diff --git "a/stack-exchange/math_overflow/shard_8.txt" "b/stack-exchange/math_overflow/shard_8.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_8.txt" +++ /dev/null @@ -1,30243 +0,0 @@ -TITLE: Is there an algorithm to find out the number of small solutions to a polynomial equation, when we vary all the coefficients? -QUESTION [6 upvotes]: Let $\Phi (z,t)$ be a polynomial given by -$$ \Phi(z,t) := z^n + A_{n-1}(t) z^{n-1} + \ldots + A_1(t) z + A_0(t).$$ -Assume that $\Phi(0,0) =0$. It is a fact that a solution $z(t)$ of the equation -$$ \Phi(z(t), t) =0 $$ -that is close to zero, can be expressed as a formal power series in $t^{1/r}$ for -some positive integer $r$. Moreover, this formal power series has a non zero -radius of convergence. This follows from the "Newton Pusieux Theorem". -My question is the following: Is there some procedure/algorithm to find -out what this $r$ is? Of course it will depend on the $A_i$. -As an example, suppose we want solutions for -$$ \Phi(z,t) = z^2 + K z + t =0$$ -where $K$ is some constant. Then a solution $z(t)$ is a power series in -$t$ if $K \neq 0$. It is a power series in $t^{1/2}$ if $K=0$. -In general is there some algorithm to find this $r$? -Everything is over the field of complex numbers $\mathbb{C}$ - -REPLY [9 votes]: This algorithm is called the Newton Polygon, and it was really invented and carefuly -described by Newton, with examples, see for example, -MR1836037 Fischer, Gerd Plane algebraic curves. Translated from the 1994 German original by Leslie Kay. Student Mathematical Library, 15. American Mathematical Society, Providence, RI, 2001. -or any other book with a similar title, or Mathematical Papers of Newton himself. -In XIX century this algorithm was a part of high school curriculum, at least in England, see, for exapmple -JFM 31.0103.05 Chrystal, G. -Algebra. Elementary textbook. 2. edition. Part II. (English) -London. 632 S. Published: 1900 -I think, for this reason, those who wrote university algebra textbooks in -XX century did not care to include it (Lang, Van der Waerden,...).<|endoftext|> -TITLE: The 2-group of extensions -QUESTION [6 upvotes]: Let $A,B$ objects of an abelian category. Then we can define the abelian group $\mathrm{Ext}^1(A,B)$ as the set of isomorphism classes of extensions $0 \to B \to E \to A \to 0$, endowed with the Baer sum. Following the principle of categorification, a finer and hopefully better invariant is the category of extensions, where morphisms are commutative diagrams as usual. Actually it is a groupoid by the Five Lemma, and I believe that the trivial extension $1$ and the Baer sum $\otimes$ make it a symmetric monoidal category in which every object has an inverse. In other words, it should be an abelian 2-group $\mathsf{Ext}^1(A,B)$, whose decategorification is the usual abelian group $\mathrm{Ext}^1(A,B)$. For this one has to find various coherence isomorphisms and check various coherence diagrams. -Question. Has this been worked out in the literature? Has this 2-group of extensions already been studied somewhere? What about specific examples? -Note that this 2-group carries more information than the group, for example the automorphism group of the unit $1$ is $\mathrm{Hom}(A,B)$. Here is an example with abelian groups: Let $p$ be a prime number. Then $\mathsf{Ext}^1(\mathbb{Z}/p,\mathbb{Z}/p)$ has $p$ non-isomorphic objects, namely the trivial extension $1$ and the extensions $X_\gamma : 0 \to \mathbb{Z}/p \xrightarrow{p \gamma} \mathbb{Z}/p^2 \xrightarrow{\text{pr}} \mathbb{Z}/p \to 0$ for $\gamma \in (\mathbb{Z}/p)^*$. We have $\mathrm{Aut}(1)=\mathbb{Z}/p$ and $\mathrm{Aut}(X_{\gamma}) = \{\alpha \in \mathbb{Z}/p^2 : \alpha \equiv 1 \bmod p\} \cong \mathbb{Z}/p$. Besides, $X_{\gamma} \otimes X_{\delta} \cong X_{\gamma + \delta}$. - -REPLY [7 votes]: A version of this, but one categorical level higher (i.e. a 3-group of (central) extensions of 2-groups) appeared explicitly in this paper of mine: -Central Extensions of smooth 2-groups and a finite dimensional string 2-group Geometry & Topology 15 (2011) 609-676. -Of course this is not the first time something like the 2-group you are interested in has appeared in the literature, but I don't have a specific reference. I don't know if there will be a good one because there is not much new information from this 2-group (see below). -Here are some observations about this "$EXT(A,B)$": - -The fastest way to see that this is an abelian 2-group satisfying all the coherence data is to view $EXT(A,-)$ as a functor from abelian groups to groupoids. As such it sends abelian group objects in Abelian groups to abelian group objects in groupoids. Then note that every abelian group B is an abelian group object. This implies that all the coherence data is there. -There is a two term chain complex of abelian groups whose homology groups are $ext^1(A,B)$ and $Hom(A,B)$. There is a well-known process which turns a chain complex of abelain groups into a category internal to abelian groups and hence into a strictly commutative abelian group object in groupoids. -There is a functor from this chain-complex 2-group into the 2-group of actual extensions which sends elements from the complex to explicit extensions. This induces a functor, which is actually an equivalence of groupoids (it is enough to check an isom on $\pi_0$ and $\pi_1$), hence it is an equivalence of abelian 2-groups. (To see it is a functor of abelian 2-groups we just need to observe that all these constructions are functorial in B). - -Either of the above two descriptions can also be used to show that all the k-invariants of this 2-group vanish. This means that this 2-group splits (non-canonically) as: -$$ EXT(A,B) \simeq ext^1(A,B) \oplus \mathcal{B} hom(A,B) $$ -where $EXT$ means the 2-group of extensions, $\pi_0 EXT = ext^1$ is the group of isomorphisms classes of extensions (thought of as a trivial 2-group), and "$\mathcal{B}$" means "shift the given abelian group up". -This means there is essentially no new information in this 2-group beyond its homotopy groups. - -REPLY [3 votes]: The paper Extensions of symmetric cat-groups by D. Bourn and E.M. Vitale defines and studies a bicategory of extensions of 2-groups (called cat-groups in the paper). In section 13, it introduces a monoidal structure that categorifies the Baer sum of extensions. -Just like Chris's answer, this is one categorical level higher then you wanted, but maybe you can decategorify it once :-)<|endoftext|> -TITLE: A Topology such that the continuous functions are exactly the polynomials -QUESTION [35 upvotes]: (I originally asked this question on Math.SE, where it received a lot of attention, but no solution.) -Which fields $K$ can be equipped with a topology such that a function $f:K \to K$ is continuous if and only if it is a polynomial function $f(x)=a_nx^n+\cdots+a_0$? Obviously, the finite fields with the discrete topology have this property, since every function $f:\Bbb F_q \to \Bbb F_q$ can be written as a polynomial. So what is with infinite fields? -I don't expect a full answer to this question, but I would even be satisfied if someone could show the existence or nonexistence of such a topology even for a single field such as $\Bbb Q, \Bbb R,\Bbb C,\Bbb Q_p$ or $ \Bbb F_q^\text{alg}$. -(Note that $(K,\tau)$ is not necessarily a topological field!) -$$ $$ -A short summary of the comments on Math.SE: Assume that you are given such a field $K$ with a topology $\tau$. Then $\tau$ is necessarily a $T_0$-space and connected. Also, the linear transformations $x \mapsto ax+b$ with $a \in K^\times$, $b \in K$ are homeomorphisms of $\tau$ (and there can be no other homeomorphisms). In the special case $K=\Bbb R$, for every $U \in \tau$ and $a \in \Bbb R$, we have $|(-\infty,a)\cap U |=|(a,\infty)\cap U |=2^{\aleph_0}$. More over, if there is an open interval $(a,b)$ and a $U \in \tau$ with $|(a,b)\cap U |<2^{\aleph_0}$, then for all $c,d \in \Bbb R$, $(-\infty,c)\cup (d,\infty) \in \tau$. As a proposal for a topology on $\Bbb R$, the coarsest topology such that all sets $p^{-1}(\Bbb Z)$ with $p \in \Bbb R[X]$ are closed, has been considered. Obviously every polynomial is continuous with respect to this topology. However, no one has given any reason why every continuous function in this topology should be a polynomial. - -REPLY [6 votes]: In the case of a topological field (edit: containing real numbers) the answer is not. Let $K$ a infinite field, with a such topology $\tau$ coherent with the algebraic field structure then $(K, +, 0, \tau)$ is a topological groups. As you said $G$ must be $T_1$ (i.e. $\{0\}$ is closed, this is equivalent to $T_0$ in a topological group), then it is completely regular, i.e. continuous functions separate closed set from points, but if $K$ is infinite the topology $\tau$ cannot be discrete or indiscrete (in these cases any function become continuous, but a polynomial can have only a finite set of zeros , then exists a lot of non polynomial functions), neither $\tau$ could be cofinite (i.e. opens are exactly complements of finite sets) because cofinite topology in a infinite set isn't $T_2$. Then exist a infinite closed $A$ and a $x\not\in A$ and a continuous function that separate these, but just because $A$ is infinite this cannot be polynomial. -I wanted to write this answer as a comment, but it was too long<|endoftext|> -TITLE: Homotopy as a general organizing principle -QUESTION [40 upvotes]: One of the realizations that led to the development of Homotopy Type Theory (HoTT) is that the ideas of homotopy theory have very broad applicability in mathematics. Indeed, Quillen model categories comprise very general ideas that arise in a variety of places. To quote Mike Shulman: - -Personally, I don’t find it especially surprising that homotopy theory has more than one application to some other subject, any more than I would find it surprising that category theory does. I think it’s becoming increasingly clear that both of them are general organizing principles of mathematics. - -I would like to build a list of interesting examples of this, especially unexpected applications of the ideas of homotopy theory in otherwise far removed areas of mathematics. - -REPLY [6 votes]: Cardinal arithmetics is usually considered far removed from homotopy theory, -so perhaps this would be an example. -Several cardinal invariants in set theory may be viewed as derived functors, in a very degenerate model category setting. is usually considered remote from homotopy theory. See Gavrilovich, Hasson, A homotopy theory for set theory, Part I, Part II. -In fact, there is also the following which appears to be an explicit attempt to -that the ideas of homotopy theory have very broad applicability in mathematics. -Gromov, In search for a structure. Part 1, On Entropy", talks entropy in terms of category theory, but perhaps not necessarily homotopy theory. He has the following to say in a postscriptum: - Apology to the Reader. Originally, Part 1 of ”Structures” was planned as about a half of an introduction to the main body of the text of my talk at the European Congress of Mathematics in Krak´w with the sole purpose to motivate o what would follow on ”mathematics in biology”. But it took me several months, instead of expected few days, to express apparently well understood simple things in an appropriately simple manner. Yet, I hope that I managed to convey the message: the mathematical language developed by the end of the 20th century by far exceeds in its expressive power anything, even imaginable, say, before 1960. Any meaningful idea coming from science can be fully developed in this language. Well..., actually, I planned to give examples where a new language was needed, and to suggest some possibilities. It would take me, I naively believed, a couple of months but the experience with writing this ”introduction” suggested a time coefficient of order 30. I decided to postpone. -There was also talks by Voevodsky (and older work by Tsvetkov(?)) who consider a category theory approach to probability. (There is an online -video talk by Voevodsky on this in Russian. I'll try to recall the name of the book by Tsvetkon(??))<|endoftext|> -TITLE: Homology of special linear group over local field -QUESTION [8 upvotes]: I am trying to compute the group -$H_1(\mathrm{SL}_2(\mathbb{Z}_2),M)$, where $\mathbb{Z}_2$ are $2$-adic integers and M is a module $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. I suppose that the group acts on $M$ by matrix multiplication. -I found a similar-looking computation in the paper of Dupont and Sah -"Homology of Euclidean groups of motion made discrete and Euclidean scissors congruences". It was shown there that $H_1(\mathrm{SO}_3(\mathbb{R}),\mathbb{R^3}) = \Omega^1_\mathbb{R}.$ -I would be very grateful for any help with computing the group or for any interpretation of its elements. - -REPLY [5 votes]: First, some general remarks on the situation for $R$ an arbitrary commutative ring. Since ${\rm diag}(-1,-1)$ acts by multiplication by $-1$ on $R^{\oplus 2}$, the homology groups ${\rm H}_i({\rm SL}_2(R),R^{\oplus 2})$ are $2$-torsion for $i\geq 1$. (This is called the center-kills-argument.) This doesn't happen in the case ${\rm SO}(3)$ because the center of ${\rm SO}(3)$ is trivial. This is a significant difference between the case of ${\rm SL}_2(R)\looparrowright R^{\oplus 2}$ in the question and the case ${\rm SO}(3)\looparrowright \mathbb{R}^3$ in the paper of Dupont and Sah. -As a particular case, if $R=K$ is a field, then the homology groups will be vector spaces over the field and therefore the homology group ${\rm H}_1({\rm SL}_2(K),K^{\oplus 2})$ can only be nontrivial in characteristic $2$. (In particular, these are trivial for $R=\mathbb{Q}_2$; the title said something about local fields.) - -In the specific situation of $R=\mathbb{Z}_2$, we can say something more about the homology group in the question, by iterated application of the Hochschild-Serre spectral sequence (well, the five-term exact sequence suffices). -First, we compute the homology ${\rm H}_1({\rm SL}_2(\mathbb{F}_2),\mathbb{F}_2^{\oplus 2})$. We have ${\rm SL}_2(\mathbb{F}_2)=S_3$, so we can use the Hochschild-Serre spectral sequence for the extension $0\to\mathbb{Z}/3\to S_3\to\mathbb{Z}/2\to 0$. To compute ${\rm H}_i(\mathbb{Z}/3,\mathbb{F}_2^{\oplus 2})$ we can use the standard resolution -$$ -\cdots\to M\xrightarrow{N} M\xrightarrow{t-1}M -$$ -where $M=\mathbb{F}_2^{\oplus 2}$, $t$ is a generator and $N=1+t+t^2$ is the norm element. In this case, $N$ is the zero map so the even homology groups are the cokernel of $t-1$ and the odd homology groups are the kernel of $t-1$. But $t-1$ is an isomorphism, so ${\rm H}_i(\mathbb{Z}/3,\mathbb{F}_2^{\oplus 2})=0$ for all $i$. The Hochschild-Serre spectral sequence then also implies that ${\rm H}_i({\rm SL}_2(\mathbb{F}_2),\mathbb{F}_2^{\oplus 2})=0$ for all $i$. -Now we can use the Hochschild-Serre spectral sequence associated to the extension $1\to\mu_2\to{\rm SL}_2(\mathbb{Z}_2)\to {\rm PSL}_2(\mathbb{Z}_2)\to 1$. The homology groups ${\rm H}_i(\mathbb{Z}/2,\mathbb{Z}_2^{\oplus 2})$ can also be computed using the standard resolution as above. The norm map is again the zero map, and $t-1$ is multiplication by $-2$. In particular, odd homology is trivial and even homology is given by $\mathbb{F}_2^{\oplus 2}$. It remains to compute the homology ${\rm H}_i({\rm PSL}_2(\mathbb{Z}_2),\mathbb{F}_2^{\oplus 2})$ with the action by left multiplication of matrices. This can be done using the extension $1\to \Gamma\to{\rm PSL}_2(\mathbb{Z}_2)\to {\rm SL}_2(\mathbb{F}_2)\to 1$. The congruence subgroup $\Gamma$ acts trivially on $\mathbb{F}_2^{\oplus 2}$. By the above computation for ${\rm SL}_2(\mathbb{F}_2)$, the contribution ${\rm H}_1({\rm SL}_2(\mathbb{F}_2),{\rm H}_0(\Gamma,\mathbb{F}_2^{\oplus 2}))\cong {\rm H}_1({\rm SL}_2(\mathbb{F}_2),\mathbb{F}_2^{\oplus 2})$ is trivial. -So we have identified -$$ -{\rm H}_1({\rm SL}_2(\mathbb{Z}_2),\mathbb{Z}_2^{\oplus 2})\cong {\rm H}_0({\rm SL}_2(\mathbb{F}_2),{\rm H}_1(\Gamma,\mathbb{F}_2^{\oplus 2})). -$$ -At this point it's not quite clear to me what the abelianization of the congruence subgroup would be (this is the relevant thing to complete the calculation). In the higher-rank cases, the abelianization is the corresponding Lie algebra over the finite field. For ${\rm SL}_2$, the abelianization would still surject onto $\mathfrak{sl}_2(\mathbb{F}_2)$ but could be bigger, the higher-rank arguments don't work because they depend on the identification of the commutator subgroups and elementary subgroups. (Note however that a direct computation shows that if the abelianization of the congruence subgroup is $\mathfrak{sl}_2(\mathbb{F}_2)$ then the coinvariants and hence the homology group in the question would be trivial.)<|endoftext|> -TITLE: History of profinite groups, when was it first mentioned? What was the original definition? -QUESTION [10 upvotes]: Searching left me hanging. One of my professors told me the definition using the topological properties was the first one but I cannot find any resources. Is that true? If not, how was it originally defined? References would be lovely. -Best regards - -REPLY [4 votes]: See also W.Krull-Galoische theorie der unendlichen algebraischen erweiterungen-Mathematische Annalen,100,1928 -The concept is there,maybee not the name<|endoftext|> -TITLE: Hartshorne-Serre's correspondence in higher codimension -QUESTION [6 upvotes]: There's a well-known correspondence (traditionally called Hartshorne-Serre) between codimension 2 smooth subvarieties $S\subset X$ of a smooth algebraic variety $X$ and certain rank two vector bundles on $X$. Is there anything similar for higher codimensional subvarieties of $X$, i.e. codimension 3 and more? - -REPLY [7 votes]: In codimension 3, you can hope for your varieties to be cut out (scheme theoretically) by the Pfaffians of an alternating map of vector bundles on $X$. -If you assume that $S$ is subcanonical, which you have to assume for Serre's correspondence (by the way, traditionnally, this is not called Hartshorne-Serre, but only Serre's correspondence) that $X = \mathbb{P}^N$ and some small extra-assumptions on $S$, then it is a theorem of Walter that $S$ is cut out by such Pfaffians (see : Walter, Pfaffians subschemes, Journal of Algebraic Geometry, 1996). -If $X$ is not the projective space, you can still hope for some structure theorems, but they are more difficult to explain (see : http://arxiv.org/pdf/math/9906170.pdf). -In codimension $4$ there is this : http://arxiv.org/pdf/1304.5248.pdf (which seems to me even much more complicated) and after (codimension $5$ or bigger) I guess nothing is known.<|endoftext|> -TITLE: PDEs involving measures; where to begin? -QUESTION [5 upvotes]: If I want to learn about existence of weak solutions to PDEs of the form -$$u_t + Au = f$$ -or -$$Au = f$$ -where $A$ is elliptic and $f$ is a measure, where do I start? I know the Galerkin method for parabolic PDEs with right hand side in $L^2(0,T;V')$ or $L^2(0,T;H)$ (where $V \subset H$). But now I want to a step deeper into PDE theory and consider measures instead. -In the literature I have come across -1) $f$ = Radon measure -2) $f$ = Young measure -... -and I don't really what how to begin. Can anyone recommend me a text/source to learn this kind of stuff, or give me a quick overview about this area? I am particularly interested in $f = \text{Dirac delta}$ but maybe that can come later. I would appreciate a book where as little measure theory as required is there since it is not my strong point. Thank you. - -REPLY [5 votes]: Weak solutions for PDEs with Radon measures as right-hand sides can be obtained by a duality technique, which roughly proceeds as follows. Assume that the adjoint operator $A^*$ acts as isomorphism from $W^{1,q}_0(\Omega)$ to $W^{-1,q}(\Omega)$ for some $q>n$ with $\Omega\subset\mathbb{R}^n$. (This is the case if the coefficients of $A$ and the domain $\Omega$ is sufficiently smooth.) The closed range theorem and reflexivity of the spaces then implies that $A$ is an isomorphism from $W^{1,q'}_0(\Omega)$ to $W^{-1,q'}(\Omega)$ for $1/q'+1/q = 1$, i.e., $q'< n/(n-1)$. Since $W^{1,q}_0(\Omega)$ is continuously and densely embedded into $C_0(\Omega)$ for $q>n$, we have the dual embedding of $C_0(\Omega)^*$ (which can be identified by the Riesz theorem with the space of Radon measures) into $W^{-1,q'}(\Omega)$. Together, this yields for every Radon measure $f\in C_0(\Omega)^*\hookrightarrow W^{-1,q'}(\Omega)$ a unique solution $u\in W^{1,q'}(\Omega)$ to $Au=f$. -If $A^*$ lacks this maximal regularity, the solution is no longer unique, and you need to either extend the space of test functions (leading to solutions in the sense of Stampacchia) or restrict the space of solutions (leading to solutions in the sense of Boccardo–Gallouët); both approaches are equivalent. All this is very nicely explained in -Meyer, C., Panizzi, L., and Schiela, A. (2011). Uniqueness criteria for solutions of the adjoint equation in state-constrained optimal control. Numerical Functional Analysis and Optimization 32.9, pp. 983–1007. -For parabolic and nonlinear equations, see, e.g., -Boccardo, L., and Gallouët, T. (1989). Non-linear elliptic and parabolic equations involving measure data. Journal of Functional Analysis 87, pp. 149–169.<|endoftext|> -TITLE: preservation of $\aleph_2$-c.c. in CS iterations -QUESTION [7 upvotes]: It seems that considerable care is taken in the literature to ensure that a countable support (CS) iteration of proper forcings preserves $\aleph_2$. Can you give an example, assuming CH, of a CS iteration of countably closed forcings such that each iterand is forced to have the $\aleph_2$-c.c. but the full iteration does not? - -REPLY [7 votes]: One of the standard (counter)examples here has to do with uniformization at $\omega_2$, and it is dealt with in Section 3 of the Appendix to Shelah's Proper and Improper forcing: -Let $S=\{\delta<\omega_2:\rm{cf}(\delta)=\omega_1\}$, and for each $\delta\in S$ let $\eta_\delta:\omega_1\rightarrow\delta$ be the increasing enumeration of a closed unbounded subset of $\delta$, and let $\bar{\eta}=\langle \eta_\delta:\delta\in S\rangle.$ -Let $\bar{C} =\langle c_\delta:\delta\in S\rangle$ be a collection of functions where $c_\delta$ maps $\omega_1$ to $\{0,1\}$. We say that $\bar{C}$ can be uniformized if there is a function $f:\omega_2\rightarrow\{0,1\}$ such that for every $\delta\in S$ the set -$$\{\alpha<\omega_1: c_\delta(\alpha)\neq f(\eta_\delta(\alpha))\}$$ -is bounded in $\omega_1$. -Shelah proved that if CH holds, then for any such $\bar{\eta}$ there is a family of colorings $\bar{C}$ as above that cannot be uniformized. Assaf Rinot has a nice write-up of this result on his blog. -Shelah points out in Discussion 3.4A on page 984 of Proper and Improper forcing that for any given $\bar{C}$, there is a very nice $\aleph_1$-closed $\aleph_2$-cc (even $\aleph_2$-centered) forcing that adds a function uniformizing $\bar{C}$. -The relevance to your question is that an iteration of such uniformizing forcings that takes care of all potential $\bar{C}$ for a given $\bar{\eta}$ must collapse $\aleph_2$ as otherwise we contradict the theorem above, and therefore the limit of the iteration certainly cannot have the $\aleph_2$-cc. -The $\aleph_2$-pic (properness isomorphism condition) from Chapter VIII of Proper and Improper Forcing is a strong version of the $\aleph_2$-cc that says (roughly) that our forcing isn't doing something like "uniformizing", and it is one way to ensure that the limit of a countable support iteration possesess the $\aleph_2$-cc.<|endoftext|> -TITLE: Is there a general theory of "representation theorems"? -QUESTION [10 upvotes]: Let $V$ and $W$ be classes of algebraic structures, and suppose we have some canonical way of constructing objects of $W$ from objects of $V$. Let's call this construction $C$, so that for all $A\in V$, $C(A)\in W$. -Then by a representation theorem, I mean a way of associating to each $B\in W$ an object $U(B)\in V$ such that $B$ embeds into $C(U(B))$. -Examples: - -$V = \operatorname{Sets}$, $W = \operatorname{Groups}$. For $X$ a set, let $C(X) = S_X$, the symmetric group on $X$. Then for any group $G$, we can take $U(G)$ to be the underlying set of $G$, and $G$ embeds in $S_{U(G)}$. -$V = \operatorname{Abelian Groups}$, $W = \operatorname{Rings}$. For $V$ a set, let $C(V) = \text{End}(V)$, the endomorphism ring of $V$. Then for any ring $R$, we can take $U(R)$ to be the additive group of $R$, and $R$ embeds in $\text{End}(U(R))$. -$V = \operatorname{Sets}$, $W = \operatorname{Boolean Algebras}$. For $X$ a set, let $C(X) = \mathcal{P}(X)$, the power set of $X$. Then for any Boolean algebra $B$, we can take $U(B)$ to be the set of ultrafilters on $B$, and $B$ embeds in $\mathcal{P}(U(B))$. - -This setup does not seem to fit nicely into a categorical framework: In the first two examples above, the construction $C$ is not a functor, at least from the usual categories of sets and abelian groups. $C$ is a functor if we only force it to respect isomorphisms (by considering $V$ as a category with invertible morphisms as arrows), but we care about a noninvertible morphism in $W$ (the embedding $B\rightarrow C(U(B))$, so we won't be able to make $U$ into a functor to $V$. Finally, even if we could make $C$ and $U$ functors, it doesn't seem that the embeddings $B\rightarrow C(U(B))$ would cohere into a natural transformation $\operatorname{id}_W\rightarrow C\circ U$. -Of course, the third example is much better behaved, categorically speaking. I am aware of the family of examples related to this one, as described, for example, in Johnstone's book Stone Spaces. -But do we know anything about the general algebraic setup? In particular are there general conditions under which representation theorems must exist? If we know that a representation theorem exists, is there any sense in which we can compute what it must be? - -REPLY [6 votes]: I'm not aware of a general theory, but the first two examples are corollaries of the Yoneda Lemma. For the first one, consider groups as categories with one object (and observe that you get a similar representation for monoids); for the second, consider rings as linear categories with one object (and use the enriched Yoneda Lemma). You could also view partial orders as categories in the usual way and thereby embed them into power sets.<|endoftext|> -TITLE: On the convergence of the function series $\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$ -QUESTION [8 upvotes]: Let $f$ be a smooth real function defined around origin. If we -differentiate term by term the series -$\hat{f}(x):=\sum_{n=0}^\infty(-1)^n\frac{f^{(n)}(x)}{n!}x^n$, we get $\frac{d}{dx}\hat{f}(x)=0$. -\begin{eqnarray}\frac{d}{dt}\hat{f}(t)&=& -\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n+ -\sum_{n=1}^{\infty}(-1)^n\frac{f^{(n)}(t)}{(n-1)!}t^{n-1}\nonumber\\&=& -\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n-\sum_{n=0}^{\infty}(-1)^n\frac{f^{(n+1)}(t)}{n!}t^n -\nonumber\\&=&0\nonumber -\end{eqnarray} -Thus $\hat{f}(t)$ should be constant. But in fact we are not -allowed to differentiate term by term from a series. -Next, suppose that $f$ is a smooth periodic function which by -the Fourier analysis we know that it has Fourier expansion. That -is we suppose $f(t)=\sum_{m=-\infty}^\infty c_me^{im\omega t}.$ Then -it is well known that we can differentiate to get -$f^{(n)}(t)=\sum(im\omega)^n c_me^{im\omega t}.$ Thus -\begin{eqnarray}\hat{f}(t)&=&\sum_n\sum_m\frac{(-1)^n(im\omega)^{n}}{n!}c_me^{im\omega - t}t^n\nonumber\\&=& -\sum_m\sum_n(\frac{(-1)^n(im\omega)^{n}}{n!}t^n)c_me^{im\omega - t}\nonumber\\&=& \sum_me^{-im\omega t}c_me^{im\omega - t}\nonumber\\&=&\sum_{m=-\infty}^{\infty}c_m\nonumber\\&=&f(0).\nonumber -\end{eqnarray} -Thus again it seems that the series should convergence to a -constant. But in the above -we have exchanged the order of two infinite sums which are not allowed. -The function -$$f(t):=\sum_{m=0}^{\infty}\frac{1}{m!}e^{i2^mt}$$is smooth -nowhere analytic, in the sense that convergence radius of the -Taylor's series of $f$ at each point is zero and therefore -$\hat{f}(t)$ diverges for all $t\ne0$. -The function -$$f(t):=\sum_{m=1}^{\infty}\frac{1}{m!}e^{i2^{-m}t}$$is -analytic at $t=0$ whose convergence radius is infinity. Thus -$\hat{f}(t)$ converges for all $t$ to $f(0)$. -In fact one can show that -a) If $f$ is analytic at origin then the series $\hat{f}$ is -convergent uniformly to the constant $f(0)$. -b)If $f$ is nowhere analytic in the sense that the radius of -convergence of the Taylor's series is zero then of course the -series is divergent. But if $f$ is nowhere analytic in the sense -that the radius of convergence of the Taylor's series is positive -but the Taylor's series does not converge to the function $f$ then -the series $\hat{f}$ may converge. -c) About the function $f(x):=e^{\frac{-1}{x^2}},f(0)=0$ one can -show that if the series is convergent then its sum is constant. -d) There are nowhere analytic functions such that the series is -convergent in a dense subset to the constant $f(0)$ and there are -nowhere analytic functions such that the series is divergent -everywhere. -Now the main questions are. - -Is there a smooth function $f$ which is not analytic at origin -and the series $\hat{f}$ is convergent in an interval around -origin and the sum is the constant $f(0).$? - -Is there a smooth function $f$ which is not analytic at origin -and the series $\hat{f}$ is convergent in an interval around -origin and the sum is not constant.? - -If we define a linear differential operator of infinite order -$f\mapsto \hat{f}-f(0)$. Then in above we said that analytic -functions at origin are contained in the space of eigenfunctions -of the zero eigenvalue of this operator. Now the question arises -that: are there nonzero eigenvalues for this operator? - -For the function $f(x):=e^{\frac{-1}{x^2}},f(0)=0$, is the -series $\hat{f}$ convergent? Please see the preprint -arXiv:1105.2611v2 [math.GM] 5 Jun 2011 and the paper: Journal of Applied Analysis, Volume 25, Issue 2, Pages 131–139, DOI: https://doi.org/10.1515/jaa-2019-0014. - -REPLY [4 votes]: Check the definition of quasianalytic functions and Denjoy-Carleman ultradifferentiable functions - the results in this field should be helpful for you.<|endoftext|> -TITLE: What are finite homotopy types? -QUESTION [18 upvotes]: Starting from an ordinary 1-categorical point of view, there are various obvious candidate definitions for ‘finite homotopy type’: - -The homotopy type of a simplicial set that has only finitely many non-degenerate simplices. -The homotopy type of a CW complex that has only finitely many cells. -The homotopy type of the nerve of a finitely-presentable category. - -Homotopy type theory affords another candidate: - -A higher inductive type that admits a finite presentation in some syntactic sense. - -Question. Do these notions coincide? To what extent is each one the ‘right’ notion of finiteness for homotopy types? For instance, are these precisely the homotopy types $X$ such that the representable functor $\mathrm{Hom}(X, -) : \infty \mathbf{Grpd} \to \infty \mathbf{Grpd}$ preserves (homotopy) filtered colimits? - -REPLY [4 votes]: Another possible definition of "finite homotopy type" but not from a categorical point of view is suggested by the paper -SPACES WITH FINITELY MANY NON-TRIVIAL HOMOTOPY GROUPS ALL OF WHICH ARE FINITE , GRAHAM ELLIS , Topdogy Vol. 36, No. 2, pp. 501-504, 1997 -This shows that for such a connected CW-space, its homotopy type may be represented by a simplicial group such that its group of $ n$ -simplices is finite for each -$n \geqslant 0$.<|endoftext|> -TITLE: Iterating definability -QUESTION [8 upvotes]: An odd -- probably basic -- question about model theory: -For $\mathcal{M}$ a structure in a (first-order) signature $\Sigma$, let $\mathcal{M}'$ be the structure in signature $\Sigma\sqcup\lbrace U\rbrace$ -- with $U$ a unary relation -- whose reduct to $\Sigma$ is $\mathcal{M}$, and interprets $U$ as $$ U^{\mathcal{M}'}=\lbrace a: a\text{ is definable in $\mathcal{M}$}.\rbrace$$ -Up to the choice of unary relation symbol $U$, this is well defined; moreover, we can iterate this through the ordinals: $$ \mathcal{M}^{(0)}=\mathcal{M}, \quad \mathcal{M}^{(\alpha+1)}=(\mathcal{M}^{(\alpha)})', \quad \mathcal{M}^{(\lambda)}=\bigcup_{\beta<\lambda}\mathcal{M}^{(\beta)} \,\,(\lambda \text{ limit}).$$ (The union notation is technically inappropriate, but its meaning is clear.) Now, for any $\mathcal{M}$, let $$ D(\mathcal{M}, \alpha)=\lbrace a\in\mathcal{M}: a\text{ is definable in $\mathcal{M}^{(\alpha)}$}\rbrace$$ be the set of elements of $\mathcal{M}$ definable after stage $\alpha$. Let $D(\mathcal{M}, \infty)=\bigcup_{\alpha\in ON} D(\mathcal{M}, \alpha)$ be the set of all eventually definable elements, and for $a\in D(\mathcal{M}, \infty)$ let the age of $a$, $age_\mathcal{M}(a)$, be the least $\beta$ such that $a\in D(\mathcal{M}, \beta)$. Clearly for each $\mathcal{M}$ there is a least upper bound, $m_\mathcal{M}$, on the ages of elements of $D(\mathcal{M}, \infty)$. -Some very easy observations: - -If $\mathcal{M}=(M, <)$ is a well-ordering, then $D(\mathcal{M}, \infty)=M$, since the $\alpha$th element of $\mathcal{M}$ is definable by stage $\alpha$ at the latest. -Even if $\mathcal{M}$ is strongly minimal, $\mathcal{M}'$ need not be: consider $\mathcal{M}=\mathbb{N}+\mathbb{Z}$ as a linear order. Presumably other niceness properties such as stability are also not preserved, but I don't have examples yet. - -My question is, what is known about the set $D(\mathcal{M}, \infty)$, the age function $age_\mathcal{M}$, or the invariant $m_\mathcal{M}$? I've been playing around with this idea for a bit, but my model theory is not very strong; I'm sure this has been treated before, but I haven't been able to find a reference. -(In case anyone is interested, I initially thought that there would be connections with notions of rank, as long as $\mathcal{M}$ is sufficiently nice; in fact, I came up with this question after using some dubious analogies to try to explain forking and rank to a friend. As far as I can tell, this initial hope is in fact bogus, but that's where this came from.) - -There are two other questions about this that I'm especially interested in. First, what if we augment first-order logic by adding a logical unary relation $D$ whose interpretation is stipulated to always be $D(\mathcal{M}, \infty)$ -- the resulting model theory seems wild (compactness and Lowenheim-Skolem fail extremely badly), but this logic "comes from" first-order logic in a natural way; is there anything nice we can say about it? Second, this time closer to computability theory: what if we replace "definable" with $\Sigma_1$-definable? Does $m_\mathcal{M}$ now have a recursion-theoretic interpretation? I consider these as just curiosities, compared to the main question (which, though more vague, I hope is still appropriate), but if anyone has anything to say on either count I'd be extremely interested. - -REPLY [5 votes]: Your construction has a family resemblence with the iterative -theories of truth, due originally to Tarski and discussed at -length in the philosophical logic literature, by Kripke, Putnam -and many others. -Rather than adding a predicate for definability, the idea in these -theories, acknowledging Tarski's nondefinability theorem that one -cannot have a full classical theory of truth in the language with -that truth predicate, is that nevertheless one may build an -increasingly robust theory of truth, in a transfinite progression, -by successively adding truth predicates over the expanded -structure that has been defined so far. One starts with an initial -structure $\mathcal{M}$, and then add a predicate for -true-in-$\mathcal{M}$, and then a predicate for truth in that -structure, and so on transfinitely. -To be precise, if $\mathcal{M}^{(0)}$ is a structure that interprets -arithmetic, so one can refer to formulas via Gödel coding, -then we define the truth predicate $T^{(0)}$ for this model, where - $T^{(0)}(\varphi,\vec a)$ holds if and only if -$\mathcal{M}^{(0)}\models\varphi[\vec a]$, where $\varphi$ is any -formula in the language of $\mathcal{M}^{(0)}$. Adjoining this new -predicate, we form the next structure $\mathcal{M}^{(1)}=\langle -M,\ldots,T^{(0)}\rangle$, and then continue the construction just -as you did. At stage $\alpha$, we have the $\alpha^{\rm th}$ -structure $\mathcal{M}^{(\alpha)}$, and then form the truth -predicate $T^{(\alpha)}$ for satisfaction in this structure. In -particular, the predicate $T^{(\alpha)}$ concerns formulas -$\varphi$ in the language in which earlier truth predicates -$T^{(\beta)}$ appear as formal predicates. -The closely related revision theories of truth (see the book The revision theory of truth by -Gupta and Belnap) have a different tack, following ideas of -Kripke, trying to unify the theory of truth into one increasingly -stable (but partial) truth predicate. In this theory, one can -refer to things like "eventually true" or stabilized truth, and so -on. There would seem to be an analogue of this approach also in -your context.<|endoftext|> -TITLE: Why do we want maps to be measurable (in countably-additive setting) -QUESTION [10 upvotes]: When I have to explain things that I am doing to people who did not do (or even did not learn) measure-theoretical probability, I think of getting a question in the title, and I am not sure I have arguments strong enough to convince that measurability is indeed the must. -Let me focus on a particular case of stochastic optimization in discrete time, in the setting of countably-additive probabilities. Being it formulated as a dynamic programming, or a gambling problem, one requires the decision to be Borel (universally, analytically) measurable with respect to the current state. Such requirement further leads to known issues as non-existence of measurable selectors. I do understand, that if the selection is not measurable, then it is not possible to define a probability measure over the space of state trajectories in a formal way. However I am no sure whether such explanation is convincing enough, since one can further argue that formal definition of probability requires measurability just in this particular framework, and perhaps the framework is something to be fixed. -It seems that Dubins and Savage used the finitely-additive framework at least for the reason to escape the measurability issues (as it written in Section 1.3 "Gambles"). So does it mean, that this question can be reduced to whether to deal with $\sigma$-additive probability or finitely-additive one? -Edited: in the edit I wanted to address the point raised by Yuri in his answer. I do know that measurability serves at least for the two goals: to define the regularity (nice sets, nice functions) and the information/dependence structure which follows from the following theorem: - -If $X$ and $Y$ are measurable maps such that $Y$ is $X$-measurable, then there exists a measurable map $f$ such that $Y = f(X)$. - -Note, however, that $X$-measurability of $Y$ guarantees the existence of a measurable map $f$. If we are just interested in a non-meaurable one, then a necessary and sufficient condition would be that level sets (pre-images) of $Y$ are saturated w.r.t. level sets of $X$. As far as I know, the dependence/information structures e.g. in game theory may be defined just based on such partitions, and serve equally well for that goal as much as $\sigma$-algebras. On the other hand, the use of the latter comes with an expense of dealing with measurable maps exclusively, which appears to be unnecessary at least for the particular of defining information structures. -The regularity task seems to be more in the need of measurability, but the only reason I could see here is the ability to measure nice sets or, more generally, to integrate nice functions. Indeed, if we fix a $\sigma$-algebra, and define a $\sigma$-additive finite measure on it, we can integrate any bounded measurable functions (which is such a nice thing after I struggled on my first year with Riemann integrability). However, yet again - is it just a nice, neat and beautiful way of doing this, or is it the way and this is way the requirements on the measurability of maps are not just due to the (relative) simplicity of dealing with them. -Perhaps, there is also a possible reason for the need of measurability besides the information structure modeling or integrating nice functions, but I don't know what could it be. I must say that I love measure-theory a lot, and found it extremely beautiful - I just want to understand myself (and also to be able to convince others) why shall we deal with measurable structures. - -Summary: in a view of the answer by Alexander, comments of Yuri and a conversation I had with Michael Greinecker, being asked the entitling question, I would argue that indeed on can define integration and probability without the notion of measurability - but in such a case one may loose many intuitive properties (e.g. LLN) which would make the interpretation of the result harder. In a particular case of the optimal control, I guess the argument could be: applying a non-measurable strategy is permitted, yet may no be able to which value would it yield, and what does this value mean (e.g. without LLN). - -REPLY [4 votes]: Regarding measurability: Depending on the audience, it may help to say that standard limit theorems (LLN, CLT, etc.) require measurability (here's something very simple I wrote recently on what you can say in the way of LLN without measurability--basically nothing, except for what you get by applying the standard W/S LLN to the measurable minorant/majorants). Most of the practical applications of probability theory depend in some way on limit theorems: e.g., statistics and CLT, stochastic algorithm convergence, Bayesian convergence, etc. -Regarding finite additivity: On the bright side, if one replaces countable with finite additivity one gets to keep some limit theorems (at least those that give explicit bounds on convergence of distributions: just approximate r.v.'s on a finitely additive space with simple ones, and apply the standard theorems to the simple ones). -On the dark side, replacing countable additivity with finite additivity will not make all sets measurable in dimensions $n\ge 3$, due to Banach-Tarski issues. In situations where there are rotational symmetries, one will still have to have non-measurable sets, assuming AC.<|endoftext|> -TITLE: Definition of "finite group of Lie type"? -QUESTION [40 upvotes]: The list of finite simple groups of Lie type has been understood for half a century, modulo some differences in notation (and identifications between some of the very small groups coming from different Lie types). Call this collection of isomorphism classes of finite groups $\mathcal{S}$. But it's not clear to me that there is a similar consensus about the meaning of "finite group of Lie type". Probably most people would place the finite general linear groups on this list. Maybe also the Weyl group of the $E_8$ root system, which is related in the Atlas of Finite Groups to the simple group denoted $G=\mathrm{O}_8^+(2)$ as the group $2.G.2$. -The term "finite group of Lie type" comes up frequently in the literature (almost 500 times in a MathSciNet search of titles and reviews). There are two main directions in which such groups are approached, which might possibly lead to different lists. Often the starting point is a simple algebraic group over a finite field (with no nontrivial proper closed connected normal subgroups), though $\mathrm{GL}_n$ doesn't quite fit here. -(1) Steinberg's efficient organization in terms of finite fixed point groups under endomorphisms, denoted $G_\sigma$, is now often expressed in terms of "Frobenius morphisms" and their fixed points $G^F$. Basically the groups of interest then come in three flavors: split (Chevalley) groups, quasi-split (Hertzig, Steinberg, Tits) groups in types $A_n (n \geq 2), D_n, E_6$, or Suzuki/Ree groups in types $B_2, G_2, F_4$ with $p = 2, 3, 2$ as defining characteristic. -This leads to a collection $\mathcal{L}_1 \supset \mathcal{S}$ of (isomorphism classes of) Lie-type groups if one is allowed to take derived groups and to factor groups by subgroups of their centers. For this I certainly want to include general linear and unitary groups, so for type $A_{n-1}$ the collection $\mathcal{L}_1$ includes $\mathrm{GL}_n, \mathrm{SL}_n, \mathrm{PGL}_n, \mathrm{PSL}_n$ and related unitary groups. Often people specify at the outset a connected reductive group $G$ over a finite field, but it's best to assume that $G$ has a simple derived group for the current purpose. -(2) The Atlas focuses instead on the finite simple groups and then builds character tables for various related groups, as in my Weyl group example above. Here the simple group $G$ can be enlarged to a group in which $G$ is normal and the quotient is abelian, or can acquire nontrivial central extensions using its Schur multiplier. Starting with the collection $\mathcal{S}$, one obtains in this way a larger collection $\mathcal{L}_2$ of (isomorphism classes of) "groups of Lie type". Partly because of questions about general linear groups and the like, I'm left with a question: - -How close to being the same are the collections $\mathcal{L}_1$ and $\mathcal{L}_2$? (Is there a natural way to define them to ensure equality?) - -REPLY [17 votes]: I feel a bit funny posting this as an answer to someone who has written a book with "finite groups of Lie type" in the title. -But I also find the matter both confusing and interesting (and full of conflicting terminology!), so here's my outside take on this, being a topologist of training: -(My own literature references include the standard sources: Carter, Gorenstein-Lyons-Solomon3..., as well as the recent (recommendable!) book by Malle-Testerman): -I thought the most general definition of a finite group of Lie type was close to your definition $\cal L_1$: -Definition ($\cal L_3$): A finite group of Lie type $G$ is a finite group obtained as ${\mathbf G}^F$, where $\mathbf G$ is a connected reductive algebraic group over an algebraically closed field of characteristic $p$, and $F$ is a Steinberg endomorphism (aka "twisted Frobenius", a composite of a Frobenius map and an automorphism of $\mathbf G$), or a group obtained from such a group ${\mathbf G}^F$ by modding out by a central subgroup, or passing to a normal subgroup with abelian quotient group. -Groups of this form will have many "Lie-like" properties, and in particular admit a canonical BN-pair structure. (Admitting a BN-pair structure would be another possible definition!) -Some authors restrict $\cal L_3$ by assuming $\mathbf G$ is semi-simple, or even simple, and some authors don't allow central quotients or passing to subgroups contained in the commutator subgroup. (One reason to allow the central quotients and subgroups is that otherwise the construction rarely produces simple groups: $|{\mathbf G}^F|$ only depends on the isogeny class of $\mathbf G$, e.g., adjoint and simply connected form have the same order, so ${\mathbf G}^F$ itself will never be simple unless $\mathbf G$ has associated Cartan matrix of determinant $1$.) Sticking with reductive (rather than simple or semi-simple) has the advantage that Levi subgroups in finite groups of Lie type are still finite groups of Lie type, which is useful if you're doing induction. -If you (for some reason) want to avoid tori $\mathbb F_q^{\times}$ and twisted products of non-trivial groups in this definition, but keep $GL_n(\mathbb F_q)$, you can maybe add the assumption to $\mathcal L_3$ that the root datum associated to $\mathbf G$ is indecomposable, and the associated root system not a product of two non-trivial root systems, or something like that? To an outsider this seems artificial? -However, not all classical groups will be in $\mathcal L_3$, e.g., $SO_{2n}^+(\mathbb F_q)$ (with the standard (classical!) definition). The problem arises from that its not connected as an algebraic group when $q$ is a power of two, so one has to pass to an index 2 subgroup, the kernel of the "pseudo/quasi-determinant", instead of the old-fashioned determinant. In particular the Weyl group of $E_8$, which is a central extension $2\cdot SO_8^+(8)$, is also not in $\mathcal L_3$, as Derek alluded to. ($W_{E_8}$ is "even further" freakish than $SO_8^+(2)$ since the central extension is related to that the index two subgroup $Spin^+_8(2)$ has non-generic Schur multiplier $\mathbb Z/2 \times \mathbb Z/2$, not zero as usual.) -I would (perhaps naively) argue that these groups also should not be considered finite groups of Lie type since they (unless I'm mistaken) do not admit the structure of a BN-pair; the structure of their p-radical subgroups is more chaotic (the Borel-Tits theorem fails). -Interpreting your $\mathcal L_2$ to consist of groups $G$ which has a filtration by normal subgroups $1< Z < H < G$, where Z is central in $G$, $G/H$ is abelian, and $H/Z$ is a simple non-abelian subquotient of a group in $\mathcal L_3$ (or equivalently $\mathcal L_1$), we will have $\mathcal L_3 \subseteq \mathcal L_2$. -With these interpretations, we can give the following response to your question: - - -$\mathcal L_1 \subseteq \mathcal L_3 \subseteq \mathcal L_2$, with $W_{E_8}$ or $SO_{2n}^+(\mathbb F_q)$, $q$ even, being examples of groups in $\mathcal L_2$ not in $\mathcal L_3$. - - -Note that a group like $GL_n(\mathbb F_p)$ of course would be in $\mathcal L_3$. Whether it is in $\mathcal L_1$ depends a bit on which $\mathbf G$ you are willing to start with. -Now, you may want to limit the class in $\mathcal L_2$ in some way to ensure that it is not larger than $\mathcal L_3$ and avoid the "pathologies". I would be surprised if there was a good way to do this, without referring back to $\mathcal L_3$: The main problem, I think, it that there is no really good "purely group theoretic" way of e.g., specifying that you only allow extensions "on top" by diagonal automorphisms (not field or graph), without getting back into the Lie theory. This is at least what goes wrong in the $SO_{2n}^+(\mathbb F_q)$, $q$ even, case. (Derek also hinted at this.) -You may also, conversely, argue that $\mathcal L_2$ is anyway too small, since it is not closed under iteratively putting things "on top" and "on the bottom"; to be "almost" of Lie type should be a bigger class... The cleanest fix for this is to go all the way and study the groups which are virtually a finite group of Lie type, i.e., has a finite index subgroup which a finite group of Lie type. Now, this class of groups is however of course just the class of virtually trivial groups, aka all finite groups... -This is not as silly as it sounds, since many deep open problems in finite group theory, such as Alperin's conjecture and its cousins, stem from the philosophy that all finite groups behave almost like finite groups of Lie type....<|endoftext|> -TITLE: The Bombieri Vinogradov Theorem restricted to moduli divisible by $k$ -QUESTION [11 upvotes]: The Bombieri-Vinogradov Theorem states that given $A>0$, there exists $B>0$ such that for $Q=\sqrt{x}\left(\log x\right)^{-B},$ we have $$\sum_{q\leq Q}\max_{y\leq x}\max_{\begin{array}{c} -a\text{ mod q}\\ -(a,q)=1 -\end{array}}\left|\psi(y;q,a)-\frac{y}{\phi(q)}\right|\ll \frac{x}{(\log x)^A}.$$ -I was wondering what happens if a restriction is put on the $q$ so that they are all divisible by some smaller integer $k$. Are there any non-trivial bounds on the average over $q$ divisible by $k$? Specifically, suppose that $k\leq Q^{1-\epsilon},$ and that there is no Siegel zero${}^{++}$ for any $\chi$ modulo $k$. Is it true that $$\sum_{\begin{array}{c} -q\leq Q\\ -k|q -\end{array}}\max_{y\leq x}\max_{\begin{array}{c} -a\text{ mod q}\\ -(a,q)=1 -\end{array}}\left|\psi(y;q,a)-\frac{y}{\phi(q)}\right|\ll_{\epsilon}\frac{1}{k} \frac{x}{(\log x)^A}.$$ -Any references would be greatly appreciated. -Thanks for your help, -${}^{++}$ Edit: As mentioned by Terence Tao in the comments, there is an issue regarding Siegel zeros modulo $k$. The original result I was asking about would give stronger bounds on the location of a possible Siegel zero modulo $k$, and for that reason it is out of reach. -I worked out a sketch/heuristic, as suggested by Terence Tao's comment, for why we have to assume that there are no Siegel zeros modulo $k$. I wrote this mainly for my own understanding, and I have added it here for anyone who is interested. I would still like to know if the result holds assuming there are no non-trivial real zeros for $\chi$ mod $k$. - -Heuristic: Suppose that $k$ is a small power of $x$. If there is an exceptional zero $\beta$ for a quadratic character $\chi$ modulo $k$ , then for every $q$ such that $k|q,$ the induced character modulo $q,$ $\chi^{\star},$ will have the same exceptional zero, and so we expect that $$\psi(x;q,a)\approx\frac{x}{\phi(q)}-\frac{\chi^{\star}(a)}{\phi(q)}\frac{x^{\beta}}{\beta}+small$$ for each $q.$ This leads us to expect that $$\sum_{\begin{array}{c} -q\leq Q\\ -k|q -\end{array}}\max_{y\leq x}\max_{a\text{ mod }q}\left|\psi\left(y;q,a\right)-\frac{y}{\phi(q)}\right|\approx x^{\beta}\sum_{\begin{array}{c} -q\leq Q\\ -k|q -\end{array}}\frac{1}{\phi(q)}\approx\frac{x^{\beta}\log x}{k},$$ however, the desired upper bound is too strong, as $$x^{\beta}\log x\ll \frac{x}{\left(\log x\right)^{B}},$$ implies that for some $C>0$, $$\beta\leq1-C\frac{\log\log x}{\log x},$$ which is stronger than the long standing bound by Siegel $$\beta\leq1-C(\epsilon)x^{-\epsilon}.$$ In the above, we have used $x$ interchangeably with $k$ for the bounds since $k$ is a small power of $x$. - -REPLY [5 votes]: Elliott has a result in this direction. Let $A>0$ and $a \geq 2$ a fixed integer. Furthermore, let $q \leq x^{1/3-\epsilon} $ be a large power of $a$. One then has that -$$ \sum_{\substack{d \leq q^{-1}x^{1/2} \log^{-A-6}(x) \\ (d,q)=1 }} \max_{(r,qd)=1} \max_{y\leq x} \left|\pi(y,qd,r) - \frac{\text{Li}(y)}{\phi(qd)} \right| \ll_{A} \frac{x}{\phi(q) \log^{A}(x) } .$$ -Taking $d=1$ recovers the Bombieri-Vinogradov theorem. Note that while Bombieri-Vinogradov doesn't contain any information about specific arithmetic progressions with modulus larger than $\log^{A}(x)$, Elliott's inequality implies that -$$\pi(x,q,r) \sim \frac{Li(x)}{\phi(q)}$$ -for $q \leq x^{1/3-\epsilon}$ that is a large power of $a$.<|endoftext|> -TITLE: Definability of Morley rank in the theory of Compact Complex Spaces(CCS) -QUESTION [7 upvotes]: Is there a counter-example showing that Morley rank in the theory of compact complex spaces (as defined by Zilber, Pillay, Moosa, ...) is not definable in families? -Given the existence of such a counterexample in DCF (constructed by Hrushovski and Scanlon; it is a family of Abelian varieties) and the similarity between two theories from the model-theoretic point of view, I am inclined to think that there should be a counter-example in CCS. But I have difficulty "translating" (at least straightforwardly) the DCF example into the CCS world since it uses some facts, such as existence of certain definable families of polarized Abelian varieties, that I am not familiar with. - -REPLY [6 votes]: Pillay and Scanlon gave an example showing Morley rank isn't definable in elementary extensions in: Compact complex manifolds with the DOP and other properties, J. Symbolic Logic, vol. 67 (2002), pp. 737–743. -In a different direction Dale Radin showed the you do have definability of dimension in the standard model in: A definability result for compact complex spaces. -J. Symbolic Logic vol. 69 (2004), pp. 241–254.<|endoftext|> -TITLE: Relationship between triangle free graphs and their minimum degree -QUESTION [5 upvotes]: Let $T$ be a triangle-free graph on $n$ vertices with minimum degree $\delta$ (which can be $0$). How does one show that $n >2\delta -1$? It seems to be true for bipartite graphs, but I cannot see how to prove it for general triangle free graphs in general. The motivation behind this question is if this is solved, I will have a much more interesting corollary to garner from it. Thank you! - -REPLY [5 votes]: In addition to the literature mentioned in the other answers, one can try some -arguments based on counting. -Let $u$ and $v$ be two of $n$ vertices in a triangle-free graph, and further assume they -are distinct and connected by an edge, with degrees $c$ and $d$ and $c$ at most $d$. -Then their neighborhoods are disjoint, so $c - 1 + d - 1$ is at most $n-2$, giving -$2\delta \leq n$. -It might be fun to recreate the $2n/5$ result: here is a start. Take an odd -cycle from a non bipartite graph with minimal cycle length. If the cycle -length is $7$ or greater, show that three vertices will produce either a triangle, -a shorter odd length cycle, or a degree at most $2n/5$. Try a similar analysis -with a $5$-cycle. Enjoy!<|endoftext|> -TITLE: Integer-distance sets -QUESTION [18 upvotes]: Let $S$ be a set of points in $\mathbb{R}^d$; I am especially interested in $d=2$. -Say that $S$ is an integer-distance set if every pair of points in $S$ is separated -by an integer Euclidean distance. - -What are examples of maximal integer distance sets? - (Maximal: no point can be added while retaining the integer-distance property - between all pairs.) - -Of course the lattice points along any one line parallel to a coordinate axis in -$\mathbb{R}^d$ constitute a countably infinite integer-distance set. -What is an example of an infinite integer-distance set of noncollinear points? -I know that Euler established that every circle contains a dense rational-distance set. -Scaling any one circle by a large common denominator provides a rich, but finite integer-distance set. For example, these five points on a circle are all separated by integer -distances: -$$ -(1221025, 0), (781456, 586092), -(439569, 586092), -(270400, 507000), -(180625, 433500) -$$ -      -I'm sure this is all known... Thanks for enlightening me! -(This is tangentially related to my earlier question, -"Rational points on a sphere in $\mathbb{R}^d$.") - -Update1. -It turns out that determining the integer-distance sets is -fundamentally open. -What is known is nicely summarized by Robert Israel and "Daniel m3." -In particular, -via the Kreisel & Kurz paper, -it is unknown (or was unknown in 2008) -whether or not there exists an 8-point -integer-distance set in $\mathbb{R}^2$, -with no three of the points collinear and no four cocircular. -Update2. -Also open is a related problem identified by Nathan Dean: -How many non-cocircular integer-distance points can be found on a parabola, -a scaling of $y = x^2$? -Nathan proved there are infinitely many sets of three such points; -Garikai Cambell proved there are infinitely many sets of four such points. -But the existence of five such points seems open. -I just learned the parabola problem from this MSE question. -Update3 (21 Jul 2013). -I ran across this just-published paper, which explores the in-some-sense obverse of the -question I asked: What are the largest point sets in $\mathbb{R}^d$ that -avoid points an integral distance apart. - -Kurz, Sascha, and Valery Mishkin. "Open Sets Avoiding Integral Distances." Discrete & Computational Geometry (2013): 1-25. (Springer link) - -Update4 (29 Nov 2014). There is a nice article at Dick Lipton's blog -on Ulam's 70-year-old un-resolved conjecture: - -If $S$ is an rational-distance set, then it is not dense in the plane. - -And that article cites the Kurz-Mishkin paper above. - -REPLY [3 votes]: RE: Daniel's and Carl's comments above, the "famous unsolved problem" is problem D20 from -Guy, R. K. Unsolved Problems in Number Theory, 2nd ed. -New York: Springer-Verlag, p. 187, 1994. -The famous unsolved problem asks about six points, the "end of the link" mentions seven points, and currently eight points is still an open question. -I think "rational distances from the 4 vertices of a unit square" is problem D19.<|endoftext|> -TITLE: Does $(\mathbb{Z}/n\mathbb{Z})^2$ ever admit a difference set when $n$ is odd? -QUESTION [9 upvotes]: A difference set of a group $G$ is a subset $D\subseteq G$ with the property that there exists an integer $\lambda>0$ such that for every non-identity member $g$ of $G$, there exist exactly $\lambda$ ordered pairs $(a,b)\in D\times D$ such that $g=ab^{-1}$. Note that $D=G$ is a difference set with $\lambda=|G|$, and so we typically only consider nontrival difference sets. -Davis showed that $(\mathbb{Z}/n\mathbb{Z})^2$ admits a nontrivial difference set when $n$ is a power of $2$. Are there any known difference sets when $n$ is odd? Perhaps cyclotomic difference sets? -As far as I know, the Paley-type construction Douglas Zare suggests in the comments (letting $D$ be the set of nonzero perfect squares in $\mathrm{GF}(n^2)$ when $n$ is prime) is only guaranteed to work when $n^2$ is $3\bmod 4$ (which never happens). However, there are hopefully weaker sufficient conditions for $n$ to satisfy, and I think the literature discusses this in the context of "cyclotomic difference sets," but I am not familiar with these results. - -REPLY [9 votes]: Such difference sets exist. There exist (nontrivial) difference sets with -$|G| = q^{d+1}[1+(q^{d+1}-1)/(q-1)]$, -$|D| = q^d(q^{d+1}-1)/(q-1)$, -$\lambda = q^d(q^d-1)/(q-1)$, -whenever $q$ is a prime power (R. L. McFarland, A family of difference sets in non-cyclic groups, JCT A, 15 (1973), pp. 1-10). More precisely, such difference sets exist in any abelian group of order $v$ which contains an elementary abelian subgroup of order $q^{d+1}$. -Take $q=7$ and $d=1$, for instance. This shows that a nontrivial difference set in $(\mathbb{Z}/21\mathbb{Z})^2$ exists.<|endoftext|> -TITLE: quasi-homomorphisms of groups -QUESTION [13 upvotes]: Suppose that $G$ is a group and $d$ is a left-invaraint metric on $G$, e.g., the word metric (provided that $G$ is finitely-generated) or distance function determined by a left-invariant Riemannian metric on $G$ (provided that $G$ is a connected Lie group). Let $\Gamma$ be another group. -A map $f: \Gamma\to G$ is called a quasi-homomorphism if there exists a constant $A$ so that for all $\gamma_1, \gamma_2\in \Gamma$ we have -$$ -d(f(\gamma_1 \gamma_2), f(\gamma_1) f(\gamma_2))\le A. -$$ -There are two sources of unbounded quasi-homomorphisms I know: - -Group homomorphisms (and their bounded perturbations). -Quasi-morphisms, i.e., quasi-homomorphisms to ${\mathbb R}$ (with the standard metric) or, equivalently, to ${\mathbb Z}$ (again, with the standard metric). (One can modify this construction by taking quasi-morphisms with values in other abelian groups, but I will not regard this as a separate source of examples.) - -There is, by now, a considerable literature on constructing quasi-morphisms, both in geometric group theory and symplectic geometry. -One can combine 1 and 2, by taking, say, compositions and direct sums/products: I am considering this as, again a trivial modification. -Question 1: Are there are other sources of quasi-morphisms which do not reduce to 1 and 2? -If this is a bit vague, here is a sub-question: -Question 2: Suppose that $\Gamma$ is higher-rank (real rank $\ge 3$) irreducible lattice (thus, it has no unbounded quasimorphisms) and $G$ is a nonelementary hyperbolic group (to exclude homomorphisms $\Gamma\to G$ with infinite image). Are there examples of unbounded quasi-homomorphisms $\Gamma\to G$? -Is there any literature on this? I could not find anything, but maybe I was looking at wrong places. -Note that the setting of quasi-homomorphisms (in the case of nonabelian targets) is very different from the one of quasi-actions, in which case there is a substantial literature and I know most of it. - -REPLY [13 votes]: As it turned out (contrary to my expectations), all quasihomomorphisms with noncommutative target groups (which are discrete with proper left invariant metric) are obtained via some natural constructions starting from homomorphisms to groups and quasihomomorphisms to abelian groups: -This is the main result of our paper with Koji Fujiwara here. -To the list of constructions I had in my question one would have to add extension from a finite index subgroup in $\Gamma$ and a lifting construction using a bounded cohomology class in $H^2(\Gamma, A)$, where $A$ is an abelian group. In particular, the answer to Question 2 is negative.<|endoftext|> -TITLE: Dehn's algorithm for word problem for surface groups -QUESTION [15 upvotes]: For some $g \geq 2$, let $\Gamma_g$ be the fundamental group of a closed genus $g$ surface and let $S_g=\{a_1,b_1,\ldots,a_g,b_g\}$ be the usual generating set for $\Gamma_g$ satisfying the surface relation $[a_1,b_1]\cdots[a_g,b_g]=1$. Dehn proved the following famous theorem. -Theorem : Let $w$ be a reduced word in the generators $S_g$ such that $w$ represents the identity in $\Gamma_g$. Then $w$ contains a subword $r_1$ of length $k \geq 2g+1$ such that there exists a word $r_2$ of length $4g-k$ with $r_1 r_2$ a cyclic permutation of $[a_1,b_1]\cdots[a_g,b_g]$. -This leads to Dehn's algorithm for solving the word problem in a surface group. Namely, start with a word $w$ which is reduced. If $w$ does not contain a subword $r_1$ as in the Theorem, then $w$ does not represent the identity in $\Gamma_g$. Otherwise, we can replace the subword $r_1$ of $w$ with $r_2$, shortening $w$. After possibly reducing $w$, we do the above again. We stop if we have reduced $w$ to the empty word. -There are many sources for Dehn's algorithm; for instance, Stillwell's book on geometric topology and combinatorial group theory gives a proof using elementary topology, and Lyndon-Schupp give a proof based on small cancellation theory. However, many sources say that Dehn's original proof used hyperbolic geometry. I've heard people tell me that he studied the usual tessalation of the hyperbolic plane by regular $4g$-gons and applied something like Gauss-Bonnet; however, I've never managed to reconstruct this kind of proof. Can someone either spell it out or give a (modern) reference? - -REPLY [4 votes]: In the paper "The combinatorial structure of cocompact discrete hyperbolic groups(here)", Cannon gave a proof of conjugacy problem for surface groups using hyperbolic geometry (Theorem 6). It's the the same proof mentioned by Stillwell in the second paragraph of his answer.<|endoftext|> -TITLE: Where does the deterministic simulation of non-deterministic ω-Turing machines fail? -QUESTION [6 upvotes]: An $\omega$-Turing machine is just a usual Turing machine $T=(Q,\Sigma,\Gamma,\delta,q_0,F)$ where $Q$ is the finite set of states, $\Sigma$ is the input alphabet, $\Gamma\supset\Sigma$ is the tape alphabet, $\delta$ is is the transition relation, $q_0$ is the initial state and $F\subset Q$ is the set of accepting states. We will consider a special acceptance condition on $\omega$-words (elements of $\Sigma^\omega$): The language $L(T)\subset\Sigma^\omega$ recognised by $T$ is the set of all $\omega$-words such that—when the initial configuration is given by this word–there exists a run of $T$ such that every input token is reald only finitely many times and only states in $F$ get visited. An $\omega$-Turing machine is called deterministic iff $\delta$ is a function. The condition that an accepting must not visit any tape position infinitely often is non-essential in the non-deterministic case because we can transform every non-deterministic $\omega$-Turing machine into an equivalent one where every run is non-oscillating. -In 1978 Cohen and Gold have proved that non-deterministic $\omega$-Turing machines are strictly more powerful than deterministic ones. The argument goes as follows: They consider a larger class of deterministic $\omega$-Turing machines with more general acceptance conditions which can all be translated into equivalent non-deterministic $\omega$-Turing machines as defined above, but the languages recognisable by these machines are closed under complementation. However, using a diagonalisation argument (using a simulation of non-deterministic $\omega$-Turing machines) they prove that the general class of languages recognisable by $\omega$-Turing machines is not closed under complementation. -Now for me it is not clear why the standard simulation of non-deterministic Turing machines by deterministic ones does not work, let me sketch my approach, which must be wrong: We could try to simulate the non-deterministic machine step-by-step by computing all possible next configurations (where the states are still in $F$). The possible configurations can get bigger and bigger, but their number always remains finite. If this deterministic machine has an accepting run then by König’s lemma (which should be applicable because of the simple acceptance condition, for Büchi acceptance it would not work because there can be arbitrary large gaps between to occurences of an accepting state) in the simulated non-deterministic one there should be an accepting run, too. I think I am missing some subtle point, maybe related to the non-oscillation condition (?), but it might also be an obvious detail in the definition. Could anybody clarify that? - -REPLY [5 votes]: You say we can remove the condition that a run read every input only finitely many times, but I don't think that's so. As you noted, König's lemma shows that acceptance is a $\Pi^0_1$ property if we remove that condition. That means that the language of such a machine is a $\Pi^0_1$-class. -On the other hand, if we retain that condition, we can build a machine that accepts Fin, the set of all infinite binary strings with only finitely many 1s. Simply make a machine that scans right through the input until it sees a 1, then it turns that 1 into a 0, runs back to the beginning of the input and repeats. Since Fin is a properly $\Sigma^0_2$-class, this shows that we can achieve strictly more by retaining the condition. -I looked at the paper you linked. Your definition of accepting is what the authors call 1'-accepting, while their Theorem 8.6, which I believe you were referring to when you said we could remove the condition, is about 3-accepting. Now, the authors do show that every 3-accepting non-deterministic machine can be simulated by a 1'-accepting non-deterministic machine, but the 1'-machine they build is explicitly oscillating. -So you claim that any non-oscillating 1'-accepting non-deterministic machine can be simulated by a non-oscillating 1'-accepting deterministic machine. I'm fairly certain your argument is correct. There is no contradiction, because the proof that non-determinism is stronger than determinism was for 3-accepting machines; 1'-accepting non-oscillating machines are less general. -Note also: the authors' proof that non-determinism is strictly stronger than determinism is for the special case of non-oscillating deterministic machines. If you allowed oscillating deterministic machines, I think you could simulate non-deterministic machines by deterministic machines without too much work.<|endoftext|> -TITLE: Origin of the term Riesz Basis -QUESTION [8 upvotes]: The term Riesz basis is in wide use for the image of an orthonormal basis in a Hilbert space under a bounded invertible map (there are lots of equivalent definitions). But I've been unable to find any papers by either Riesz on this subject. Does anyone know a reference that justifies the name or otherwise know of a first use or something that sheds light on the origins of this term? - -REPLY [5 votes]: According to the article Riesz system in the EoM, some early references that use this term are from N.K. Bari (or Bary). The 1946 reference seems hard to locate online, but the 1951 reference in Russian can be. I can confirm that she uses the terms Bessel system, Hilbert system, Fischer-Riesz system and Riesz basis without citing any other reference, except referring obliquely to her earlier work. It is conceivable that these terms were introduced by Bari. Also, in the first line of this 1951 comment by Gel'fand, he attributes the term Riesz basis to her.<|endoftext|> -TITLE: Contractively complemented subspaces of $c_0(I)$ -QUESTION [5 upvotes]: Does every contractively complemented subspace of $c_0(I)$ is isometric to $c_0(J)$ for some $J\subseteq I$? -May be someone has a counterexample? - -REPLY [2 votes]: As you may know already, every contractively complemented subspace of $\ell_1(I)$ is isometric to $\ell_1(J)$ for some $J \subset I$ --- see for example Semadeni, Banach Spaces of Continuous Functions, 1971, p. 488. This is just the dual of your desired proposition. If $X$ is a contractively complemented subspace of $c_0(I)$ by a projection $P$, then the adjoint $P^*$ (as in Bill's answer above) is an isometric embedding of $X^*$ into $\ell_1(I)$ such that $P^*(X)$ is contractively complemented by the restriction operator, so that $P^*(X)$ is isometric to some $\ell_1(J)$. Can you then argue by w*-continuity of the isometry that $X$ must be isometric to $c_0(J)$?<|endoftext|> -TITLE: Reverse mathematics of Hilbert's Theorem 90 -QUESTION [13 upvotes]: What is known, and what is published, on the reverse mathematics of the nest of results called Hilbert's Theorem 90? - -REPLY [11 votes]: It is provable in RCA0 that if $K$ is a finite Galois extension of $F$ with Galois group $G$ then $H^1(G,L^\times)$ is trivial (every cocycle $a:G \to L^\times$ is a coboundary). -The finite Galois theory in RCA0 was discussed by Friedman, Simpson and Smith in Countable algebra and set existence axioms [APAL 25 (1983), 141–181; MR0725732]. The executive summary is that finite Galois theory works fine in RCA0. -The usual proof of Theorem 90 works in RCA0. The linear independence of characters is provable using only $\Sigma^0_1$-induction since one can check linear dependence by testing at a fixed finite basis of $L$ over $K$. It follows that if $a:G \to L^\times$ is a cocycle then there is a $x \in L^\times$ such that $y = \sum_{\sigma \in G} a(\sigma)\sigma(x)$ is nonzero. Standard algebraic manipulations then show that $a(\sigma) = y/\sigma(y)$, i.e. $a$ is a coboundary. -Infinite Galois extensions are more subtle as discussed in my paper with Jeff Hirst and Paul Shafer. For Theorem 90, it seems that the issue is a definitional one more than anything else. -If $K$ is an infinite Galois extension of $F$ such that $F$ is a subset of $K$ (as opposed to being merely embeddable in $K$) then the Galois group $G$ of $K$ over $F$ is a separably closed subgroup of permutations of $K$. Then $G$ can be represented as a complete separable metric space in the usual manner and one can make sense of the idea of a continuous cocycle $a:G \to L^\times$. In that scenario, RCA0 does prove that $H^1(G,L^\times)$ is trivial since every continuous cocycle is a lifting of a cocycle from a finite Galois subextension. -In the general case where $F$ is merely embeddable in $K$, several issues arise. Galois theory still makes sense over WKL0 but the Galois group $G$ is not necessarily separably closed. It is unclear what a continuous cocycle would be in that case. So the main issue is to define $H^1(G,L^\times)$ before proving anything about it. Note that this case is impossible in ACA0 since we can always comprehend the range of an embedding. So, assuming ACA0, $H^1(G,L^\times)$ is always well defined and always trivial.<|endoftext|> -TITLE: Failure of diamond at large cardinals -QUESTION [14 upvotes]: What is known about the failure of $\Diamond_{\kappa}$ (diamond at $\kappa$) for $\kappa$ (the least) inaccessible, (the least) Mahlo and (the least) weakly compact. -Remark. The problem of forcing the failure of diamond at a weakly compact cardinal is solved recently by Gitik's student Omer Ben-Neria (see The failure of diamond at large cardinals and GCH). This solves a longstanding open problem. -Does anyone know the main ideas of the proof? - -REPLY [10 votes]: The failure of diamond at large cardinals is an interesting and -difficult problem. Let me just mention a few references to -supplement those mentioned by Andres. -Much of the interest in diamond principles at large cardinals lies -in the case of $\Diamond_\kappa(\text{REG})$, and in this case -somewhat more is known than in the general case. - -Woodin proved the consistency of the failure of -$\Diamond_\kappa(\text{REG})$ at a weakly compact cardinal. -Kai Hauser extended this result to the -case of indescribable cardinals. -Mirna Džamonja and I extended this result further to the case of -strongly unfoldable cardinals, which can be viewed as a -transfinite extensions of the indescribability hierarchy. Our -paper is M. Džamonja, J. D. Hamkins, Diamond (on the regulars) can fail at any strongly unfoldable cardinal, Ann. Pure Appl. Logic, vol. 144, iss. 1-3, pp. 83-95, 2006.. - -To my way of thinking, the problem is deeply connected with the -Laver diamond principle, which I introduced in A class of strong diamond principles. This develops the concept of the Laver diamond for a large -variety of large cardinal concepts, where one has a Laver function $\ell:\kappa\to V_\kappa$, such that one can obtain the Laver anticipation property $j(\ell)(\kappa)=x$ for a variety of sets $x$ with embeddings $j$ witnessing that large cardinal property. (Hellsten had around the same time developed independently the case of the Laver diamond for weakly -compact cardinals, so there is some overlap between Hellsten's work -and my paper.) One of the interesting things that Mirna and I -proved was that one can have an ordinal-anticipating Laver -function for strong unfoldability, even when there is no -set-anticipating Laver function. And there can be pseudo Laver -functions for strong unfoldability, that anticipate every ground -model set on a stationary set, but still -$\Diamond_\kappa(\text{REG})$ fails. -There are some extremely interesting open questions about this, -such as: - -Can there be a $\theta$-supercompact cardinal having no -$\theta$-supercompactness Laver function? -Can there be an uplifting cardinal with no uplifting Laver -function?<|endoftext|> -TITLE: Trimble's theory of operads -QUESTION [6 upvotes]: I have two questions about this note by T. Trimble, which I warmly invite to answer. -First question -I am trying to prove that these two functors are mutually inverses: - -The functor $U\colon \mathbf C\to [\mathbf{Sets}^\mathbf{Fin},\mathbf C]$ which sends $c\in\bf C$ to the functor $F\mapsto \int^nF(n)\cdot c^n$ (given two cmc categories (see Trimble's note), we denote $[\bf C,D]$ the category of product preserving and cocontinuous functors $\bf C\to D$); -The functor $V\colon [\mathbf{Sets}^\mathbf{Fin},\mathbf C]\to \bf C$ which evaluates a functor $\Phi\colon \mathbf{Sets}^\mathbf{Fin}\to \mathbf C$ in $\hom(1,-)\colon \bf Fin\to Sets$. - -It is quite obvious that $VU(c)\cong c$ since -$$ -c\stackrel{VU}{\longmapsto}\int^n \hom(1,n)\cdot c^n\stackrel{\text{Yoneda}} {\cong}c^1=c -$$ -I am stuck in proving that the composition $U\circ V$ is isomorphic to ${\rm id}_{[\mathbf{Sets}^\mathbf{Fin},\mathbf C]}$: following Trimble's notations -$$ -\Phi\stackrel{UV}{\longmapsto} \left(F\mapsto \int^n F(n)\cdot \Phi(\hom(1,-))^n\right) -$$ -now $\int^n F(n)\cdot \Phi(\hom(1,-))^n$ can be simplified since $\Phi$ is product preserving, but I'm not able to find out why it should be equal to $\Phi(F)$. -Second question -Unless a mysterious duality is involved, I think that this construction gives a co-monad. Where am I wrong? - -This question was originally posted on math.SE; Martin Brandenburg pointed out that the equivalence ${\rm Cocont}(\mathbf{Sets}^\mathbf{Fin},\mathbf C)\cong {\rm Fun}(\mathbf{Fin}^\text{op},\mathbf C)$ restricts to an equivalence between the sole product-preserving functors, which is sort of the reason this equivalence exists. The fact is that even if it works, I think that the aim of Trimble's note was to give an explicit definition of the functors realizing the equivalence: Instead of finding a formal reason for the equivalence to hold (which is exactly the Universal Property of PSh(C), as Martin Brandenburg said), I would rather know how to manipulate that coend in order to get $\Phi(F)$ back. - -REPLY [11 votes]: For the first question, we have natural isomorphisms -$$U V(\Phi)(F) = \int^n F(n)\cdot \Phi(\hom(1,-))^n \cong \int^n F(n)\cdot \Phi(\hom(1,-)^n) \cong \int^n \Phi(F(n)\cdot \hom(1,-)^n)$$ -$$\cong \Phi(\int^n F(n)\cdot \hom(1,-)^n) \cong \Phi(\int^n F(n) \cdot \hom(n, -)) \cong \Phi(F)$$ -where the first isomorphism uses the fact that $\Phi$ preserves products, the second and third use cocontinuity of $\Phi$ (I think Buschi Sergio pointed out that the notation $S \cdot G$ indicates a coproduct of copies of $G$ indexed over a set $S$), and the last uses the Yoneda lemma. Being natural in $F$, this shows we have an isomorphism $UV(\Phi) \cong \Phi$ (natural in $\Phi$). -It might help to compare Kelly's theory of operads, where a launching point is a universal property of Day convolution: - -If $M$ is a symmetric monoidal category, then the Yoneda embedding $y: M \to Set^{M^{op}}$ is a symmetric monoidal functor (where the codomain has been equipped with the Day convolution product induced from the product on $M$), and is universal among symmetric monoidal functors from $M$ to symmetric monoidally cocomplete categories (meaning the monoidal product preserves colimits in each factor). - -Here what we are essentially doing is making the observation/calculation that if the monoidal product on $M$ is the cartesian product, then the induced Day convolution is also the cartesian product, and if we replace "symmetric monoidal" by "cartesian monoidal" throughout the displayed statement, the result again holds. Then, parallel to what Kelly does, we observe that $\text{Fin}^{op}$ (the category opposite to finite sets) is the free cartesian monoidal category on one object (i.e., on the terminal category), analogous to the fact that the finite permutation category $\mathbf{P}$ is the free symmetric monoidal category on one object. -So the idea is just to develop a little theory of cartesian operads parallel to Kelly's approach to ordinary permutative operads, and then observe the essential equivalence between cartesian operads and (finitary) Lawvere theories. -By the way, I wouldn't call this "my theory" -- it's my feeling that all the observations I'm making are already well-known to a lot of people, but it seems to be very hard to find a source where it's all written down. More on this in a moment. - -For the second question, it might help to remember that on $Cat$ (2-category of small categories, say), the functor $X^-: Cat^{op} \to Cat$ for a given category $X$ reverses the direction of 1-cells but preserves the direction of 2-cells. (This is an easy exercise.) So given say a 2-cell $F G \to H$, we would get an induced 2-cell $X^G X^F \to X^H$. Similarly, if $M$ is a monad on $C$ with multiplication $m: MM \to M$, we get an induced monad (not comonad) on $X^C$ with multiplication $X^m: X^M X^M \to X^M$. -On similar grounds, given a cartesian operad $M$, inducing a (cocontinuous product-preserving) monad $- \odot M: \text{Set}^{\text{Fin}} \to \text{Set}^{\text{Fin}}$, we get an induced monad -$$[- \odot M, C]: [\text{Set}^{\text{Fin}}, C] \to [\text{Set}^{\text{Fin}}, C]$$ -for any cartesian monoidally cocomplete $C$. - -Finally, I'd like to say what spurred me to start writing the note referred to in the OP. It was really this question of Martin Brandenburg, where my answer involved a coend formula for monads $T$ based on a Lawvere theory $\theta$. Martin asked for a proof of the coend formula (which he said he knew in the case where $C = \text{Set}$, but not for general cartesian monoidally cocomplete $C$). -I had tried to explain what was behind the coend formula in a string of comments that appealed to an analogy with ordinary permutative operads -- that there is a completely parallel development for cartesian operads and Lawvere theories. Based on Martin's lack of response to that comment string, it seems my arguments didn't convince him -- and unfortunately I couldn't refer to the literature because I don't know of a place where this is spelled out. (Even though I believe it has to be well-known to people who have studied Kelly's work on operads. There's an article by Hyland and Power which comes pretty close to being a reference for what Martin wants, but not quite.) -Thus, I began writing all this stuff down, out of a belief that it's all very soft and conceptual and ought to be recorded. I just hadn't announced it to anyone at MO yet. -Now, spurred by yet another MO question, I may get back to that note (which was left in a slightly incomplete state)!<|endoftext|> -TITLE: Analytic uniformization -QUESTION [7 upvotes]: Suppose I am given a subset of $2^\omega\times\omega^\omega$ of some bounded Borel rank. Can I get an analytic uniformization of this set? - -REPLY [5 votes]: There is an arithmetical set $A\subseteq 2^{<\omega}\times \omega^{<\omega}$ so that for any $x\in 2^{\omega}$, $A(x)=\{\sigma\mid \exists n(x|n,\sigma)\in A\}$ is an $x$-recursive tree which has an infinite path but no infinite path hyperarithmetic in $x$. -Now let $B$ be an arithmetical set so that $(x,y)\in B$ if and only if $y$ is an infinite path through $A(x)$. $B$ has the following property: -(1). For any $x$, there is some $y$ so that $(x, -y)\in B$; -(2) No analytic function uniformizing $B$. Suppose not, then there is a real $z$ and $\Sigma^1_1(z)$-function $f$ uniformizing $B$. Let $x\geq_h z$, then $f(x)\leq_h x\oplus z\leq_h x$, a contradiction.<|endoftext|> -TITLE: The singular values of the Hilbert matrix -QUESTION [15 upvotes]: The $n\times n$ Hilbert matrix $H$ is defined as follows -$$H_{ij} = \frac{1}{i+j-1}, \qquad 1\leq i,j\leq n$$ -What is known about the singular values $\sigma_1 \geq \cdots \geq \sigma_n$ of $H$? -For example, it is known that the matrix is very ill-conditioned, i.e., [1] -$$\dfrac{\sigma_1}{\sigma_n} = \mathcal O \left( \frac{1+\sqrt{2})^{4n}}{\sqrt{n}} \right)$$ -But are there estimates for $\sigma_k$ for $2\leq k\leq n-1$? Or is there a bound on $1\leq k\leq n$ such that $\sigma_k > \epsilon$ for some $\epsilon>0$? I am interested in this problem because I would like to know the numerical rank of $H$. - -[1] J. Todd, "The condition of the finite segments of the Hilbert matrix." Contributions to the solution of systems of linear equations and the determination of eigenvalues, 39 (1954), pp. 109-116. - -REPLY [7 votes]: I came back to this a few months ago and I can now answer my own question. I hope it is appropriate to answer my own question given the length of time. -Bernhard Beckermann and I just submitted a paper [1] that shows that if $AX-XB = F$ with $A$ and $B$ normal matrices, then the singular values of $X$ can be bounded as -$$ -\sigma_{1+\nu k}(X) \leq Z_k(\sigma(A),\sigma(B))\|X\|_2, \qquad \nu = {\rm rank}(F), -$$ -where $\sigma(A)$ and $\sigma(B)$ are the spectra of $A$ and $B$, respectively, and $Z_k(E,F)$ is the so-called Zolotarev number. More precisely, -$$ -Z_k(E,F) = \inf_{r\in\mathcal{R}_{k,k}} \frac{\sup_{z\in E}|r(z)|}{\inf_{z\in F}|r(z)|}, -$$ -where $\mathcal{R}_{k,k}$ is the space of degree $(k,k)$ rational functions. This is called the third Zolotarev problem. -In the case of the Hilbert matrix, let $A = {\rm diag}(1/2,3/2,\ldots,n-1/2)$ and $B = -A$ then -$$ -AH - HB = \begin{bmatrix}1\\\vdots\\1 \end{bmatrix}\begin{bmatrix}1&\cdots&1 \end{bmatrix}. -$$ -Hence, $\nu=1$ and -$$ -\sigma_{1+k}(H) \leq Z_k([-n+1/2,-1/2],[1/2,n-1/2])\|H\|_2. -$$ -Fortunately, bounds on the Zolotarev numbers are well-studied (in [1] we give asymptotically tight bounds) and we conclude that -$$ -\sigma_{1+k}(H) \leq 4\left[{\rm exp}\left(\frac{\pi^2}{2\log(8n-4)}\right)\right]^{-2k}\|H\|_2. -$$ -If we seek the numerical rank of $H$ for some $0<\epsilon<1$, i.e., $k = {\rm rank}_\epsilon(H)$ if $k$ is the smallest integer such that $\sigma_{k+1}(H)\leq \epsilon\|H\|_2$, then from the bound on the singular values of $H$ we conclude that -$$ -{\rm rank}_\epsilon(H) \leq \Bigg\lceil \frac{\log(8n-4)\log(4/\epsilon)}{\pi^2} \Bigg\rceil. -$$ -This shows that Hilbert matrices are not only exponentially ill-conditioned with $n$, but its singular values decay geometrically to zero too. This methodology extends to any matrix with displacement structure such as Pick, Cauchy, Loewner, real Vandermonde, and positive definite Hankel matrices. For more details, see [1].<|endoftext|> -TITLE: Partition of a group into small subsets -QUESTION [6 upvotes]: A nonempty subset $S$ of a group $G$ is called small if there is an infinite sequence of elements $g_n$ in $G$ such that the translated sets $g_nS$ are pairwise disjoint. -Question: Is there a group which is a (disjoint) union of three small subsets, but it is not a union of two small subsets? -Remark: Such a group must be non-amenable (clear) and must not contain a copy of the non-abelian free group (in fact it is an exercise to see that the groups which are a union of two small sets are exactly those containing the free group). -Bonus question: Is every non-amenable group a finite union of small subsets? - -REPLY [3 votes]: I believe the answer to the bonus question is negative, that is, there exist non-amenable (finitely generated) groups which are not representable as a union of finitely many small subsets. This is directly related to the problem of constructing non-amenable groups -with arbitrarily large Tarski numbers discussed here. Here is a sketch of the -argument. -Using a variation of the method that Mark Sapir has described in the above post, one can construct a finitely generated non-amenable group G with the following property: -For every n there is a finite index subgroup G(n) of G such that every n-generated subgroup of G(n) is finite. -Now suppose such G is a union of k small subsets S_1,..., S_k. Then each finite index subgroup H of G is also a union of k subsets, namely the intersections of S_1,...,S_k -with H, which are still small (as subsets of H). This easily implies that H must have a non-amenable subgroup generated by at most k^2 elements, which is a contradiction.<|endoftext|> -TITLE: When are maps between topological spaces homotopic? -QUESTION [7 upvotes]: I wanted to ask if there is any known mehod to quantify 'how many' homotopy classes of maps there are between two given topological spaces $X$, $Y$ (CW-complexes, say). -So far I had the following idea: -If two maps $f,g:X\rightarrow Y$ are homotopic, then they induce the same maps on all homotopy groups. Maybe there's a class of spaces for which the converse is also true. Then I would have a characterization via Homomorphisms between homotopy groups. -Does anybody know if there is a class of spaces for which this is true? - -REPLY [5 votes]: Vidit's comments are relevant to a paper of Graham Ellis -Homotopy classification the J.H.C. Whitehead way. -Exposition. Math. {6} (1988) 97--110. -The more general result given here is that if $C$ is a crossed complex and $X$ is a CW-complex then there is a bijection of homotopy classes -$$[X,BC] \cong [\Pi X_*, C]$$ -where $BC$ is the classifying space of the crossed complex, and $\Pi X_*$ is the fundamental crossed complex of the skeletal filtration of $X$. A cubical version of the proof is given in the EMS Tract Vol 15 Nonabelian Algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids, where a pdf is available. This book also contains our generalisation of the work in CHII on the relation between crossed complexes and chain complexes with a group of operators. He writes in essence that the former have better realisation properties, and the latter are better for calculation. -Note that the construction $BC$ generalises Eilenberg-Mac Lane spaces, including the local system case. -This has been generalised to the equivariant case, -Brown, R., Golasinski, M., Porter, T. and Tonks, A. -Spaces of maps into classifying spaces for equivariant - crossed complexes. II. The general topological group case. -$K$-Theory 23 (2001) 129--155. arxiv 9808111 -There is a lot in Whitehead's CHII ! One example is his result on free crossed modules. I wrote up his proof ([30] in my publication list) and the referee wrote that: "The theorem is not new, the proof is not new, and the paper should be published." This theorem is sometimes stated in texts on algebraic topology but rarely proved. Note that Whitehead's "homotopy systems" are our "reduced free crossed complexes". -To go back to the original question, it must be said that the homotopy groups are but a pale shadow of the homotopy type, and indeed crossed complexes give only a "linear" perspective, though linear methods are often useful as an approximation. -The fist displayed equivalence is part of a weak equivalence -$$ B(CRS(\Pi X_*, C )) \to (BC) ^X$$ -which does give information on homotopies in this case. See the above mentioned book, Theorem 11.4.19.<|endoftext|> -TITLE: Surgery diagram for the Seifert-Weber space -QUESTION [11 upvotes]: In every reference I see, the Seifert-Weber space is presented as an identification space (specifically identify opposite faces of a dodecahedron by a 3/10ths twist). -What I can't seem to find is a surgery diagram for this manifold. From its homology we know it doesn't arise as surgery on a knot. Its volume is approximately 11.1991. Checking with SnapPy there are a lot of candidate links in this range: -In [14]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:13.0])) -Out[14]: 1103 -In [15]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:12.0])) -Out[15]: 365 -In [16]: len(filter(lambda x: x.num_cusps() >= 3,HTLinkExteriors[11.19:12.5])) -Out[16]: 620 - -so my attempts at guessing have been unsuccessful. -One could calculate a surgery diagram by-hand from the Heegaard diagram that comes from the identification space and then try to simplify the result. This seems daunting to do by hand, but I'm also unaware of software for the task. -Edit -I accepted Daniel's answer before working through the details. The process certainly gives the Seifert-Weber space a surgery on a manifold. The problem is that the branched-covers of $S^3$ over the trivial 2-component unlink are not themselves link complements, as in the case of a 1-component unlink, so the method in Rolfsen doesn't yield surgery on a link in $S^3$, but somewhere else. That somewhere else has non-separating 2-spheres in it, illustrated by this picture: - -The cut disks are indicated by red and blue, the two light grey wireframe balls are the complements of the two knot components after cutting along the disks. As long as at least two of these gadgets are glued together light-to-dark, the yellow curve will pierce the solid grey 2-sphere only once. - -REPLY [7 votes]: So Gates Rudd, Dunfield, and Obeidin algorithmically obtained a surgery diagram in their preprint Computing a link diagram from its exterior. Here's Figure 21 from Section 9.2. -Actually I had made one back in 2019 based on the earlier answers of Moskovich and Ruberman and Agol's comment. For whatever reason, I had forgotten about it. Some simplification could be done, but then 5-fold reverberation gets lost. - -Edit: So I went ahead and made a Sketches of Topology post giving some details about how this diagram was obtained.<|endoftext|> -TITLE: Algebraic numbers and the complex projective line minus three points -QUESTION [7 upvotes]: Deligne's monograph “Le Groupe Fondamental de la Droite Projective Moins Trois Points” begins by remarking that when X is the projective line over the complex numbers, minus three points: "every finite cover of X can be described by equations with algebraic number coefficients." -see http://www.math.ias.edu/files/deligne/GaloisGroups.pdf -Is the proof something like the Hilbert irreducibility theorem? -I mean is it like the following? For any cover given by a complex polynomial in two variables, the finitely many complex coefficients can be regarded as variables which can then be specialized to algebraic values which meet whatever rational polynomial conditions as the originals did while avoiding finitely many others, to give an isomorphic cover. Or will I waste my time if I try to formulate such conditions? - -REPLY [3 votes]: The comments are all correct. Deligne is citing one half of what is called Belyi's theorem, though experts feel this half was actually clear much earlier due to Weil. Weil used a version of the insight described above by JSE. B Kock gives a nice exposition in "http://arxiv.org/abs/math/0108222".<|endoftext|> -TITLE: Non-iterative modal logics -QUESTION [6 upvotes]: Let S be a propositional modal logic system (extension of K, or even E) with a single unary modal operator and defined by a single non-iterative axiom (i.e. of modal degree 1). -Is it true that for such a system S, every theorem that is a non-iterative formula has a proof consisting only of non-iterative formulas? -If yes, can anyone provide a reference where I can find this result? -If no, can you provide a counterexample? -EDIT: Adding clarification of where this comes from. -Denote by P the property in question. A slightly stronger property P' requires all the formulas in the proof to also not exceed N propositional variables, where N is the maximum between the number of variables in the non-iterative axiom and the number of variables in the non-iterative theorem to prove. So basically we are trying to prove a theorem without increasing the modal degree nor the number of propositional variables during the proof (with a possible exception for PC formulas; for example one may use $p \rightarrow p \lor q$ to infer that $\square p \rightarrow \square p \lor \lozenge p$ and this would not count as an increase in the number of variables). -It looks like for a system defined by a combination of iterative and non-iterative axioms these properties do not hold (universally). For example, see Hughes and Cresswell pp 58, in order to prove $\square p\rightarrow \square \square p$ in S5, one must use formulas of modal degree 3 during the proof. The same thing appears to be the case when proving that T+B+S4 implies S5. I also suspect that increasing the number of variables does not help in avoiding the increase in modal degree in these cases, therefore the distinction between P and P'. -However, P and even P' seem to hold for non-iterative systems. Given that these systems are relatively simple (i.e. they have FMP and a number of other "nice" properties), I was thinking that P and/or P' may be known results, but I cannot find them. -An algebraic explanation would be interesting for me too. (But I do not see a tag for algebraic-logic.) - -REPLY [4 votes]: $\let\ET\bigwedge\let\LOR\bigvee\let\EQ\Leftrightarrow$The property is true for extensions of K (i.e., normal modal logics). You didn’t really describe the proof system you are interested in, but based on the discussion in the question, I will assume it is a Hilbert-style proof system with the rules of modus ponens and necessitation, and substitution instances of a fixed set of axioms, including a complete axiomatization of classical propositional logic, and the distributivity axiom of K. (It seems that you also allow substitution to be used as a rule. I prefer the formulation I gave with no substitution rule, but axioms schemata closed under substitution, because it has nicer structural properties. Of course, one can simulate schemata in the other system by an explicit application of substitution on the base form of the axiom.) -First, note that the “stronger property” putting a bound on the number of variables is trivially equivalent to the original formulation: given a proof of a formula $A$, you can uniformly substitute a fixed formula (e.g., $\bot$ if it’s in the language, or a variable from $A$) in all formulas in the proof for each variable which does not occur in $A$, obtaining a proof of $A$ which only involves variables from $A$. -Let me define a Boolean substitution to be a substitution $\sigma$ such that $\sigma(p)$ is a Boolean formula (i.e., $\Box$-free) for every variable $p$. - -Theorem: Let $S\cup\{A\}$ be a set of formulas of modal degree $\le1$. If $\vdash_{\mathrm K\oplus S}A$, then $A$ has a derivation using - -(degree-$1$ instances of) classical propositional tautologies, and the rule of modus ponens, -axioms $\Box B$, where $B$ is a $\Box$-free classical tautology, and Boolean substitution instances of $\Box(p\to q)\to(\Box p\to\Box q)$, -Boolean substitution instances of $B\in S$. - -Moreover, all formulas in the proof use only variables occurring in $A$. - -Proof: -Assume that the conclusion fails. Unless stated otherwise, all formulas below are required to use only the variables from $A$. By the completeness of classical propositional logic, there exists a Boolean assignment $v$ to variables and boxed Boolean formulas such that $v(A)=0$, but $v(B)=1$ for every axiom of type (2), (3). We will construct a Kripke model based on an $S$-frame where $A$ is false. -Let $\{p_0,\dots,p_{n-1}\}$ be an enumeration of all variables occurring in $A$, and if $e$ is any Boolean assignment $e\colon\{p_i:i -TITLE: How does the work of a pure mathematician impact society? -QUESTION [48 upvotes]: First, I will explain my situation. -In my University most of the careers are doing videos to explain what we do and try to attract more people to our careers. -I am in a really bad position, because the people who are in charge of the video want me to explain what a pure mathematician does and how it helps society. They want practical examples, and maybe naming some companies that work with pure mathematicians, and what they do in those companies. All this in only 5 or 10 minutes, so I think that the best I can do is give an example. -Another reason that I am in a bad position: In my University we have the career "Mathematical engineering" and they do mostly applications and some research in numerical analysis and optimization. (*) -I know that pure mathematics is increasing its importance in society every year. -Many people in my country think that mathematics has stagnated over time and now only engineers develop science. -I think that the most practical thing I can do is give some examples of what we are doing with mathematics today (since 2000). -If some of you can help me, I need the following: - -A subject in mathematics that does not appear in (*). Preferably dynamical systems, logic, algebraic geometry, functional analysis, p-adic analysis or partial differential equations. -A research topic in that subject. -Practical applications of that research and the institution that made the application. - -Extra 1. If you know an institution (not a University) that contracts with pure mathematicians and you know what they do there, please tell me also. -Extra 2. If you have a very good short phrase explaining "what a mathematician does" or "how mathematics helps society" I will appreciate it too. -Thanks in advance. - -REPLY [3 votes]: Of course there are many companies hiring pure mathematicians. But they do not work as pure mathematicians, they become programmers, consultants, or statisticians… That is the truth and prospective students deserve to know it. - -REPLY [2 votes]: Clifford algebras have a wide range of applications inside and outside of Mathematics, including Differential Geometry, Computer Vision, Robotics, Theoretical Physics, Computer Science. See for instance: http://www.amazon.com/Geometric-Computing-Clifford-Algebras-Applications/dp/3642074421<|endoftext|> -TITLE: $K_0$ group of graph underlying an approximately finite (AF) C* algebra -QUESTION [7 upvotes]: Say we have an AF C* algebra $A$ described by some Bratteli diagram $E$. If $M_\infty (A)=\displaystyle{\lim_\rightarrow M_n(A)}$ and $P(A)$ are the projections in this algebra, we know that $K_0(A)^+=P(A)/\sim$ where $\sim$ is the von Neumann equivalence relation, and the Grothendieck group construction gives us $K_0(A)$. -My question is: is there an analogous description of projectors and equivalence at the level of the graph $E$ itself which yields the same $K_0$ group as the AF algebra it describes? I have thought about this question in terms of the path algebra of $E$, but really don't know much about this area. - -REPLY [5 votes]: Kerov's thesis "Asymptotic representation theory of the symmetric group" has a nice discussion of these ideas. I think I can remember an answer to your question in that book, but I'll have to check later. -As you probably know, the ordered abelian group $K_0(A)$ is a complete invariant of the algebra $A$. Here is a construction that recovers not only $K_0(A)$, but the positive cone as well. -Consider a certain class of integer-valued functions on $E$ which obey the same branching rule as the dimension function: -$$ f : E \longrightarrow \mathbb{Z}$$ -$$ \sum_{\lambda \nearrow \Lambda} f(\lambda) = f(\Lambda),$$ -Where the notation $\lambda \nearrow \Lambda$ indicates that $\Lambda$ covers $\lambda$ in the Bratteli diagram. -We don't actually require $f$ to be defined on all of $E$, just almost everywhere (past some finite stage). The condition is only required to hold in cases where $f$ is defined. -Such functions form a group under addition. The nonnegative cone is functions which are nonnegative in their domain. I think it is straightforward to check that this construction produces $K_0(A)$ as an ordered abelian group.<|endoftext|> -TITLE: $\infty$-categorical interpretation of type theory -QUESTION [18 upvotes]: One can read at several places that Martin-löf type theory should be the internal language of a locally Cartesian closed infinity category, and that the univalence axiom should distinguished infinity topos among locally Cartesian closed infinity categories. This is generally presented as a commonly accepted fact but not proven yet, for example, in the introduction of the HoTT book, one can read "in particular questions of coherence and strictness remains to be addressed" -Roughly,I would like to understand what are the difficulties for proving this (it seems to be an extremely natural result), or more precisely what does mean the sentence I quoted from the book ? -Thank you ! - -REPLY [12 votes]: More generally, the issue with such interpretation is that substitution in type theory is interpreted by pullback in category theory, and substitution in ordinary type theory preserves all type-theoretic operations strictly and functorially, so we need some model for an $(\infty,1)$-category in which pullback has these properties. This is the purpose of the various flavors of categorical models of dependent types that type theorists have developed, so what one needs is a coherence/strictification theorem for $(\infty,1)$-categories making them into such a structure. -A natural approach (for lots of reasons) is to start with a locally presentable $(\infty,1)$-category and present it by a suitable model category, take the display maps to be the fibrations, and then strictify it somehow. It's not too hard to show that a right proper Cisinski model category will give rise to the appropriate structure on these display maps, up to isomorphism, to model type theory. Cisinski and Gepner-Kock have shown that any locally presentable, locally cartesian closed $(\infty,1)$-category can be presented by some right proper Cisinski model category. Finally, a recent coherence theorem of Lumsdaine-Warren (still unpublished) applies to strictify this structure as necessary. -Thus, putting it all together, type theory (without universe types) admits a model in any locally presentable locally cartesian closed $(\infty,1)$-category. And it should probably be possible to remove the local presentability condition by passing to presheaf categories and then restricting the model to the representables. As Urs said, what remains to be done is to handle univalent universes, and there we have partial results only. (Higher inductive types are also not covered by the above sketch, but they should be handled by a forthcoming paper of Lumsdaine and myself.) -I would personally stop short of claiming yet that it is proven that homotopy type theory is "exactly the internal language of" locally cartesian closed $(\infty,1)$-categories, because for that I would want to have a complete functorial semantics in place. In particular, I would want it to be the case that the syntactic model of type theory is initial among locally cartesian closed $(\infty,1)$-toposes, and I don't think we know anything of that sort yet.<|endoftext|> -TITLE: How many principal bundles are there over a given base? -QUESTION [5 upvotes]: I'm currently considering principal bundles and classifying spaces in the context of gauge theory and a crucial question came to my mind: -Is there a way to say how many (isomorphism classes of) principal bundles there are over a given base space? -For finite gauge groups the answer is yes: They correspond to elements of $\mathrm{Hom}(\pi_1(M),G)/G$ (where $M$ is the base and $G$ the gauge group). -Is there a similar characterization for arbitrary compact Lie groups $G$? - -REPLY [3 votes]: Let me add a few further remarks to the above very good answer. I think that the problem of classifying principal bundles is one of the most fundamental questions and applications of algebraic topology. -The basic reason why the classification of principal bundles for $G$ a compact Lie group is so much more complicated than for a finite group is that the compact Lie group is not discrete and has higher homotopy. For example, the principal $G$-bundles over $S^n$ are classified by $\pi_n(BG)\cong \pi_{n-1}(G)$. In particular, the complete classification of bundles is not even known over spheres of arbitrary dimension. -The question simplifies a bit if you stabilize, i.e., you look at $O(\infty)$-bundles or $U(\infty)$-bundles. In these cases, the homotopy sets $[M,BO]$ or $[M,BU]$ are topological K-groups and can in principle be computed using long exact sequences and such things. -However, in general, the classification of principal bundles over finite CW-complexes is going to be more and more complicated with growing dimension. To give you some flavour of the sort of results to expect, you might want to have a look at some of the following papers: -A.Dold and H.Whitney. Classification of oriented sphere bundles over a $4$-complex. Ann. of Math. (2) 69 (1959), 667-677. -I.M.James and E.Thomas. An approach to the enumeration problem for non-stable vector bundles. J. Math. Mech. 14 (1965), 485-506. -F.P.Peterson. Some remarks on Chern classes. Ann. of Math. (2) 69 (1959), 414-420. -L. Smith. Complex 2-plane bundles over $\mathbb{CP}(n)$. Manuscripta Math. 24 (1978), 221-228. -R.M.Switzer. Complex 2-plane bundles over complex projective space. Math. Z. 168 (1979), 275-287. -... and for something more recent (look at the progress in dimension)... -M. Cadek and J. Vanzura. On oriented vector bundles over CW-complexes of dimension 6 and 7. Comment. Math. Univ. Carolin. 33 (1992), 727-736. -B. Antieau and B. Williams. On the classification of oriented 3-plane bundles over a 6-complex. arXiv:1209.2219. -This does not even say anything about the classification of principal bundles with exceptional structure groups.... -[Edit:] I should have said that in all the above cases, the results are proved using obstruction theory. The answer then classifies bundles in terms of characteristic classes in suitable cohomology theories, together with additional data like compatibilities with Steenrod operations etc. This is a standard procedure in algebraic topology. Look at Hatcher's algebraic topology book for an introduction to Postnikov towers and obstruction theory.<|endoftext|> -TITLE: A few standard results (on metrizability and relative separation strength) without Choice? -QUESTION [6 upvotes]: I've been going back over some results from Munkres's Topology, and I'm curious about some things. (I originally posted this on M.SE, but I think it is probably a better fit here.) -I know that Choice principles have some connection to the separation axioms (in ZF, at least)--for example: - -Locally compact Hausdorff spaces are Baire if and only if the Axiom of Dependent Multiple Choices holds. [Due to Fossy and Morillon, I believe.] -Complete pseudometric spaces are Baire if and only if the Axiom of Dependent Choices holds. - -Still, it seems likely that compactness (or the weaker condition of completeness) and "Baireness" may be playing a substantial part, here. -I know that the Urysohn Metrization Theorem--which states that regular, Hausdorff, second-countable spaces are metrizable (or equivalently, that a space is separable and metrizable if and only if it is regular, Hausdorff, and second countable)--holds in ZF, though the Urysohn Lemma--which states that a space is normal Hausdorff if and only if any two disjoint closed sets can be separated by a continuous function--does not. That's a nice result. -First, the usual proof of the Urysohn Lemma (cf. Munkres) uses Dependent Choice, and it has been suggested to me that Dependent Multiple Choices may do the trick. Does anyone know a proof for this? -Second, I wonder what is known about how much Choice is needed to prove the following metrization theorems, or whether any of them are known to be equivalent in ZF: - -Nagata-Smirnov Metrization Theorem: A topological space $X$ is metrizable if and only if it is $T_3$ and has a basis that is countably locally finite [= $\sigma$-locally finite base]. -Smirnov Metrization Theorem: A topological space is metrizable if and only if it is paracompact Hausdorff and locally metrizable. -Bing Metrization Theorem: A topological space $X$ is metrizable if and only if it is $T_3$ and has a basis that is countably locally discrete [= $\sigma$-discrete base]. - -REPLY [5 votes]: To prove Urysohn's Lemma using Dependent Multiple Choice (DMC), proceed as in the usual proof, using Dependent Choice (DC), with the following modification. At each stage of the usual inductive construction, you need to find some open sets (usually $2^n$ of them at stage $n$) subject to requirements that look like this: You already have open sets $U$ and $V$ with the closure $\overline U$ of $U$ included in $V$, and you want an open $W$ with $\overline U\subseteq W\subseteq \overline W\subseteq V$. Normality of thespace gives you the existence of such a $W$, but you use DC to choose appropriate $W$'s for all the stages together. Using DMC instead of DC, you get, instead of a particular $W$ for given $U$ and $V$, a finite set of such $W$'s. But then you can replace those finitely many $W$'s with their intersection (or their union) to get a single appropriate $W$. That's the essential idea of the proof; one still needs to be careful about some technical details, but, if I remember correctly, they're rather straightforward.<|endoftext|> -TITLE: Lower bounds on matrix eigenvalues -QUESTION [6 upvotes]: Let $A$ be a real $n\times n$ matrix and let $\mu_1, \dots, \mu_n$ the (generalized, complex) eigenvalues of $A$. Assume that -$$ 0 < \alpha < \mathrm{Re}(\mu_1) < \dots < \mathrm{Re}(\mu_n).$$ -I am interested in a lower bound on the eigenvalues of $A^t A$ in relation to $\alpha$. -If $A$ is symmetric, then the lowest eigenvalue of $A^tA$ is bigger than $\alpha^2$. But if $A$ is non-symmetric, then this is not true. -Question: Is there a positive number $C(\alpha)$ of $\alpha$ such that we have: If for a matrix $A$, the above inequality holds, then the smallest eigenvalue of $A^tA$ is bigger than $C(\alpha)$? And what is the best such $C(\alpha)$? -\Edit: For example, for the matrix -$$A := \begin{pmatrix} \alpha & 1 \\ 0 & \alpha \end{pmatrix},$$ -the matrix $A^t A$ has the smallest eigenvalue $\alpha^2 + \frac{1}{2}\bigl( 1 - \sqrt{4\alpha^2 + 1}\bigr)$, which is always smaller than $\alpha^2$. -\Edit: Fixed a typing error. - -REPLY [5 votes]: No, there is no such $C(\alpha)$, even for fixed $n>1$. -Indeed, consider the matrix $A = \begin{pmatrix} \alpha & x \\ 0 & \alpha\end{pmatrix}$. -The product of the eigenvalues of $A^t A$ does not depend on $x$ (it is equal to $det(A)^2=\alpha^4$), whereas their sum goes to $\infty$ (it is $Tr(A^tA)=2\alpha^2+x^2$). Hence one the the eigenvalues goes to $\infty$, and the other to $0$.<|endoftext|> -TITLE: Exact sequence for divisor class groups -QUESTION [6 upvotes]: Let $X$ be a either a projective scheme or a compact complex space. Then one has an exact sequence $$ (1) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}),$$ -where $\textrm{Pic}(X)$ and $\textrm{Cl}(X)$ denote the groups of Cartier (resp. Weil) divisors on $X$ up to linear equivalence. -The last arrow is in general not surjective, as can be shown by simple examples. -Now, in the paper by J.Birgener and U. Storch -Zur Berechnung der Divisorenklassengruppen kompletter lokaler Ringe, Nova Acta Leopoldina 52 Nr. 240, 7-63 (1981) -page 11, it is claimed that in the algebraic case one actually has the sequence -$$ (2) \quad 0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{x \in \textrm{Sing}(X)} \textrm{Cl}(\mathcal{O}_{X,x}) \to H^2(X, \mathcal{O}_X^{\ast}) \to 0.$$ -My problem is that $(2)$ seems to me not true. For instance, take a smooth cubic threefold $X \subset \mathbb{P}^4$. Then the inclusion $\textrm{Pic}(X) \to \textrm{Cl}(X)$ is an isomorphism, so $(2)$ would imply $ H^2(X, \mathcal{O}_X^{\ast})=0$. However, by the exponential sequence one finds $H^2(X, \mathcal{O}_X^{\ast})=H^3(X, \mathbb{Z})= \mathbb{Z}^{10}$, and this is a contradiction. -So my question is: - -am I missing something? If not, can one correct $(2)$ in some way? - -Any answer or reference to the existing literature will be appreciated. Thank you! - -REPLY [11 votes]: The exponential sequence does not quite work that way for the algebraic cohomology $H^q(X,\mathcal{O}_{X}^*)$ for $q>1$. Of course the analytic cohomology, $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$, does satisfy the exponential sequence. However, GAGA does not directly apply for $q>1$. Of course it does apply for $q=1$, because both the analytic and algebraic cohomology groups classify invertible sheaves, and GAGA gives an equivalence between these groups. But for $q=2$, one only has that the torsion subgroup of $H^q(X,\mathcal{O}_X^*)$ (and this group is a torsion group in "typical" cases) equals the torsion subgroup of $H^q(X^{\text{an}},\mathcal{O}_{X^{\text{an}}}^*)$. For a cubic threefold, there is no torsion in $H^3(X^{\text{an}};\mathbb{Z})$. Thus $H^2(X,\mathcal{O}_X^*)$ is zero. -Edit. I should make clear that "algebraic cohomology" above really means etale cohomology.<|endoftext|> -TITLE: Terminology question for poset maps -QUESTION [8 upvotes]: Is there a standard name for order-preserving maps $f\colon P\to Q$ of posets with the property that the image of a lower set is a lower set, or equivalently if $q\leq f(p)$ then there exists $p'\leq p$ with $f(p')=q$? -If you view the poset as a category, then this condition says that the functor associated to $P$ is surjective on in-stars. -In my research I need to consider the category of posets with these kinds of morphisms and I would like to know their name. For example, if $P$ and $Q$ are face posets of regular CW complexes, then this property says that the image of each closed cell is a closed cell. - -REPLY [4 votes]: I think one name for this is a simulation.<|endoftext|> -TITLE: Is there an irreducible, noncompact commuting, nonnormal operator, with spectrum strictly continuous? -QUESTION [6 upvotes]: Let $H$ be an infinite dimensional separable Hilbert space. -Definition: The commutant $\mathcal{S}'$ of a subset $\mathcal{S} \subset B(H)$ is $ \{A \in B(H) : AB=BA \ , \ \forall B \in \mathcal{S} \} $. -Definitions : An operator $A \in B(H)$ is : - -Irreducible (Halmos 1968) if its commutant $\{ A\}'$ does not contain projections other than $0$ and $I$ ($A \ne A_{1} \oplus A_{2}$, $A$ generates $B(H)$ as von Neumann algebra : $\{A,A^{*}\}''=B(H)$). -Nonnormal if $\{ A\}'$ does not contain $A^{*}$ (i.e. $AA^{*} \ne A^{*}A$). -Noncompact commuting if $\{ A\}'$ does not contain a compact operator. - -Definition : The spectrum $\sigma(A)$ of $A \in B(H)$ is $\{\lambda \in \mathbb{C} : A - \lambda I \text{ not bijective} \}$. - It decomposes as follows: - -Point spectrum: $\sigma_{p}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \text{ not injective} \}$ -Continuous spectrum: $\sigma_{c}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{injective, dense nonclosed range} \}$ -Residual spectrum: $\sigma_{r}(A) = \{\lambda \in \mathbb{C} : A - \lambda I \ \text{ injective, nondense range} \}$ - -The spectrum of $A$ is strictly continuous if $\sigma(A) = \sigma_{c}(A)$. -Examples: - -Let $S$ be the bilateral shift defined on $H = l^{2}(\mathbb{Z})$ by $S.e_{n} = e_{n+1} $. -Its spectrum is strictly continuous : $\sigma(S) = \sigma_{c}(S) = \mathbb{S}^{1}$. -It's also a unitary operator ($SS^{*} = S^{*}S = I$), so a fortiori a normal operator. -It is noncompact commuting and reducible. -Let $T$ be the unilateral shift defined on $H = l^{2}(\mathbb{N})$ by $T.e_{n} = e_{n+1} $. -Its spectrum is not strictly continuous because $0 \in \sigma_{r}(T)$. -It's a nonnormal operator because $[T^{*},T].e_{0} = e_{0}$. -It is noncompact commuting and irreducible. -Let $V$ the Volterra operator defined on $H= L^{2}[0,1]$ by $(V.f)(t)=\int_0^tf(x)dx$. -Its spectrum is strictly continuous $\sigma(V) = \sigma_{c}(V) = \{ 0\}$. -It is compact, irreducible and nonnormal (see here). -Let $p$ be a non-constant polynomial (see here). -Then $p(V)$ is nonnormal, compact commuting and irreducible. -Its spectrum is strictly continuous $\sigma(p(V)) = \sigma_{c}(p(V)) = \{ p(0)\}$. -It's compact commuting, nonnormal and irreducible. -Let $S \oplus V$ defined on $l^{2}(\mathbb{Z}) \oplus L^{2}[0,1]$. -It is reducible, compact commuting, nonnormal and with spectrum strictly continuous. - -If you find a mistake, thank you let me know in comment. - -The main question: Is there an irreducible, noncompact commuting and nonnormal operator, with spectrum strictly continuous ? - -Bonus questions : How classify these operators ? - -REPLY [6 votes]: Essentially I think weighted shifts should be a sufficiently rich class of operators. Consider, for instance, the following example. -Take the doubly infinite sequence -$$ w_k=\left\{\begin{array}{ll} 2 & \text{if } k<0 \\ 1 & \text{if } k\geq 0 \end{array}\right. $$ -and let $W$ be the weighted shift on $\ell^2(\mathbb{Z})$ defined by -$$ We_j=w_je_{j+1} \qquad \forall j\in\mathbb{Z}\qquad (\text{here }(e_j)\text{ is the usual basis}).$$ -Then - -W has strictly continuous spectrum. This is an almost identical argument the usual bilateral shift. -W is not normal. The adjoint is given by $W^\ast e_j=w_{j-1}e_{j-1}$ and so $$W^\ast We_j=w_j^2e_{j},\quad \text{but}\quad WW^\ast e_j = w_{j-1}^2e_j.$$ -$W$ is irreducible. This follows from Corollary 2 of -R. Gellar, Operators commuting with a weighted shift, Proc. Amer. Math. Soc. 23 (1969), 538-545. -I do not seem to be able to show $W$ is non-compact-commuting. See Bill's comment below.<|endoftext|> -TITLE: Winning sets of full measure (Schmidt's game) -QUESTION [6 upvotes]: A quick reminder of the definition of Schmidt's game: - -Let ${X}$ be a metric space and ${S\subset X}$ be a subset. Let - ${0<\alpha,\beta<1}$ be constants. Bob chooses any open ball - ${B_0\subset X}$ with radius ${\rho_0}$. Then Alice chooses a ball - ${B_1\subset B_0}$ with radius ${\rho_1=\alpha\rho_0}$. Then Bob - chooses a ball ${B_2\subset B_1}$ with radius ${\rho_2=\beta\rho_1}$, - then Alice chooses a ball ${B_3\subset B_2}$ with radius - ${\rho_3=\alpha\rho_2}$ and so on. Let ${x}$ be the (single) point in - the intersection of all balls ${B_n}$. If ${x\in S}$ then Alice wins - the game. Otherwise Bob wins. If Alice can force a victory, then the - set ${S}$ is called ${(\alpha,\beta)}$-winning. $S$ called - $\alpha$-winning if it's $(\alpha,\beta)$-winning for all $0 < \beta < 1$. One can also define $windim(S)$ to be the least upper bound on all $\alpha$ such that $S$ is $\alpha$-winning. - -It's easy to see that $S$ need to be dense to be $(\alpha,\beta)$-winning. It's suprising though, that some sets of lebesgue measure $0$ are $\alpha$-winning (badly approximated numbers for $\alpha < \frac{1}{2}$). -Does some criterion exist for the inverse claim: sets of full measure (in a sense that $\mu(S^c)=0$) that are not $\alpha$-winning for some $\alpha$'s? It seems like the complement of the ternary Cantor set might be an example of that, but I couldn't find a good reasoning. - -REPLY [3 votes]: I'm not sure if you're more interested in general conditions for this or an example with a simple proof. I can give a simple example and explanation. Let $S$ be the set of numbers normal in base $b$. $S$ is a set of full measure (follows by SLLN, Birkhoff ergodic theorem, etc.) W. Schmidt proved that $S^c$ is $1/2$-winning. So $S^c$ is $\alpha$-winning for all $\alpha \in (0,1/2]$. But the property of being $\alpha$-winning is preserved under countable intersections. So if $S$ were $\alpha$-winning, then $S \cap S^c=\emptyset$ would also be $\alpha$-winning. Does that answer your question? There are many other examples along these lines in some of the more recent literature on Schmidt games that I can give if you're interested.<|endoftext|> -TITLE: Origins of Axiomatic Reasoning -QUESTION [5 upvotes]: Is there any evidence that axiomatic reasoning has been used prior to Thales of Milet (624-547BC), who is generally credited for the "invention" of axioms. -In this context I understand axioms in the original interpretation as checkable facts, whose truth is beyond doubt. -The motivation for my question is the following: -Siddharta Gautoma (563-483BC) based his religion (Buddhism) on "4 Noble Truths" of which the first three are related to observable "truths beyond doubt". -So, assuming that Thales of Milet and Siddharta Gautoma developed their ideas independently, it seems likely that the idea of basing arguments on checkable "truths beyond doubt", has been around for some time and may also have been in use in other cultures. -To clarify: I am not interested in religious debates; therefore I did not list the Noble Truths of Buddhism. - -REPLY [5 votes]: You'll find an extensive overview of early uses of the axiomatic method in Axiomatic Philosophy, by Pritam Sen (1996). One of the earliest sources is the Upanishad of the Hindu faith, written in the period 1000-500 BC.<|endoftext|> -TITLE: Is this a known combinatorial problem? -QUESTION [8 upvotes]: Let $M\subset\mathbb{N}$ be a finite set. For every positive integer $n$ set -$$D_n(M)=\{W\subset \mathbb{N} \text{ finite }|\ \forall\ i=0,\ldots,n-1\ \exists\ w\in W: w\in M+i\},$$ where $M+i=\{m+i|\ m\in M\}$. I would like to understand -$$d_n(M)=Min\{|W|\ | \ W\in D_n(M)\}.$$ -I am mainly interested in the asymptotic behaviour of the $d_n(M)$'s. So I would like to know $$d(M)=\lim_{n\to\infty}\frac{d_n(M)}{n}.$$ -For example, for $M=\{1,2\}$ we have $(d_1(M),d_2(M),\ldots)=(1,1,2,2,3,3,\ldots)$ and so $d(M)=\frac{1}{2}$. -My question is, if this is (related to) a known combinatorial problem? It seems a fairly natural problem to me, so I could well imagine that it has been treated in the literature. I encountered it when trying to explicitly compute some value in an algebraic-geometric example. -I would like to know things like, is the sequence of first differences $\Delta d_n(M)=d_n(M)-d_{n-1}(M)$ always periodic? - -REPLY [14 votes]: Yes this is a known combinatorial problem. You should look into the literature of covering or tiling groups by translates of a finite set. Your function $d(M)$ is commonly known as the "covering density". Newman called it "codensity" in his article "Complements of finite sets of integers", Michigan Math J 14 (1967), 481–486. -In particular it is true that there is a periodic covering achieving the covering density, from which I believe your observation follows. This was proved in - -W. M. Schmidt and D. M. Tuller, "Covering and packing in $\mathbb Z^n$ and $\mathbb R^n$ I", Monatsh Math 153 - (2008), 265–281 - -and also in section 5 of - -B. Bollobas, S. Janson, O. Riordan, "On covering by translates of a set", Random Structures Algorithms 38:1-2 (2011), 33–67. - -The sets which achieve $d(M)=\frac{1}{|M|}$ are the ones which tile $\mathbb Z$ by translations, and recognizing such sets is a notorious problem. I hope this helps.<|endoftext|> -TITLE: What are Moschovakis cardinals? -QUESTION [13 upvotes]: The question is exactly that of the title: what are Moschovakis cardinals? -Background. In a recent answer to the question, "Are there examples of statements that have been proven whose consistency proofs came before their proofs?," user14111 posted (Are there examples of statements that have been proven whose consistency proofs came before their proofs?) an answer involving "Moschovakis cardinals," a large cardinal notion which was shown to be inconsistent at some point in time. Now, googling for Moschovakis cardinals reveals nothing besides that answer and this (http://mathforum.org/kb/thread.jspa?forumID=13&threadID=22263&messageID=59655#59655) Math Forum Discussion post, which seems(?) to be responding to a post which was then deleted. -According to user14111, the notion of a Moschovakis cardinal arose in an unpublished manuscript circulated around the late 60s; given the timing, my current guess is that "Moschovakis cardinal" is just a synonym for "Reinhardt cardinal," but I'll admit there is no real basis for my guess. -Why I'm interested. (Assuming these aren't just Reinhardts in disguise) I'm always interested in large cardinal axioms inconsistent with ZFC; in particular, can Moschovakis cardinals survive in ZF? Also, on a purely historical level, it would be interesting to know about. -Even if Moschovakis=Reinhardt, I'm still intrigued: why would that name be used? I've heard Reinhardt cardinals called Kunen cardinals before, since Kunen proved their inconsistency; but Moschovakis seems to have no relation to the subject that I'm aware of. - -REPLY [11 votes]: The following quotations are from "A survey of recent results in set theory" by A. R. D. Mathias, a preprint dated July 1968, and stated to be "a draft of a survey to be published in the Proceedings from the UCLA Set Theory Institute." (I found this today while doing some housecleaning.) -The first quotation begins on p. 28, column 2, and continues onto p. 29, column 1. - -D 2001 (Moschovakis) Let $\ \kappa,\lambda,\nu$ be cardinals, $\mu$ an ordinal. - $\kappa\underset{\lambda}\to(\mu)_{\nu}^{\lt\omega}$ iff for every $f$ with domain the set of sequences of - length $\lambda$ of finite subsets of $\kappa$ and range a subset of $\nu$ (in - symbols, $f:([\kappa]^{\lt\omega})^{\lambda}\to\nu$), there is a sequence $\langle x_{\eta}\ |\eta\lt\lambda\rangle$ of - subsets of $\kappa$, each of order type $\mu$, (i.e. $\langle x_{\eta}, \in\restriction x_{\eta}\rangle\cong\mu$), - with the property that for every $s$ and $t$ in the domain of $f$, - if for all $\eta\lt\lambda,\ s_{\eta}\subseteq x_{\eta},t_{\eta}\subseteq x_{\eta}$, and $\overline{\overline s}_{\eta}=\overline{\overline t}_{\eta}$, then - $f(s)=f(t)$.The case $\lambda=1$ is the Erdös-Rado property $\kappa\to(\mu)_{\nu}^{\lt\omega}$. The case - $\nu=2$ will be written $\kappa\underset{\lambda}\to(\mu)^{\lt\omega}$.$\kappa$ is a $\underline{\text{Ramsey}}$ cardinal iff $\kappa\to(\kappa)^{\lt\omega}$. $\quad\kappa$ is a $\underline{\text{Moschovakis}}$ cardinal - iff $\kappa\underset{\omega_1}\to(\omega_1)^{\lt\omega}$ - -From p. 30, column 2: - -Little is known about the size of Moschovakis cardinals relative to - other large numbers; Silver has made the following simple observation:T 2011 $\text{ZF}+\text{AC}\vdash$ If there are both Moschovakis and strongly compact - cardinals, then the first Moschovakis cardinal is smaller than - the first strongly compact. - -From p. 51, what must have been the motivation for considering such awful things: - -T 3025 (Moschovakis; Prikry) $\text{ZF}+\text{AC}\vdash$ If there is a Moschovakis - cardinal, then every $\underset{\sim}\Sigma^1_2$ game is determined. - -This is all pretty much Greek to me but I suppose it will mean something to you, if I didn't botch the typing too badly.<|endoftext|> -TITLE: Can one use Brownian motion to prove that two manifolds are not conformally equivalent? -QUESTION [10 upvotes]: Let me start by a very simple example; consider the following question: -"Let D1 be a square and D2 a rectangle (boundary included). View them -as subsets of the complex plane. Does there exist a conformal map (extending -to the boundary) taking D1 to D2?" -Of course, the answer is no, but I want to point out an unusual "proof" -of this assertion. Suppose the answer was yes. I think we can assume that -the center gets mapped to the center. Start a brownian motion from the center -of D1. The probability that this brownian motion hits any of the four sides -is equal. However the probability that a brownian motion hits any of the -four sides starting from the center of of D2 is not equal. And this is -a contradiction, because brownian motion is conformally invariant (which -is a non trivial fact, but its true). -I believe this "proof" can be made rigorous. My question is the following: -Can this same idea be used to show for instance two complex manifolds are -not biholomorphic to each other? Of course there maybe a simpler proof using -more direct methods, but I am still curious to know if the idea of using -brownian motion can be used to answer such a question (ie are two manifolds conformally equivalent). - -REPLY [11 votes]: An easy example is the proof that the open disk is not conformally equivalent to the plane, since the tail sigma-field of the Brownian motion on the disk is nontrivial (it contains information about the boundary point) whereas the tail sigma-field of the BM in the plane is trivial. I guess this observation should have generalizations in terms of boundary theory of Markov processes.<|endoftext|> -TITLE: Conditions for a topological group to be a Lie group -QUESTION [7 upvotes]: In flipping through the Springer lecture notes on Serre's 1964 'Lie Algebras and Lie Groups' lectures at Harvard, I found this pair of suprising results (page 157): -Let $G$ be a locally compact group. Then - - -(Gleason-Montgomery-Zippin-Yamabe) G is a real Lie group iff it does not contain arbitrarily small subgroups (i.e., there exists a neighbourhood of the identity containing no nontrivial subgroup). -(Lazard) G is a $p$-adic Lie group iff it contains an open subgroup $U$ such that $U$ is a finitely generated pro-$p$-group with $[U,U] \subset U^{p^2}$. - - -Are there further results that tell us when $G$ is a Lie group over $K$, $K = \mathbb{C}$ or $[K: \mathbb{Q}_p] < \infty$? - -REPLY [5 votes]: Every locally compact and locally contractible topological group is a Lie group -(Hofmann-Neeb arXiv:math/0609684).<|endoftext|> -TITLE: Metrics for lines in $\mathbb{R}^3$? -QUESTION [17 upvotes]: I seek a metric $d(\cdot,\cdot)$ between pairs of (infinite) lines in $\mathbb{R}^3$. -Let $s$ be the minimum distance between a pair of lines $L_1$ and $L_2$. -Ideally, I would like these properties: - -If $L_1$ and $L_2$ are parallel, then $d(L_1,L_2) = s$. -$d(L_1,L_2)$ increases with the degree of skewness between the lines, i.e., the angle $\theta$ between their projections onto a plane orthogonal to a segment that realize $s$, where $\theta=0$ for parallel lines and $\theta=\pi/2$ for orthogonal lines. - -A natural definition (suggested in an earlier MSE question) is $d(L_1,L_2) = s + |\theta|$. -But this is not a metric. -For example, a sufficiently large value of $a$ below ensures the triangle inequality will be violated: -      -I would be interested to learn of metrics defined on lines in space, and whether or not any such metric satisfies the properties above. -Perhaps the properties cannot be achieved by any metric? - -Update. -Here is my current understanding of the rich variety of the erudite answers provided. -Apologies in advance if my summary is inaccurate. -First, there is no such metric, interpreting my second condition as (naturally) demanding invariance under Euclidean motions, as convincingly demonstrated by Robert Bryant, Vidit Nanda, and Pierre Simon. -Second, a looser interpretation requires only that if we fix $L_1$, then $d(L_1,L_2)$ is monotonic with respect to $\theta$ as $L_2$ is spun "about their intersection point in the plane that contains them [Yoav Kallus]." Then, Will Sawin's metric satisfies this condition. Here is an example of why this metric fails the more stringent condition—it depends on the relationship between the lines and the origin: -    -The right lines could be further apart than the left lines (depending on $a$ and $b$). - -REPLY [6 votes]: Such a metric does exist. Simply let $d(L_1,L_2)$ equal the angle between $L_1$ and $L_2$ plus the distance between $P_1$ and $Q_1$, where $P_1$ is the closest point in $L_1$ to $0$ and $Q_2$ is the closest point in $L_2$ to $0$. -This is a metric because it is a sum of two pseudometrics and the distance between distinct lines is $0$. We easily verify that the two bullet points are satisfied, because parallel lines have their closest point to $0$ in the plane which is perpendicular to the lines and passes through $0$. -It dodges Vidit's proof because it is not, as Robert notes, invariant under the group of Euclidean motions.<|endoftext|> -TITLE: Tiling the square with rectangles of small diagonals -QUESTION [16 upvotes]: For a given integer $k\ge3$, tile the unit square with $k$ rectangles so that the longest of the rectangles' diagonals be as short as possible. Call such a tiling optimal. The solutions are obvious in the easy cases when $k$ is the square of an integer and for a few small values of $k$ only (unpublished). In each of the solved cases, the sides of all rectangles turn out to be rational and their diagonals are equal. -Question. In an optimal tiling, must the sides of all rectangles be rational and their diagonals be equal? -The analogous question for tiling the $n$-dimensional cube with rectangular boxes can be asked for every $n\ge3$ as well. - -REPLY [3 votes]: In a new related question I give a conjecture for the unit square which agrees with the $k=5$ and $k=8$ solutions here. -If $s^2 \lt k \lt (s+1)^2$ then the optimal solution has $s$ or $s+1$ rows (depending on which square is closer) each with $s$ or $s+1$ rectangles. More specifically: -If $k=s^2+t$ with $0 \lt t \le s$ then the optimal solution has $s$ rows with $s-t$ rows having $s$ rectangles $b \times \frac1s $ and $t$ rows of $s+1$ rectangles $a \times \frac1{s+1}$ where $b^2+\frac{1}{s^2}=a^2+\frac1{(s+1)^2}$ and $(s-t)b+ta=1$ -But if $k=s^2+t$ with $s \le t \lt 2s+1$ then the optimal solution has $s+1$ rows with $2s+1-t$ rows having $s$ rectangles $a \times \frac1s $ and $t-s$ rows of $s+1$ rectangles $\frac1{s+1} \times b$ where $a^2+\frac{1}{s^2}=b^2+\frac1{(s+1)^2}$ and $(2s-t+1)a+(t-s)b=1$ -Note that in case $k=s^2+s,$ either description gives all rectangles $\frac1s \times \frac1{s+1}.$<|endoftext|> -TITLE: $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat (right) $\mathbb{Z}G$-module? -QUESTION [14 upvotes]: Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by-finite group. -Denote the (Banach) space (in fact an algebra under convolution) $$\ell^1(G)=\left\{\sum_{g\in G}\lambda_g g\;\Big|\; \sum_{g\in G}|\lambda_g|<\infty, \,\lambda_g\in \mathbb{C}\right\}.$$ -Note that $\mathbb{Z}G\subset \ell^1(G)$ and we can consider $\ell^1(G)$ as a right $\mathbb{Z}G$-module. My question is: - -$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat $\mathbb{Z}G$-(right) module? - -If it is true, any reference (maybe for related topics)? Otherwise, any counterexample? Especially, is it true for the Heisenberg group? - -RK: I have asked a general one in MSE but no answer appeared, so I asked here a more specific one. - -REPLY [4 votes]: The general answer is no for my question. -For example, for $G=\mathbb{Z}^2\rtimes\mathbb{Z}$ with exponential growth rate, $\ell^1(G)$ is not flat over $\mathbb{Z}G$, the proof is based on rather elementary calculation. But it is too long to be present here.. -The point is to find $f\in\mathbb{Z}G\cap\ell^1(G)^{\times}$, find $h\in\mathbb{Z}G$, and do calculation to show for any $t\in\mathbb{Z}G$ with $th\in\mathbb{Z}Gf$, $t\not\in\ell^1(G)^{\times}$. Then we have $0\to\frac{\mathbb{Z}Gf+\mathbb{Z}Gh}{\mathbb{Z}Gf}\to\frac{\mathbb{Z}G}{\mathbb{Z}Gf}$ fails the flatness of $\ell^1{G}$. -It is open when $G$ is the Heisenberg group. -Motivation behind this question is proposition 2.1, theorem 3.1 and conjecture 3.6 in the paper here. Note that conjecture 3.6 is false ingle general, but for the most interesting case, i.e., $G$ is Heisenberg group, it is still open.<|endoftext|> -TITLE: certain trigonometric homeomorphisms -QUESTION [8 upvotes]: Are there any simple characterizations of rational functions $f(x,y)$ with real coefficients such that $\theta\mapsto f(\cos\theta,\sin\theta)$ is a homeomorphism from $\mathbb R\bmod 2\pi$ to $\mathbb R\cup\{\infty\}$ (the last set being the one-point compactification)? -(The fact that $\theta\mapsto\sec\theta+\tan\theta$ is such a homeomorphism despite the fact that each term separately is two-to-one onto its image and the first does not map onto the codomain, is what first made me think of this.) - -REPLY [17 votes]: The magic words are $\tan(\theta/2).$ That substitution reduces your question to asking which rational functions $\mathbb{R} \rightarrow \mathbb{R}$ are homeomorphisms. Those are precisely the functions whose derivative does not change sign, so differentiating our function we get a rational function which does not change sign. This is true if and only if both the numerator and denominator do not change sign, so let's just say they stay positive (or non-negative if you want homeo instead of diffeo). A polynomial is nonnegative on the real line if and only if it is the sum of (two) squares. -So, there you have it. -sum of two squares to express a nonnegative polynomial as a sum of two squares, factor it over $\mathbb{C}.$ It will have either real roots of even multiplicity or conjugate complex roots. A quadratic polynomial with no real roots can be written as a sum of two squares ("completing the square"), an even power of a polynomial can be written as a sum of two squares (one of which is $0$), and now note that the product of sums of two squares is likewise a sum of two squares (think of norms) -sigh to complete the answer, note that if the rational function has a pole, it can only have one simple pole (otherwise, the map is not $1$-$1$ at infinity), furthermore, the limits at $\pm \infty$ have to be equal and finite. Suppose that pole is at $x=a.$ Precompose your function by a linear fractional $\phi$ which sends infinity to $a$ - $x \mapsto \frac{a x}{x+ a + 1}$ is my personal favorite. Now, you have a rational function which sends infinity to infinity, so you can apply what I had said before.<|endoftext|> -TITLE: Converse of Poincaré-Hopf theorem -QUESTION [14 upvotes]: Let $M$ be a connected, compact, oriented manifold of dimension $n<7$. If any two maps $M \to M$ having equal degrees are homotopic, must $M$ be diffeomorphic to the $n$-sphere? - -REPLY [8 votes]: To complete @RyanBudney's suggestion (and @Misha's comment), there are 3-manifolds in the snappea census which have trivial isometry group, so in particular have no non-trivial degree 1 self-maps, and which have trivial first betti number. If the map is degree $>0$, then this follows from Gromov's proof of Mostow rigidity (using the Gromov norm, the degree must $=\pm 1$, and the map is homotopic to an isometry). If a degree zero map from a 3-manifold to itself is homotopically non-trivial, then it must surject the fundamental group of some covering space. If this covering space is finite-index, then since the group is co-finitely Hopfian, this map must induce an isomorphism on fundamental group, and thus must be degree $\pm 1$, a contradiction. Thus, the image group must be infinite index. These groups have positive betti number, a contradiction.<|endoftext|> -TITLE: Stabilizer Action on vector bundle on a stack -QUESTION [5 upvotes]: Suppose you have a Deligne Mumford stack $X$ and a geometric point $x:Spec{k}\rightarrow X$ with stabilizer group $Stab(x)$.Let $F$ be a locally free sheaf on $X$. -How is the action of $Stab(x)$ on $F\otimes {k}$ defined? -Also, suppose there is a surjective map $f:X\rightarrow Y$ to another DM stack such that $f_{*}F$ is locally free. What is the induced action of $Stab(fx)$ on $f_{*}F\otimes{k}$? -Let $y$ be a geometric point $y: Spec{k}\rightarrow Y$ with closed image. Is it true that every action of $Stab(y)$ on $f_{*}F\otimes{k}$ comes from an action of $Stab(x)$ on $F$, where $x$ is a closed geometric point $x:Spec{k'}\rightarrow X$?(Under the surjectivity assumption I think that the map is surjective on closed points and on their stabilizers, but if not, assume $f$ is also a map which satisfies this condition) - -REPLY [6 votes]: For the first question the short answer is that the map $x$ factorises -canonically through the classifying stack $BG$ with $G = Stab(x)$ and the -category of quasi-coherent modules on $BG$ is equivalent to the category -of $G$-vectorspaces over $k$. -Explicitly, you can see the action if you view $F$ as a sheaf on big fppf site on $X$. This site has the category $Sch/X$ of schemes over $X$ as underlying category. The map $x$ is given by a functor $x:Sch/k \to Sch/X$ and $Stab(x)$ is the sheaf of groups defined by $u \mapsto Aut(x(u))$, where -$u:T \to Spec\ k$ is an object in $Sch/k$. Now $F \otimes k = x^{-1} F$ is just the sheaf given by the composition $F \circ x$. For any $u$ as above, we have an action of $Aut(x(u))$ on $F(x(u))$, simply because $F$ is a functor. This gives $F \otimes k$ the structure of a sheaf of $G$-sets. In fact, the construction gives a functor from sheaves of sets on $X$ to sheaves of $G$-sets on $Spec\ k$. We also get a functor to the sheaves of $O_X$-modules to sheaves of $k[G]$-modules which preserves the property of being quasi-coherent. -Note that the above description is very general. For instance, you could take any scheme instead of $Spec\ k$ and any, not necessarily algebraic, stack instead of $X$. -I don't think it is possible to give a general answer to the second question as you stated it. It might be a good idea to have a look on the definition of the push-forward functor. You could start with the push-forward of set-valued sheaves (http://stacks.math.columbia.edu/tag/00XF). A good baby example to consider is to take a group homomorhism $G \to H$ and letting $X = BG$ and $Y = BH$. Verify that pulling back a sheaf from $Y$ to $X$ corresponds to restricting the group action and pushing forward from $X$ to $Y$ corresponds to induction. In particular, if $H$ is trivial, you get the sheaf of invariants. This example also works for quasi-coherent sheaves. Then pull-back and push-forward corresponds to restriction and induction of $G$-representations. In general, the push-forward functor for set-valued sheaves does not coincide with the push-forward functor for quasi-coherent sheaves. In the example with the classifying stacks it works since the map $f\circ x$ is flat. -Surjectivity does not imply surjectivivty on stabilisers. In the example above, the maps are surjective regardless of choice of map $G \to H$ since the underlying topological spaces of classifying stacks are just one point sets. But there are of course group homomorphisms which are not surjective. I'm not sure what you mean by that every action on $f_*F\otimes k$ "comes from" an action on $F$. There is no action on $F$. Just on $F \otimes k$, and it is uniquely determined by $F$.<|endoftext|> -TITLE: A colorful fractal structure on a graph provided by Wilson's algorithm: any explanation? -QUESTION [5 upvotes]: Consider a big finite rescaled piece of $\mathbb{Z}^2$, i.e. consider a unit square with a thick grid. Famous Wilson's method allows to generate a colored spanning tree of such a graph in a uniform way by popping out the cycles: if the mesh-size tends to 0, then the interesting fractal structure occurs: one can read about this in R.Lyons, Y. Peres Probability on Trees and Networks. It is said there, that this fractal structure is not yet explained. Could someone point to me some works connected to this question, if there are any? - -REPLY [5 votes]: Is this the fractal structure you have in mind? - -    -Image from Russell Lyons' book page. Figure 4.6 in the book. Caption: -    -"The distances in the tree to the path between opposite -corners in a uniform spanning tree in a 200x200 grid."<|endoftext|> -TITLE: Is the preimage of a nonreduced subscheme nonreduced? -QUESTION [6 upvotes]: Say $X \to Y$ is a surjective map of algebraic varieties, and $Z \subset Y$ is nonreduced. Then is the preimage $Z \times_Y X$ also nonreduced? - -REPLY [11 votes]: In Allen's notation, take: -$R = k[t]$, $X = \operatorname{Spec} R$ -$S = k[x,y]/(y^2 - x^2(x-1) )$, $Y= \operatorname{Spec} S$. -with the map defined by: -$y = t(t^2+1)$ -$x=(t^2+1)$ -$I=(x)$, $Z = \operatorname {Spec} S/I$. -$S/I$ contains a nilpotent, $y$ so $Z$ is non-reduced. $X \to Y$ is a surjective map of varieties. $X \times_Y Z = \operatorname {Spec} R/RI = \operatorname{Spec} k[t]/(t^2+1)$, which is reduced.<|endoftext|> -TITLE: Function transformation of exponentials -QUESTION [5 upvotes]: I came across the following function transformation: -$$ -\sum_{j=-\infty}^{\infty} e^{(-j^2\cdot t)} = \sqrt{\frac{\pi}{t}} \cdot \sum_{j=-\infty}^{\infty} e^{(-\frac{\pi^2}{t}\cdot j^2)} -$$ -where $ j \in \mathbb{Z}$ (i.e. integers). -Can anyone help me to understand why this relation is true? Thanks! - -REPLY [5 votes]: This is the functional equation for the theta function. A nice proof (using Poisson summation) can be found here.<|endoftext|> -TITLE: A game on sets of reals -QUESTION [18 upvotes]: A 2 player game on $\mathcal{P}(\mathbb{R})$: Players take turns playing uncountable sets of reals. Each play must be a subset of the previously played set. Player 1 wins if the intersection of all the sets is nonempty. Otherwise player 2 wins. -Assuming AC, player 2 has a winning strategy. On can prove in ZF that 2 wins the analogous game on any well-ordered set. -What might happen without AC? I heard a false argument that 1 wins assuming the perfect set property, so be careful. - -REPLY [16 votes]: This is a great question! I've now managed to eliminate the use of countable choice. -Theorem. Without using any choice principle, it follows that player I can have no winning strategy in the game. -Proof: Working in ZF only, suppose toward contradiction that player I has a winning strategy $\tau$. Look at the set $A$ played by player I on the first move, using strategy $\tau$. Note that $A$ is uncountable, but it can have no uncountable well-orderable subset, since in this case player II could play that subset and move to the well-orderable case, which according to what you have said would put player II in a winning position and therefore contradict that $\tau$ is winning for player I. -Let us now argue that there is a choice function on the uncountable subsets of $A$. To see this, fix any uncountable $B\subset A$ and let us have player II respond with $B$ on player II's first move, with player I continuing subsequently to play according to $\tau$. At each subsequent stage $n$, let us have player II follow the (losing) strategy, where when confronted with a set $E\subset A$, he will find the first rational interval $I$ (in some canonical enumeration) of width $\frac 1n$ such that $E\cap I$ is uncountable, and then play $E\cap I$. -We could deduce that such an interval $I$ exists if we had countable choice, since otherwise $E$ would be a countable union of countable sets. But I claim that we do not need to assume any choice principle in order to make this conclusion. If the set $E$, which was played by player I according to the strategy $\tau$, is the countable union of countable sets $\bigcup_n E_n$, then this is now a winning position for player II, since every uncountable subset of $E$ must contain points from infinitely many $E_n$, and so player II can always subsequently play on his $k^{th}$ step so as to remove the sets $E_i$ for $i\leq k$ from play (a countable removal). Since $\tau$ is winning for player I, however, it must be that the set $E$ is not a countable union of countable sets and so such a rational interval $I$ exists. -Continuing with the main line of argument, what we have is that player II is playing in a prescribed manner (without using any AC), but ensuring that the diameters of the sets in the play of the game go to $0$. Since $\tau$ is winning, the final intersection set will therefore be a singleton subset of $B$, and thus we have provided a means to choose an element from any such $B$. -Using this choice function, we can now choose elements from uncountable subsets of $A$. It follows that we can define an injection of $\omega_1$ into $A$: we first choose a point $a_0\in A$, and then choose a point from $A-\{a_0\}$ and so on. At each stage $\beta$, we choose a point in $A-\{a_\alpha\mid\alpha\lt\beta\}$, which is an uncountable subset of $A$. The set of $a_\alpha$ is an uncountable well-orderable subset of $A$, which contradicts our assumption that there was no such subset. So player I cannot have such a winning strategy after all. QED -This theorem does not seem to imply that player II must have a winning strategy, however, since perhaps the game is undetermined. -Following the link in Andres's comment, we see that Asaf Karagila had had essentially the same argument over at math.SE. And the trick to eliminate the use of countable AC uses the same idea as in Andres's answer over at math.SE.<|endoftext|> -TITLE: Standard (special) spines and hyperbolic structure on 3-manifolds -QUESTION [7 upvotes]: My question relates to constructing angled triangulations or hyperbolic triangulations for $3$--manifolds. Briefly, an angle triangulation can be considered as an assignment of a real number (called an angle) to each edge of a tetrahedron such that when we glue the tetrahedron up to obtain the $3$--manifold, around each edge the sum of the angles are $2\pi$, and around each vertex the sum of angles is $\pi$ (we also require that opposite edges of each tetrahedron have the same angle). In short, an angle structure corresponds to the linear part of Thurston's gluing equations -- it corresponds to a weak hyperbolic structure in that the induced metric on the manifold may not be complete. -Now I have assumed the topological conditions required to obtain a hyperbolic structure on the manifold via Thurston's hyperbolization theorem (i.e., the manifolds I am looking at are compact, irreducible, atoroidal with torus boundary) and I am able to explicitly construct special spines for these. Dual to these special spines are ideal triangulations (this comes from the work of Matveev). -My question is whether there are well known conditions for a special spine to be 'geometric'. That is given a special spine, we can look at the special spine as a special polyhedron onto which the manifold collapses. Does it follow that if I look at the special spine as a hyperbolic polyhedron, and I know that the special spine when thickened gives me a $3$--manifold, does this imply that the $3$--manifold is a hyperbolic $3$--manifold. Any discussion, questions and ideas will be appreciated. This is my first time posting here so apologies in advance if I have not followed protocol. - -REPLY [4 votes]: If an ideal triangulation (or its dual spine) admits an angle structure (with positive angles), then the manifold must admit a complete hyperbolic metric of finite volume (see Theorem 10.2, an observation of Casson). However, when one straightens these tetrahedra in the complete hyperbolic metric, they might not be positively oriented. Casson also observed (after Igor Rivin) that the volume function on the space of angle structures is maximized at a complete hyperbolic structure; he made some attempt to find a different proof of geometrization of cusped hyperbolic manifolds using this approach (see Futer-Gueritaud for an exposition). However, as far as I know, there is no nice characterization of when an ideal triangulation with a positive angle structure admits a positive solution to Thurston's gluing equations. -One slight correction: an angle structure (with positive angles) gives rise to an incomplete hyperbolic metric on the complement of the 1-skeleton.<|endoftext|> -TITLE: A simple proof for a theorem of Szekeres and Turán -QUESTION [11 upvotes]: Szekeres and Turán found in 1937 a formula for the sum of the squares and the sum of the fourth powers of determinants of all $n$ by $n$ matrices with $\pm 1$ entries. (The sum of squares case follows easily from Cauchy-Binet identity.) Later Turán published a simpler proof for the sum of the fourth powers but in Chinese. I vaguely remember that there are simpler probabilistic proofs for both cases. -My question is about simple proofs for these identities, especially the one for 4th powers. -Is there a formula for the 6th power? - -REPLY [5 votes]: This is an answer, but my insight is hardly a mathematical one. Nonetheless, here we go: -The 1955 paper of Turán is, in fact, published in both English and Chinese in the same journal. The Chinese version is pages 411-417 in the journal; the English language version is pages 417 - 423. -Turán writes: "I am pretty sure that on the way one can evaluate $M_n(6)$ or $S_n(6)$ after a little longer calculations and also further." But he doesn't give a formula.<|endoftext|> -TITLE: Parameter estimation for stochastic differential equation from discrete observations -QUESTION [9 upvotes]: Suppose we have a time-series $x(t_i)$ at discrete times $t_i$ and we want to estimate the parameters of an underlying SDE corresponding to this time-series: -$$dx_t = f(x_t,\theta)dt + \sigma(x_t,\theta)dB_t$$ -I have read that there exists a lot of different numerical methods to approach this problem, but I have not found a comprehensive and comparative survey about this problem. - -What is the best method (according to the type of data/model) ? - -Is there a Python or Matlab toolbox doing the job ? I have had a look at SDE Toolbox for Matlab but it is not clear to me how to use it, so it would be appreciated to provide an example. - - -I think it can be useful to have an overview of this topic since it may be interesting for many people. Thank you. - -REPLY [2 votes]: Part 1: Methods -There is no best method, but most methods end up in quite a similar area. Here's an old review which is still quite good, though some of the language and specific methods need to be updated. The idea is either to develop some loss function against your data or take a Bayesian route. Let's look at the two separately. -Loss functions are an optimization-based approach. You essentially choose properties about the data you want to match. For example, method of moments and generalized method of moments are simply the process: - -Solve N times -Compare the difference of some average quantity of your Monte Carlo solution against the data. Make the difference your scalar loss function $L(y)$ where $y$ is your numerical solution. -Use whatever optimization algorithm to find the $y$ that minimizes the function $L(y)$ - -Then you mix and match. Use whatever SDE discretization method you like with your chosen loss function and local/global optimization algorithm, and that's a potential parameter estimation method. The simplest is Euler-Maruyama with $L_2$ loss on the expected value: -$$ L(y) = \sum_i \Vert E[y(t_i)] - E[d_i] \Vert $$ -for discrete data points $d_i$ at time $t_i$, and then throw that into an optimization package. Inside of that is a parameter $N$ for how many times you need to solve the SDE, and the higher $N$ is the more accurate your expected value is. Doing this on the means is generally referred to as the method of moments approach, where you could also do some $E[h(y(t_i))]$ which is then generalized method of moments. -Very much related to method of moments approaches is likelihood-based approaches. In this case, you assume that your data goes according to some distribution, solve your SDE $N$ times, compare the two distributions, and then use the distance of the distributions (KL-divergence, etc.) or the likelihood to calculate the fit. In this case you get another $L(y)$ which you put into an optimizer. To do this, you have to have some pre-specified likelihood as your input data, and usually this is done by assuming that the data has a distribution and performing a maximum-likelihood fit. For example, you can assume that your data is normally distributed at each $t_i$, then find the mean and variance (those are the sufficient statistics for MLE of the normal distribution), and so then you assume that at time $t_i$ you must have $N(\mu_i,\sigma_i)$, and calculate the loss against this. This approach can make it much easier to incorporate distributional assumptions about the errors, and allows correlations to be taken into account by using multivariate distributions. In some sense, it's a super-set of the previous approach. For ODEs, minimizing the $L_2$-loss is equivalent to maximizing the log-liklihood under the assumption of constant $\sigma$. While I don't know of a result for SDEs, I assume there's something similar here. -Let me make a quick remark that you don't have to sample $N$ solutions. Instead, you can sample from the solution's distribution directly using "exact" methods (see this link or by solving PDEs for the distribution directly, though these methods can have trouble scaling to high-dimensional problems. But yes, mix and match these techniques with choices for loss functions and optimizers and it covers a large amount of what people have done. -The other approach is Bayesian. In this case, you again need to assume some likelihood on your data, but now you assume some prior distribution on your parameters, and then use an MCMC technique where you - -Sample parameters from the prior -Calculate the likelihood for those parameters (solving the equations $N$ times, etc) -Use the MCMC rules to choose whether to reject the sample -Choose new parameters to step to, and repeat. - -Just like with ODEs, this process is much more expensive but instead of getting a point estimate out you get posterior distributions for your possible parameters. -While there are quite a few published methods, when you dig into the details it's usually some variant of how to make the choices for the steps in these overarching ideas. -Part 2: Toolboxes -I do not know of a toolbox in MATLAB or Python for this. Using what I described above you can put together some SDE solver toolbox, such as @horchler's or JiTCSDE, with an optimization routine. However, there are much more developed tools in Julia (including a more substantial selection of SDE solvers, and the built-in parameter estimation tooling from DifferentialEquations.jl currently handles the optimization-based approaches (here's an example), with the Bayesian approaches slated as coming soon.<|endoftext|> -TITLE: An analogue of the Bass-Quillen conjecture with power or Laurent series -QUESTION [12 upvotes]: The famous Quillen-Suslin theorem (formerly known as Serre's problem/conjecture) states that every projective module over $k[x_1,\dots, x_n]$ is free for $k$ a field. Replacing $k$ by a more general ring, we get the Bass-Quillen conjecture: - -Let $R$ be a regular ring and $P$ a projective module over $R[x_1,\dots, x_n]$, then $P \cong Q \otimes_R R[x_1,\dots, x_n]$ for a projective $R$-module $Q$. - -This has been proven in many cases, for example for $R$ of Krull dimension $\leq 2$ or if $R$ is a localization of an affine $k$-algebra for $k$ a field. -Now one could put forward similar conjectures replacing polynomial variables by power or Laurent series variables: - -(Power) Let $R$ be a regular ring and $P$ a projective module over $R[[x_1,\dots, x_n]]$, then $P \cong Q \otimes_R R[[x_1,\dots, x_n]]$ for a projective $R$-module $Q$. -(Laurent) Let $R$ be a regular ring and $P$ a projective module over $R((x_1,\dots, x_n))$, then $P \cong Q \otimes_R R((x_1,\dots, x_n))$ for a projective $R$-module $Q$. - -If I understand the (affine) Horrocks Theorem correctly, then the Laurent series version of the conjecture actually implies the original Bass-Serre conjecture for a ring $R$ if $n=1$. Actually, Horrocks proves the Laurent series version for $R$ regular local either of dimension $\leq 1$ or of dimension $\leq 2$ and containing a field already in 1964. My question is now: - -What is known about the power and Laurent series versions of the Bass-Quillen conjecture? - -REPLY [2 votes]: I believe it should be possible to remove the "Noetherian" hypothesis in Mohan's answer, using the following slight generalization of the Nakayama-style fact: -Lemma: Let $f : R \to A$ be a ring map admitting a section $g : A \to R$ and such that the inclusion $A^{\times} \subseteq g^{-1}(R^{\times})$ is an equality. Let $K$ be a finitely presented $A$-module such that $K \otimes_{A,g} R = 0$. Then $K = 0$. -Proof: Let $$ A^{\oplus s} \stackrel{\varphi}{\to} A^{\oplus r} \to K \to 0 $$ be a presentation of $K$ as a $A$-module, and let $$ M \in \mathrm{Mat}_{r \times s}(A) $$ be the matrix corresponding to $\varphi$. Since $\varphi \otimes_{A,g} R$ is surjective, there exists $N \in \mathrm{Mat}_{s \times r}(R)$ such that $g(M) \cdot N = \operatorname{id}_{r}$, hence $g(M \cdot f(N)) = \operatorname{id}_{r}$, which means $M \cdot f(N)$ is invertible (since $A^{\times} = g^{-1}(R^{\times})$). -Remarks: I guess in this question we only consider finitely generated projective modules. To check that $K,C$ are finitely presented, we may use e.g. part (4) of [SP, 0519].<|endoftext|> -TITLE: Large cardinals and mild extensions -QUESTION [5 upvotes]: It is known that for many large cardinals $\kappa$ (like weakly compact, measurable,...) , $\kappa$ remains large of the same type after forcings of size $<\kappa$. Now the questions are: -Question (1): Are there any large cardinals for which this result is unknown? -Question (2): In particular is the above result true for some small large cardinals like "reflecting cardinals", "unfoldable cardinals", "ineffable cardinals", "subtle cardinals" and ...? -Question (3): Is there any large cardinals whose consistency with $GCH$ is not known? - -REPLY [6 votes]: These are sweeping questions, whose full answer would constitute several book chapters. So let me just sketch some general information. (If you want, please ask a more focussed question about specific notions.) -The basic situation is that the Levy-Solovay phenomenon is extremely widespread, and holds for nearly all of the usual large cardinal notions. For large cardinal properties witnessed by embeddings $j:V\to M$ with critical point $\kappa$, where the forcing has size less than $\kappa$, it is easy to see that the embedding lifts to the forcing extension $j:V[G]\to M[j(G)]$ and still generally witnesses the large cardinal property in the extension. This kind of reasoning applies to measurable cardinals, supercompact, partially supercompact, strongly compact, strong, huge, almost huge, etc. etc. By considering smaller embeddings $j:M\to N$, the same idea works with weak compactness, unfoldability, strong unfoldability, uplifting cardinals, extendible cardinals, etc. etc. If one lifts $V_\kappa\prec V_\theta$ to $V_\kappa[G]\prec V_\theta[G]$, then one makes the argument work for reflecting cardinals, etc. -Nevertheless, there are a few variant large cardinal notions that are not preserved by small forcing. One large class of such cardinals are the Laver indestructible large cardinals of various types, such as Laver indestructible supercompact cardinals or Laver indestructible measurable cardinals, indestructible weakly compact cardinals and so on. The main results of my papers, Small forcing makes any cardinal superdestructible and Superdestructibility: a dual to Laver indestructibility, show that every Laver indestructible supercompact cardinal is destroyed by small forcing. The general conclusion is that small forcing generally ruins indestructibility. -I suppose that is isn't clear that we should regard the notion of a "Laver indestructible supercompact cardinal" as a large cardinal notion, per se, but I believe that it is sensible to do so. After all, set theorists are often interested in the forcing absoluteness of their concepts, and to make a large cardinal absolute by certain kinds of forcing is a natural strengthening of the concept. But this argument does show that the answer to the question depends on exactly what we count as a large cardinal notion. -Meanwhile, there is also the downward version of the Levy-Solovay theorem, where one asks whether a cardinal can become newly large after small forcing. This also is known in almost all the standard cases, but there remain a few open cases. For example, in my paper Small forcing creates neither strong nor Woodin cardinals, the question remains open whether small forcing can ever increase the degree of strongness of a cardinal (and the only open case is that perhaps a $\lt\lambda$-strong cardinal becomes $\lambda$-strong for a limit $\lambda$ of small cofinality). -The situation with GCH is similar. Almost all the standard large cardinal notions are known to be consistent with the GCH. Indeed, most are provably preserved by the canonical forcing of the GCH. For the smaller large cardinals, one may alternatively appeal to the canonical inner models, which have the large cardinals and the GCH. -Meanwhile, again one may form counterexamples, if one has a more expansive concept of what counts as a large cardinal notion. For example, the existence of a non-measurable weakly measurable cardinal implies $2^\kappa\gt\kappa^+$.<|endoftext|> -TITLE: Fundamental units of imaginary quartic fields -QUESTION [8 upvotes]: Let $F/{\mathbb Q}$ be an imaginary quartic extension (i.e. the degree $[K:{\mathbb Q}]=4$ and no embedding of $K$ in ${\mathbb C}$ has its image inside the real numbers). Then the unit group of the integer ring ${\mathcal O}_K$ is infinite cyclic up to the roots of unity in $K$ and one can pick a generator $\varepsilon_F$ with absolute value $>1$ (say we have chosen an embedding into ${\mathbb C}$). -I am interested in the asymptotics of $|\varepsilon_F|$; more precisely my question is the following: if we fix an imaginary quadratic extension $F_D={\mathbb Q}(\sqrt{-D})$ of ${\mathbb Q}$ then for any fundamental discriminant $d$ in the ring of integers ${\mathcal O}_D$, $F=F_D(\sqrt d)/{\mathbb Q}$ is an imaginary quartic extension, and it is known that $|\varepsilon_F|$ tends to infinity as $d$ does. I would be interested in knowing whether this convergence is uniform in $D$ or not, that is whether if given $M>1$ there is a $N\ge 0$ such that for any $D\in{\mathbb Z}_{>0}$ there are at most $N$ fundamental discriminants $d\in {\mathcal O}_D$ such that $|\varepsilon_F|\le M$. If this turns out not to be the case then I would be interested in the asymptotics of the numbers of $d$ with $|\varepsilon_F|\le M$ as (square-free) $D\to +\infty$. -It is well-known that one can reformulate this in terms of Pell-like equations: the units in such a $F$ are given by $1/2(t+u\sqrt d)$ where $(t,u)$ is an integer solution of $t^2-u^2d=4$. -My motivation for asking this question comes from geometry: the norms of fundamental units in quadratic extensions of $F_D$ correspond to the lengths of closed geodesics on the associated Bianchi orbifold, and the question amounts to asking if the number of such lengths which are less than $e^M$ is bounded when $D$ varies. - -REPLY [8 votes]: There's certainly some uniform bound, as a special case of the -theorem that for each $n$ and $M$ there are only finitely many -algebraic integers $\epsilon$ of degree $n$ each of whose conjugates -has absolute value at most $M$. Here $n=4$, and since $\epsilon$ -is a unit conjugate to $\pm\epsilon^{-1}$ (once $D \lt -4$), the proof -leads to the estimate $N = O(M^2)$ with an effective (and reasonably small) -implied constant. -Write the minimal equation of $\epsilon$ as -$0 = (x-\epsilon) (x\mp\epsilon^{-1}) = x^2 + ax \pm 1$ -when $\epsilon$ is an algebraic number of degree $2$ with norm $\pm 1$, -and as -$$ -0 = (x-\epsilon) (x-\bar\epsilon) (x\mp\epsilon^{-1}) (x\mp\bar\epsilon^{-1}) -$$ $$ -= (x^2 - 2{\rm Re}(\epsilon) + 1) (x^2 \mp 2{\rm Re}(\epsilon^{-1}) + 1) -$$ $$ -= x^4 + ax^3 + bx^2 \pm ax + 1 -$$ -when $\epsilon$ has degree $4$. -In each case the coefficients $a$ or $a,b$ are integers of size $O(M)$. -Thus there are $O(M^2)$ possibilities in all, so you can take $N = O(M^2)$ -as claimed. -As with Pell equations, we expect that the actual count is asymptotically smaller, -but $N$ must still grow as some power of $M$ because of families such as -$d = 4(t^2 \mp 1)$, $\epsilon = t + \sqrt{t^2 \mp 1}$.<|endoftext|> -TITLE: Are there better arithmetics on ordinals? -QUESTION [6 upvotes]: First fix the following notations:‎ -‎‎ -‎ -$‎\mathcal{L}_{AR}:=‎$‎‎The ‎first order language ‎‎$‎\lbrace ‎\overline{0},\overline{S},\overline{+},\overline{‎\times}, ‎\sqsubset‎‎ ‎‎\rbrace‎$‎‎ ‎which ‎‎$‎‎‎\overline{0}‎$ ‎is a‎ ‎constant ‎symbol‎, $‎\overline{S}‎$ is a ‎unary ‎function ‎symbol, $‎‎‎\overline{+}‎$ , ‎‎$\overline{‎\times‎}‎$ ‎are ‎binary ‎function ‎symbols ‎and‎ ‎‎‎‎$‎\sqsubset‎$‎ ‎is a binary ‎relation ‎symbol. -‎ -In the usual set theoretic literature, ordinal numbers are generalization of natural numbers and ‎the proper class of all ordinals ($‎‎Ord$) as an amorphous accumulation is very "‎similar" ‎to the set of all natural numbers (‎‎$‎\omega‎$). But we need some "bones" to mold these bodies and complete their similarity by equipping them with a same structure. In this way we know one of the most important‎ structures of ‎$‎\omega‎$ is its well ordered Peano arithmetic $‎‎‎\langle \omega , 0, S, +, ‎\times, ‎<‎ ‎\rangle‎‎$ in the language ‎‎‎of $‎\mathcal{L}_{AR}‎$ ‎with ‎the usual ‎interpretations of ‎symbols‎. In order to build some structural similarity between ‎$‎\omega‎$ ‎and ‎$‎Ord‎$ ‎‎‎‎we usually ‎endow ‎‎$‎Ord‎$ ‎by some special‎ ‎generalizations ‎of ‎‎natural ‎operations on ‎‎$‎‎‎\omega‎$. This method gives ‎us ‎a‎ ‎(proper class)‎ $‎\mathcal{L}_{AR}‎$ - ‎‎‎structure‎ $‎‎‎\langle Ord , 0_{ord}‎, S_{ord}, ‎+_{ord}‎, ‎\times_{ord}, ‎<_{ord}‎ ‎\rangle‎$ by the following interpretations: ‎ -‎‎ -‎ -‎$<_{ord}\subseteq Ord‎\times ‎Ord‎‎‎$‎ -‎ -‎$<_{ord}:= ‎\lbrace ‎(‎\alpha , ‎\beta )\in Ord ‎\times ‎Ord~|~ ‎\alpha ‎\in \beta ‎‎‎‎\rbrace‎$‎‎‎ -‎ -‎$‎‎‎‎‎‎‎0_{ord}\in Ord‎$‎‎‎‎ -‎‎$0_{ord}:=‎\emptyset‎$‎ -‎‎ -‎‎‎ -$S_{ord}:Ord\longrightarrow Ord‎$‎ -‎$S_{ord}(‎‎\alpha):= ‎\alpha‎ ‎\cup ‎‎\lbrace ‎\alpha ‎‎\rbrace‎‎‎$‎‎‎‎‎‎ -‎��� -‎$‎‎+_{ord}:Ord\times Ord\longrightarrow Ord‎$‎ -‎$‎\alpha +_{ord} ‎\beta ‎:=‎ ordertype\langle\alpha ‎\times ‎\lbrace 0 ‎‎\rbrace ‎\cup ‎‎‎\beta ‎\times ‎\lbrace 1 ‎‎\rbrace , ‎<_{train~‎order}‎‎‎‎\rangle‎‎‎‎‎$‎ -‎ -‎ -$‎‎‎\times‎‎_{ord}:Ord\times Ord\longrightarrow Ord‎$‎ -‎$‎\alpha \times_{ord} ‎\beta ‎:=‎ ordertype ‎\langle \beta ‎\times ‎\alpha‎ , ‎<_{lexographic~‎order}‎ ‎‎‎\rangle‎‎‎‎‎‎‎$‎ -‎ -‎ -Now the following facts are clear by ‎definitions‎‎: ‎‎ -‎‎‎ -‎‎ -Fact (1): On ‎natural ‎numbers,‎$‎0_{ord}‎$‎,$S_{ord}‎‎$‎,‎$+_{ord}‎‎$‎,$\times _{ord}‎‎$,‎‎$‎<_{ord}‎$ ‎are ‎equal ‎to ‎‎$‎0‎$‎,$‎S‎$‎,‎$‎+‎$‎,$‎\times‎$,$‎<‎$‎. -‎ -Fact ‎(2):‎$‎‎‎\langle \omega , 0, S, +, ‎\times ,‎ ‎<‎ ‎\rangle ‎\subseteq‎ ‎‎‎\langle Ord,0_{ord}‎, S_{ord},‎+_{ord}‎,\times_{ord},‎<_{ord}‎\rangle‎$‎‎‎ -‎ -By the fact (1) ordinary ordinal arithmetic "extends" the natural arithmetic of natural numbers. And the fact (2) says that this ordinal arithmetic has "‎primitive" (quantifier free) properties of natural number arithmetic. But something is uncomplete because we have the following facts:‎ -‎ -‎ -‎Fact ‎(3):‎ ‎‎$‎‎‎\langle \omega , 0‎‎\rangle ‎\prec ‎\langle ‎‎Ord, 0_{ord}\rangle‎‎‎$‎ -‎‎‎ -‎‎ -Proof:‎ Easy induction on formulas. -‎‎ -Fact (4): $‎‎‎\langle \omega , S‎‎\rangle ‎\nprec ‎\langle ‎‎Ord, S_{ord}\rangle‎‎‎$‎ -‎‎ -Proof: ‎Consider the sentence: ‎$‎\exists‎ x~‎\exists‎ y~(\neg (x=y) ‎‎\wedge ‎\forall‎ z~(\neg (‎\overline{S}(z)=x‎) ‎\wedge ‎\neg (‎\overline{S}(z)=y)‎))‎‎‎‎$ ‎which ‎is ‎true ‎in ‎‎$‎‎\langle ‎‎Ord, S_{ord}\rangle‎$ ‎but ‎false ‎in‎ $‎‎‎\langle \omega , S‎‎\rangle‎$‎. -‎‎ -‎ -Fact (5):‎ $‎‎‎\langle \omega , +‎‎\rangle ‎\nprec ‎\langle ‎‎Ord, +_{ord}\rangle‎‎‎$‎ -‎‎ -‎ -Proof:‎‎ Consider the sentence: ‎$‎‎‎\forall ‎x~‎\forall ‎y~(x ~‎\overline{+}~y=y~‎\overline{+}~x‎)‎‎$ ‎which ‎is ‎true ‎in ‎‎$‎‎‎‎\langle \omega , +‎‎\rangle‎$ ‎but ‎false ‎in ‎‎$‎‎\langle ‎‎Ord, +_{ord}\rangle‎‎‎$. -‎‎ -Fact (6):‎ $‎‎‎\langle \omega , ‎\times‎ ‎‎\rangle ‎\nprec ‎\langle ‎‎Ord, ‎\times‎_{ord}\rangle‎‎‎$‎ -‎ -‎‎‎ -Proof:‎‎ Consider the sentence: ‎$‎‎‎\forall ‎x~‎\forall ‎y~(x ~‎\overline{‎\times‎}~y=y~‎\overline{‎\times‎}~x‎)‎‎$ ‎which ‎is ‎true ‎in ‎‎$‎‎‎‎\langle \omega , ‎\times‎‎‎\rangle‎$ ‎but ‎false ‎in ‎‎$‎‎\langle ‎‎Ord, ‎\times‎_{ord}\rangle‎‎‎$.‎ -‎ -Fact (7): $‎‎‎\langle \omega , <‎‎\rangle ‎\nprec ‎\langle ‎‎Ord, <_{ord}\rangle‎‎‎$‎ -‎‎ -‎ -Proof:‎ Consider the sentence: $‎\exists ‎x‎~‎\exists y‎~(\neg (x=y)‎\wedge‎ ‎\forall ‎z‎~(z‎\sqsubset‎ x ‎\longrightarrow ‎‎\exists ‎t~(z\sqsubset t ‎\wedge ‎z\sqsubset x)‎‎‎)‎\wedge‎ ‎\forall u‎~(u‎\sqsubset ‎y‎ ‎\longrightarrow ‎‎\exists v~(u\sqsubset v ‎\wedge v\sqsubset y)‎‎‎))‎‎‎‎$ ‎which ‎is ‎true ‎in ‎‎$‎‎\langle ‎‎Ord, <_{ord}\rangle‎$ ‎but ‎false ‎in‎ $‎‎‎\langle \omega , <‎‎\rangle‎$‎. -‎ -‎ -Obviously by the facts we have: $‎‎‎\langle \omega ,0,S,+,\times,‎‎<‎‎\rangle ‎\nprec‎‎ ‎‎‎\langle Ord, 0_{ord}‎,S_{ord},‎+_{ord}‎,\times_{ord},‎‎<_{ord}‎ ‎\rangle‎$‎‎‎. -So we can observe that ordinary ordinal arithmetic hasn't "all" (first order) properties of standard arithmetic on natural numbers and some properties are "missed". Now there are some natural questions here:‎ -‎ -‎ -Question (1): ‎Is ‎this "‎incompleteness" of ‎ordinal ‎arithmetic ‎fundamental? ‎In ‎other ‎words, are there some ‎interpretations ‎‎$S^{*}‎‎$‎, ‎$+^{*}‎‎$‎, ‎$\times^{*}‎‎$‎, ‎$<^{*}‎‎$‎ for ‎‎$‎\mathcal{L}_{AR}‎‎‎$ ‎symbols ‎such that $‎‎‎\langle \omega , 0, S, +, ‎\times ,‎ ‎<‎ ‎\rangle ‎\prec‎‎ ‎‎‎\langle Ord , 0‎, S^{*}, ‎+^{*}‎, ‎\times^{*} ,‎ ‎<^{*}‎ ‎\rangle‎$? -(We call this interpretation, a "good" arithmetic on ordinals.) -‎ -Remak (1): Note that‎‎‎ possitive ‎answer ‎of above ‎question means that we can find a "good" ordinal arithmetic which "extends" natural number arithmetic and satisfies "all" of its properties too. So it seems that ‎‎‎‎we ‎must ‎"desert" ‎the ‎"classic" ordinal ‎arithmetic ‎and build a ‎‎"modern" ‎set ‎theory based on this (these) "well behavior" ordinal arithmetic(s) which will be a "renaissance" in set theory! ‎‎ -‎ -‎ -Now consider the following questions in two cases dependent on the answer of question (1):‎ -‎ -‎ -If the answer of question (1) be negative: -‎ -Question (2): Is there any non trivial sub language ‎$‎‎‎‎\lbrace ‎\overline{0}‎ ‎‎\rbrace ‎‎\varsubsetneq ‎‎‎\mathcal{L} ‎\varsubsetneq ‎\mathcal{L}_{AR}‎$‎ ‎such ‎that‎ ‎the answer of question ‎(1) ‎be ‎positive ‎up ‎to ‎‎$‎‎‎\mathcal{L}‎$‎? If yes, what are the maximal languages between ‎$‎‎‎‎‎‎\lbrace \overline{0}‎ ‎\rbrace$ and ‎$‎‎‎\mathcal{L}_{AR}$‎‎? -‎‎ -‎ -If the answer of question (1) be possitive: -‎ -‎ -Question (3): ‎Is ‎the ‎"good" ordinal arithmetic ‎in ‎question ‎(1) ‎unique?‎ -‎‎ -‎ -Question (4): How much is the degree of (first order) "unifiability" between the collections ‎$‎‎‎\omega‎$ ‎and ‎‎$‎‎Ord$? ‎Precisely ‎is ‎the following ‎sentence ‎true?‎ -‎ -"For all first order language $‎\mathcal{L}‎$‎ and for all $‎\mathcal{L}‎$ - structure ‎$‎‎‎\mathcal{M}‎$ with ‎$‎‎Dom(‎\mathcal{M}‎)=‎\omega‎$, there is an $‎\mathcal{L}‎$ - structure ‎$‎‎‎\mathcal{N}‎$ with $‎‎Dom(‎\mathcal{N}‎)=Ord‎$ which ‎$‎\mathcal{M} \prec ‎\mathcal{N}‎$ ‎‎‎‎‎"‎. -‎ -‎ -Question (5): Let ‎$‎‎‎\langle Ord , 0 , S^{*} ,‎+^{*}, ‎\times^{*}, ‎<^{*}‎ ‎‎\rangle‎‎$ ‎‎be a‎ ‎"good" ‎arithmetic ‎on ‎‎$‎Ord‎$‎. ‎Define‎ $‎‎‎‎\mathbb{N}^{ord}:=‎‎‎\langle Ord , 0 , S^{*} ,‎+^{*}, ‎\times^{*}, ‎<^{*}‎ ‎‎\rangle‎$ ‎and build ‎‎$‎‎‎‎\mathbb{Z}^{ord}‎$, ‎$‎‎‎‎\mathbb{Q}^{ord}‎$, ‎$‎‎‎‎\mathbb{R}^{ord}‎$ and ‎$‎‎‎‎\mathbb{C}^{ord}‎$ from ‎$‎‎‎\mathbb{N}^{ord}‎$ ‎in ‎the usual ‎methods ‎which we produce ‎‎$‎\mathbb{Z}‎$, ‎‎‎‎‎$‎‎‎\mathbb{Q}‎$‎, ‎‎$‎\mathbb{R}‎$ and ‎$‎‎‎\mathbb{C}‎$ ‎from ‎‎$‎‎‎\mathbb{N}=‎‎‎\langle ‎‎‎\omega , 0 , S ,‎+, ‎\times, ‎<‎ ‎‎\rangle‎‎$‎. Is there any well behavior real and complex "analysis" on $‎\mathbb{R}^{ord}‎$ and ‎$‎‎‎\mathbb{C}^{ord}‎$? In other words, is the behavior of "continous line of infinities" (‎‎$‎‎‎‎\mathbb{R}^{ord}‎$‎) and "complex plane of infinities" (‎$‎‎‎\mathbb{C}^{ord}‎$‎) as same as ‎$‎‎‎\mathbb{R}‎$ ‎and ‎‎$‎‎‎\mathbb{C}‎$? More precisely, do we have ‎$‎‎‎\mathbb{R}\prec ‎\mathbb{R}^{ord}‎$ ‎and‎ ‎‎$‎‎‎\mathbb{C}\prec ‎\mathbb{C}^{ord}‎$‎ in the language of (ordered) fields?‎‎‎ -‎ -‎ -Remak (2):‎ ‎Note ‎that we can produce ‎‎$‎‎‎‎\mathbb{Z}^{ord}‎$, ‎$‎‎‎‎\mathbb{Q}^{ord}‎$, ‎$‎‎‎‎\mathbb{R}^{ord}‎$ and ‎$‎‎‎‎\mathbb{C}^{ord}‎$ by the usual structure $‎‎‎‎\mathbb{N}^{ord}:=‎‎‎\langle Ord , 0 , S_{ord} ,‎+_{ord}, ‎\times_{ord}, ‎<_{ord}‎ ‎‎\rangle‎$ but the arithmetics on ‎‎$‎‎‎‎\mathbb{Z}^{ord}‎$, ‎$‎‎‎‎\mathbb{Q}^{ord}‎$, ‎$‎‎‎‎\mathbb{R}^{ord}‎$ and ‎$‎‎‎‎\mathbb{C}^{ord}‎$ will be very bad and complicated. So at first we need to fix our arithmetic on ‎$‎‎Ord$ ‎by choosing ‎the ‎best ‎one. ‎Anyway, i‎f ‎the ‎answer ‎of ‎above ‎question ‎be ‎possitive, ‎we ‎can ‎go ‎beyond ‎current infinitary "combinatorics" (number theory) ‎and ‎build a ‎"‎‎goo‎d behavior" infinitary real and complex "analysis" which could be an extra revolutionary development in our set theory and mathematics. -‎ -Question (6): Assume the answer of question (5) be possitive, what is the interpretation of "transcendental" infinitary real or "imaginary" infinitary complex numbers? For example what is the meaninig of ‎$\sqrt[\beth_{1}‎]{‎\aleph_{‎\omega‎}}‎$ ‎or ‎$‎‎i^{\aleph_{1}}$‎? ‎Is ‎ther‎e any fundamental transcendental infinitary real numbers such as ‎$\pi^{ord}‎‎$ and ‎‎$e^{ord}‎‎$, "hidden" between "integer" infinitary numbers such as ‎$‎‎\aleph_{2}$ ‎and ‎‎$‎‎\aleph_{4}$‎ ‎with a‎ ‎fundamental ‎relation similar to ‎‎$‎‎e^{i\pi}+1=0$‎? -‎ -‎ -Question (7): ‎What ‎is ‎the ‎answer ‎of ‎above ‎questions ‎in ‎logics with ‎more "expression power" than first order, such as higher order or infinitary logics? -‎‎‎ - -REPLY [8 votes]: I think you'll be interested in the surreal numbers -- see Philip Ehrlich's paper, "The absolute arithmetic continuum and the unification of all numbers great and small" (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bsl/1327328438). If my understanding is right, the surreals form the "maximal" elementary extension of the ordered field $\mathbb{R}$. In particular, in the surreal numbers many odd expressions like those in your Question 6 are meaningful. -Of course, since things are class-sized now, how the surreals behave -- or in general, how anything you're looking for behaves -- will depend on the ambient set theory, to some extent; I don't know of any particular point of interest, but it's worth keeping in mind. -One final remark, towards your Question 7: many logics stronger than first-order -- including higher-order logic (with the full, as opposed to Henkin, semantics) and infinitary logic -- can characterize $\mathbb{N}$ up to isomorphism, so the answer is trivially "no." You'd probably want to restrict your attention to compact logics extending first-order logic; there are some interesting ones here, but they tend to be a little odd. Saharon Shelah has a number of (difficult) papers on the subject; as examples, the logic gotten from first-order logic by allowing quantification over automorphisms of definable fields is compact, as are some logics with messier generalized quantifiers: see shelah.logic.at/files/375.ps and http://arxiv.org/pdf/math/0009080.pdf, page 3, respectively. It's worth noting that there is no logic stronger than first-order logic which is compact and has the Downward Lowenheim-Skolem property (this is Lindstrom's Theorem). -If you're interested in the area of logics beyond first-order, I heartily recommend the book "Model-Theoretic Logics" ed. Barwise and Feferman, which is chapter-by-chapter available at ProjectEuclid: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pl/1235417263. - -REPLY [5 votes]: As Joel commented, you're asking for a class-sized elementary extension of the structure $\langle\omega,0,S,+,\times,<\rangle$. Such things exist (under reasonable assumptions about proper classes), but they have essentially nothing to do with ordinal numbers. In particular, the order relation $<$ cannot be a well-ordering. -There are reasonable operations on ordinals that look somewhat more like the familiar operations on natural numbers. For example, the "natural sum" of two ordinals, $\alpha\oplus\beta$, is defined as the largest ordinal that can be partitioned into a subset of order-type $\alpha$ and a subset of order-type $\beta$. Unlike genuine ordinal addition, $\oplus$ is commutative. If you're interested in this sort of thing, look up "natural sum" and "natural product" or the equivalent names "Hessenberg sum" and "Hessenberg product". But don't expect to find elementary extensions of the natural numbers in this way. These Hessenberg operations are still compatible with the well-ordering of the ordinals.<|endoftext|> -TITLE: Combinatorics of resultants -QUESTION [8 upvotes]: This is a crosspost of https://math.stackexchange.com/questions/446470/combinatorics-of-resultants which received no answer. [EDIT: I deleted the initial copy of the question on MathSE]. -Let $f(z)=\sum_{i=0}^{D_f}x_iz^i$ and $g(z)=\sum_{i=0}^{D_g}y_iz^i$ be two polynomials. I would like to know the number of monomials (in the variables $x_i$ and $y_i$) in the resultant in $z$ of $f$ and $g$. Equivalently, this is the number of monomials in the determinant of the Sylvester matrix: -$$ -\left( -\begin{array}{ccccccccc} -x_0 & x_1 & x_2 & \cdots & x_{D_f} & 0 & 0 & \cdots & 0\\ -0 & x_0 & x_1 & \cdots & x_{D_f-1} & x_{D_f}& 0 & \cdots & 0 \\ -\vdots & \ddots & \ddots &\cdots &\ddots &\ddots & \ddots &\cdots &\vdots\\ -y_0 & y_1 & y_2 & \cdots & y_{D_g} & 0 & 0 & \cdots & 0\\ -0 & y_0 & y_1 & \cdots & y_{D_g-1} & y_{D_g}& 0 & \cdots & 0 \\ -\vdots & \ddots & \ddots &\cdots &\ddots &\ddots & \ddots &\cdots &\vdots\\ -\end{array} -\right). -$$ -I think there is a classical answer to this problem (I even seem to recall having seen it once) but I can't find a pointer to it. Can anyone point me either to a closed form or even better a tight and simple upper bound on this number of monomials ? - -REPLY [5 votes]: Yes, this (like everything else under the Sun) was studied, but not a lot. The reference is: -M. Kalkbrener, An upper bound on the number of monomials in the Sylvester resultant.<|endoftext|> -TITLE: Coequalizers in stable (infinity,1)-categories -QUESTION [10 upvotes]: I have read it claimed in several places that in a stable $(\infty,1)$-category, the coequalizer of parallel maps $f,g:X\to Y$ can be identified with the cokernel of $f-g$ (i.e. the pushout of the map $X\to 0$ along $f-g$). How is this proven? Is it written down anywhere? - -REPLY [2 votes]: I think you prove this pretty much as one does in an ordinary abelian category. -First note that the coequalizer $f,g: X \rightarrow Y$ is also given by the pushout of $(f, 1)$ and $(g, 1)$, both maps $X \oplus Y \rightarrow Y$. (This is proven exactly as in the ordinary case, by checking universal properties). -So it suffices to identify this pushout with the other. For this, note that the pushout of $X \rightarrow 0$ along the map $X \rightarrow X \oplus Y$ given by $(1, -g)$ is $Y$ together with the map $X \oplus Y \rightarrow Y$ given by $(g, 1)$ (since $g-g = 0$). Now use the pasting lemma for pushouts (which works in an $\infty$-categorical context, Lemma 4.4.2.1 in HTT). -Here's a few more details if you don't appreciate how reckless I'm being: For the first step it suffices by Lemma 4.2.4.3 of HTT to show that the map $\text{Fun}(\text{Fork}, \mathcal{C}) \rightarrow \text{Fun}(\Delta^1 \vee \Delta^1, \mathcal{C})$ given by the construction I've outlined above, is fully faithful. That way the object representing the functor assigning to an object the space of maps out of a fork diagram or a deleted square corresponding to it will be represented by equivalent objects, by Yoneda. To prove this, note that we have an inverse functor defined on the essential image which is an equivalence by the universal property of a coproduct, then the result follows by 2 out of 3. -To show that the pushout of $X \rightarrow 0$ by $X \rightarrow X \oplus Y$ is as described we will use another pasting argument. First note that the cofiber of the anti-diagonal $$X \stackrel{(1,-1)}\rightarrow X \oplus X$$ -is just $X$ where the map $X \oplus X \rightarrow X$ is addition. Using this pushout square and the fact that the pushout of $+: X \oplus X \rightarrow X$ along $$X \oplus X \stackrel{(1, g)}\rightarrow Y$$ -is $Y$ with the desired map, we can apply the pasting lemma and we're done. -I think I can make this last part a little more rigorous using the characterization of the additive inverse of a morphism given by Lemma 1.1.2.10 in Higher Algebra. If you like I can add that in later...<|endoftext|> -TITLE: Is there a constant $c>0$, such that every natural number $n>1$ is the sum of primes, each with size at least $cn$? -QUESTION [7 upvotes]: I am sure that this is well-known, but I looked around for the last half hour and couldn't see an answer. I just wondered whether it's possible to insist on taking all primes to be large in Vinogradov type results? -Thanks in advance for any help. - -REPLY [16 votes]: The usual proof of Vinogradov's result can be modified to show that every sufficiently large odd $n$ has $\asymp n^2/(\log n)^3$ representations as a sum of three primes with each prime exceeding $cn$, provided $c>0$ is sufficiently small. This gives (easily) a positive answer to your original question. -The best unconditional result of this sort seems to be by Baker and Harman ( R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 356 (1998), 763–780.). They show that every sufficiently large odd $n$ can be written as a sum of three primes from $[\frac{n}{3}-n^{4/7},\frac{n}{3}+n^{4/7}]$.<|endoftext|> -TITLE: Busy Beaver modulo 2 -QUESTION [24 upvotes]: There is well-known Rado's "Busy Beaver" sequence — the maximal number of marks which a halting Turing machine with n states, 2 symbols (blank, mark) can produce onto an initially blank two-way infinite tape. It begins: - -1, 4, 6, 13, ... - -The sequence grows faster than any computable function and so is non-computable itself. Calculating the next (5th) term is widely believed to be a very difficult task (at least as difficult as solving the Collatz problem). -Let's consider a sequence of remainders of the "Busy Beaver" sequence elements modulo 2: - -1, 0, 0, 1, ... - -Is anything known about computability of this latter sequence? - -REPLY [12 votes]: I will not answer the question since I think -it is not completely defined without completely specifying the machine model. -In the following I explain why. -Note that the exact running time of a machine is not natural mathematically, -it heavily depends on the particular model of the machine we use -(that is why in complexity theory -we get rid of such constants using asymptotic notations like $O$). -Therefore, -the question of computing the running time modulo some value is not natural either. -We can easily consider a machine model -where all halting machines halt in even number of steps and -another model where all halting machines halt in odd number of steps, -and in both cases your question is trivially computable. -Even more artificially, -we can define a machine model where -a halting machine with size $k$ halts in even number of steps iff -a particular machine halts in $k$ steps (which would make the problem uncomputable). -These are all valid models of computation though artificial ones. -The function is trivially computable in one model of computation, -and uncomputable in another model of computation. -That tells us that the question is not a natural one. -Note that this is not the case regarding $BB$, -it is uncomputable in all acceptable models of computation -(having same computational power as Turing machines). -The answer doesn't depend on the particular model of computation we use, -the numbers will be different, -but the question about computability has the same answer in all of them. -Your question about $BB(k) \mod 2$ for -a particular fixed completely specified model of computation is well defined, -but IMHO it is not a very interesting one: -it is too much dependent on the particularities of the machine model. -Unless you have some reasonable restrictions on the models such that -the answer will be the same on all such models -the question would not be very interesting IMHO. -In a sense, this is similar to asking -if a particular Diophantine equation has any integer answers -without explaining why that particular equation is of any interest. -Here is the same, from mathematical perspective, -the answer to the question is not a natural one and -too much dependent on the particular encoding of the machines -and I don't see any way of getting around this dependence. -(Also from practical perspective, -I also don't see any use for computability of this function.)<|endoftext|> -TITLE: random walk and Brownian motion on Riemannian manifold -QUESTION [12 upvotes]: As we know, the random walk on $\mathbb{Z}/n$ will converge(in some sense) to the Brownian motion on $\mathbb{R}$ when $n\to\infty$. I would like to know is there some higher dimensional analogy result. -Edit: As pointed by Nate Eldredge, there is a generalization of Donsker's theorem on manifold. But, I am interested in the following more topological generalization. -For a compact Riemannian manifold $X$, if a triangulation is given. Is there a canoncal way to defined a random walk $W$ on the vertex of the triangulation, such that $W^n$ the random walk defined in this way after $n$-th barycentric subdivision will converge to the Brownian motion on $X$. - -REPLY [7 votes]: Nicolas Th. Varopoulos, Brownian motion and random walks on manifolds, Annales de l'Institut Fourier 34(2) (1984), 243-269. -Abstract: We develop a procedure that allows us to “discretise” the Brownian motion on a Riemannian manifold. We construct thus a random walk that is a good approximation of the Brownian motion.<|endoftext|> -TITLE: Reference request: (co)limits in Eilenberg--Moore (V-)categories -QUESTION [7 upvotes]: The following result seems to be well known: - -If T is a (V-)monad on a (V-)category C, then the forgetful functor $U^T \colon C^T \to C$ creates - -any limits that exist in C, and -any colimits that exist in C and are preserved by T. - - -But I don't know of a published proof for general V. Toposes, Triples and Theories proves (1) for V = Set and leaves (2) for the special case of coequalizers as an exercise. Kelly's book doesn't mention monads at all. Lack, in Codescent objects and coherence, says that 'of course' (2) is true for V = Cat, but doesn't give a citation, while Blackwell, Kelly & Power, in 2-dimensional monad theory, say that (1) is 'well known' but don't give a reference either. And so on. -My questions are: - -Has a complete proof of this result (assuming it's true as stated) been published? -Has its bicategorical analogue been treated? - -REPLY [4 votes]: A result slightly weaker than your point 1. (creation of limits) for general enriched categories can be found in -Eduardo J. Dubuc, Kan extensions in Enriched Category Theory, Lecture Notes in Mathematics, Volume 145, 1970, Chapter II -where a few special cases are proved. Proposition II.4.7, Dubuc proves that powers (also known as cotensors) are created, and Proposition II.4.8 covers conical limits and ends. (The proof of the first proposition is very detailed, while the second is mostly left as an exercise.) -Taken together, these results show that the category of algebras is complete if the base is. I do not know a reference for creation of a general weighted limit (which might exist even if the category doesn't have copowers, products, and ends), nor for the creation of colimits that are preserved by $T$.<|endoftext|> -TITLE: On free limits of iterations -QUESTION [8 upvotes]: Shelah isolates the notion of "$\aleph_1$-free iteration" in the first two sections of Chapter IX from Proper and Improper Forcing, and he proves there that properness is preserved by this sort of iteration. He mentions in his "On what I do not understand" paper that this was his third proof of the preservation of properness, and he goes on to pose a few questions about "nep forcings" (non-elementary proper) and preservation of properties in free limits. -My real questions are: "What are $\aleph_1$-free iterations good for?" and "Has anyone other than Shelah worked with them?", but I'll settle for an answer to the question "Are there situations where the use of $\aleph_1$-free iteration rather than countable support iteration has been critical?" - -Edit: I did have a chance to talk with Shelah about this. As far as he is aware, no one else has worked with the idea, and it is still unknown if free limits have "real applications" other than making some proofs of iteration theorems run a little smoother. - -REPLY [4 votes]: This is more of a comment than an answer, but as I couldn't post a comment, I'm writing here. -If I understood correctly the definitions in chapter IX from Proper and Improper Forcing, it seems that you can find some variants of free limits (not $\aleph_1$-free, though) in the following work of Shelah: -In the paper "On CON($\mathfrak{d}_{\lambda}>cov_{\lambda}(meagre)$)" (Sh:945) in page 29 there is a construction that resembles the free limit, which is used in order to establish a result that is somewhat analogous to a basic result of Judah and Shelah (JdSh:292) on FS iterations of Suslin forcing (namely the complete embeddability of iterations along a subset of the original ordinal). -It also seems that free-like constructions appear in Shelah's paper "Properness without elementaricity" (this is definition 5.7 in Sh630), but I don't know what is it good for, as I'm still reading this paper.<|endoftext|> -TITLE: Constructing Metrics for specific Topological Spaces, and Refinements of the Cantor-Space in particular -QUESTION [7 upvotes]: I have a Problem in general, given some some Topological Space $(X, \tau)$ from which I know it is metrisable, how can I find a metric (that is at best in some sence constructive and easy, at the very best could be easily programmed on a Computer). -It came up to me because I am trying to construct metrics for Refinements of the Cantor-Space. The Cantorspace is the sets of all one-sided-infinite words over $X = \{ 0,1 \}$, i.e. the set $X^{\mathbb N}$. The Standard-Topology is given by the Basis $\mathcal B = \{ w \cdot X^{\mathbb N} : w \in X^{*} \} \cup \{ \emptyset \}$ and is also the Product-Topology on $X^{\mathbb N}$ where $X$ is equipped with the discrete topology. A metric which induces this Topology could be easily found, it is -$$ - \operatorname{d}(u,v) = 2^{-r} ~ \textrm{ with } ~ r = \min \{ n : u_n \ne v_n \} -$$ -with the convention $\min \emptyset = \infty$ and $2^{-\infty} = 0$. Details could be found for example in the Book Infinite Words by Perrin/Pin (see here). -Ok, now one refinement of the above Topology is given by the base -$$ - \mathcal B_A := \{ F : F\subseteq X^{\mathbb N} \land F \textrm{is a regular $\omega$-language and closed in Cantor space} \}. -$$ -Details on these Topology and the form of the closure and interior operator for it could be found in the Paper Topologies Refining the Cantor Topology on $X^{\mathbb N}$ -Now the Space $X^{\mathbb N}$ equipped with the Topology generated by $\mathcal B_A$ is normal and second-countable, so according to Urysohn Metrisation Theorem is metrisable. So I have gone through the proof of Urysohn's Lemma and Urysohns Metrisation Theorem. And despite it's "constructive" charakter I am unable to derive any usable metric for the space with Baisis $\mathcal B_A$ from it. -Now at this point, does anybody know examples where something similar is desired or could point me to papers which could help me? Even some hints if the Urysohn functions look simpler on these spaces? -To get more specific, for the construcition of the Urysohn function I looked at Munkres, Topology. And there it is constructed by successively selecting open sets which are nested in each other, I have no idea how to derive a simple representation of such a function form this. Furthermore, I looked at the original paper Zum Metrisationsproblem. In it he starts by enumerating the Basis sets -$$ - \{ U_1, U_2, U_3, \ldots \} -$$ -and then selecting those pairs $\pi_n = (U_i, U_k)$ with $\overline{U_i} \subseteq U_k$, guess it would help me to give a simple characterisation of these sets, but I didn't succeeded. Then for each $n$ and pair $\pi_n$ he defines the continuous Urysohn function -$$ - f_n(x) = 0 ~ \textrm{ on } \overline{U_i}, \quad f_n(x) = 1 ~ \textrm{ on } X \setminus U_k. -$$ -I would be really glad for hints or references on how I could tackle this Problem of constructing easy and handy metrics, thanks! - -REPLY [4 votes]: Some aspects of the standard metrization theorems are non-constructive, so there is probably no general solution for the problem you pose. -However, for your space with a basis consisting of closed regular ω-languages, there is a simpler way since this is a zero-dimensional space. A second-countable Hausdorff zero-dimensional space $E$ with clopen basis $(B_i)_{i\in\mathbb{N}}$ can always be embedded into Cantor space via the map $f:E \to 2^{\mathbb{N}}$ where each $f(x)$ is the characteristic function of the set $\{i \in \mathbb{N} : x \in B_i\}$ (i.e. $f(x)(i) = 1$ iff $x \in B_i$). Then, $d(f(x),f(y))$, where $d$ is the usual metric on Cantor space, gives a metric for $E$.<|endoftext|> -TITLE: What is the Twisted Trace Formula? -QUESTION [16 upvotes]: I am studying the trace formula using "An Introduction to the Trace Formula" by James Arthur. I would like to understand the twisted trace formula, but unfortunately I never came across a good article that mainly deals with it. For example, in Section 26 of the paper mentioned above, Arthur mentions how the formula looks like in the case of compact quotient, but that is it. I appreciate greatly if you could answer any of the following questions. - -How is the kernel modified? -How do the coarse geometric and spectral expansions look like? How about the fine expansions? -How can the twisted trace formula be useful? Is there something that the twisted trace formula can detect, but the regular trace formula can not? - -Thank you! - -REPLY [9 votes]: For simplicity assume that $G$ is a reductive $\mathbb{Q}$-group that is anisotropic. -Assume that it admits an automorphism $\theta$. -Let $f \in C_c^\infty(G(\mathbb{A}))$. -One has the usual kernel -$$ -K_f(x,y):=\sum_{\gamma \in G(F)}f(x^{-1}\gamma y). -$$ -The trace formula is an expression for -$$ -\mathrm{tr}R(f):=\int_{G(\mathbb{Q}) \backslash G(\mathbb{A})}K_f(x,x)dx -$$ -The twisted trace formula is an expression for -$$ -\mathrm{tr} R(\theta^{-1}\circ f):=\int_{G(\mathbb{Q}) \backslash G(\mathbb{A})}K_f(x,{}^{\theta}x)dx. -$$ -The point is that if one expands $\mathrm{tr} R(\theta^{-1} \circ f)$ in terms of cuspidal automorphic representations, then only those representations that are isomorphic to their $\theta$ conjugates contribute, for the same reason that the trace of an $n \times n$ matrix associated to a cyclic permutation of order $n$ vanishes. Thus the twisted trace formula isolates only part of the spectrum. One then tries to compare this part of the spectrum to spectra of other groups and thereby deduce cases of Langlands functoriality. This is the ultimate goal of what is known as twisted endoscopy. -The morning seminar (arXiv:1204.2888), as reproduced by Labesse and Waldspurger, is a high-level reference for this.<|endoftext|> -TITLE: Generating functions with all non-zero coefficients equal to one -QUESTION [6 upvotes]: I have been wondering if there are any useful generating functions with all non-zero coefficients equal to one. Obviously, the trivial generating function $\frac{1}{1-x}$ has significant applications, as do monomial symmetric functions but for the purposes of this question, we should ignore them. (Edit: As Graham has commented, indicator functions also fall into this category. The best formulation I can think of for why these should be ignored is that they are most interesting for the purpose of taking products of generating functions, rather than being directly used for computation.) -More specifically, such a generating function would (in my mind) have to be multi-variate, enumerate some object of interest and facilitate computations related to that object in a situation where direct computation is not straightforward. As an example of what I would consider cheating, by specializing many variables to one, the generating function in the aforementioned question allows for computing joint distributions of entries in a reduced word with great ease. However, to compute the generating function, it seems to me one would have to enumerate all such words anyways, hence no labor is saved. Were this not the case, this function would be an excellent example. -What would be an example of such a generating function where computation is assisted without being embedded in constructing the generating function? Even if the generating function serves as a useful book keeping device, that would be okay. -Please comment below with any suggested improvements for what should define a "useful" generating function. Explanations for why such a function cannot exist are welcome as well. - -REPLY [7 votes]: Although you say you're not interested in examples with indicator functions, I think there are some examples with indicator functions of polytopes that fit your criteria. I'm writing this answer because I don't think they have the same flavor which you are trying to rule out. -To every lattice point one may associate a Laurent monomial and therefore to every polytope we may associate the sum of all such monomials corresponding to lattice points inside the polytope. This (possibly huge) Laurent polynomial is very important, for example, when one cares about weighted enumeration of lattice points inside polytopes and other facts. -Brion's formula allows one to write this Laurent polynomial as the sum of generating functions of the lattice points of the vertex cones of the polytope. Even though generating functions for cones feel easier than computing this indicator function for the polytope itself, it is not apriori obvious that this makes the computation faster. -However Barvinok's theorem uses Brion's formula together with a decomposition of cones into unimodular cones to conclude that this computation can be done in polynomial time. You can read Barvinok's "A polynomial time algorithm for counting integral points in polyhedra when the dimensionis fixed", Math. Oper. Res. 19 (1994), 769–779. -This is in my opinion one of the nicest results that uses generating functions with 0/1 coefficients in addition to some geometric fact to help compute a combinatorial function very fast. There are of course applications in Ehrhart theory etc.<|endoftext|> -TITLE: Help identify this generalized sign of real representations -QUESTION [7 upvotes]: Let $V$ be a real representation of a finite group $G$. -Define $\mathbb Z[I]_{I\leq G}$ to be the ring over the integers generated by subgroups of $G$ with multiplication corresponding to intersection of subgroups. -Associate an element $\sigma_V\in \mathbb Z[I]_{I\leq G}$ so that -$$\sigma_V= \sum_{I\leq G} n_II$$ -and for all subgroups $H$ of $G$, the following equality holds: -$$\sum_{I\geq H}n_I=(-1)^{dim V^H}$$ -Question: Is $\sigma_V$ some standard representation theoretic thingy? (My knowledge of representation theory doesn't go much further than Wikipedia...) -I am using $\sigma_V$ as a kind of generalized sign of the determinant of complex conjugation on $V\oplus iV$. I think that it has the following nice properties: -$$\sigma_{V\oplus W}=\sigma_V\sigma_W$$ -$$\sigma_V^2=1G$$ -If $I\leq G$ is normal, then $n_I$ is an integer divisible by the index of $I\leq G$. -A second question: is $n_{\{1\}}$ always $0$ or $\pm \lvert G\rvert$? This is the case in the few examples I have computed, but I have no general reason to expect it to be true. -For example: if $G=S_n$, and $\rho$ is the representation which permutes the coordinates of $\mathbb R^n$, then $n_I=(-1)^k k!$ if $I$ is a subgroup of $S_n$ corresponding to partitioning $\{1,\dotsc,n\}$ into $k$ nonempty subsets, and $n_I=0$ otherwise. - -REPLY [3 votes]: I think the answer to your second question is "no". Note first that by the M\"obius inversion formula (see for example Chapter 3 of Richard Stanley's ``Enumerative Combinatorics, Volume 1"), for any real $G$-module $V$, -$$ -n_{\{1\}}=\sum_{H \leq G}\mu(1,H)(-1)^{\dim V^H}, -$$ -where $\mu$ is the M\"obius function on the subgroup lattice of $G$. (As an aside, it follows from this that if all the dimensions of the fixed point subspaces have the same parity, then $n_{\{1\}}=0$.) -Now let $V$ be the regular module for $G$. Then, for each $H \leq G$, -$$ -\dim V^H=[G:H]. -$$ -Let ${\mathcal S}_2(G)$ be the set of all subgroups of $G$ that contain at least one Sylow $2$-subgroup of $G$ (that is, have odd index in $G$). Since the sum of $\mu(1,H)$ over all subgroups $H$ of $G$ is zero, -$$ -n_{\{1\}}=-2\sum_{H \in {\mathcal S}_2(G)}\mu(1,H). -$$ -Now one can take $G=PSL_2(31)$. A Sylow $2$-subgroup of $G$ is dihedral of order $32$ and is maximal in $G$. By results of Philip Hall (the paper is called "The Eulerian functions of a group", and can be found in Hall's collected works), $\mu(1,G)=2|G|$ and, for any Sylow $2$-subgroup $D$ of $G$, $\mu(1,D)=0$. So, -$$ -n_{\{1\}}=-4|G|. -$$ -A similar but slightly more complicated argument argument should yield the same answer if $31$ is replaced by any prime $p$ satisfying $p \equiv \pm 1 \bmod 16$ and $p \equiv \pm 1 \bmod 5$. (One has to handle dihedral overgroups of Sylow $2$-subgroups, but these will satisfy $\mu(1,D)=0$.)<|endoftext|> -TITLE: Reference Request: Probability and (Nonlinear) PDEs -QUESTION [8 upvotes]: I'm a graduate student interested in learning about probability and (mostly evolutionary) PDEs, just for fun (and as an excuse to learn some probability). I'm mostly interested in things along the lines of: what can we (almost surely) prove for randomized (if that's the right word) initial data which fails (or has yet to be proven) in the deterministic setting. I'm less interested in, e.g., stochastic PDE. I'm also not entirely opposed to there being some elliptic theory, but I'd like the focus to be primarily on evolution equations. -Ideally I'd like to read a book (or (series of) paper(s)) that covers a variety of topics and gives (with proof) "representative" results without getting too bogged down in details (e.g., in the name of maximal generality); something like Tao's Nonlinear Dispersive Equations (but for the probabilistic setting). I've learned standard linear and nonlinear (deterministic) PDE theory and (consequently) real analysis/measure theory, but no probability. The PDE book wouldn't have to cover probability, but if it doesn't, I'd appreciate a suggested probability book to learn what I'd need to understand the PDE material. Thanks! - -REPLY [2 votes]: Caveat: I am not an expert on the subject. My knowledge of it consists of having read a paper or two, private conversations with three or four actual experts a few years ago, and sitting through five or six lectures/seminars on the topic. So the summary below is very biased and incomplete. -Caveat 2: I hope I am interpreting correctly what you meant by "randomised initial data". - -I am not aware of any books that cover what you are interested in. The field is still quite active and I am not aware of anyone actually sitting down and summarising the development in the past 20 years or so. -A somewhat recent review article that is sufficiently introductory was written by Burq and Tzvetkov. You should also consult its references. -A paper that everyone refers to is Lebowitz, Rose, and Speer; it will give you at least the initial point of view of studying invariant Gibbs measures for nonlinear dispersive equations. In terms of the pure mathematics, much of the early breakthroughs seem to be due to Bourgain 1 2 3. He gave a quick review of the state of the art in 2000. -Since then, there has been much more development for construction of the invariant Gibbs measure and almost sure existence of solution below critical regularity. Some of the names of people whom I know worked on, or are working on, the subject include (in no particular order): Jean Bourgain, Tadahiro Oh, Nicolay Tzvetkov, Aynur Bulut, Nicolas Burq, Kay Kirkpatrick... Anyone of those people would be more qualified than I to answer this question (unfortunately none of them appears to be on MathOverflow). You should check the arXiv for their papers on the subject, and I am pretty sure Tadahiro and Aynur at least (whom I know better personally) will be happy to answer some of your questions by e-mail. -Last but not least, you may want to check out Staffilani's slides on the subject: 1 2. (The second link I just stumbled upon now while looking for the first one; it looks like a nice introduction!)<|endoftext|> -TITLE: Intuition behind Pincus' "injectively bounded statements" -QUESTION [7 upvotes]: In - -David Pincus, Zermelo-Fraenkel Consistency Results by Fraenkel-Mostowski Methods, - The Journal of Symbolic Logic, Vol. 37, No. 4 (Dec., 1972), pp. 721-743 - -Pincus introduces the notion of injectively bounded statements, which he proves are sentences which can be transferred from a (permutation) model of ZFA to a (symmetric) model of ZF. I have no intuition for what these statements are, and he only gives a couple of examples (allow me to ignore the special case of projectively bounded statements, I am after generality here). -I would like, if possible, a more structural explanation of what it means for a statement to be injectively bounded, rather that something that looks like a mess of codings via ordinals. - -REPLY [5 votes]: (Note: this isn't something I really know, so this might be wildly off base.) -To start with, let's look at a weaker transfer principle: the Jech-Sochor Embedding Theorem. -Jech-Sochor says that sentences depending only on a bounded amount of the cumulative hierarchy above the set $A$ of atoms can be "passed over" to a model of genuine $\mathsf{ZF}$. More precisely, Jech-Sochor states that - -Let $V$ be a transitive model of $\mathsf{ZFA}$ with a set $A$ of atoms and let $\gamma\in \mathsf{Ord}^V$. Then there is a model $W$ of $\mathsf{ZF}$ and an embedding $i: V\rightarrow W: x\mapsto \tilde{x}$ such that $(P_\gamma(A)^V, \in^V)\cong (P_\gamma(\tilde{A})^W, \in^W)$. - -So statements of "fixed depth" can be transferred. For example, "non-well-orderability" can be preserved by setting $\gamma=\omega+2$ (really, we just care about the powerset, but we also want to talk about maps from $\omega$ so we need to go up $\omega+1$ many levels to get $\omega$ into $P_\gamma(A)$, and then one more level to get the desired maps). -In Pincus, this property of the truth of $\phi(X)$ only depending on some fixed level of the cumulative hierarchy over $X$ is called boundability; so, for example, on page 722 Pincus phrases the Jech-Sochor theorem as: - -"A boundable statement is transferable." - -The question is whether we can improve this result to transfer statements that don't necessarily depend just on $P_\gamma(X)$ for some fixed $\gamma$, but are still "locally determined" in some sense. This "locally determined" is his condition that - -$\vert x\vert\le\sigma(y)$ - -in a (sur/in)jectively boundable statement. So now we've switched from caring about the number of powersets required to reach a set, to caring about its "cardinality" being small. Note, though, that the bound on the size of $x$ itself must be boundable, so the idea of Jech-Sochor isn't really going away. -The requirement that each element of $x$ have no intersection with the transitive closure of $y$ seems more technical, and I'm not sure if there's a clean intuition behind it. His example 2B1 shows why I feel okay not caring about this part of the definition too much - in the end, we take some class of potential counterexamples (field expansions that might be algebraic closures) but which don't satisfy this disjointness condition, and just slide them over in an appropriately definable manner (in this case, $x\mapsto \lbrace (w, y): w\in x\rbrace$). I suspect that in general something like this will be possible without much difficulty (although I am not sure on this point). -So what we're left with is that a (sur/in)jectively bounded sentence is essentially a $\Sigma_2$-sentence where - -the universal quantification is taken over sets of boundable size, and - -the matrix of the sentence is boundable in the original Jech-Sochor sense. - - -(Of course, this isn't really $\Sigma_2$, since this "matrix" might well have quantifiers, but oh well.) What's really new here is this universal quantifier. -At this point it would be nice to see a injectively boundable statement which is not boundable. I think the clearest example is Pincus' 2B6 on page 724 (actually, he uses this as an example of an injectively boundable statement which is not surjectively boundable, but that distinction seems less intuitively crucial to me). The statement here is - -"Every infinite partially ordered set has either an infinite chain or an infinite antichain but there is an infinite, Dedekind-finite set." - -This sentence is a conjunction $\Phi\wedge\Psi$, where - -$\Phi\equiv$ "Every infinite poset has a chain or antichain," and - -$\Psi\equiv$"There is a strictly Dedekind-finite set." - - -Now $\Psi$ already transferable by Jech-Sochor (since "is a Dedekind-finite set" depends just on the powerset). $\Phi$, however, is a bit trickier, since it involves quantifying over the class of all posets! And this certainly can't be done with Jech-Sochor. -Instead, we use a trick. First, we can rewrite $\Phi$: $$\Phi\equiv "\forall x(\vert x\vert_-\le\omega\implies (\text{ if $x$ is an infinite poset, then $x$ has an infinite chain or antichain})) "$$ since if $\vert x\vert_->\omega$ then we can already build a chain or antichain without choice. Now the conclusion of this implication is a boundable formula of $x$, since it really only talks about the powerset! So even though the whole sentence $\Phi$ wasn't boundable, by massaging it a bit we got it to the point where the universal quantifier causing all the trouble was just over sets of small "size," and this was enough for it to be injectively boundable. -Note that here, the specific notion of size we use is crucial: $\Phi$ isn't surjectively boundable, since $\vert x\vert^-$ can behave more weirdly on Dedekind-finite sets. So this is what should motivate injective boundability: it's the broadest obvious way to push up the strength of Jech-Sochor to allow some non-powerset-bounded universal quantification. - -Hopefully, this helps. Tl;dr: injectively-bounded properties will generally look like "This simple property $(\neg\Psi)$ does not always $(y)$ have small witnesses $(\forall x[\vert x\vert_-< . . .])$."<|endoftext|> -TITLE: Reference for tetrahedral Coxeter group -QUESTION [6 upvotes]: Let G be the group with 4 generators, each of order 2, such that the product of any 2, say ab, has order 3 (i.e., ababab=e). That is, this is an infinite reflection group with Coxeter diagram a tetrahedron. I am looking for references for this group... - -REPLY [3 votes]: The group itself shall be the group generated by reflections in the sides of a regular ideal tetrahedron, whose dihedral angles are all $\pi/3$. For a reference, there are many Coxeter diagrams listed in a paper by Johnson, Kellerhals, Ratcliffe and Tschantz, -called "The size of a hyperbolic simplex". -Please mind the fact that you may have to look at the barycentric subdivision of your simplex before you find its counterpart (a simplex from the subdivision, which is an orthoscheme) in their table. -Definitely, the ideal simplex reflection group contains interesting subgroups, which are manifold groups. There should be the eight-knot group in there, since the figure-eight complement comes from glueing two regular ideal tetrahedra. I suppose that Neil's references shed more light on this kind of questions. Hope my reply describes in more geometric detail the group you were interested in.<|endoftext|> -TITLE: The fundamental group of a $3$-manifold with a boundary of genus $>0$ -QUESTION [7 upvotes]: Let $M$ be an orientable $3$-manifold with connected boundary $\Sigma_g$, a surface of genus $g>0$. -I would like to find a reference to the following two statements. -1) $\pi_1(M)\ne 0$. -2) $\pi_1(M)\ne \pi_1(\Sigma_g)$. -2)' If 2) is too hard I would be happy just to know that the map $\pi_1(\Sigma_g)\to M$ -is not an isomorphism. -I think I can prove the first statement by contradiction. If $\pi_1(M)$ were trivial it would stay so after gluing a handlebody to $\Sigma_g$. In the resulting simply-connected manifold one can chose loops inside $M$ that have non-zero linking number with loops inside the handlebody hence they are not null-homologous in $M$. Hence we get a contradiction with $\pi_1(M)=0$. -But I don't see how to prove 2), and this might be hard. I would be grateful for some tips. - -REPLY [8 votes]: The magic words are "half lives, half dies". See Hatcher's notes, Lemma 3.5 (and read the rest of the notes, while you are at it :))<|endoftext|> -TITLE: historical antecedents of mathematical talks -QUESTION [8 upvotes]: Is there a general reference of how mathematical talks, say academic talks, evolve in history? Before the International Congress of mthematics, is there any antecedent of todays talks? - -REPLY [4 votes]: Well, if you are interested to learn how Euler lectured on math and physics to a lay audience, you might want to take a look at the Letters of Euler to a German Princess. This book collects the lectures on elementary science that Euler gave in Berlin to the Princess of Anhalt Dessau, in the early 1760's. Dominic Klyve gives a nice overview of "Euler as Master Teacher". -For a more technical/advanced talk, the lecture which Riemann delivered on June 10, 1854 has been preserved here. Dedekind writes about this lecture: - -Gauss sat at the lecture, which surpassed all his expectations and on the way back from the colloquium - meeting he spoke to Wilhelm Weber, with the greatest appreciation, and - with an excitement rare for him, about the depth of the ideas - presented by Riemann.<|endoftext|> -TITLE: What's a noncommutative set? -QUESTION [22 upvotes]: This issue is for logicians and operator algebraists (but also for anyone who is interested). -Let's start by short reminders on von Neumann algebra (for more details, see [J], [T], [W]): -Let $H$ be a separable Hilbert space and $B(H)$ the algebra of bounded operators. -Definition: A von Neumann algebra is a *-subalgebra $M \subset B(H)$ stable under bicommutant: -$M^{*} = M$ and $M'' = M$. -Theorem: The abelian von Neumann algebras are exactly the algebras $L^{\infty}(X)$ with $(X, \mu)$ a standard measure space. They are isomorphic to one of the following: - -$l^{\infty}(\{1,2,...,n \})$, $n \geq 1$ -$l^{\infty}(\mathbb{N})$ -$L^{\infty}([0,1])$ -$L^{\infty}([0,1]\cup \{1,2,...,n \})$ -$L^{\infty}([0,1]\cup \mathbb{N})$ - -Noncommutative philosophy: There are various schools of noncommutative philosophy, here is the school close to operator algebras. This issue is not about philosophy, so I will explain it quickly (for more details see for example the introduction of this book). First an intuitive idea : in the same way as there are classical physics and quantum physics, there are classical mathematics and quantum mathematics. What does it mean in practice ? It means the following : in the classical mathematics there are many different structures, for example, the measurable, topological or Riemannian spaces. The point is to encode each structure by using the framework of commutative operator algebras. For the previous examples, it's the commutative von Neumann algebras, C$^{*}$-algebras and spectral triples. Now if we take these operator algebraic structures and if we remove the commutativity, we obtain what we call noncommutative analogues : noncommutative measurable, topological or Riemannian spaces. -This school explores noncommutative analogues of more and more structured objects, it goes in one direction. My point is to question about the other direction (back to the Source) : -What's the noncommutative analogue of a set (called a noncommutative set) ? - -What is a noncommutative set? - -The von Neumann algebras of the standard measure space $[0,1]$, $[0,1]\cup \{1,2,...,n \}$ and $[0,1]\cup \mathbb{N}$ are not isomorphic, but as sets, these spaces are isomorphic (i.e., same cardinal). -Is there a natural equivalence relation $\sim$ on the von Neumann algebras, forgetting the measure space but remembering the set space, on abelian von Neumann algebras? -Remark: If $M \sim N$, we could say that they are isomorphic as noncommutative sets. -The equivalence class could be called the quantum cardinal (a link with cyclic subfactor theory?). -Are there noncommutative analogues of the ZFC axioms ? -What I'm looking for seems different of what is called quantum set in the literature... - -REPLY [2 votes]: Andre Kornell wrote a new paper entiteld Quantum sets, available on arxiv from yesterday: arXiv:1804.00581. The first sentence of the paper is the following: - -This paper concerns the quantum generalization of sets, in the sense - of noncommutative geometry. - -Robin Giles and Hans Kummer defined (in this paper published in 1971) a quantum set as an atomic von Neumann algebra (i.e. whose projection lattice is an atomic lattice, which means that every projection contains a minimal projection), in other words, a $\ell^{\infty}$-direct sum of type ${\rm I}$ factors. -The above paper of Andre Kornell takes a smaller class: there, a quantum set is a direct sum of type ${\rm I}$ finite factors, i.e. a direct sum of matrix algebras $M_n(\mathbb{C})$. -These notions should be called atomic noncommutative sets (of type ${\rm I}$).<|endoftext|> -TITLE: Class number of real maximal subfield of cyclotomic fields -QUESTION [9 upvotes]: Let $p$ be a prime number and $h_p^+$ the class number of $\mathbb{Q}(\zeta_p + \zeta_p^{-1})$. What is known about the values of $p$ for which $h_p^+ = 1$? -Are there infinitely many? Finitely many? Something else? -(As usual, $\zeta_p$ denotes a primitive $p$-th root of unity.) - -REPLY [5 votes]: I thought it should be mentioned that van der Linden's result (which depends on Odlyzko's discriminant bounds) has recently been improved upon by a more careful accounting of the contribution of prime ideals to the explicit formula of the Dedekind zeta function of the Hilbert class field of $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$. -The value of $h_p^+$ is now known up to 151 (and up to 241 under GRH). Please see http://arxiv.org/abs/1407.2373.<|endoftext|> -TITLE: Are dual spaces barreled? -QUESTION [6 upvotes]: Let $X$ denote a topological affine space (with no additional assumptions). Let $X^*$ denote its dual space of continuous affine functionals, equipped with the weak-$*$ topology. It is easy to see that $X^*$ is a topological vector space, since it has a zero functional; it is also locally convex. -Is the dual space necessarily a barreled space? If so, why? If not, could you provide a counterexample? -The Wikipedia page says, "locally convex spaces which are Baire spaces are barrelled." Is the dual space necessarily a Baire space? - -REPLY [9 votes]: No, it need not be barreled. -Let $X$ be an infinite-dimensional normed vector space. The closed unit ball $B_{X^\ast}$ in the dual space $X^\ast$ is a barrel in the weak-$\ast$ topology: it is compact, convex, balanced and absorbing. It is not a weak-$\ast$ neighborhood of zero because its interior is empty: every basic open set in the weak-$\ast$ topology contains an affine subspace of finite codimension. -This also shows directly that the dual space of $X$ is not Baire: $X^\ast = \bigcup_{n=1}^\infty n B_{X^\ast}$ shows that $X^\ast$ is a countable union of closed nowhere dense sets.<|endoftext|> -TITLE: Kakeya and Nikodym maximal functions -QUESTION [7 upvotes]: I've been working through part of Terry Tao's 1999 article "The Bochner-Riesz Conjecture Implies the Restriction Conjecture." (It appeared in the Duke Mathematical Journal.) A little more specifically, I care about the proof of Theorem 4.10, see below. -Here's the necessary background. For $0 < \delta \ll 1$, and $f$ a compactly supported function defined on $\mathbb{R}^{n}$, the Kakeya and Nikodym maximal functions of $f$, denoted $f_{\delta}^{*}$ and $f_{\delta}^{**}$, are defined as follows. First, $f_{\delta}^{*}$ is the function on $S^{n - 1}$ given by -$$ -f_{\delta}^{*}(\omega) -= \sup_{T} \frac{1}{|T|} \int_{T} | f(y) | \, dy, -$$ -where the supremum is taken over all right cylindrical tubes $T$ having length $1$, radius $\delta$, and axis parallel to $\omega$. Next, $f_{\delta}^{**}$ is the function on $\mathbb{R}^{n}$ given by -$$ -f_{\delta}^{**}(x) -= \sup_{T} \frac{1}{|T|} \int_{T} | f(y) | \, dy, -$$ -where this time the supremum is over all right cylindrical tubes $T$ having length $1$, radius $\delta$, and which contain $x$. (In both definitions, $|T|$ denotes the Lebesgue measure of $T$.) -Having defined these maximal functions (or maximal operators), what we're interested in are the size of the $L^{p}$ bounds for them. Tao uses $K(p, \alpha)$ to denote the estimate -$$ -\| f_{\delta}^{*} \|_{p} -\leq C_{n, p, \alpha} \delta^{- (n/p - 1) - \alpha} \| f \|_{p}, -$$ -and likewise he uses $N(p, \alpha)$ to denote the estimate -$$ -\| f_{\delta}^{**} \|_{p} -\leq C_{n, p, \alpha} \delta^{- (n/p - 1) - \alpha} \| f \|_{p}. -$$ -It's conjectured that $K(p, \alpha)$ and $N(p, \alpha)$ hold for all $1 \leq p \leq n$ and all $\alpha > 0$, although this is not presently known. The idea here is that $K(p, \alpha)$ and $N(p, \alpha)$ give estimates for the norms of the two maximal operators as $\delta \to 0$, i.e. as we consider thinner and thinner tubes. -Now on to what I'm specifically interested in. Part of Tao's Theorem 4.10 is that $K(p, \alpha)$ implies $N(p, \alpha)$, and this is what I'm struggling with. He first makes a reduction to showing a "frozen" estimate for the Nikodym maximal function, namely -$$ -\| f_{\delta}^{**}(\underline{x}, 0) \|_{p} -\leq C_{n, p, \alpha} \delta^{- (n/p - 1) - \alpha} \| f \|_{p}, -$$ -where $\mathbb{R}^{n}$ has been parameterized by $x = (\underline{x}, x_{n})$. I'm okay with this reduction, as well as with another reduction whereby we assume that $f$ is supported in the "slab" $0 < x_{n} \leq 1$. Tao makes one further reduction which I do not understand the argument for: that we may assume $f$ is supported in the slab $1/2 < x_{n} \leq 1$. He says, "the condition $\alpha < (n + 1) / p$ and scaling considerations ensure that the other contributions are more favourable than this main term." That is, if $f$ is supported in $0 < x_{n} \leq 1$, the contribution to $f_{\delta}^{**}$ from the portion of $f$ supported in $0 < x_{n} \leq 1/2$ is in some less than the contribution from the portion of $f$ supported in $1/2 < x_{n} \leq 1$. -I'm guessing that the argument for this would involve a dyadic decomposition of $f$ into the slabs $2^{-k - 1} < x_{n} \leq 2^{-k}$ and some sort of re-scaling to take these thinner slabs to the slab $1/2 < x_{n} \leq 1$, but I've tried a couple different ideas and neither of them worked for me. So what, more precisely, is the argument reducing a function on $0 < x_{n} \leq 1$ to a function $1/2 < x_{n} \leq 1$? - -REPLY [9 votes]: Yes, the argument is dyadic decomposition followed by rescaling. I think I forgot to mention in the paper one initial reduction, which is to only consider the portion of the Nikodym maximal function coming from tubes which make an angle of at most 1/10 (say) with the basis vector $e_n$; note that one can reduce to this case by a finite partition of unity of the unit sphere. Once one restricts to such nearly-vertical tubes, the scaling argument should go through (averaging on $\delta$-tubes centred at a point $(\underline{x},0)$ applied to a function supported on the slab $\{ 2^{-k-1} < x_n \leq 2^{-k} \}$ will rescale to an average on $2^k \delta$-tubes centred at $(2^k \underline{x},0)$ supported on the slab $\{ 1/2 < x_n \leq 1 \}$). (It may be helpful to first consider the extreme case $2^{-k} = \delta$, which is an easy case of the Nikodym conjecture; the remaining cases are an interpolation between this easy case and the most difficult case $k=0$, which is intuitively why the rescaling is guaranteed to be favorable.)<|endoftext|> -TITLE: An example of a proof that is explanatory but not beautiful? (or vice versa) -QUESTION [29 upvotes]: This question has a philosophical bent, but hopefully it will evoke straightforward, mathematical answers that would be appropriate for this list (like my earlier question about beautiful proofs appropriate for high school level.) -The background is this: we routinely distinguish between proofs that explain and proofs that demonstrate. This distinction has been around at least since Aristotle's time, but it is an open question, for instance in contemporary philosophy of mathematics, what explanation really means. One recent suggestion has been that explanation might be related to beauty. It seems reasonable that explanatory proofs are nicer than non-explanatory ones in SOME way, but are they necessarily more beautiful? And similarly, is beauty necessary for explanation? It seems a good way to attempt to answer these question is to look at a bunch of good examples. And it seems a good way to get examples is to ask mathematicians, which is why I post the question here. -To clarify: the question is not at all to discuss the nature of explanation or beauty (if you want to discuss, I can give you my email address and we can chat offline). The purpose to is collect some good mathematical examples that help understand the relation between beauty and explanation in mathematics. -Examples that would be relevant: beautiful proofs that are not explanatory, explanatory proofs that are not beautiful. -Thanks in advance. - -REPLY [3 votes]: In the game of 'chomp', the first player always has a winning strategy, by a simple and (I think) beautiful argument (he can 'steal' a possible winning strategy from the second player). However, as far as I know, apart from special cases, there is no general known way of describing the winning strategy (which would maybe count as more explanatory)<|endoftext|> -TITLE: A double cover of $\mathbb{P}^n \times \mathbb{P}^n$ -QUESTION [8 upvotes]: Let $X$ be a double cover of $\mathbb{P}^n \times \mathbb{P}^n$ branching along bi-degree $(d,e)$ hypersurface. Is it possible to write $X$ as a complete intersection of hypersurfaces in some weighted projective space (for certain $d,e$)? -This is motivated by the fact that a double cover of $\mathbb{P}^n$ branching along degree $2d$ hypersurface can be written as a hypersurface in $\mathbb{P}^n_{(1^{n+1},d)}$. - -REPLY [11 votes]: The answer is no. The reason is that the Picard group of a complete intersection of dimension $\ge 3$ in a weighted projective space is of rank 1 and the Picard group of the double cover is of rank $\ge 2$.<|endoftext|> -TITLE: Research-level mathematical bookstores -QUESTION [30 upvotes]: I'm interested in compiling a list of bookstores around the world that stock a good selection of high-level mathematical books. The aim is so that a mathematician travelling, or on holiday, can easily check the list to see if there is a book store nearby that she or he could happily spend an hour browsing in. -Some particulars: - -It must be a physical bookstore. -It should have a good selection of research level books. This criterion is designed to exclude stores that stock primarily popular books and undergraduate textbooks, such as some university and chain bookshops. - -REPLY [10 votes]: In Paris, there's Gibert (26-34 Boulevard Saint-Michel). It's a giant bookstore with several floors of books including one whole floor devoted to scientific books. A sizable part of that floor is dedicated to maths, and there are enough books that they're sorted by subject (number theory, algebra, analysis, ...). During Springer's yellow sale, there's a big table overflowing with cheap Springer books, including many tomes of Bourbaki. If you go, there's a high risk of running into somebody you know.<|endoftext|> -TITLE: Two infinite dimensional algebras such that the center of their tensor product is bigger than the tensor product of their centers -QUESTION [15 upvotes]: I am searching for two infinite dimensional algebras such that the center of their tensor product is bigger than the tensor product of their centers. Who knows of such examples? Thanks a lot. - -REPLY [5 votes]: It seems to me that a simple example is as follows: Let $R = \mathbb{Z}$, let $A$ be the noncommutative $\mathbb{Z}$-algebra on two generators $x$, $y$, obeying $xy=yx+2$ and let $B = \mathbb{Z}/2 \mathbb{Z}$. -Note that $A$ is basically the Weyl algebra and its center is $\mathbb{Z}$. Here is a more detailed argument: Let $W = \mathbb{Q}\langle x,y \rangle / (xy-yx-2)$; $W$ is isomorphic to the Weyl algebra. There is an obvious map $A \to W$. We claim that it is injective. Proof: By a PBW like argument, $x^i y^j$ spans $A$ over $\mathbb{Z}$, and these monomials are linearly independent in $W$. $\square$ So $x^i y^j$ is a $\mathbb{Z}$-basis of $A$, and we can compute in it in the usual way to see that $\mathbb{Z} = Z(A)$. -So $Z(A) \otimes_R Z(B)$ is $\mathbb{F}_2$, but $A \otimes_R B$ is $\mathbb{F}_2[x,y]$.<|endoftext|> -TITLE: Class numbers of orders -QUESTION [8 upvotes]: Consider an order $R$ in a number field $L$. Let $C_R$ be the set of $R$-fractional ideals modulo $L^\times$. Let $O$ be the maximal order in $L$, and $C_O$ be the class group of $O$. -My question: Is there a formula that relates $\# C_R$ with $\# C_O$, that involves perhaps the conductor of $R$? -A note: Beware that in the definition of $C_R$ I really consider fractional ideals: I do not ask that they are non-singular, so $C_R$ is not the Picard group of the order $R$ (In fact, it is not even a group!). - -REPLY [6 votes]: The ideal class semigroups mentioned in the question got studied in this setting (orders of number fields) and in other and more general ones by various authors in recent years. -A starting references is: - -Zanardo, P.; Zannier, U. The class semigroup of orders in number fields. - Math. Proc. Cambridge Philos. Soc. 115 (1994), 379–391. - -They show that this semigroup is a Clifford semigroup for orders of quadratic fields, but for degree greater 2, there always is some order such that the respective class semigroup is not a Clifford semigroup. -There is also an older paper on this theme (that it appears was overlooked in recent literature for some time but reapperas in still more recent one) - -Dade, E. C.; Taussky, O.; Zassenhaus, H. On the theory of orders, in paricular on the semigroup of ideal classes and genera of an order in an algebraic number field. Math. Ann. 148 (1962) 31–64. - -Starting from these two papers and looking for papers that quote them in MathSciNet for example one will find several more recent contributions. Investigating these semiclass groups; but it seems (but I donot have a good overview) the emphasis is more to generalize to more general structures (say, other domains than just orders) than more details in the number-theoretic setting. -Yet, in particular the early papers (I do not know for the recent ones) contain also some explicit information on these semigroups, but as commented earlier I would not know of in some sense simple descriptions (but again my knowledge is superficial here).<|endoftext|> -TITLE: von neumann algebras and measurable spaces -QUESTION [8 upvotes]: I've read some pages on links between von neumann (VN) algebras and measurable spaces (Spectra of $C^*$ algebras and Non-commutative geometry from von Neumann algebras?), but I can't get the following: - -VN algebras are C*-algebras -C*-algebras are equivalent to compact separated topological spaces -VN are equivalent to measurable spaces - -How come then we can find measurable spaces which are not topological spaces if any VN algebra is a C*-algebra? I suspect my question to be silly but I don't have the answer. -Thanks for your help. - -REPLY [8 votes]: I highly recommend Segal's original paper Equivalences of Measure Spaces -[American Journal of Mathematics -Vol. 73, No. 2 (1951), pp. 275-313], where he introduced localizable spaces, since this was before the terminology took off. -In it he shows that an arbitrary measure space has maximal abelian (i.e. strongly closed) $L^\infty$ algebra if and only if it is localizable. -So there do exist measure spaces which for which $L^\infty(X)$ is not a von Neumann algebra. But (and it's a big but), if you're at all interested in integration, then the class of localizable measure spaces is really as large a class of interesting measure spaces as there is. The following excerpt explains why: -The class of measure spaces with these properties (we call such spaces "localizable") constitutes in some ways a more natural generalization of the $\sigma$-finite measure spaces, than the class of arbitrary measure spaces. In particular, for a measure space to be localizable is equivalent to the validity for the space of the conclusion of the Radon-Nikodym theorem, or alternatively to the conclusion of the Riesz representation theorem for continuous linear functionals on the Banach space of integrable functions. Every measure space is metrically equivalent (by which we mean there is a measure-preserving isomorphism between the $\sigma$-finite measure rings - roughly speaking this means the spaces are equivalent as far as integration over them is concerned) to a localizable space, and this latter space is essentially unique. -The point of view being held here is that measure theory is not about defining odd pathological measure spaces, but about all the cool stuff you can build on the nicer spaces: integration theory, probability theory, dynamical systems, stochastic processes, ergodic theory. And all that interesting stuff can be done, or is being worked out, in von Neumann algebras. -So von Neumann algebras capture the integration bit of measure theory. -Regarding your point in the comments: "since in vNT all what distinguishes measure theory from topology (mainly measurable spaces not being topological) is absent (we are restricted to LMS, which are topological)?" The category of localizable measure spaces and measurable maps is equivalent to the category of hyperstonean topological spaces and hyperstonean maps - this is not a subcategory of topological spaces and continuous maps - the maps between these spaces have to preserve an extra structure, namely a family of normal measures. -(Just like the category of topological spaces is not a subcategory of sets, because again they have different morphisms, but there is a forgetful functor Top-->Set).<|endoftext|> -TITLE: Weitzenböck Identity for $\Delta_{\bar{\partial}_E}$ -QUESTION [9 upvotes]: This question is related to this MO question and this MSE question. - -Let $E$ be a hermitian holomorphic vector bundle over a hermitian manifold $X$. The bundle $\bigwedge^{\bullet,\bullet}X\otimes E$ has an induced hermitian metric. As $E$ is holomorphic, there is a Dolbeault operator $\bar{\partial}_E$ which acts on -$$\Omega^{\bullet,\bullet}(X, E) = \Gamma\left(X, \bigwedge\nolimits^{\!\bullet,\bullet}X\otimes E\right),$$ -and it has adjoint $\bar{\partial}_E^*$. Both $\bar{\partial}_E$ and $\bar{\partial}_E^*$ are first order differential operators and -$$\Delta_{\bar{\partial}_E} = \bar{\partial}_E\bar{\partial}_E^* + \bar{\partial}_E^*\bar{\partial}_E$$ -is a second order differential operator. Furthermore, $2\Delta_{\bar{\partial}_E}$ is a generalised Laplacian; that is, -$$\sigma_2(2\Delta_{\bar{\partial}_E})(\xi) = - \|\xi\|^2\operatorname{id}_{E_x}$$ -where $x = \pi(\xi)$ and $\pi : TX \to X$ is the natural projection map. - -Proposition 9.1.27 (Nicolaesu's Lectures on the Geometry of Manifolds) -Let $L$ be a generalised Laplacian on $F$. There is a unique metric connection $\nabla$ on $F$ and $\mathfrak{R} \in \operatorname{End}(F)$ such that -$$L = \nabla^*\nabla + \mathfrak{R}.$$ - -I want to know what the connection is for the generalised Laplacian $2\Delta_{\bar{\partial}_E}$ under the assumption that $X$ is Kähler. If I'm not mistaken, the following result is what I need. - -Proposition 3.67 (Berline, Getzler, & Vergne's Heat Kernels and Dirac Operators) -Let $\mathcal{W}$ be a holomorphic vector bundle with hermitian metric on a Kähler manifold $X$. The tensor product of the Levi-Civita connection with the canonical connection of $\mathcal{W}$ is a Clifford connection on the Clifford module $\bigwedge(T^{0,1}X)^*\otimes\mathcal{W}$, with associated Dirac operator $\sqrt{2}(\bar{\partial}_E + \bar{\partial}_E^*)$. - -First of all, $\sqrt{2}(\bar{\partial}_E+\bar{\partial}_E^*)$ is a Dirac operator for $2\Delta_{\bar{\partial}_E}$, i.e -$$\left(\sqrt{2}(\bar{\partial}_E+\bar{\partial}_E^*)\right)^2 = 2\Delta_{\bar{\partial}_E}.$$ -If $\mathcal{W} = \bigwedge(T^{1,0}X)^*\otimes E$ where the first factor has the metric induced by the metric on $X$ and $E$ is equipped with a metric such that the tensor product metric agrees with the metric on $\mathcal{W}$. Then -$$\bigwedge(T^{0,1}X)^*\otimes\bigwedge(T^{1,0}X)^*\otimes E \cong \bigwedge\nolimits^{\!\bullet, \bullet}X\otimes E.$$ -The connection on $\bigwedge(T^{0,1}X)^*$ is the Levi-Civita connection, and the connection on $\bigwedge(T^{1,0}X)^*$ is the Chern connection, but as the manifold is Kähler, the Chern connection coincides with the Levi-Civita connection. Therefore on -$$\left(\bigwedge(T^{0,1}X)^*\otimes\bigwedge(T^{1,0}X)^*\right)\otimes E \cong\bigwedge\nolimits^{\!\bullet, \bullet}X\otimes E$$ -the connection is given by the Levi-Civita connection on the first factor and the Chern connection on the second factor. - - -Is this the metric connection from Proposition 9.1.27 for the generalised Laplacian $2\Delta_{\bar{\partial}_E}$? -If so, is it also the metric connection for the generalised Laplacian $2\Delta_{\partial_E}$? -Can any of this be generalised to the case where $X$ is just a hermitian manifold? - - -I'm still getting used to the whole language involved in this setup, so if it seems that I've missed something, please let me know. - -Added later: Thanks to the two answers, several of my questions have been answered. If $X$ is a Kähler manifold, then on $\Omega^{p,q}(X, E)$ we have $\Delta_{\bar{\partial}_E} = \nabla^*\nabla + \mathfrak{R}_1$ and $\Delta_{\partial_E} = \nabla^*\nabla + \mathfrak{R}_2$ where, in both cases, $\nabla$ is the connection induced by the Chern connection. Combining these two decompositions, we have $\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + (\mathfrak{R}_1 - \mathfrak{R}_2)$. As $\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + [iF_{\nabla}, \Lambda]$ we see that $\mathfrak{R}_1 - \mathfrak{R}_2 = [iF_{\nabla}, \Lambda]$. -In the case that $X$ is a hermitian manifold, I am struggling to see how to generalise the above consideration. I know that $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + [iF_{\nabla}, \Lambda] + [\bar{\partial}_E, [L, \Lambda_{\bar{\partial}\omega}]] - [\partial_E, [L, \Lambda_{\partial\omega}]].$$ Combining the decompositions $\Delta_{\bar{\partial}_E} = \nabla_1^*\nabla_1 + \mathfrak{R}_1$ and $\Delta_{\partial_E} = \nabla_2^*\nabla_2 + \mathfrak{R}_2$, we have $$\Delta_{\bar{\partial}_E} = \Delta_{\partial_E} + (\nabla_1^*\nabla_1 - \nabla_2^*\nabla_2) + (\mathfrak{R}_1 - \mathfrak{R}_2).$$ Therefore $$(\nabla_1^*\nabla_1 - \nabla_2^*\nabla_2) + (\mathfrak{R}_1 - \mathfrak{R}_2) = [iF_{\nabla}, \Lambda] + [\bar{\partial}_E, [L, \Lambda_{\bar{\partial}\omega}]] - [\partial_E, [L, \Lambda_{\partial\omega}]].$$ What I can't deduce is the correspondence between the terms. On the right hand side, the first term is order zero, whilst the other two terms are order one. On the left hand side, the first bracket is at most order two, but I'm guessing they have the same principal part so it is actually order one. Is $\nabla_1 \neq \nabla_2$ or are they equal so that only $\mathfrak{R}_1 - \mathfrak{R}_2$ represents the right hand side. Surely this can't be the case as $\mathfrak{R}_1$ and $\mathfrak{R}_2$ are order zero, aren't they? - -REPLY [4 votes]: The answer for both questions 1, 2 is yes, the metric connection in Proposition 9.1.27 is the Levi-Civita induced connection if the background manifold is Kahler. When $X$ is only a Hermitian manifold, the result is not true. One such example is discussed in great detail in Bismut's paper - -A local index theorem for non-Kahler manifolds, Math. Ann., 284(1989), 681-699.<|endoftext|> -TITLE: Is there always a zero between consecutive local extrema of $\Re \zeta(1/2+it)$ (or $\Im \zeta(1/2+i t)$ -QUESTION [8 upvotes]: Based on limited numerical evidence, I am inclined to suspect that -there is always zero of $\Re \zeta(1/2+it)$ between consecutive local -extrema of $\Re \zeta(1/2+it)$ -(and the same for $\Im \zeta(1/2+i t)$). -There are very short intervals having two zeros. -For Siegel $Z$ function, RH implies this for $t$ large enough. - -Is it true (maybe conditionally)? -Counterexamples? (Please check for 2 zeros in a short interval). - -REPLY [8 votes]: This is a nice problem. On the Riemann Hypothesis this is true, for both the real and imaginary parts. Here's the proof for $v(t)=\text{Im}(\zeta(1/2+i t))$. -Write $\zeta(1/2+i t)=Z(t)\exp(-i\theta(t))$, where $Z(t)$ is the Hardy function and -$\theta(t)=\text{Im}(\log \Gamma (1/4+ i t/2)-(t \log \pi)/2).$ -We have -$ -v(t)=-Z(t)\sin(\theta(t))$. So the zeros of $v(t)$ are precisely the zeros of $\zeta(s)$ on the critical line (i.e., zeros of $Z(t)$), and the zeros of $\sin(\theta(t))$, which are known as Gram points. ($g_k$ such that $\theta(g_k)=\pi k$.) -Take derivatives to see that -$$ -v^\prime(t)=0\Leftrightarrow \frac{Z^\prime(t)}{Z(t)}= --\cot(\theta(t))\theta^\prime(t). -$$ -Now on RH, $Z^\prime(t)/Z(t)$ is monotone decreasing between the zeros of $Z(t)$, see Edwards book "Riemann's Zeta Function". -On the other hand, Stirling's formula gives that -$$ -\theta(t)=t\log(t/2\pi)/2-t/2-\pi/8+O(1/t) -$$ -and -$$ -\theta^\prime(t)=\log(t/2\pi)/2+O(1/t^2) -$$ -so $\theta(t)$, $\theta^\prime(t)$ are both monotone increasing. So is $-\cot(x)$ between consecutive multiples of $\pi$. Thus $-\cot(\theta(t))\theta^\prime(t)$ is monotone increasing between consecutive Gram points. -The figure below shows the graph of $Z^\prime(t)/Z(t)$ (in blue) and the graph of $-\cot(\theta(t))\theta^\prime(t)$ (in red) in a typical situation where the Gram points and the (imaginary part of the) Riemann zeros alternate. These are the so called 'good Gram points'. The blue and red vertical lines show the poles at the Riemann zeros and Gram points. Between a Riemann zero and a subsequent Gram point, $Z^\prime(t)/Z(t)$ decreases from $+\infty$ to a finite value, while $-\cot(\theta(t))\theta^\prime(t)$ increases from a finite value to $+\infty$. By the Intermediate Value Theorem, there is precisely one intersection. Between a Gram point and a subsequent Riemann zero, the situation is reversed. -      - (source) -'But wait', you say, 'not all Gram points are good. What about the bad Gram points?' The second and third figures together (note different vertical and horizontal scales) show what happens at the first bad Gram point ($k=126$). Between two consecutive Gram points without an intervening Riemann zero, $Z^\prime(t)/Z(t)$ decreases from one finite value to another, while $-\cot(\theta(t))\theta^\prime(t)$ increases from $-\infty$ to $+\infty$. Again by the Intermediate Value Theorem, there is precisely one intersection. Between two consecutive Riemann zeros without an intervening Gram point, the situation is reversed. -     - (source) -     - (source) -Thus in all scenarios, between two consecutive zeros of $v(t)$ there is precisely one zero of $v^\prime(t)$. The zeros are simple as long as no Riemann zero is a Gram point. -The situation for $u(t)=\text{Re}(\zeta(1/2+i t))$ is similar; instead of Gram points we consider $h_k$ such that $\theta(h_k)=(k+1/2)\pi$.<|endoftext|> -TITLE: Shape of axioms in algebraic structures -QUESTION [5 upvotes]: When defining algebraic structures (like monoids, groups, etc...), are there some constraints on the shape of the axioms, for the structure to have good properties that we implicitly use in many proofs (like behaving well with respect to morphisms and quotients)? -For instance, is the following axiom acceptable, in a structure equipped with a unary function $f$ and a binary operator $\circ$: -$$\forall x, \text{ if }x\circ x=x\text{ then } f(x)=x.$$ -More specifically, does it make sense to study varieties generated by (subclasses of) such classes of objects, even if the class of structures so defined is not a variety, since this axiom is not an equation? - -REPLY [4 votes]: x = "preservation". -Usually one is motivated to study a structure because it -serves as a model for something of interest. Likewise a -certain statement or class of statements may have nice -consequences for its class of models. It is uncommon -to say 'I want to build this structure, but only if I can -characterize it with regular identities or in iambic -pentameter." Preservation theorems are the family -of theorems that relate the shape of a characterizing -language/theory to the shape of the constructions -preserved. -If you need a special kind of suit to dress up a class -of models of which your structure is a member, -you may not get a precise fit: some members of the -class are excluded, or perhaps some additional members -are included. Whether this is desirable depends on the -party for which one is dressing up. -The short answer is : it depends. If you are looking -at alternative axiomatizations for your class, then why? -Do you need a term rewriting system to work on it? -Is recursive axiomatizability sufficient? Do you need -a mixed presentation of a second order axiom and -one or two modifying first order axioms? Are you -trying to fit your class in some poset of defined classes? -Shape does matter. I can't tell you why before you tell -me enough motivation.<|endoftext|> -TITLE: Structure theorem for finite dimensional $C^*$-algebras and their representations -QUESTION [12 upvotes]: I would like a source for some Artin-Wedderburn type facts about these algebras which seem to have easy proofs, and are probably written somewhere. -Let $\mathcal{A} \subset M_n(\mathbb{C})$ be an algebra which is closed under taking adjoints. (That is, $X \in \mathcal{A} \Rightarrow X^*\in \mathcal{A}$) -Then, - $\mathcal{A}\cong \bigoplus_i M_{n_i}(\mathbb{C})$ for some finite sequence of integers $n_i.$ Furthermore if $\pi:\bigoplus_i M_{n_i}(\mathbb{C}) \rightarrow \mathcal{A}$ is a $*$-isomorphism, then there is a unitary $U$ and integers $m_i$ such that for every $ \bigoplus_i X_i\in\bigoplus_i M_{n_i}(\mathbb{C}),$ we can write the homomorphism via the formula -$\pi(\bigoplus_i X_i) = U^* \bigoplus_i (X_i \otimes I_{m_i}) U.$ - -REPLY [9 votes]: $\def\CC{\mathbb C}$Suppose $A$ is a $*$-closed subalgebra of $M_n(\CC)$. $\CC^n$ is a left $M_n(\CC)$-module in a canonical way, and therefore also an $A$-module. Using the fact that $A$ is $*$-closed it is easy to see that if $S\subseteq\CC^n$ is an $A$-submodule, then $S^\perp$ is also an $A$-submodule. This implies that $\CC^n$ is a semisimple $A$-module. -$\CC^n$ is then a semisimple faithful $A$-module: it follows at once that $A$ itself is a semisimple algebra (because the Jacobson radical acts by zero on every semisimple module). The decomposition you want then follows from the usual A-W theorem. - -REPLY [7 votes]: If your main concern is to have a reference to quote, then 1) is Theorem III.1.1 and 2) is essentially Corollary III.1.2 or Lemma III.2.1 in Davidson's book, as mentioned by Mike Jury. -This answer just aims at giving a direct (not using the larger compact case as in Davidson's book) self-contained and elementary operator algebraic approach. Hopefully this is correct... - -Your first statement can be seen as the finite-dimensional case of the structure of a type I von Neumann algebra which somehow generalizes Artin-Wedderburn over $\mathbb{C}$. Your second statement is the root of Brattelli diagrams. Both can be proved by elementary manipulations of the projections of the algebra. - -Note that if a finite-dimensional $*$-subalgebra $A$ of $B(H)$ does not contain $\mbox{Id}_H$, still the unital $*$-algebra $A\oplus \mathbb{C}\mbox{Id}_H$ is spanned by its projections by functional calculus (or just diagonalization of normal operators), so $A$ is also spanned by its projections (projections of $A\oplus \mathbb{C}\mbox{Id}_H$ are $p\oplus 0$ and $-p\oplus \mbox{Id}_H$ for projections $p$ of $A$). Whence the joint of all the projections of $A$ is a unit for $A$. So a finite-dimensional $*$-subalgebra of $B(H)$ is always a unital von Neumann algebra. Up to reducing $H$, we can assume that this unit is $\mbox{Id}_H$. - -1) If $A$ is a finite-dimensional $*$-subalgebra of $B(H)$, then $A$ is unitarily equivalent to $\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$. - -In short: a finite-dimensional von Neumann algebra must be of type $\rm{I}$, whence unitarily equivalent to $\sum_{\alpha}^\oplus A_\alpha \otimes B(H_\alpha)$ with $A_\alpha$ Abelian (Theorem I.27 in Takesaki I). In the finite-dimensional case, this gives the assertion. -Here is how the above goes in the finite-dimensional case directly. First, considering the minimal projections of the center $A\cap A'\simeq \mathbb{C}^r$ which decompose $A$ into blocks, we can restrict to the factor case $A\cap A'=\mathbb{C}1$. By compactness of the unit sphere of $A$, there are finitely many Murray-von Neumann equivalence classes of projections since two close enough projections ($\|p-q\|<1$) must be homotopic, whence equivalent. Therefore, there are minimal projections in particular. Since the order on projections is total in a factor, all the minimal projections are equivalent. Finally, take $p_1\oplus \ldots \oplus p_n$ a maximal orthogonal set of minimal projections. By maximality, we must have $p_1\oplus \ldots \oplus p_n=1$. Since these minimal projections must be pairwise equivalent, the partial isometries implementing these equivalences yield a unitary equivalence between $A$ and $M_n(\mathbb{C})\otimes p_1 Ap_1$. And by minimality of $p_1$, $p_1Ap_1\simeq \mathbb{C}$. -Decomposing the range into blocks, your second question follows from the following fact. - -2) If $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$, and if $\pi:A\longrightarrow B(H)$ is a $*$-homomorphism, then $\pi$ is unitarily equivalent to an orthogonal direct sum of possibly zero diagonal embeddings $x_j\longmapsto x_j\otimes 1_j$ where $x=(x_1,\ldots,x_r)\in A$. - -Without loss of generality, we can assume that $\pi$ is unital, up to reducing $H$. -The key point is that a unital $*$-homomorphism $\pi:A\longrightarrow B(H)$ sends projections to projections (self-adjoint idempotents), and partial isometries to partial isometries. -First, if $A=\bigoplus_{j=1}^r M_{n_j}(\mathbb{C})$, then the corresponding orthogonal decomposition of the unit $1=\bigoplus p_j$ is sent to an orthogonal decomposition of the unit of $B(H)$. This allows us to restrict to the case $A=M_n(\mathbb{C})$, and to assume $\pi$ unital again. -Now the matrix units $e_{ij}$ with $1$ in $(ij)$-th position and $0$ elsewhere can be interpreted as partial isometries implementing the Murray-von Neumann equivalence of the projections $e_{ii}$ and $e_{jj}$. This is preserved under the unital $*$-homomorphism $\pi$. It follows that $\pi$ is, up to unitary equivalence, a diagonal embedding. -The case $A=M_2(\mathbb{C})$, with weaker assumptions, is used and explained with somewhat lengthy details in this MSE answer. It works the exact same way for $A=M_n(\mathbb{C})$.<|endoftext|> -TITLE: Non-amenable groups with arbitrarily large Tarski number? -QUESTION [85 upvotes]: Just out of curiosity, I wonder whether there are non-amenable groups with arbitrarily large Tarski numbers. The Tarski number $\tau(G)$ of a discrete group $G$ is the smallest $n$ such that $G$ admits a paradoxical decomposition with $n$ pieces: $\exists A_1,\ldots,A_k,B_1\ldots,B_l\subset G$, $\exists g_1,\ldots,g_k,h_1\ldots,h_l\in G$ such that $k+l=n$ and $$G = \bigsqcup_{i=1}^k A_i\sqcup\bigsqcup_{j=1}^l B_j = \bigsqcup_{i=1}^k g_iA_i = \bigsqcup_{j=1}^l h_jB_j \quad\mbox{(disjoint unions)}.$$ Tarski's theorem says that $\tau(G)<\infty$ iff $G$ is non-amenable. It is known that $\tau(G)=4$ iff $G$ contains a non-abelian free subgroup. (See a survey paper by Ceccherini-Silberstein, Grigorchuck, and de la Harpe) -If $G$ is a non-amenable group such that every $m$ generated subgroup of it is amenable, then it satisfies $\tau(G)>m+2$ (because one may assume $g_1=e=h_1$ in the paradoxical decomposition). Such $G$ probably exists, but I do not know any examples even for $m=2$. - -REPLY [205 votes]: It is indeed an open problem, as Misha said. But here is a solution. In E. Golod, Some problems of Burnside type. 1968 Proc. Internat. Congr. Math. (Moscow, -1966) pp. 284-289. Izdat. ”Mir”, Moscow, Golod announced, for every $m$ an infinite finitely generated torsion group all of whose $m$-generated subgroups are finite. A proof can be found in Ershov, Mikhail, Golod-Shafarevich groups: a survey. Internat. J. Algebra Comput. 22 (2012), no. 5, 1230001, 68 pp (Theorem 3.3). The proof starts with a Golod-Shafarevich group $G$. If you assume that $G$ has property (T) (such groups exist by Ershov, see the survey), the resulting group will have property (T). Thus there exists a finitely generated infinite property (T), hence non-amenable, group with arbitrary large Tarski number. - Correction. Misha Ershov sent me two corrections. - -It is easier to deduce the answer from Theorem 3.3 and Ershov's theorem that any GS group is non-amenable, it also can be found in the survey (this does follow from existence of property (T) GS group). - -There exists a generalized GS group with property (T) -and all m-gen. subgroups finite (for every fixed m). Thus a property (T) group with arbitrary large Tarski number also exists.<|endoftext|> -TITLE: Explicit description of the oplax limit of a functor to Cat? -QUESTION [10 upvotes]: The nCatLab Grothendieck construction page gives an explicit description of the oplax colimit of any functor to Cat. Can someone give me a similarly explicit description (the objects and morphisms) of an oplax limit of any functor to Cat (or a link to a page which describes it)? (I've found Reedy model structures on oplax limits, but that leaves unspecified the "'obvious' coherence conditions".) -Additionally, is there a name for such a category, analogous to "Grothendieck construction" or "category of elements"? -Context: The reason I'm interested in this is because I'm trying to formulate the categorical dependent sum and dependent product in Coq. I think the oplax (co)limit are the dependent sum/product pushed across a Yoneda-like transformation (though I'm not entirely sure that it's Yoneda). Coq's dependent sum and product are more similar to the oplax (co)limit formulation, and while nCatLab has good pages on dependent sum and dependent product, it doesn't seem to have such a page on oplax limits. -Edit: I'm looking for a description of the objects and morphisms in this category, possibly together with the composition law. Here is my guess at what the objects and morphisms are: Given a functor $F : \mathcal C \to \text{Cat}$, and letting $F_0$ denote its action on objects and $F_1$ denote its action on morphisms, - -Objects consist of the following components - -For each object $r \in \mathcal C$, an object $x_r \in F_0(r)$ -For all objects $s, d \in \mathcal C$ and each morphism $m \in \text{Hom}_{\mathcal C}(s, d)$, a morphism $f_m \in \text{Hom}_{F_0(d)}((F_1(m))_0(x_s), x_d)$ (Note: This doesn't agree with Reedy model structures on oplax limits, but I can't figure out how to typecheck what's there; I've added a comment to that effect.) -For all $r \in \mathcal C$, a proof that $f_{\text{id}_r} = \text{id}_{x_r}$ (well, actually, that $f_{\text{id}_r}$ is equal to the isomorphism induced by the proof that $(F_1(\text{id}_r))_0(x_r) = x_r$) -For all $p, q, r \in \mathcal C$ and all morphisms $m_0 \in \text{Hom}_{\mathcal C}(q, r)$ and $m_1 \in \text{Hom}_{\mathcal C}(p, q)$, a proof that $f_{m_0 \circ m_1} = (F_1(m_1))_1(f_{m_0}) \circ f_{m_1}$ - -Morphisms from $(x, f)$ to $(x', f')$ consist of the following components: - -For each object $r \in \mathcal C$, a functor $g_r : x_r \to x'_r$. -Some coherence condition I haven't managed to phrase yet, corresponding to the commutativity square for natural transformations. - - -Did I get anything wrong? (In particular, is the second component of objects right? Also, what changes for lax vs. oplax?) - -REPLY [8 votes]: The oplax limit is the category of sections for the functor from the Grothendieck construction to the base category. -The strong limit is the category of cartesian sections (every arrow in the base category gets mapped to a cartesian one). -Notice how this goes along very well with the interpretation as dependent product and as $\forall$: The set theoretic product is just the set of sections into the disjoint union. -[edit] -Given a strong functor $F:X\to\mathrm{Cat}$ we denote the Grothendieck construction by -$$\mathrm{Gr}(F).$$ -There is a canonical functor $\pi:\mathrm{Gr}(F)\to X$. Sections of this functor are functors -$$s:X\to\mathrm{Gr}(F)$$ -such that $s\circ\pi=\mathrm{id}$. -So even more explicitely a section amounts to -forall $x\in X$ an object $s_x\in F(x)$ and ...<|endoftext|> -TITLE: Foliations by holomorphic curves on complex surfaces -QUESTION [11 upvotes]: On a complex surface, does there exist a non-singular foliation by holomorphic curves that is NOT a holomorphic foliation, i.e. a transversally holomorphic foliation? -The surface should be compact and can admit holomorphic foliations. - -REPLY [2 votes]: This is a question which is related to Lagrangian foliations on hyperkaehler manifolds, but much of the results are conjectures, or unpublished. Let $M$ be a K3, and $\omega$ a cohomology class on a boundary of a Kaehler cone, satisfying $\omega^2=0$ (such clases are called nef). There are quite a few of such classes on K3; for a general non-algebraic K3, for all classes with $\omega^2=0$, either $\omega$ or $-\omega$ is nef. -It follows from the general compactness arguments that $\omega$ is represented by a closed, positive (1,1)-current. If this current is continuous, it has exactly one positive and one vanishing eigenvalue, hence its kernels give a foliation by holomorphic curves. -The tangent bundle of a general non-algebraic K3 has no rank 1 -coherent subsheaves, hence this K3 has no holomorphic foliations. -However, in Serge Cantat's thesis (unpublished) he showed an existence -of a Holder continuous positive closed (1,1)-form representing a class -with $\omega^2=0$, using McMullen's exotic holomorphic automorphisms on -non-algebraic K3. This should give such a non-holomorphic foliation by holomorphic curves. -I think that such foliations (without relation to K3) -were also discussed in Marco Brunella's paper -Brunella, Marco -On Kähler surfaces with semipositive Ricci curvature. -Riv. Math. Univ. Parma (N.S.) 1 (2010), no. 2, 441–450. -but it is not available anywhere, so I cannot check.<|endoftext|> -TITLE: Groups as Union of Proper Subgroups: References -QUESTION [11 upvotes]: There are interesting theorems about groups as union of proper subgroups. The first result in this subject is the theorem of Scorza(1926): "a groups if union of three proper subgroups if and only it has quotient $C_2\times C_2$." In 1959, Haber and Rosenfeld proved interesting theorems on the groups as union of subgroups. Then, in 1994, J. H. E. Cohn proved some interesting theorems about groups as union of few proper subgroups, and made conjectures. -While reading these three papers, which have large gaps in the publishing years, I couldn't find other initial references on "Groups as union of subgroups". -It will be a great pleasure, if one provides a list of references on the subject "Groups as union of proper subgroups", from 1926 to 1959 and from 1959 to 1994. -Especially, it is known that a non-cyclic $p$-group can not be union of $p$-proper subgroups, and if it is union of $p+1$ proper subgroups, then all the subgroups are maximal, and theire intersection has index $p^2$ in $G$. I would like to get original references for this theorem also. -Thanks in advance!! - -REPLY [2 votes]: To be complete, there is another reference here. -Theorem (Bruckheimer, Bryan and Muir) A group is the union of three proper subgroups if and only if it has a quotient isomorphic to $V_4$. -The proof appeared in the American Math. Monthly $77$, no. $1 (1970)$. As remarked by @soluble, the theorem seems to be proved earlier by the Italian mathematician Gaetano Scorza, I gruppi che possono pensarsi come somma di tre loro sottogruppi, Boll. Un. Mat. Ital. $5 (1926), 216-218$.<|endoftext|> -TITLE: A question about the $C^{2,\alpha}$ regularity of concave fully nonlinear uniformly elliptic equation -QUESTION [6 upvotes]: While reading Theorem 6.6 of Chapter Six of "Fully nonlinear elliptic equation" by Luis A. Caffarelli and Xavier Cabre in the American mathematical society colloquium publications vol. 43, I get two problems as follow. -The theorem 6.6 of this chapter is to prove the $C^{2,\alpha}$ regularity of the viscosity solution of the concave fully nonlinear uniformly elliptic equation $F(D^2u)=0$, where $F$ is just a concave function defined on the symmetric matrices and F is not required to be differentiable. -My first problem is that the uniform linear elliptic operator is special concave uniform elliptic, and in order to get $C^{2,\alpha}$ of the solution, we need the $C^{0,\alpha}$ differentiability of the coefficients of the operator. But here the theorem states that we can get the $C^{2,\alpha}$ regularity of the viscosity solution without any differentiability of the equation. I guess the reason may be that we consider the viscosity solution here, but I'm not sure whether we can get the $C^{2,\alpha}$ regularity of the classical solution of the concave uniformly elliptic equation. (This may be obvious wrong, but I want to know if the reason is just that we consider the viscosity solution.) -The second problem is the proof given by the book. In order to prove the theorem, the author applied the Evans-Krylov theorem to the $C^{1,1}$ viscosity solution of the equation $F(D^2 u)=0$, and the $C^{1,1}$ regularity of such an equation is obtained in the proof. While applying the process of proving the Evans-Krylov theorem to the $C^{1,1}$ viscosity solution, we can just get the $C^{2,\alpha}$ regularity in the subset $B_{\frac{1}{2}}\cap A$ of $B_{\frac{1}{2}}$, where $B_{\frac{1}{2}}-A$ is of measure zero. I wonder how can we get the regularity of the viscosity solution in $B_{\frac{1}{2}}$ without any other assumption. In the paper of Evans, the smoothness of $F$ is needed in order to use the method of continuity, and Evans considered the $C^{2,\alpha}$ regularity of the classic solution. -For the second problem, I have tried to use the smooth concave uniformly elliptic operator $F_k$ to approximate $F$, but in order to get a convergent viscosity solution of $F$ from the limit of $u_k$ that is the viscosity solution of $F_k$, we need a uniform bound for $u_k$, which is an obstruction for me now. - -REPLY [6 votes]: A linear uniformly elliptic equation of the form $F(D^2u) = 0$ can only be $\Delta u = 0$ (up to an affine transformation of the solution) since the only inputs are the second derivatives. Equations like $a^{ij}(x)u_{ij} = 0$ are of the form $F(D^2u,x) = 0$. If $F(.,x_0)$ is concave for each $x_0$ (corresponding to freezing the coefficient in the linear setting) and $F$ is Holder continuous in $x$, then we get $C^{2,\alpha}$ regularity by applying Evans-Krylov and a perturbation argument (see ch. 8 of Caffarelli-Cabre). This is a generalization of Schauder estimates. -A more interesting concave equation "built out of" linear ones is the Bellman equation, -$$\inf_{\alpha}a^{ij}_{\alpha}u_{ij} = 0$$ -where $a^{ij}_{\alpha}$ is a collection of uniformly elliptic constant-coeffient matrices. -For the second question, the proof actually gives $C^{2,\alpha}$ regularity for a $C^{1,1}$ viscosity solution in all of $B_{1/2}$. Indeed, one shows the oscillation decay of $D^2u$ (as an $L^{\infty}$ map) -$$diam(D^2u(B_{\delta^k}(x))) < 2^{-k}\|u\|_{C^{1,1}(B_1)}$$ -for some universal $\delta$. It follows that $D^2u$ is Holder continuous up to modification on a set of measure $0$. By using quadratic approximations to $u$ at nearby points it is straightforward to show that $u$ is in fact $C^{2,\alpha}$ everywhere. (Think for example that the function $\chi_{\{0\}}$ cannot be the derivative of any differentiable function, since if it were it would lie close to a line of slope $1$ near $0$ and violate being locally constant away from 0). -As a side note, it is an interesting result of Armstrong, Silvestre and Smart that for any uniformly elliptic equation $F(D^2u) = 0$ with $F \in C^1$ we get $C^{2,\alpha}$ regularity off of a set of Hausdorff dimension $n-\epsilon$ for some small universal $\epsilon$.<|endoftext|> -TITLE: Growth of powers of non-negative integer matrices -QUESTION [8 upvotes]: In what I am currently doing, there naturally appears the following question: let $A$ be a square matrix with non-negative integer entries. Let $a_n$ be the sum of all entries of $A^n$. -Question: How the sequence $\{a_n\}_{n\geq 1}$ can grow? -Of course if $A$ is positive, then Perron-Frobenius Theorem tells us the answer, but in the general case of non-negative matrices, it can be difficult to guess the asymptotics of the sequence $\{A^n\}_{n\geq 1}$. So, I thought may be there is something known for this case, when we have actually integer matrices. Any references and comments would be appreciated. - -REPLY [4 votes]: I am not sure I understand the question. Any matrix $A$ (integer or not, positive or not) has a Jordan canonical form $A = MJM^{-1},$ whereupon $A^n = M J^n M^{-1}.$ If $A$ is integer and nonsingular, the biggest eigenvalue is at least $1$ in modulus (since the determinant is at least $1$ in absolute value). If it IS equal to $1$ in modulus, the sum of the elements will be polynomial, if it is greater than one, it will be exponential -- if there is a single eigenvalue of maximal modulus, it will be really exponential, otherwise at least there will be a positive density subsequence of $n$ for which it is. Which numbers can occur as eigenvalues of nonnegative integer matrices was answered by Doug Lind in: -Lind, D. A.(1-WA) -The entropies of topological Markov shifts and a related class of algebraic integers. -Ergodic Theory Dynam. Systems 4 (1984), no. 2, 283–300. -58F11 (15A48 28D20) -MR1149738 (92m:11117) Reviewed -Lind, Douglas(1-WA) -Matrices of Perron numbers. -J. Number Theory 40 (1992), no. 2, 211–217. -11R06 (15A48 58F03)<|endoftext|> -TITLE: Collision polynomials -QUESTION [6 upvotes]: Consider $P_n(x)$ polynomials defined through the recurrence relations -$$P_n(x)=2(1-x)P_{n-1}(x)-(1+x)^2P_{n-2}(x),$$ with $P_0(x)=1$ and $P_1(x)=1-3x$. -In fact, the explicit solution of these recurrence relations is given by the formula -$$P_n(x)=\frac{1}{2}\left[ (1+i\sqrt{x})^{2n+1}+(1-i\sqrt{x})^{2n+1}\right].$$ -What condition(s) should satisfy the number $n$, the polynomial $P_n(x)$ to be reducible? I have checked, that $P_n(x)$ is reducible for $n=4,7,10,12,13,16,17,19,\ldots$ What is special about these numbers? -Another question: given a number $n$, for what numbers $0 -TITLE: Is the fundamental group of $II_{1}$ factors invariant under a relation? -QUESTION [7 upvotes]: In order to define the equivalence relation, let's first recall the Tomita-Takesaki modular theory and conditional expectation for von Neumann algebras. -Let $H$ be a separable Hilbert space and $B(H)$ the algebra of bounded operators. -Definition: A von Neumann algebra is a *-subalgebra $M \subset B(H)$ stable under bicommutant: $M^{*} = M$ and $M'' = M$. -Modular theory : Let $M \subset B(H)$ be a von Neumann algebra. Let $\Omega \in H$ be a cyclic and separating vector (i.e., $M.\Omega$ and $M'.\Omega$ are dense in $H$). Let $S : H \to H$ be the closure of the anti-linear map $x\Omega \to x^{*}\Omega$. Then, $S$ admits a polar decomposition $S = J\Delta^{1/2}$, with $J$ anti-linear unitary and $\Delta$ positive. Then, $JMJ = M'$ and $\Delta^{it} M \Delta^{-it} = M$. - Let $\sigma_{\Omega}^{t}(x) = \Delta^{it} x \Delta^{-it}$ the modular action of $\mathbb{R}$ on $M$. -Conditional expectation (Takesaki 1972) : Let $N \subset M$ be an inclusion of von Neumann algebra, then there is a conditional expectation of $M$ onto $N$ with respect to $\Omega$ (cyclic and separating) if $N$ is invariant under the modular action (i.e., $\sigma_{\Omega}^{t}(N) = N)$. -Notation : if $\exists \Omega$ verifying the previous conditions, we note $N \subset_{e} M$. -Remark : The modular theory is trivial for $M = L(\Gamma) \subset B(H)$, with $\Gamma$ a discrete group and $H = l^{2}(\Gamma)$ (because $\Delta = I$). In particular, it's trivial for the abelian von Neumann algebras. -As a consequence, in this case: $N \subset M$ $\Leftrightarrow$ $N \subset_{e} M$. -Notation : Let $N$ and $M$ be two von Neumann algebras. - If $\exists P \simeq N$ such that $ P \subset_{e} M$, we note $N \hookrightarrow_{e} M$. - -Equivalence relation : $M \sim N$ if $N \hookrightarrow_{e} M \hookrightarrow_{e} N$. - -Philosophy : $M \sim N$ could significate they are isomorphic as noncommutative sets (see here). -Examples : - -Among $l^{\infty}(\{1,2,...,n \})$, $l^{\infty}(\mathbb{N})$ -and $L^{\infty}([0,1])$ none is equivalent to another. -$L^{\infty}([0,1])$, $L^{\infty}([0,1]\cup \{1,2,...,n \})$ -and $L^{\infty}([0,1]\cup \mathbb{N})$ are pairwise equivalent, -because $L^{\infty}([0,1]) \subset L^{\infty}([0,1] \cup \{2,3,...,n\}) \subset L^{\infty}([0,1] \cup \mathbb{N}_{\geq 2}) \hookrightarrow L^{\infty}(\mathbb{R})$ -and $L^{\infty}([0,1]) \simeq L^{\infty}(\mathbb{R})$ -Obviously $L^{\infty}([0,1]) \not\sim B(H)$. -Let $R \subset B(H)$ be the hyperfinite $II_{1}$ factor, $R_{\infty} = R \otimes B(H)$ the hyperfinite $II_{\infty}$ factor. $ B(H) \hookrightarrow_{e} R_{\infty} \hookrightarrow_{e} B(H \otimes H)$ and $B(H) \simeq B(H \otimes H)$. So, $R \not\sim B(H) \sim R_{\infty}$. -Let $\Gamma$ be a non-amenable ICC discrete group. Then $L(\Gamma) \not\hookrightarrow_{e} B(H)$ and $L_{\infty}(\Gamma) = L(\Gamma) \otimes B(H) \not\hookrightarrow_{e} B(H \otimes H) $ so $L(\Gamma) \not\sim B(H) \not\sim L_{\infty}(\Gamma)$. -Let $\mathbb{F}_{2} = \langle a,b \vert \ \rangle $ and $\mathbb{F}_{\infty} = \langle a_{1},a_{2},... \vert \ \rangle $. -Then $\mathbb{F}_{2} \hookrightarrow \mathbb{F}_{n} \hookrightarrow \mathbb{F}_{\infty} \hookrightarrow\mathbb{F}_{2} $ (the last injection is given by $a_{n} \to b^{-n}ab^{n}$). -Consequence : $L(\mathbb{F}_{2}) \sim L(\mathbb{F}_{n}) \sim L(\mathbb{F}_{\infty}) $ - - -Fundamental group (see here) : The fundamental group of a type $II_{1}$ factor is the set of numbers $t > 0$ for which its - amplification by $t$ is isomorphic to itself: $\mathcal{F}(M) = \{t>0 \ \vert \ M^{t}\simeq M \}$. - -Examples: - -There is a semi-direct product $ \Gamma = \mathbb{Z}^{2} \rtimes SL(2,\mathbb{Z})$ such that $\mathcal{F}(L(\Gamma)) = \{1\}$ -It's countable for $II_{1}$ factors with property (T). -$\mathcal{F}(R) = \mathcal{F}(L(\mathbb{F}_{\infty})) = \mathbb{R}_{+}^{*}$ -Open : $\mathcal{F}(L(\mathbb{F}_{2})) = \{1\}$ or $\mathbb{R}_{+}^{*}$, but we still do not know which it is. -This is a reformulation of the free group factor isomorphism problem: $L(\mathbb{F}_{2}) \simeq L(\mathbb{F}_{\infty}) $ ? - - -Question: Is the fundamental group $\mathcal{F}(M)$ of a $II_{1}$ factor $M$ - invariant under $\sim$ ? - -Remark : an affirmative answer would solve the free group factor isomorphism problem. -Because this problem is very difficult, if this question admits an affirmative answer, I do not expect that the proof will be given here without a colossal work, but I would be interested to know if (in your opinion) this way seems promising. If it admits a negative answer, then in addition to a possible counter-example, I would be interested to know if you see a manner to reformulate the question for becoming open. - -REPLY [11 votes]: Here is a counterexample. I don't see any easy way to augment the question to something more natural. -Let $Q$ be a $w$-rigid II$_1$ factor with trivial fundamental group, e.g., $Q = L( \mathbb Z^2 \rtimes SL_2(\mathbb Z) )$. Let $\mathcal S \subset \mathbb R_+^*$ be a non-trival subgroup. Set $M = *_{s \in \mathcal S} Q^s$, and $N = Q * M$. (We may take $M$ and $N$ separable if we take $\mathcal S$ countable, and $Q$ separable.) Clearly we have $M \hookrightarrow N$, and by a result of Dykema and Rădulescu (Theorem 1.5 from http://www.ams.org/mathscinet-getitem?mr=1735079) we have $M \cong M * L(\mathbb F_\infty)$ from which it follows easily that $N \hookrightarrow M * M \hookrightarrow M$. -(Note that by Umegaki's Theorem http://www.ams.org/mathscinet-getitem?mr=68751 there always exist normal conditional expectations for von Neumann subalgebras of II$_1$ factors.) -Corollary 6.5 in my paper with Ioana and Popa http://www.ams.org/mathscinet-getitem?mr=2386109 shows that $\mathcal F(M) = \mathcal S$, while $\mathcal F(N) = \{ 1 \}$.<|endoftext|> -TITLE: Andreev's Theorem and Thurston's hyperbolization theorem -QUESTION [6 upvotes]: I am attempting to get to grips with Thurston's hyperbolization theorem for Haken $3$--manifolds. In particular I was looking at the section related to gluing up along hierarchy surfaces in Otal, Jean-Pierre (1998), "Thurston's hyperbolization of Haken manifolds" -From what I understand we take hierarchy for a Haken $3$--manifold $M$ with corners (a decomposition of the manifold with corners along incompressible surfaces into a collection of $3$--balls), use something like Andreev's theorem to put a hyperbolic structure on the $3$--balls, and use an inductive argument to re-glue the manifold along the hierarchy surfaces in such a way that the original manifold carries a complete hyperbolic metric. I am speaking very loosely here since I can't say I understand all the details. -My question is as follows. Is it possible to re-formulate the this construction in terms of manifolds with boundary pattern (in the spirit of Johannson), and if so, if the boundary pattern on the collection of $3$--balls corresponds to the cuts along hierarchy surfaces, does the boundary pattern carry all the data needed for re-gluing? - -REPLY [11 votes]: The sorts of hierarchies that Johannson makes use of are called "simple hierarchies", and go back to Waldhausen (used by him in the solution of the word problem and homotopy rigidity of Haken 3-manifolds). The boundary patterns associated -to these hierarchies don't contain all of the information one needs to glue up the hierarchy. As an example, consider the first cut of a Haken manifold, say along a separating incompressible surface. The boundary pattern here is trivial, consisting of the entire boundary, which is two copies of the first surface. Then the second cut will be along a (boundary)-incompressible surface (possibly disconnected), which we may assume meets both components of the boundary, and the resulting boundary pattern will be the second surface and the first surface cut up along the boundary of the second surface. To reglue the manifold, we may glue the 2nd boundary pattern by any automorphism of this surface (with boundary), and then glue the 1st stage by any of the automorphisms of the 1st surface. Notice that the boundary pattern associated to one copy of the 1st surface does not ``see" the boundary pattern associated to the other side, which is partly why the pattern is not enough to recover the gluing. For a discussion of simple hierarchies, see the paper of Aitchison-Rubinstein and references therein. -One remark (since you mention Andreev's theorem): Thurston told me that he originally came up with a new proof of Andreev's theorem for hyperbolic reflection polyhedra by the sort of inductive argument used in the proof of geometrization of Haken 3-manifolds. It turns out that reflection orbifolds are either tetrahedral or Haken. The theory here is much simpler, because compactification of the hyperbolic structure on non-compact reflection groups is simpler (the bending laminations of the boundary of the convex hull are finitely many arcs). After succeeding in this case, he worked on the general case of Haken manifolds, and in fact incorporated the reflection group techniques in his "orbifold trick" to make the gluing problem have no boundary.<|endoftext|> -TITLE: Word evaluating to a group element and its inverse with different frequency -QUESTION [34 upvotes]: I'm supervising an undergraduate research project. Among other things, I've got the student to look at this paper of Gene Kopp and John Wiltshire-Gordon. This question arose from a missing complex conjugate in something the student wrote. -Let $g$ be an element of a finite group $G$, and $w$ a word in $n$ variables. If you evaluate the word on all $n$-tuples of elements of $G$, does it give $g$ and $g^{-1}$ the same number of times? -I thought the answer must be "no", but found it frustratingly difficult to come up with an example. After tapping local knowledge it seems that I was right. There's a recent paper of Alexander Lubotzky that proves that if $G$ is a finite simple group then the only restriction on a subset $A\subseteq G$ for it to be the image of the word map for some word in 2 variables is that it contains the identity and is fixed by $\operatorname{Aut}(G)$. Since there are finite simple groups (e.g., the Mathieu group $M_{11}$) with elements that are not sent to their inverses by any automorphism, this answers the question. -However, my real question is whether there's a relatively simple example? -Lubotzky's paper doesn't give an explicit word, although it does show that, for $M_{11}$, there's a word that works with length at most about $1.7\times 10^{244552995}$. -Presumably one can do a bit better than that? -There are obvious restrictions on $g\in G$ and $w$ that rule out really small examples. There can't be any automorphism of $G$ sending $g$ to $g^{-1}$ or any automorphism of the free group $F_n$ sending $w$ to $w^{-1}$. - -REPLY [19 votes]: Yes, one can do much better than $1.7 \times 10^{244552995}$ -(not surprisingly, because we're asking less than Lubotzky: -one of the two counts must be less than the other, -but not necessarily zero). -In fact a word of length $10$ suffices. -I tried $G = M_{11}$ and $g$ an element of order $11$, and took $n=2$, -which makes exhaustive computation easily feasible (the first variable -can be assumed to lie in one of the $10$ conjugacy classes so -there's only $10 \, |M_{11}| = 79200$ -group elements to compute given $w$). -None of the words $w(x,y) = x^a y^b x^c y^d$ seems to work, -but several of the form $w(x,y) = x^a y^b x^c y^d x^e$ solve the problem. -The first one (in lexicographic order) with all exponents at most $3$ is -$(a,b,c,d,e) = (1,2,1,3,3)$, i.e. $w(x,y) = x y^2 x y^3 x^3$, -for which $w(x,y) = g$ has $7491$ solutions -but $w(x,y) = g^{-1}$ has only $7458$.<|endoftext|> -TITLE: Name and/or uses for the spectrum of a stalk? -QUESTION [9 upvotes]: Suppose that $X$ is a scheme and $x\in X$ is a point. The stalk of $X$ at $x$ is a (local) ring and we can form its spectrum $Y_x=\rm{Spec}(\mathcal{O}_{X,x})$. -There is a canonical map $Y_x\to X$. We can define it by fixing an affine neighborhood $x\in U\cong \rm{Spec}(R)$, making $x$ as a prime ideal in $R$ and $\mathcal{O}_{X,x}\cong R_x$ a localization. The quotient $R\to R_x$ then induces the map of schemes $Y_x\to U \subseteq X$. -My question is this: is there a name for this construction? Are there familiar methods or theorems where it arises? - -REPLY [13 votes]: Topologically the scheme $\rm{Spec}(\mathcal{O}_{X,x})$ is exactly the intersection of all neighbourhoods of $x$ and algebraically it contains every infinitesimal neighbourhood of $X$. -Although technically it is not the germ of$X$ at $x$, it seems to me that it contains so much information about that germ that it could be considered a materialization of that germ . -Also: technically it is not a subscheme of $X$ (since it is not locally closed) but I can imagine a world where a broader notion of subscheme would allow the monomorphism $j: S=\rm{Spec}(\mathcal{O}_{X,x})\hookrightarrow X$ to be called a subscheme, an almost open subscheme if you will. -One argument for that broader point of view is that the induced canonical morphism $j^{-1}\mathcal O_X=\mathcal O_X\mid S \stackrel {\cong}{\to} \mathcal O_S$ of sheaves over $S$ is an isomorphism, just as if $S\subset X$ were open. -In particular, given a morphism of rings $A\to B$ , the corresponding morphism of affine schemes $\phi:\rm{Spec} (B)\to \rm{Spec} (A)$ and a prime ideal $\mathfrak p\subset A$, the morphism $\rm{Spec} (B_{\mathfrak p}) \to \rm{Spec}(A_{\mathfrak p} )$ is a pleasant thickening of the genuine fiber $\phi^{-1}(\mathfrak p)=\rm{Spec}(B\otimes_A \kappa(\mathfrak p))$ of $\mathfrak p$. -Considerations of such almost germs $\rm{Spec}(A_{\mathfrak p} )$ of $\rm{Spec}(A)$ at $\mathfrak p$ may be of help when following reasonings in commutative algebra. -My first contact with one of these almost germs (or almost open subschemes) was in Mumford's Red Book, Chapter Two, §1, Example F, where he describes an example as "... a startling way to make a scheme out of the non closed points in the plane" and draws one of his celebrated pictures to illustrate the notion. - -REPLY [12 votes]: In EGA1, 2.4 this is called the local scheme of $X$ at $x$.<|endoftext|> -TITLE: Smooth projective varieties with infinite abelian fundamental group and finite $\pi_2$ -QUESTION [7 upvotes]: Let $X$ be a smooth projective complex algebraic variety of general type. Suppose that the (topological) fundamental group of $X$ is an infinite abelian group and that $\pi_2(X^{an})$ is finite. -What can we say about $X$? -I am mainly interested in the case where $\dim X = 2$, and $\Omega^1_X$ satisfies some positivity properties such as ampleness. -What is "known" about surfaces $X$ with infinite abelian fundamental group, finite $\pi_2$ and ample $\Omega^1_X$? - -REPLY [7 votes]: With regard to surfaces: it is a simple theorem that $\pi_2$ of any closed $4$-manifold is torsion free; so one doesn't need any complex geometry to see that there are no such projective surfaces. The argument I will give was observed by the late Jerry Levine and me in 2004 after some correspondence with Gurjar following his paper referred to in ulrich's answer. The idea is to apply the argument that follows to the universal cover of a closed 4-manifold. It seems that the more interesting question, addressed by Gurjar for complex surfaces, is whether $\pi_2$ is actually free; our attempts to resolve this ran into some well-known problems related to group cohomology and ends of manifolds. -Let X be a simply connected $4$-manifold with empty boundary. Then $H_2 (X)$ is isomorphic to -$H^2_c (X)$, cohomology with compact supports. This is the direct -limit of $H^2 (X,X-C)$, where $C$ ranges over compact subsets of X. we -have an exact sequence -$$ -0 \to H^1 (X-C) \to H^2 (X,X-C) \to H^2 (X) -$$ -Now $H^1 (X-C)$ is torsion-free because $H^1$ always is. $H^2 -(X)=Hom(H_2 (X),Z)$ is also torsion-free since $X$ is simply connected, and so $H^2 (X,X-C)$ is -torsion-free. -But now it is easy to see that the direct limit of a family of -torsion-free groups is torsion-free.<|endoftext|> -TITLE: Covering convex polygons with inscribed disks -QUESTION [5 upvotes]: The following problem came up when discussing mapping software (e.g., Google maps) with computer scientists. By $B(c,r)$ I mean the planar disk (open or closed, it doesn't matter) of radius $r$ around the point $c$. -Consider a finite collection of points in the plane whose convex hull we call $P$. Here's the question: - -Given $\epsilon, \delta > 0$ and the vertices of $P$, can we efficiently produce a finite collection of balls $B_j = B(c_j,r_j)$ so that - (1) each $B_j$ is contained completely in $P$, - (2) the area of $P$ not contained in any of the balls is smaller than $\epsilon$, and - (3) the area covered by more than one ball is smaller than $\delta$? - -Basically, the problem requires an efficient way to cover a convex polytope by inscribed disks so that we simultaneously maximize the covered area and minimize the multiply-covered area. I would be happy with any approach, greedy or otherwise, which achieves such an optimization If this is hopeless, can we make progress by dropping requirement (3) entirely? -Note also that the naive/greedy approach which starts by finding the largest inscribed disk $B_P \subset P$ and focusing attention on $P' = P \setminus B_P$ immediately puts us in an unfriendly setting: $P'$ is neither convex nor a polytope. I can't see how to make a greedy recursive approach work without essentially constructing a polygonal approximation of $B_P$. - -REPLY [5 votes]: Here is a slight variation of the nice algorithm of Yoav. -(1) Pack $P$ with disks with Eppstein's $O(n \log n)$ algorithm (where $n$ is the number -of vertices of $P$). -(2) Modify as described in this paper, - -Marshall Bern, Erik Demaine, David Eppstein, and Barry Hayes, “A Disk-Packing Algorithm for an Origami Magic Trick”, in Proceedings of the International Conference on Fun with Algorithms (FUN'98), Isola d'Elba, Italy, June 18–20, 1998, pages 32–42. (link here) - -so that all gaps are either 3-gaps or 4-gaps, i.e., bounded by three or four arcs, as in the figure below. -         -(3) Proceed as in Yoav's algorithm. -The advantage is that from now on, all gaps are 3- or 4-gaps and are easier to fill -than arbitrarily shaped gaps. -One other point. -You could stop the filling at some stage above $\epsilon$ coverage, -and enlarge the disks -keeping their centers fixed, so that they overlap by $\le \delta$. -This would reduce the number of disks needed. -If you follow this enlarging, you should start with $P$ offset inward to accommodate -the later enlarging.<|endoftext|> -TITLE: Easy proof of topological property of Zoll manifolds -QUESTION [8 upvotes]: It is known that the cohomology ring of a Zoll manifold---a riemannian manifold all of whose geodesics are periodic with the same minimal period---must be the same as the cohomology ring of a compact rank one symmetric space (see Besse's book Manifolds all of whose geodesics are closed for references). -Is there a simple and elementary proof of the following much weaker property? -The first Betti number of a Zoll manifold is equal to zero. -Addendum. -The comment by Thomas Richard got me thinking and here is something that should lead to a proof that the fundamental group of a Zoll manifold is either trivial or isomorphic to $\mathbb{Z}_2$: -Any two prime closed geodesics in a Zoll manifold are homotopic. Indeed, if $v_x$ is a unit vector tangent to a geodesic $\gamma$ and $w_y$ is a unit vectors tangent to a geodesic $\sigma$, then a continuous path on the unit tangent bundle joining these two unit vectors, taken as the initial conditions of prime closed geodesics, will define a homotopy between $\gamma$ and $\sigma$. -Note that there is at least one closed geodesic representing each non-trivial homotopy class of loops, but the geodesic doesn't have to be prime. Still ... - -REPLY [8 votes]: As asked by @alvarezpaiva, I repost my remark as an answer. -After his addendum and answer showing that the fundamental group (if non-trivial) has one generator, given by a prime closed geodesic, you just have to observe that this geodesic is homotopic to itself with reversed orientation. -Hence the fundamental group has order at most $2$.<|endoftext|> -TITLE: looking for proof or partial proof of determinant conjecture -QUESTION [18 upvotes]: Math people: -I am looking for a proof of a conjecture I made. I need to give two definitions. For distinct real numbers $x_1, x_2, \ldots, x_k$, define $\sigma(x_1, x_2, \ldots, x_k) =1$ if $(x_1, x_2, \ldots, x_k)$ is an even permutation of an increasing sequence, and $\sigma(x_1, x_2, \ldots, x_k) =-1$ if $(x_1, x_2, \ldots, x_k)$ is an odd permutation of an increasing sequence. For example, $\sigma(2, 1, 10, 8) = 1$ because $(2,1,10,8)$ is an even permutation of $(1,2,8,10)$, and $\sigma(2, 1, 8, 10) = -1$ because $(2,1,8,10)$ is an odd permutation of $(1,2,8,10)$. For real $B$, $n\geq 1$ and distinct real numbers $\mu_1, \mu_2, \ldots, \mu_n, \gamma_1, \gamma_2, \ldots, \gamma_n$, let $M(B;\mu_1,\mu_2,\ldots,\mu_n;\gamma_1,\gamma_2,\ldots,\gamma_n)$ be the $n$-by-$n$ matrix defined by -$$ M(B;\mu_1,\mu_2,\ldots,\mu_n;\gamma_1,\gamma_2,\ldots,\gamma_n)_{i,j}=\frac{\exp(-B\gamma_j)}{\mu_i+\gamma_j}+\frac{\exp(B\gamma_j)}{\mu_i-\gamma_j}. $$ -My conjecture is the following: if $n \geq 1$, $B \geq 0$, and $\mu_1, \mu_2, \ldots, \mu_n, \gamma_1, \gamma_2, \ldots, \gamma_n$ are distinct positive numbers with -$0<\mu_1 < \mu_2 < \cdots < \mu_n$ and $0<\gamma_1 < \gamma_2 < \cdots < \gamma_n$ , then -$$\operatorname{sgn}(\operatorname{det}(M(B;\mu_1,\mu_2,\ldots,\mu_n;\gamma_1,\gamma_2,\ldots,\gamma_n))) = (-1)^{\frac{n(n+1)}{2}} - \sigma(\mu_1, \mu_2, \ldots, \mu_n, \gamma_1, \gamma_2, \ldots, \gamma_n). $$ -Of course $\operatorname{sgn}(x)$ is the sign of $x$, which is $1$, $-1$, or $0$. I have proven this is true for $n=1$ and $n=2$. For $n$ between $3$ and $20$, I have run thousands of experiments in Matlab using randomly generated $\mu$'s and $\gamma$'s. In a set of one thousand experiments, the conjectured equation will typically hold every single time, or might fail once or twice, with the determinant (with the wrong sign) being extremely small, so perhaps roundoff error is the culprit. -UPDATE: let $d(B)$ be the determinant of the matrix, where the other parameters should be clear from context. $d(B)$ is an analytic function of $B$. It suffices to show that $\frac{\partial^m d}{\partial B^m}$ has the desired sign at $B=0$ for all $m \geq 0$. Unfortunately, the determinant of $\frac{\partial M}{\partial B}$ is not the same thing as $\frac{\partial^m d}{\partial B^m}$ (if it were, properties of Cauchy matrices would yield the desired conclusion). Since the conjecture is true for $n=1$, that means that the displayed formula for $M_{i,j}$ above, and all its derivatives with respect to $B$, have the same sign as $\mu_i - \gamma_j$ at $B=0$, and $M_{i,j}$ has that sign for all positive $B$. I proved the conjecture for $n=2$, by computing the determinant of $M$ and its derivatives with respect to $B$ at $B=0$, and looking at the six possible orderings of $\mu_1, \mu_2, \gamma_1$, and $\gamma_2$ given the restrictions $\mu_1 < \mu_2$ and $\gamma_1 < \gamma_2$. I had some help from Maple multiplying out, simplifying and factoring algebraic expressions. I am trying to prove the general case by induction on $n$, expanding the determinant along the last row or column, but the determinants of the $n-1$-by-$n-1$ minors don't seem to necessarily have the ``right'' signs. -Thanks to some comments provided below, unless I am confused, the conjecture can be proven for $B=0$ and large positive $B$, for any $n$, using properties of Cauchy matrices. - -REPLY [8 votes]: I have a proof if all the $\mu$'s are larger than all of the $\gamma$'s. I've spent a while trying to figure out how modify this proof to work for other orderings, but I'm giving up. This argument is inspired by a computation of the Caucy determinant by Doron Zielberger (skip to the 25 minute mark). -We start with the identity -$$e^{\mu B} \int_B^{\infty} e^{- \mu t} (e^{\gamma t} + e^{-\gamma t}) dt = \frac{e^{-B \gamma}}{\mu+\gamma} + \frac{e^{B \gamma}}{\mu-\gamma}.$$ -Note that we need $\mu > \gamma$ in order for the integral to converge; that is why I can't extend this proof to any other ordering of the $\mu$'s and $\gamma$'s -So the determinant is -$$\det \left( e^{\mu_i B} \int_{t_i=B}^{\infty} e^{-\mu_i t_i} (e^{\gamma_j t_i} + e^{- \gamma_j t_i}) d t_i \right).$$ -Note that we are using $n$ different integration variables -- one for each row of the matrix. By the multilinearity of the determinant, this is -$$\exp\left(\sum B \mu_i \right) \times \int_{t_1=B}^{\infty} \int_{t_2=B}^{\infty} \cdots \int_{t_n=B}^{\infty} \exp\left( - \sum \mu_i t_i \right) \det \left( e^{\gamma_j t_i} + e^{- \gamma_j t_i} \right) dt_1 dt_2 \cdots dt_n.$$ -Note that permuting $(t_1, \ldots, t_n)$ only changes $ \det \left( e^{\gamma_j t_i} + e^{- \gamma_j t_i} \right)$ by a sign, but changes $\exp(\mu_i t_i)$ to $\exp(\mu_{\sigma(i)} t_i)$ for some permutation $\sigma$. Lumping together all $n!$ reorderings of the $t_i$, the integral is -$$\int_{B \leq t_1 \leq \cdots \leq t_n} \sum_{\sigma \in S_n} (-1)^{\sigma} \exp(\sum t_i \mu_{\sigma(i)}) \det \left( e^{\gamma_j t_i} + e^{- \gamma_j t_i} \right) dt_1 \cdots dt_n $$ -$$= \int_{B \leq t_1 \leq \cdots \leq t_n} \det(e^{\mu_j t_i}) \det \left( e^{\gamma_j t_i} + e^{- \gamma_j t_i} \right) dt_1 \cdots dt_n.$$ -We claim that the integrand is positive for $0 < t_1 < \cdots < t_n$ so the integral is positive, as desired. So we are done once we prove: -Lemma If $\mu_1 < \mu_2 < \cdots < \mu_n$ and $\gamma_1 < \gamma_2 < \cdots < \gamma_n$ and $t_1 < t_2 < \cdots < t_n$, then -$$\det \left( e^{\mu_j t_i} \right) \ \mbox{and} \ \det \left( e^{\gamma_j t_i} + e^{- \gamma_j t_i} \right)$$ -are positive. -Proof It is enough to show that these determinants don't vanish anywhere in this range, since they then must have constant sign and it is easy to check that the correct sign is positive. If the first determinant vanished, then there would be some nonzero function $\sum a_j z^{\mu_j}$ which vanished at $z = e^{t_1}$, $e^{t_2}$, ... $e^{t_n}$. But then this is a "polynomial with real exponents" having $n$ nonzero terms, and $n$ postive roots, contradicting Descartes rule of signs. (I wrote out a proof of Descartes' rule of signs for real exponents here.) -Similarly, if the second determinant vanishes, then we get a real-exponent-polynomial $\sum b_j (z^{\gamma_j} + z^{-\gamma_j})$ with $2n$ terms and roots at the $2n$ points $e^{\pm t_i}$. Again, a contradiction.<|endoftext|> -TITLE: Is the infimum of Salem numbers > 1? -QUESTION [9 upvotes]: BACKGROUND -A Salem number is an algebraic integer $\theta$ such that all the Galois conjugates of $\theta$ are $\leq 1$ in absolute value, and at least one of them lies on the unit circle. Their importance is derived for example from the fact that the minimal polynomial of a Salem number, -$$ -P(x) = x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1 -$$ -is conjectured to minimize the Mahler measure ($M(P) = 1.17...$) over all $P \in \mathbb{Z}[x]$ with $M(P) > 1$. -The closely related Pisot numbers are algebraic integers $\theta > 1$ such that all the Galois conjugates of $\theta$ are of absolute value $< 1$. Their set is closed and in particular there exists a smallest Pisot number. This has been found by Siegel to be the plastic constant, $\theta_0 = 1,32471\ldots$, a root of $g(x) = x^3 - x - 1$. It is known that for any monic, non-reciprocal polynomial $P$ we have $M(P) \geq M(g) = \theta_0$. This is Smyth's theorem. -THE QUESTION -I am currently reading ``Conjecture de Lehmer et petits nombres de Salem" by Bertin and Pathiaux-Delefosse. The book is from 1989 and it is stated in it that it is still not known if $\inf T > 1$ where $T$ is the set of all Salem numbers. Has there been any developments since then? Is this conjecture still open? -Precisely: Has it been proved or disproved that $\inf T> 1$? - -REPLY [3 votes]: Just wanted to add: the conjecture that Salem numbers are bounded away from one (the "Salem Conjecture") is wide open as the other answers state, and it's equivalent to part of what is known as the "Short Geodesic Conjecture" for hyperbolic orbifolds: that there is a positive universal lower bound for the lengths of geodesics in arithmetic hyperbolic 2-orbifolds. The case of 3-orbifolds implies the Salem Conjecture. You can find this information in Maclachlan and Reid's book The Arithmetic of Hyperbolic 3-Manifolds, Section 12.3, along with some comments on the Salem Conjecture.<|endoftext|> -TITLE: Why stationary sets were named such? -QUESTION [6 upvotes]: My question is about terminology: -Do you know why stationary sets were named such? -Going over the following MO question about the intuition behind stationary sets, the only compelling argument I can think of is Fodor's lemma. -Is this the reason? - -REPLY [4 votes]: In Infinite Combinatorics, in: Handbook of the History of Logic, 6. Sets and Extensions in the Twentieth Century, p 226, footnote 214, Jean Larson states that the term was first used in G. Bloch: Sur les ensembles stationnaires de nombres ordinaux et les suites distinguees de fonctions regressives, Comptes Rendus Acad. Sci Paris, 236(1953), 265-268. The reason for the name was probably Neumer's theorem, a weaker and earlier form of Fodor's theorem.<|endoftext|> -TITLE: Riemann hypothesis and Kakeya needle problem -QUESTION [7 upvotes]: The question may be a bit vague. I noticed an analogy that both Riemann hypothesis and Kakeya needle problem has been proved in finite fields. Can somebody shed light on why finite field analogues are "easier"? - -REPLY [24 votes]: The Riemann hypothesis is proved over function fields (like the fraction field of F_q[t]), not finite fields, and the "real version" is a question about the integers. Kakeya is proved over finite fields, and the "real version" is a question about, well, the reals. So the situations are quite different. I'd say that the truth of RH over function fields really does make me feel more confidence that RH is true over number fields, because the analogy between function fields and number fields is in many ways a very close one. On the other hand, the truth of Kakeya over finite fields does not tell me very much about the truth of the real Kakeya problem. For one thing, what's true over finite fields -- that a Kakeya set has measure bounded away from 0 -- is totally false over the real numbers! (See: Besicovich set.) An intermediate multiple-scale case is that of a power series ring over a finite field -- in this case, measure-0 Kakeya sets exist by a theorem of Dummit and Hablicsek, but we don't know the answer to the Kakeya problem, i.e. we don't know whether a Kakeya set over F_q[[t]] has full dimension. -One reason (but not the only reason) the Kakeya problem is easier over finite fields than it is over the real numbers is that finite fields only have one scale, while the real numbers have multiple scales. -One reason (but not the only reason) the Riemann Hypothesis is easier over function fields than it is over number fields is that the zeta function in the function field case has an interpretation in terms of the etale cohomology of a variety over a finite field. In the number field case there is no such interpretation at present, despite the best efforts of the F_un-ologists.<|endoftext|> -TITLE: Mordell-Weil of an elliptic surface after adjoining a nontorsion section: as small as possible? -QUESTION [6 upvotes]: Let $k$ be an algebraically closed field of characteristic $0$, let $C_{/k}$ be a nice (smooth, projective, geometrically integral curve), let $K = k(C)$, and let $\overline{K}$ be an algebraic closure of $K$. Let $E_{/K}$ be an elliptic curve with $j(E) \notin k$. Let $P \in E(\overline{K}) \setminus E(K)$ be a point of infinite order. Let $K(P)$ be the field of definition of $P$ (equivalently, the field obtained by adjoining to $K$ the coordinates of $P$ in a Weierstrass equation for $E$). Then -$\langle P, E(K) \rangle \subset E(K(P))$. -Must we have equality? -Comments: -1) For my application, I may assume that $E_{/K}$ is semistable, so please feel free to address the question under that additional hypothesis if it helps. (But I don't see how it does...) -2) I am not able to assume anything about $C$. -3) For my application, I have already dealt with the case in which $P$ has finite order, and in that case I could assume that $C = X_m(n)$ -- the elliptic modular curve parameterizing $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ torsion structures -- and $E$ is the universal elliptic curve over $C$ (the four pairs $(m,n) \in \{(1,1), (1,2), (1,3), (2,2)\}$ in which there is no universal elliptic curve are excluded). In this case the result follows from work of Shioda and Cox-Parry. -4) If this is true, it seems to be closely related to Shioda's landmark 1972 paper on elliptic modular surfaces. I confess that I am asking this question before I have fully absorbed this important paper: I have several collaborators who would be happy if my plate were cleaner. -Added: If I may, I'll try one variant of the question: what if $[n]P \in E(K)$ for some $n \in \mathbb{Z}^+$? - -REPLY [5 votes]: I will sketch a counterexample for the modified question. The idea behind the construction is similar to the counterexample for the original question. Only the geometric details of this construction are trickier and does not integrate so well in my previous answer. For this reason I will post this a new answer. -For this example we take $k=\mathbb{C}$ and $C=\mathbb{P}^1$. -Take two polynomials $a(t), b(t)$ of degree 2 and 4 and consider the elliptic curve $E_{a,b}: y^2=x(x^2+a(t)x+b(t))$. This is a rational elliptic surface, with Mordell-Weil group $\mathbb{Z}^4 \times \mathbb{Z}/2\mathbb{Z}$, provided that $a$ and $b$ are sufficiently general, e.g., $a$ and $b$ have no common zero and both $b$ and $a^2-4b$ are squarefree. -Take a point $Q\in E_{a,b}(K(t))$ of infinite order. Let $T$ be a point of order 2, different from $(0,0)$ and take for $P$ the point $Q+T$. Then $2P\in E_{a,b}(\mathbb{C}(t))$ and if $a^2-4b$ is not a square then $P$ is not in $E_{a,b}(\mathbb{C}(t))$. -As in my previous answer, it suffices to find a $(a,b)$ such that the twist of $E_{a,b}$ over $K(P)=K(\sqrt{a^2-4b})$ has positive rank. I will sketch two arguments why such an $(a,b)$ exists. -Let $U \subset \mathbb{C}[t]_2\times \mathbb{C}[t]_4$ be the locus of polynomials such that $E_{a,b}$ defines a rational elliptic surface with positive Mordell-Weil rank and precisely one point of order $2$. (The complement of $U$ is the union of the locus where $a$ is a multiple of $b^2$, together with the locus where $a^2-4b$ is a square, and finitely many codimension 4 components parametrizing the Mordell-Weil rank $0$ surfaces.) -For any integer $m$ consider the locus $L_m$ of $(a,b)$ such that the twisted elliptic curve $(a^2-4b)y^2=x(x^2+ax+b)$ has a section of infinite order which intersects the zero section in $m$ points. -Using Hodge theory, Lefschetz $(1,1)$ and the fact that the corresponding elliptic surface has $h^{2,0}=2$ it follows that this locus is locally given by two equations. This implies that $L_m$ is either contained in the complement of $U$ or $L_m$ contains a component of codimenion at most two in $U$. -Now the complement of $U$ has only one component of codimension two. Since for most $m,m'$ we have that $L_m\neq L_{m'}$ we find at least one component of some $L_m$ which is not contained in $U$. Actually, there is a standard argument in Noether-Lefschetz theory to check whether the union of $L_m$ is dense in the analytic topology on $U$ and I expect that this argument applies also here. -A second way to construct an explicit example. It is rather easy to find equations for the loci $L_m$. For $L_0$ you have $19$ equations in $25$ variables. Since the equations are of small degree I expect that with some help from a computer you can find an explicit example. Moreover, modern computer algebra software should be able to check whether $L_0$ is in the complement of $U$ or not. -Edit: If you want that $E_{a,b}$ is semistable then you need to do a bit more work. In this case you need to exclude all $(a,b)$ with a common zero from $U$. This defines a codimension one locus. Hence, depending on the strategy you need to show that $\cup L_m$ is dense or that $L_0$ is not contained in the complement of $U$.<|endoftext|> -TITLE: Correlation of a quadratic character with the Mobius function -QUESTION [5 upvotes]: Let $p$ be a prime number and consider the sum $S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)$. For how small an $x$ in terms of $p$ is it known that $S(x)=o(x)$? -I am especially interested in unconditional results. - -REPLY [13 votes]: Wirsing's Theorem tells us that if $f$ is multiplicative and each $f(p)=-1,0 $ or $ 1$ then -$\sum_{n\leq x} f(n) = o(x)$ as $\sum_{p\leq x} (1-f(p))/p \to \infty$ (and one cannot do much better). Moreover one can get an explicit upper bound: -$ \sum_{n\leq x} f(n) \ll x \exp( -.32\sum_{p\leq x} (1-f(p))/p)$. -In your case $f(n)=\mu(n)(n/p)$ so that -$\sum_{p\leq x} (1-f(p))/p = 1/p+ 2\sum_{q\leq x, (q/p)=1} 1/q $. -Therefore to get the bound $o(x)$ you need that a significant number of the $q\leq x$ satisfy $(q/p)=1$. So your question becomes: For what $x$ can we guarantee this? -Or, in other words, is it possible that $(q/p)=-1$ for "most" of the primes $\leq x$ (as may well be the case of one has a Siegel zero)? -If we use quadratic reciprocity, then $(q/p)=-1$ is equivalent to demanding $(p/q)=$ something fixed, and we can find such $p$ for which this holds for all but one prime $q\ll \log p$, by Dirichlet's Theorem. But then, by smooth number estimates, one knows that for almost all such $p$ one has $\sum_{n\leq x} \mu(n)(n/p) \gg \rho(A)x$ for $x=(\log p)^A$ (for each $A$). -So we have "proved" that for any fixed $A>0$, the estimate $\sum_{n\leq x} \mu(n)(n/p) = o(x)$ does not hold uniformly for $x=(\log p)^A$. -The same ideas give, assuming GRH, that $\sum_{n\leq x} \mu(n)(n/p) = o(x)$ does hold uniformly provided $\log x/\log\log p \to \infty$ as $p\to \infty$. -These ideas can be found in my paper "Large Character Sums" with Soundararajan, though there we looked at character sums $\sum_{n\leq x} \chi(n)$; it should not take much to modify those ideas for this situation.<|endoftext|> -TITLE: prime divisor of elements of $A+B$ -QUESTION [9 upvotes]: Consider two infinite subsets $A$ and $B$ of natural numbers. I am positive that the set of prime divisors of elements of $A+B$ is infinite but I can not prove it. Maybe this is a known result. If anybody has an idea or a reference please let me know. (by $A+B$ I mean $\lbrace a+b \; | \; a\in A, b\in B\rbrace$) - -REPLY [12 votes]: This was proved by Erdős and Turán when they were undergraduate students, see Theorem III here. A stronger result was proved by Győry, Stewart, and Tijdeman, see Theorem 1 here.<|endoftext|> -TITLE: Diffeology as a sheaf on the site of smooth manifolds -QUESTION [9 upvotes]: Souriau's definition of diffeology may be phrased as defining a concrete sheaf on the category $\mathsf{Open}$ of open subsets of Euclidean/coordinate spaces. It seems to me, unless I am missing something, that any such sheaf extends to a sheaf on the site $\mathsf{Man}$ of all smooth manifolds. Is there a proof written down somewhere? -More generally it seems to me that the inclusion $\mathsf{Open} \hookrightarrow \mathsf{Man}$ should induce an equivalence of 2-categories between the 2-category of stacks on $\mathsf{Man}$ and the stacks on $\mathsf{Open}$. Is this true? If so, is there a reference? (Metzler seems to mentions something like this in passing in here: arXiv:math/0306176 [math.DG] .) - -REPLY [6 votes]: Actually, one doesn't need the comparison lemma in this case. As it turns out, $\mathbf{Man}$ is the Karoubi envelope of $\mathbf{Open},$ (see the Examples section of http://ncatlab.org/nlab/show/Karoubi+envelope), which implies that if $$i:\mathbf{Open} \hookrightarrow \mathbf{Man}$$ is the canonical inclusion, the induced restriction functor $$i^*:Psh_n\left(\mathbf{Man}\right) \to Psh_n\left(\mathbf{Open} \right)$$ between their categories of presheaves of $n$-groupoids for any $n$ is already an equivalence. -Edit: In this question (Proof that the category of presheaves on a category $C$ is equivalent to the category of presheaves on its Karoubi envelope) it discusses that presheaves of sets on the Karoubi envelope of a small category is equivalent to presheaves on the original category, and gives a reference. -Another Edit: First I gave the following argument, but it seems to have a gap (skip over this to get to the direct answer): -Now, consider the functor $i_!:Psh_\infty\left(\mathbf{Open} \right) \to Psh_\infty\left(\mathbf{Man}\right)$ which is left adjoint to $i^*.$ Consider the composite $i^*i_!$ which is colimit preserving. It also restricts to an equivalence on $0$-truncated objects, by the above (wait: Why shoudl $i_!$ send $0$-truncated objects to $0$-truncated objects?). If $F$ is an arbitrary presheaf on $\mathbf{Open}$, then $F$ can be represented as a simplicial presheaf, hence there exists a simplicial diagram $c_F:\Delta^{op} \to Psh_\infty\left(\mathbf{Open} \right)$ for which each ${c_F}_n$ is $0$-truncated and such that the colimit of $c_F$ is $F.$ Since $i^*i_!$ is colimit preserving, it must send $F$ to itself. A similary argument works using the composite $i_!i^*$, and one concludes that $i_!$ and $i^*$ form an equivalence of $\infty$-categories. In particular, they restrict to an equivalence between $n$-truncated objects for any $n$, hence the induced map $$i^*:Psh_n\left(\mathbf{Man}\right) \to Psh_n\left(\mathbf{Open} \right)$$ is an equivalence for all $n$. -Unfortunately (see the bold wait) I don't see how to show that $i_!$ preserves $0$-truncated objects (i.e. agrees with the 1-categorical left Kan extension when restricted to presheaves of sets) until I show its an equivalence, so here's another way to finish the argument: -(Start reading again here if you skipped): Consider the restriction functor $$i^*:Psh\left(\mathbf{Man}\right) \to Psh\left(\mathbf{Open} \right)$$ of presheaves of sets, which has a right adjoint $i_*$. (Explicitly, by the Yoneda lemma, one has that $i_*(F)(M)\cong Hom(i^*y(M),F)$ where $y$ is the Yoneda embedding.). Since $i^*$ also has a left adjoint $i_!,$ $i^*$ is left exact, and since $i$ is full and faithful, so is $i_*.$ So we have that the adjunction $(i^*,i_*)$ exhibits $Psh\left(\mathbf{Open} \right)$ as a left exact localization of $Psh\left(\mathbf{Man}\right)$. Hence, there is a unique Grothendieck topology $J$ on $\mathbf{Man}$ for which $Sh_J\left(\mathbf{Man}\right)\cong Psh\left(\mathbf{Open}\right)$ (more precisely, such that the localization induced by sheafification agrees with the one above). However, since we know that $i^*$ is an equivalence, it implies that this Grothendieck topology must be the trivial one. Notice that we also have an induced adjunction $(i^*,i_*)$ between $\infty$-presheaves. (Here there is no danger of the abuse of notation, since both functors are left exact, so preserves $n$-truncated objects for all $n$, so their restriction to presheaves of sets agree with the ones above). This adjunction is still a left exact localization, and since $Psh_\infty\left(\mathbf{Man}\right)$ is $1$-localic, it must again correspond to a unique Grothendieck topology. This left exact localization factors uniquely as a topological localization (one coming from a Grothendieck topology), followed by a cotopological one (one for which the only monos sent to equivalences are equivalences). The covering sieves of the topology corresponding to the topological part of the localization correspond exactly to those monos $f:S \to y(M)$ such that $i^*(f)$ is an equivalence. However, subobjects of representable objects in $Psh_\infty\left(\mathbf{Man}\right)$ are the same as subobjects in $Psh\left(\mathbf{Man}\right),$ so one sees the resulting class of covering sieves must be the same as for the $1$-categorical case, which we have argued only gave the maximal sieves (the trivial Grothendieck topology). It follows that the localization must be cotopological. Since the $\infty$-category of $\infty$-groupoids is hypercomplete and colimits are computed pointwise in presheaves, $Psh_\infty\left(\mathbf{Man}\right)$ is hypercomplete, and hence the localization must be trivial. Hence $$i^*:Psh_\infty\left(\mathbf{Man}\right) \to Psh_\infty\left(\mathbf{Open} \right)$$ is an equivalence, and by restriction to $n$-truncated objects, -$$i^*:Psh_n\left(\mathbf{Man}\right) \to Psh_n\left(\mathbf{Open} \right)$$ is as well.<|endoftext|> -TITLE: Contractible topological groups -QUESTION [14 upvotes]: Does there exist a Hausdorff topological group which is contractible and of finite covering dimension but which is not homeomorphic to $\mathbb{R}^n$ for some $n$? - -REPLY [17 votes]: If a topological group is contractible, then it is locally contractible (using the group operation produce a contraction which does not move the unit of the group). By a classical result of [A. Gleason, R. Palais, On a class of transformation groups, Amer. J. Math. -79 (1957), 631–648], a locally path-connected finite-dimensional topological group is a Lie group and being contractible, is homeomorphic to an Euclidean space.<|endoftext|> -TITLE: Is there a survey of recent work relating to the Hausdorff dimension of sets defined through some restriction of digits? -QUESTION [5 upvotes]: I am familiar with the work of Helmut Cajar, but his book is thirty years old and it's clear that there has been substantial progress since then. I have been spending a lot of time looking through mathscinet citations and am honestly a bit overwhelmed. Also, I'm not just interested in sets defined through restrictions on digits of their decimals expansions, but also similar results pertaining to continued fraction expansion, Cantor series expansions, the Luroth series expansion, Engel series expansion, etc. - -REPLY [2 votes]: I would guess you'll find many recent pointers in the publication list of Lars Olsen, and in that of Godofredo Iommi, for example: - -Applications -of multifractal divergence points to sets of numbers defined by their -N-adic expansion (2004) -The frequency of -digits in the Lüroth expansion (2009) -Hausdorff Dimension of Cantor -Series (2009)<|endoftext|> -TITLE: Estimate term in Ramanujan Lost Notebook (classic analytic number theory) -QUESTION [12 upvotes]: This is the fault of Igor Rivin, who asked about sums of divisor functions. I will put in links eventually. What I would like to know is the size of the right hand side in Ramanujan's formula (381), see original and a rendition FROM HERE in the second jpeg below. The quotes from Ramanujan are in (including some sign corrections) Nicolas and Robin, The Ramanujan Journal, Volume 1, Issue 2, June 1997, pages 119-153, Highly Composite Numbers by Srinivasa Ramanujan. -The questions are about the surprising size of the thing. For $s = 1/2$, formula 380, as a function of $N$ it is still larger than any power of $\log N$. Note that the construction "generalised superior highly composite numbers" always gives something smaller than any root $N^\alpha$ for $\alpha > 0$. -Question (A): is there a subinterval of the given $1/2 < s < 1$ for which the right-hand side of (381) has comparable growth to some $(\log N)^\beta$ for some $0 < \beta < \infty$? -Question (B): how do we get down to the really tiny growth for $s=1$, dominant term $\log \log N? $ Note that this was later improved in Robin's Criterion. -Between (330) and (332) we find -$$ S_s(x) = - s \sum \; \frac{x^{\rho - s}}{\rho (\rho - s)} $$ -where the sum is ``over the complex roots of $\zeta.$'' -With that in mind, the right hand side of (381) is -$$ | \zeta(s) | \exp \left\{ \mbox{Li} \left( (\log N)^{1-s} \right) - \frac{2s(2^{1/(2s)} -1)}{2s-1} \frac{(\log N)^{\frac{1}{2}-s}}{\log \log N} \right\} -+ \frac{S_s(\log N)}{\log \log N} + O \left\{ \frac{(\log N)^{\frac{1}{2}-s}}{(\log \log N)^2} \right\} $$ - -REPLY [2 votes]: EDIT, Monday: while specific constants depend on RH, the following is unconditional: for $s=1$ the "ratio" grows as $\log \log n,$ but for fixed $0 < s < 1$ the "ratio" grows faster than any power of $\log n.$ -I heard from Jean-Louis Nicolas. For fixed $\frac{1}{2} < s < 1$ and $n \rightarrow \infty$ the growth of Ramanujan's "generalised superior highly composite numbers" is still swifter than any power of $\log n.$ My own fiddling with formula (381) shows that the single dominant term is an awfully good estimate. Note that $\zeta(3/4) \approx -3.441285$ -================================== - - t 0.75 - - ratio = sigma_{3/4}(n) / n^{3/4} - - Ram = exp( Li ( (log n)^{1/4} ) ) - -================================== - - ratio 1.594603558 2 bump 2^1 - ratio 2.294142325 log n 1.791759469 6 bump 3^1 - ratio 2.802796526 log n 2.48490665 12 bump 2^2 - ratio 3.641028199 log n 4.094344562 60 bump 5^1 - ratio 4.033928726 log n 4.787491743 120 bump 2^3 - ratio 4.573537136 log n 5.886104031 360 bump 3^2 - ratio 5.63628118 log n 7.832014181 2520 bump 7^1 - ratio 5.962699514 log n 8.525161361 5040 bump 2^4 - ratio 6.949884201 log n 10.92305663 55440 bump 11^1 - ratio 7.965010441 log n 13.48800599 720720 bump 13^1 - ratio 8.224276385 log n 14.18115317 1441440 bump 2^5 - ratio 8.649956395 log n 15.27976546 4324320 bump 3^3 - ratio 9.245517654 over Ram 3.028600308 log n 16.88920337 21621600 bump 5^2 - ratio 10.34983665 over Ram 3.165867659 log n 19.72241672 367567200 bump 17^1 - ratio 11.48711898 over Ram 3.083261736 log n 22.6668557 6983776800 bump 19^1 - ratio 12.58086187 over Ram 3.172163117 log n 25.80234991 160626866400 bump 23^1 - ratio 12.81668461 over Ram 3.231623923 log n 26.49549709 321253732800 bump 2^6 - ratio 13.84228269 over Ram 3.095238829 log n 29.86279292 bump 29^1 - ratio 14.4487679 over Ram 3.230853497 log n 31.80870307 bump 7^2 - ratio 15.54855733 over Ram 3.101387246 log n 35.24269028 bump 31^1 - ratio 15.88423195 over Ram 3.168342458 log n 36.34130256 bump 3^4 - ratio 16.94303437 over Ram 3.19811678 log n 39.95222048 bump 37^1 - ratio 17.9887252 over Ram 3.217053505 log n 43.66579254 bump 41^1 - ratio 18.18553099 over Ram 3.084784213 log n 44.35893972 bump 2^7 - ratio 19.26852148 over Ram 3.103456721 log n 48.12013984 bump 43^1 - ratio 20.34195462 over Ram 3.114010326 log n 51.97028744 bump 47^1 - ratio 21.37753969 over Ram 3.113314008 log n 55.94057936 bump 53^1 - ratio 21.78937675 over Ram 3.173291822 log n 57.55001727 bump 5^3 - ratio 22.81291905 over Ram 3.163514705 log n 61.62755471 bump 59^1 - ratio 23.85808075 over Ram 3.152929072 log n 65.73842858 bump 61^1 - ratio 24.87685838 over Ram 3.135532889 log n 69.9431212 bump 67^1 - ratio 25.46187993 over Ram 3.209270267 log n 72.34101647 bump 11^2 - ratio 26.50287028 over Ram 3.188425241 log n 76.60369635 bump 71^1 - ratio 26.67341265 over Ram 3.208942324 log n 77.29684353 bump 2^8 - ratio 27.74144999 over Ram 3.187825713 log n 81.58730297 bump 73^1 - ratio 28.78835877 over Ram 3.162018062 log n 85.95675082 bump 79^1 - ratio 29.05524627 over Ram 3.191332103 log n 87.05536311 bump 3^5 - ratio 30.11185937 over Ram 3.163394697 log n 91.47420372 bump 83^1 - ratio 31.15104963 over Ram 3.132063376 log n 95.96284009 bump 89^1 - ratio 31.73094398 over Ram 3.190368501 log n 98.52778944 bump 13^2 - ratio 32.75755091 over Ram 3.154083607 log n 103.1025004 bump 97^1 - ratio 33.78573385 over Ram 3.253082917 log n 107.7176209 bump 101^1 - ratio 34.83070774 over Ram 3.213503967 log n 112.3523499 bump 103^1 - ratio 35.87765418 over Ram 3.173480786 log n 117.0251788 bump 107^1 - ratio 36.94119501 over Ram 3.267554004 log n 121.7165266 bump 109^1 - ratio 38.00705978 over Ram 3.224796113 log n 126.4439145 bump 113^1 - ratio 38.15155234 over Ram 3.237055915 log n 127.1370616 bump 2^9 - ratio 39.16001407 over Ram 3.188810947 log n 131.9812487 bump 127^1 - ratio 40.17133587 over Ram 3.271163165 log n 136.8564461 bump 131^1 - ratio 41.1745089 over Ram 3.21940471 log n 141.776427 bump 137^1 - ratio 41.57610945 over Ram 3.250805563 log n 143.7223371 bump 7^3 - ratio 42.60313855 over Ram 3.200038518 log n 148.6568111 bump 139^1 - ratio 43.60210727 over Ram 3.275073797 log n 153.6607574 bump 149^1 - ratio 44.61432694 over Ram 3.22071367 log n 158.6780372 bump 151^1 - ratio 45.18291626 over Ram 3.261760202 log n 161.5112505 bump 17^2 - ratio 46.20162379 over Ram 3.206931917 log n 166.5674964 bump 157^1 - ratio 47.21440711 over Ram 3.277230899 log n 171.6612466 bump 163^1 - ratio 48.23074282 over Ram 3.220290278 log n 176.7792404 bump 167^1 - ratio 49.24183179 over Ram 3.287799086 log n 181.932532 bump 173^1 - ratio 50.24805522 over Ram 3.228541517 log n 187.1199178 bump 179^1 - ratio 51.26631908 over Ram 3.293967076 log n 192.3184148 bump 181^1 - ratio 51.82404921 over Ram 3.205575163 log n 195.2628538 bump 19^2 - ratio 52.83273596 over Ram 3.267967455 log n 200.5151272 bump 191^1 - ratio 53.04563115 over Ram 3.281136081 log n 201.6137395 bump 3^6 - ratio 54.07005962 over Ram 3.220955494 log n 206.8764297 bump 193^1 - ratio 55.09832965 over Ram 3.282209579 log n 212.1596334 bump 197^1 - ratio 56.13824653 over Ram 3.22181287 log n 217.4529382 bump 199^1 - ratio 56.45557713 over Ram 3.240024693 log n 219.0623761 bump 5^4 - ratio 57.47533017 over Ram 3.298549027 log n 224.4142343 bump 211^1 - ratio 57.60476235 over Ram 3.305977229 log n 225.1073815 bump 2^10 - ratio 58.60299034 over Ram 3.241380692 log n 230.5145532 bump 223^1 - ratio 59.6050657 over Ram 3.296806323 log n 235.9395033 bump 227^1 - ratio 60.61759257 over Ram 3.232422062 log n 241.3732253 bump 229^1 - ratio 61.63403256 over Ram 3.286623539 log n 246.8242637 bump 233^1 - ratio 62.64799568 over Ram 3.221820568 log n 252.3007273 bump 239^1 - ratio 63.67221839 over Ram 3.27449363 log n 257.7855242 bump 241^1 - ratio 64.68192381 over Ram 3.209099212 log n 263.3109771 bump 251^1 - ratio 65.68962805 over Ram 3.259094987 log n 268.8600532 bump 257^1 - ratio 66.69547057 over Ram 3.308998395 log n 274.4322072 bump 263^1 - ratio 67.69958251 over Ram 3.241375141 log n 280.0269186 bump 269^1 - ratio 68.71316482 over Ram 3.289904251 log n 285.6290375 bump 271^1 - ratio 69.28195086 over Ram 3.317137047 log n 288.7645317 bump 23^2 - ratio 70.30232697 over Ram 3.24929488 log n 294.3885492 bump 277^1 - ratio 71.32665713 over Ram 3.296638274 log n 300.0269038 bump 281^1 - ratio 72.36039885 over Ram 3.229428647 log n 305.6723507 bump 283^1 - ratio 73.382162 over Ram 3.275029711 log n 311.3525233 bump 293^1 - ratio 74.38270722 over Ram 3.319683824 log n 317.0793711 bump 307^1 - ratio 75.38709562 over Ram 3.249769843 log n 322.819164 bump 311^1 - ratio 76.400164 over Ram 3.293440965 log n 328.5653672 bump 313^1 - ratio 77.41711457 over Ram 3.224374375 log n 334.324269 bump 317^1 - ratio 78.41473649 over Ram 3.265924703 log n 340.1263873 bump 331^1 - ratio 79.41169082 over Ram 3.307447227 log n 345.9464703 bump 337^1 - ratio 80.39941875 over Ram 3.236182228 log n 351.7957951 bump 347^1 - ratio 81.39513097 over Ram 3.276260954 log n 357.650867 bump 349^1 - ratio 82.39459553 over Ram 3.316490715 log n 363.517335 bump 353^1 - ratio 83.39362406 over Ram 3.244889968 log n 369.4006574 bump 359^1 - ratio 84.38818935 over Ram 3.283589029 log n 375.3060193 bump 367^1 - ratio 85.38244953 over Ram 3.322276218 log n 381.2275977 bump 373^1 - ratio 86.37645598 over Ram 3.24984013 log n 387.1651339 bump 379^1 - ratio 87.37414754 over Ram 3.28737742 log n 393.1131689 bump 383^1 - ratio 87.49088072 over Ram 3.291769406 log n 393.8063161 bump 2^11 - ratio 88.48973144 over Ram 3.329350308 log n 399.7698954 bump 389^1 - ratio 89.48467856 over Ram 3.256298948 log n 405.7538317 bump 397^1 - ratio 90.48327589 over Ram 3.292637363 log n 411.7477931 bump 401^1 - ratio 91.47816766 over Ram 3.328840935 log n 417.7615083 bump 409^1 - ratio 92.46594017 over Ram 3.255165011 log n 423.7993792 bump 419^1 - ratio 93.46081907 over Ram 3.290188663 log n 429.842012 bump 421^1 - ratio 94.01493491 over Ram 3.309695722 log n 433.2093079 bump 29^2 - ratio 95.00882623 over Ram 3.23649366 log n 439.2754159 bump 431^1 - ratio 96.00974315 over Ram 3.270590084 log n 445.3461537 bump 433^1 - ratio 97.01081885 over Ram 3.304691917 log n 451.4306531 bump 439^1 - ratio 98.01547483 over Ram 3.338915714 log n 457.5242229 bump 443^1 - ratio 99.02034486 over Ram 3.264797057 log n 463.6312457 bump 449^1 - ratio 100.0221593 over Ram 3.297827853 log n 469.7559291 bump 457^1 - ratio 101.0275168 over Ram 3.330975468 log n 475.8893272 bump 461^1 - ratio 102.0396879 over Ram 3.257021972 log n 482.0270542 bump 463^1 - ratio 103.0554254 over Ram 3.289443466 log n 488.1733835 bump 467^1 - ratio 103.2368708 over Ram 3.295235055 log n 489.2719958 bump 3^7 - ratio 103.6295806 over Ram 3.307770024 log n 491.6698911 bump 11^3 - ratio 104.6417008 over Ram 3.34007606 log n 497.8415917 bump 479^1 - ratio 105.2050828 over Ram 3.251655773 log n 501.2755789 bump 31^2 - ratio 106.2199051 over Ram 3.283021679 log n 507.463843 bump 487^1 - ratio 107.2382498 over Ram 3.314496454 log n 513.6602871 bump 491^1 - ratio 108.2539705 over Ram 3.240575218 log n 519.8728932 bump 499^1 - ratio 109.2731903 over Ram 3.271085492 log n 526.0934834 bump 503^1 - ratio 110.292897 over Ram 3.301610344 log n 532.3259314 bump 509^1 - ratio 111.3042884 over Ram 3.331886276 log n 538.5816814 bump 521^1 - ratio 112.3220256 over Ram 3.257208542 log n 544.8412629 bump 523^1 - ratio 113.3233319 over Ram 3.286245266 log n 551.1346822 bump 541^1 - ratio 114.3252422 over Ram 3.315299503 log n 557.439131 bump 547^1 - ratio 115.3223698 over Ram 3.344215048 log n 563.7616962 bump 557^1 - ratio 116.320144 over Ram 3.268343789 log n 570.0949758 bump 563^1 - ratio 117.3185812 over Ram 3.296397707 log n 576.4388563 bump 569^1 - ratio 118.3229419 over Ram 3.324618066 log n 582.7862455 bump 571^1 - ratio 119.3279906 over Ram 3.249339042 log n 589.1440878 bump 577^1 - ratio 120.3285981 over Ram 3.276585902 log n 595.5191126 bump 587^1 - ratio 121.3299295 over Ram 3.303852473 log n 601.904307 bump 593^1 - ratio 122.3319989 over Ram 3.331139143 log n 608.2995686 bump 599^1 - ratio 123.3398218 over Ram 3.255529759 log n 614.6981635 bump 601^1 - ratio 124.3484052 over Ram 3.282151114 log n 621.1066923 bump 607^1 - ratio 125.3577623 over Ram 3.308792893 log n 627.5250572 bump 613^1 - ratio 126.370361 over Ram 3.335520231 log n 633.9499263 bump 617^1 - ratio 127.3886645 over Ram 3.259857576 log n 640.3780315 bump 619^1 - ratio 128.4004973 over Ram 3.285750232 log n 646.8253374 bump 631^1 - ratio 129.4084107 over Ram 3.311542589 log n 653.2883669 bump 641^1 - ratio 130.4218652 over Ram 3.337476745 log n 659.7545116 bump 643^1 - ratio 131.4385169 over Ram 3.261534446 log n 666.2268579 bump 647^1 - ratio 132.4560248 over Ram 3.286783033 log n 672.708435 bump 653^1 - ratio 133.4743996 over Ram 3.312053134 log n 679.1991585 bump 659^1 - ratio 134.4982745 over Ram 3.337459713 log n 685.6929124 bump 661^1 - ratio 135.5161752 over Ram 3.261385529 log n 692.2046577 bump 673^1 - ratio 135.8203463 over Ram 3.268705832 log n 694.1505679 bump 7^4 - ratio 136.8436943 over Ram 3.293334126 log n 700.6682391 bump 677^1 - ratio 137.4137237 over Ram 3.307052677 log n 704.279157 bump 37^2 - ratio 137.5227394 over Ram 3.309676293 log n 704.9723042 bump 2^12 - ratio 138.5520797 over Ram 3.334448801 log n 711.4987991 bump 683^1 - ratio 139.5801067 over Ram 3.258552335 log n 718.0369389 bump 691^1 - ratio 140.6046611 over Ram 3.282470958 log n 724.5894468 bump 701^1 - ratio 141.6279895 over Ram 3.306360962 log n 731.1533023 bump 709^1 - ratio 142.6479948 over Ram 3.330173386 log n 737.7311637 bump 719^1 - ratio 143.6668556 over Ram 3.353959091 log n 744.3200902 bump 727^1 - ratio 144.6866876 over Ram 3.277153174 log n 750.9172359 bump 733^1 - ratio 145.7074983 over Ram 3.300274535 log n 757.5225338 bump 739^1 - ratio 146.7313577 over Ram 3.323464947 log n 764.1332298 bump 743^1 - ratio 147.754163 over Ram 3.346631486 log n 770.7546355 bump 751^1 - ratio 148.0025434 over Ram 3.352257301 log n 772.3640734 bump 5^5 - ratio 149.0280708 over Ram 3.275506404 log n 778.9934366 bump 757^1 - ratio 149.4258815 over Ram 3.284249934 log n 781.558386 bump 13^3 - ratio 150.457187 over Ram 3.306917125 log n 788.1930194 bump 761^1 - ratio 151.4874976 over Ram 3.32956245 log n 794.8381103 bump 769^1 - ratio 152.5208351 over Ram 3.352274302 log n 801.4883894 bump 773^1 - ratio 153.5473095 over Ram 3.275430973 log n 808.1566176 bump 787^1 - ratio 154.5709524 over Ram 3.297267055 log n 814.8374723 bump 797^1 - ratio 155.5899343 over Ram 3.319003711 log n 821.5332712 bump 809^1 - ratio 156.613736 over Ram 3.340843181 log n 828.2315393 bump 811^1 - ratio 157.6348459 over Ram 3.264122817 log n 834.9420624 bump 821^1 - ratio 158.6607396 over Ram 3.285365855 log n 841.6550186 bump 823^1 - ratio 159.2299653 over Ram 3.297152732 log n 845.3685906 bump 41^2 - ratio 160.2624786 over Ram 3.318532842 log n 852.0863953 bump 827^1 - ratio 161.2998063 over Ram 3.340012641 log n 858.8066155 bump 829^1 - ratio 162.3345013 over Ram 3.361437928 log n 865.5388262 bump 839^1 - ratio 163.3629889 over Ram 3.284177807 log n 872.2875857 bump 853^1 - ratio 164.3943674 over Ram 3.304912187 log n 879.0410237 bump 857^1 - ratio 165.4304444 over Ram 3.325741025 log n 885.7967926 bump 859^1 - ratio 166.4694248 over Ram 3.34662823 log n 892.5572073 bump 863^1 - ratio 167.5023879 over Ram 3.269806211 log n 899.3337143 bump 877^1 - ratio 168.5382193 over Ram 3.290026627 log n 906.1147719 bump 881^1 - ratio 169.5786852 over Ram 3.310337514 log n 912.8980971 bump 883^1 - ratio 170.6220317 over Ram 3.330704631 log n 919.6859421 bump 887^1 - ratio 171.1931289 over Ram 3.341852992 log n 923.4471422 bump 43^2 - ratio 172.2289407 over Ram 3.362073026 log n 930.2572846 bump 907^1 - ratio 173.2675862 over Ram 3.284840511 log n 937.0718275 bump 911^1 - ratio 174.3056659 over Ram 3.304520627 log n 943.8951137 bump 919^1 - ratio 175.3415227 over Ram 3.324158601 log n 950.7292224 bump 929^1 - ratio 176.3768557 over Ram 3.343786646 log n 957.5719057 bump 937^1 - ratio 177.4149801 over Ram 3.363467609 log n 964.4188488 bump 941^1 - ratio 178.4542487 over Ram 3.286142674 log n 971.2721479 bump 947^1 - ratio 179.4946651 over Ram 3.305301405 log n 978.1317628 bump 953^1 - ratio 180.5297637 over Ram 3.324362209 log n 985.0059613 bump 967^1 - ratio 181.5676132 over Ram 3.343473671 log n 991.8842878 bump 971^1 - ratio 182.6066178 over Ram 3.362606403 log n 998.7687744 bump 977^1 - ratio 183.6467807 over Ram 3.285267457 log n 1005.659384 bump 983^1 - ratio 184.6865287 over Ram 3.303867566 log n 1012.558098 bump 991^1 - ratio 185.7274403 over Ram 3.322488492 log n 1019.462849 bump 997^1 - ratio 186.7648676 over Ram 3.341047087 log n 1026.379564 bump 1009^1 - ratio 187.8049988 over Ram 3.359654051 log n 1033.300235 bump 1013^1 - ratio 188.8463003 over Ram 3.282372532 log n 1040.226812 bump 1019^1 - ratio 189.8918367 over Ram 3.300545195 log n 1047.15535 bump 1021^1 - ratio 190.9355046 over Ram 3.318685379 log n 1054.093635 bump 1031^1 - ratio 191.9833843 over Ram 3.336898771 log n 1061.033857 bump 1033^1 - ratio 193.0324483 over Ram 3.355132747 log n 1067.979871 bump 1039^1 - ratio 194.0796943 over Ram 3.278040755 log n 1074.935464 bump 1049^1 - ratio 195.1311187 over Ram 3.295799501 log n 1081.892961 bump 1051^1 - ratio 196.1807578 over Ram 3.313528093 log n 1088.859928 bump 1061^1 - ratio 197.2345536 over Ram 3.331326893 log n 1095.828779 bump 1063^1 - ratio 198.289547 over Ram 3.349145919 log n 1102.803258 bump 1069^1 - ratio 198.8724667 over Ram 3.358991537 log n 1106.653405 bump 47^2 - ratio 199.9229823 over Ram 3.281810856 log n 1113.644582 bump 1087^1 - ratio 200.9761418 over Ram 3.299098865 log n 1120.639432 bump 1091^1 - ratio 202.0333959 over Ram 3.316454089 log n 1127.636114 bump 1093^1 - ratio 203.093304 over Ram 3.333852877 log n 1134.636448 bump 1097^1 - ratio 204.1544227 over Ram 3.35127154 log n 1141.642237 bump 1103^1 - ratio 205.2167544 over Ram 3.368710113 log n 1148.653451 bump 1109^1 - ratio 206.2788728 over Ram 3.291415684 log n 1155.671853 bump 1117^1 - ratio 207.3422073 over Ram 3.308382405 log n 1162.695612 bump 1123^1 - ratio 208.4067603 over Ram 3.325368567 log n 1169.724699 bump 1129^1 - ratio 209.4614029 over Ram 3.342196597 log n 1176.773086 bump 1151^1 - ratio 210.5200032 over Ram 3.359087778 log n 1183.823208 bump 1153^1 - ratio 211.577085 over Ram 3.375954729 log n 1190.881966 bump 1163^1 - ratio 212.6340265 over Ram 3.298354947 log n 1197.94758 bump 1171^1 - ratio 212.7979734 over Ram 3.300898072 log n 1199.046192 bump 3^8 - ratio 213.8542558 over Ram 3.317283006 log n 1206.120309 bump 1181^1 - ratio 214.9117545 over Ram 3.333686807 log n 1213.199493 bump 1187^1 - ratio 215.9704714 over Ram 3.350109503 log n 1220.28372 bump 1193^1 - ratio 217.0290842 over Ram 3.366530584 log n 1227.37463 bump 1201^1 - ratio 218.0849831 over Ram 3.289163364 log n 1234.475481 bump 1213^1 - ratio 219.1434026 over Ram 3.305126474 log n 1241.579626 bump 1217^1 - ratio 219.2466955 over Ram 3.30668434 log n 1242.272773 bump 2^13 - ratio 220.3068355 over Ram 3.322673399 log n 1249.381835 bump 1223^1 - ratio 221.3681987 over Ram 3.338680907 log n 1256.495791 bump 1229^1 - ratio 222.4333755 over Ram 3.354745931 log n 1263.611373 bump 1231^1 - ratio 223.4997818 over Ram 3.370829498 log n 1270.731817 bump 1237^1 - ratio 224.5635702 over Ram 3.29345261 log n 1277.861916 bump 1249^1 - ratio 225.6260484 over Ram 3.30903493 log n 1284.999989 bump 1259^1 - ratio 226.6822481 over Ram 3.324525172 log n 1292.152258 bump 1277^1 - ratio 227.7421474 over Ram 3.34006967 log n 1299.306092 bump 1279^1 - ratio 228.8045116 over Ram 3.355650319 log n 1306.463048 bump 1283^1 - ratio 229.8681032 over Ram 3.371248969 log n 1313.62467 bump 1289^1 - ratio 230.9353971 over Ram 3.293907657 log n 1320.787842 bump 1291^1 - ratio 232.0039242 over Ram 3.309148411 log n 1327.955652 bump 1297^1 - ratio 233.074919 over Ram 3.324424362 log n 1335.12654 bump 1301^1 - ratio 234.1496189 over Ram 3.339753162 log n 1342.298965 bump 1303^1 - ratio 235.2267951 over Ram 3.355117282 log n 1349.474454 bump 1307^1 - ratio 236.3015346 over Ram 3.370446645 log n 1356.659084 bump 1319^1 - ratio 237.3799583 over Ram 3.293283751 log n 1363.845228 bump 1321^1 - ratio 238.4596279 over Ram 3.308262515 log n 1371.035904 bump 1327^1 - ratio 239.0477073 over Ram 3.316421218 log n 1375.006196 bump 53^2 - ratio 240.114527 over Ram 3.331221707 log n 1382.222171 bump 1361^1 - ratio 241.1825781 over Ram 3.346039283 log n 1389.442545 bump 1367^1 - ratio 242.251862 over Ram 3.360873961 log n 1396.667298 bump 1373^1 - ratio 243.3212169 over Ram 3.375709625 log n 1403.897861 bump 1381^1 - ratio 244.3849109 over Ram 3.298208543 log n 1411.141374 bump 1399^1 - ratio 245.4475631 over Ram 3.312550053 log n 1418.39201 bump 1409^1 - ratio 246.5069511 over Ram 3.326847509 log n 1425.652532 bump 1423^1 - ratio 247.568674 over Ram 3.341176477 log n 1432.915862 bump 1427^1 - ratio 248.6338503 over Ram 3.355552052 log n 1440.180592 bump 1429^1 - ratio 249.7013693 over Ram 3.369959244 log n 1447.448117 bump 1433^1 - ratio 250.7701174 over Ram 3.292696017 log n 1454.719821 bump 1439^1 - ratio 251.8389861 over Ram 3.306730623 log n 1461.997069 bump 1447^1 - ratio 252.9101907 over Ram 3.320795899 log n 1469.277077 bump 1451^1 - ratio 253.9848409 over Ram 3.334906418 log n 1476.558463 bump 1453^1 - ratio 255.060727 over Ram 3.349033165 log n 1483.843969 bump 1459^1 - ratio 256.1345535 over Ram 3.363132868 log n 1491.137667 bump 1471^1 - ratio 257.2074353 over Ram 3.377220167 log n 1498.43814 bump 1481^1 - ratio 258.2837212 over Ram 3.299876046 log n 1505.739962 bump 1483^1 - ratio 259.3623296 over Ram 3.313656527 log n 1513.044478 bump 1487^1 - ratio 260.4443511 over Ram 3.327480615 log n 1520.350338 bump 1489^1 - ratio 261.5287026 over Ram 3.341334471 log n 1527.658881 bump 1493^1 - ratio 262.6142983 over Ram 3.355204224 log n 1534.971435 bump 1499^1 - ratio 263.6979009 over Ram 3.369048512 log n 1542.291961 bump 1511^1 - ratio 264.7795384 over Ram 3.382867694 log n 1549.620399 bump 1523^1 - ratio 265.8613535 over Ram 3.305462762 log n 1556.954075 bump 1531^1 - ratio 266.9412466 over Ram 3.318889108 log n 1564.295559 bump 1543^1 - ratio 268.0223746 over Ram 3.332330807 log n 1571.640924 bump 1549^1 - ratio 269.1057836 over Ram 3.345800867 log n 1578.988868 bump 1553^1 - ratio 270.1904307 over Ram 3.359286319 log n 1586.340668 bump 1559^1 - ratio 271.275277 over Ram 3.372774248 log n 1593.697586 bump 1567^1 - ratio 272.3623985 over Ram 3.295729417 log n 1601.057054 bump 1571^1 - ratio 273.4497264 over Ram 3.308886662 log n 1608.421601 bump 1579^1 - ratio 274.5393257 over Ram 3.322071391 log n 1615.788678 bump 1583^1 - ratio 275.6260663 over Ram 3.335221528 log n 1623.16456 bump 1597^1 - ratio 276.7150635 over Ram 3.348398972 log n 1630.542943 bump 1601^1 - ratio 277.8053005 over Ram 3.361591417 log n 1637.925068 bump 1607^1 - ratio 278.8988123 over Ram 3.37482349 log n 1645.308436 bump 1609^1 - ratio 279.9945861 over Ram 3.297853938 log n 1652.694287 bump 1613^1 - ratio 280.4154458 over Ram 3.302810941 log n 1655.5275 bump 17^3 - ratio 281.5141146 over Ram 3.315751367 log n 1662.917064 bump 1619^1 - ratio 282.6160672 over Ram 3.328730471 log n 1670.307863 bump 1621^1 - ratio 283.7192721 over Ram 3.341724324 log n 1677.702356 bump 1627^1 - ratio 284.8217054 over Ram 3.35470909 log n 1685.102977 bump 1637^1 - ratio 285.9183887 over Ram 3.367626129 log n 1692.515741 bump 1657^1 - ratio 287.0163142 over Ram 3.380557801 log n 1699.932119 bump 1663^1 - ratio 287.6212267 over Ram 3.387682628 log n 1704.009656 bump 59^2<|endoftext|> -TITLE: Consequences of a bound on possible counterexamples to Riemann hypothesis -QUESTION [7 upvotes]: The Riemann hypothesis has many strong consequences in number theory. The question is: would a bound on the number of zeros of Riemann zeta-function in the critical strip with real part not equal 1/2 have significant consequences? For example, what would the statement that a number of such zeros is finite imply? - -REPLY [12 votes]: Such bounds are generally called zero-density estimates (for the Riemann zeta function or more general $L$-functions), and they have significant consequences. Chapter 10 of Iwaniec-Kowalski's Analytic number theory is devoted to this topic. -A famous example of such a result and application is Huxley's theorem (Inventiones Mathematicae 15 (1972), 164-170): for $x^{7/12+\epsilon}0$. -More generally, one has that if $\zeta(s)$ has no zeros in the complex halfplane $H_{\sigma} =\{z \colon \Re(z)> \sigma\}$ then -$$ -\pi(x)- \operatorname{Li}(x) = O(x^{\sigma + \epsilon}) -$$ -for every $\epsilon>0$. -Now if the number of zeros were finite it would in particular follow that there is some $\sigma_0<1$ such that $H_{\sigma_0}$ contains no zeros of $\zeta$ (recall that all zeros are known to have real part less than $1$ and if their number is finite take the max of the real parts), and we had -$$ -\pi(x)- \operatorname{Li}(x) = O(x^{\sigma_0 + \epsilon}) -$$ -for every $\epsilon >0$. -This would be a lot better than the current error term of the form $O(x \exp(-c (\log x)^{3/5}) )$.<|endoftext|> -TITLE: Do maximal polyhedra have algebraic volume? -QUESTION [10 upvotes]: Is it possible to prove that for every $n > 3$ the maximal possible volume of a convex polyhedron having $n$ vertices inscribed in a sphere of unit radius is an algebraic number? - -Update: What can be said about degrees of these algebraic numbers for different $n$? - -REPLY [16 votes]: Yes. Let $\mathbb{R}^{alg}$ be the field of real algebraic numbers. This is a real closed field which means that, for any statement of first order logic, using the symbols $0$, $1$, $+$, $\times$, $=$, $<$, that statement is true in $\mathbb{R}^{alg}$ if and only if it is true in $\mathbb{R}$. -The statement "The volume of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is $V$" is a first order statement in the coordinates of the $x_i$ and $V$. Proof sketch There are finitely many simplicial complexes on the abstract vertex set $1$, $2$, ..., $n$. For each of these simplicial complexes, the statement that this simplicial complex is a triangulation of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is finitely many polynomial inequalities and, if the simplicial complex is such a triangulation, then the volume of $\mathrm{Hull}(x_1, x_2, \ldots, x_n)$ is a polynomial in the coordinates of the $x$'s. So we can encode $\mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n)) = V$ as - -$\Delta_1$ encodes a triangulation of the convex hull and $V=\cdots$ - OR $\Delta_2$ encodes a triangulation of the convex hull and - $V=\cdots$ OR ... - -So I can talk about $\mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$ in the first order theory of ordered fields. Consider the statement: - -There exists an $A$ such that there are $n$ points on the unit sphere - with $A = \mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$ and such - that, for any points $y_1$, \dots, $y_n$ on the unit sphere, we have - $A \geq \mathrm{Vol}(\mathrm{Hull}(x_1, \ldots, x_n))$. - -This is a first order statement by the above discussion. It is true when all the variables range over $\mathbb{R}$, because $(S^2)^n$ is compact. So it is also true when all variables range over $\mathbb{R}^{alg}$. Let $A^{alg}$ be the maximum volume of a sphere with real algebraic coordinates; we have just shown that this number exists. Let $A$ be the maximum volume of a sphere with real coordinates. -It remains to check that $A = A^{alg}$. Let $f(t)$ be the minimal polynomial of $A^{alg}$ over $\mathbb{Q}$. Let the real roots of $f$ be $x_1$, $x_2$, ...., $x_n$ with $A^{alg} = x_r$. Consider the statement - -The number $A$ in the previous paragraph obeys $f(A) = 0$. There are - $n-1$ other numbers $x_1$, $x_2$, ..., $x_{r-1}$, $x_{r+1}$, ..., - $x_n$ with $f(x_1)=f(x_2) = \cdots = f(x_n)=0$ and $x_1 < x_2 < \cdots -> < x_{r-1} < A < x_{r+1} < \cdots < x_n$. - -This is a first order statement which has been constructed to be true in $\mathbb{R}^{alg}$. So it is also true in $\mathbb{R}$, and we see that $A$ is the $r$-th root of $f$ in $\mathbb{R}$. Thus, $A = A^{alg}$, as claimed.<|endoftext|> -TITLE: Does there always exist a line bundle whose Chern class represents an integer symplectic form? -QUESTION [8 upvotes]: Let $(M, \omega, J)$ be a compact symplectic manifold with a -compatible almost complex structure $J$, such that the symplectic -form determines an integer cohomology class, ie -$$ [\omega] \in H^2(M, \mathbb{Z}).$$ -Does there exist an "almost homolorphic line bundle" $L \rightarrow M $ -such that its first Chern class is $[\omega]$? -By "almost holomorphic line bundle", I mean a complex line bundle, whose -transition data is pseudo holomorphic (ie differential commutes with J). -Note that, if M was a Kahler manifold (ie J is integrable), the answer is yes. -$\textbf{Modified Question:} $ It seems the answer to the above question is no. -Here is a weaker question: Suppose $(M,\omega)$ is a compatc symplectic manifold, -with integral symplectic class. Does there exist an $\omega$-compatible almost complex structure $J$, such that the answer to the -above question is yes (ie there will exist an "almost homolorphic" line bunde (wrt to the J) whose first Chern class is $\omega$)? -It seems there are examples of $(M, \omega)$ such that for a generic compatible $J$ the answer is no. But still there -could be some compatible $J$ for which the answer is yes. - -REPLY [10 votes]: The answer is already 'no' in dimension $4$. The generic almost complex structure compatible with a symplectic structure in dimension $4$ does not admit any pseudoholomorphic functions (in your sense) other than the constants, so, for such data $(M,\omega,J)$, you are asking whether there is a line bundle $L$ that can be described by constant transition functions such that its first Chern class is $[\omega]$. -To see that this is not always possible, consider the special case in which $M$ is simply-connected, and choose $J$ generically compatible with $\omega$ such that the sheaf of $J$-pseudoholomorphic functions is the constant sheaf. Then any $L$ that can be built with constant transition functions will inherit from that construction a flat connection $\nabla$ with holonomy in $\mathbb{C}^\ast$. Since $M$ is simply connected and nabla is flat, the holonomy will be trivial, so $L$ will have a nonvanishing parallel section and hence be the trivial line bundle, so its first Chern class will be zero.<|endoftext|> -TITLE: Minimum requirements for a Kähler manifold to be hyperkähler -QUESTION [11 upvotes]: In 'panoramic view of Riemmannian geometry' when introducing hyperkähler manifolds, Berger states, informally, that a hyperkähler manifold is a Riemmannian manifold which is Kähler for more than one different almost complex structures. -I was wondering whereas this was a theorem or just a 'catchphrase'. In other words, my question is : if a Riemannian manifold is Kähler for two different (linearly independent) almost complex structures, is it hyperkähler ? -Of course on has to ask the metric to be irreducible, since things such as $\mathbb{S}^2\times M$ where $M$ is hyperkähler has at least 6 (!) independent almost complex structures. And being $4n+2$ dimensional, they can't be hyperkähler ! -I think this is true in (real) dimension 4 for the following reason : the holonomy group $G$ of a Kähler $(M^4,g)$ is contained in $U(2)$. It leaves the Kähler form invariant. If there are two almost complex structures, there are two Kähler forms, so $G$ must act trivially on a 2-dimensional subspace of $\Lambda^2T_p^*M$. Since the holonomy representation of $U(2)$ only preserves the Kähler form, $G$ is a strict subgroup of $U(2)$ and is therefore contained in $SU(2)=Sp(1)$, hence $M$ is hyperkähler. -My knowledge of representation theory is too scarce to see if this carry on in higher dimensions. - -REPLY [10 votes]: Suppose that the metric on $M^n$ has irreducible holonomy, is simply connected (or, slightly more generally, that the restricted holonomy $H^0$ acts irreducibly), and that there exist two independent parallel complex structures. Since there is at least one parallel complex structure $I$, the holonomy group $H^0$ is a subgroup of $\mathrm{U}(n/2)$. -If $H^0=\mathrm{U}(n/2)$ then $I$ generates the ring of parallel endomorphisms of the tangent bundle, and the only other parallel complex structure is $-I$, so the restricted holonomy must be a proper subgroup of $\mathrm{U}(n/2)$. -By Berger's classification (since it acts irreducibly), $H^0$ must be either $\mathrm{SU}(n/2)$ or $\mathrm{Sp}(n/4)$ or else the metric must be locally symmetric. -When $n=4$, one has $\mathrm{SU}(n/2)=\mathrm{Sp}(n/4)$, so if $H^0=\mathrm{SU}(2)$, the metric is hyperKähler. If $n>4$ and $H^0=\mathrm{SU}(n/2)$, then, again, $I$ generates the ring of parallel endomorphisms of the tangent bundle, and the only other parallel complex structure is $-I$, contrary to hypothesis, so the restricted holonomy can't be $\mathrm{SU}(n/2)$. -Finally, if the metric were locally symmetric, it would have to be locally isometric to one of the known irreducible Hermitian symmetric spaces, but looking at the list of these, one sees again that the ring of parallel endomorphisms of the tangent bundle is generated by $I$, so none of these admit a linearly independent parallel complex structure. -Thus, the only possibility is that the restricted holonomy is $\mathrm{Sp}(n/4)$, i.e., the metric is (locally) hyperKähler.<|endoftext|> -TITLE: A game on equiangular polygons -QUESTION [5 upvotes]: Let $n\geq 3$ be a natural number and Consider the following game: -Correspond an integer to each vertices of an equiangular polygon -(at least two of the numbers are unequal). -(1) Replace the number of each vertices with the number obtained by the sum of -its neighborhood numbers minus its own number. (Edit: Do this for all vertices at once, not one by one) -This gives us a new equiangular polygon with an integer corresponded to each of its vertices. -Again do as in (1) and Continue the progress. -Let $A$ be the set containing the numbers corresponded to vertices of this polygon -in this (infinite) process. Now the question is: -what are the possible values of $n$ if we want $A$ to be a bounded set of integers!? - -REPLY [5 votes]: For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded. -The question is more subtle if you ask for which $n$ it is possible to find a non-constant integer choice of labels for which the set of numbers $A$ is bounded. The answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$. -$$ *$$ -The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$ (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated according to their multiplicity, which is either $1$ or $2$) -$$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$ -In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $ in the spectral decomposition, implying that $A=\{L^k v\}_{k\ge0}$ is unbounded. Note that $v$ is certainly not a constant vector, as required. Also note that for even $n$, $\lambda_{n/2}=-3$ corresponds to the example in Barry Cipra's comment. -For what $n$ it is possible to find non-constant integer labels $v=(v_1,\dots,v_n)$ of the vertices for which $A=\{L^k v\}_{k\ge0}$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue $-1$. -If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits. -Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ -that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$ for some $m$. So $1$ is one of the eigenvalues of $L^m$, which are of the form $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or $\lambda_k=1$. In the former case, $k=0$ and $v$ is a simple eigenvalue of $L$ with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.<|endoftext|> -TITLE: How $a+b$ can grow when $a!b! \mid n!$ -QUESTION [13 upvotes]: Let $a,b,n$ be natural numbers such that $a!b! \mid n!$. I am looking for a (somehow best) upper bound of $a+b$ in terms of $n$ (for large values on $n$). For example it is clear to see that we must have $a+b \leq 2n$. But unfortunately I am looking for much smaller bound ! Any idea would be helpful. - -REPLY [11 votes]: This is really a comment, but given all the interest in the question it seemed a good idea to write it as an answer. Apparently what was discussed earlier (in the answers of Seva and GH from MO) is an old result of Erdos! The reference is Aufgabe 557, Elemente Math. (1968) 111-113, which may be found here: http://retro.seals.ch/digbib/view?rid=elemat-001:1968:23::117&id=browse&id2=browse4&id3= . I found this reference in an interesting paper of Erdos, Graham, Ruzsa and Strauss from 1975, which you can find here (see bottom of page 90): http://math.ucsd.edu/~ronspubs/75_03_prime_factors.pdf -Erdos, Graham, Ruzsa and Strauss also comment that there is a constant $c>0$ such that for all $n$ except a set of density zero we have -$$ -\frac{(2n)!}{n!(n+[c\log n])!} \in {\Bbb N}. -$$ -Thus the bound of $2n+O(\log n)$ is best possible. -This is also in Erdos's exercise linked above. Note on the history: Erdos proposed as an exercise to show the upper bound $a+b\le n+C\log n$ (part 1) and that this is best possible (part 2). Part 1 was solved by P. Bundschuh (same solution as in GH from MO's response). The solution printed for Part 2 was that sent by the proposer Erdos. -They also raise a "curious problem" in this paper: Namely it is possible that $n!/(a!b!)$ is not an integer for $a+b \ge n+c\log n$ only because of small primes. In other words for every $c$ it is asked whether there exists a $k$ such that for infinitely many (or even all sufficiently large $n$) there are suitable $a$ and $b$ with $a+b>n+c\log n$ and such that $n!/(a!b!)$ has no prime factor $>k$ in its denominator.<|endoftext|> -TITLE: The number of relevant scales for a finite metric space -QUESTION [11 upvotes]: For an $n$-element metric space $X=\{x_1,\dots,x_n\}$ with metric -$d$ we introduce an array containing $\frac{n(n-1)}2$ numbers -$d(x_i,x_j)$, $i d_1+\sum_{i -TITLE: Information from Moment Polytopes -QUESTION [12 upvotes]: Let $T$ be a compact real torus, and $X$ a Hamiltonian $T$-manifold (which you may take to be a smooth complex projective variety) with moment map $\mu:X\rightarrow\frak{t}^*$. If $\dim(T)=\frac{1}{2}\dim(X)$, then $X$ is classified up to $T$-equivariant symplectomorphism by its moment polytope $\mu(X)$. If we relax this dimension condition, then what information about $X$ might we hope to obtain from an explicit description of its moment polytope? For instance, is the $T$-equivariant cohomology of $X$ nicely related to $\mu(X)$? My question is somewhat vague, so please feel free to answer in any fashion you deem to be acceptable. - -REPLY [8 votes]: Here's two natural things to ask about any compact group action on a compact manifold: (1) what are the (finitely many) conjugacy classes of stabilizer groups? Assume our group is a torus, so we can omit "conjugacy classes of". (2) For each such stabilizer, what are the (finitely many) components of its fixed point set? -So, we have a poset of such submanifolds, and a torus subgroup for each. Now comes the additional question in the Hamiltonian situation: (3) what's the moment polytope for each of these submanifolds? -In actual examples, (1-3) are easy to figure out, and it's silly to answer (3) only for the whole manifold (or one might say, only for the fixed points). -You ask if $H^*_T(X;\mathbb Q)$ can be computed from this sort of data alone. In the case that the minimal strata in (2) are all $S^2$s, $X$ is called a GKM space (for Goresky-Kottwitz-MacPherson), and indeed it can; it's easy to find references with that keyword.<|endoftext|> -TITLE: Patterns among integer-distance points -QUESTION [28 upvotes]: Mark each point of $\mathbb{N}^2$ ($\mathbb{N}$ the natural numbers) if its -Euclidean distance from the origin is an integer. One obtains a plot like this, symmetric about the $45^\circ$ diagonal. - -There are obvious patterns here. The straight lines through the origin are derived -from scalings of Pythagorean triples: the $(3,4,5)$ triangle, the $(5,12,13)$ triangle, the $(8,15,17)$ triangle, etc. -But other patterns are discernible, some of which I am perhaps hallucinating: - -Do these patterns reflect Diophantine curves dense in integer-distance points? -To what extent is, in some sense, this entire plot (extended indefinitely) understood -as a union of such curves? -Or do there remain unknowns lurking in here, i.e., there are sporadic -points with no as-yet apparent logic behind their appearance? -(Tangentially related to the MO question, "Integer-distance sets".) - -Addendum. It is now clear (after looking at plots extending further) -that the blue and tan curves are actually one, crossing the diagonal, -and not pinched off as I drew them. - -REPLY [2 votes]: Here are the various curves from my answer:<|endoftext|> -TITLE: Lex $\infty$-colimits -QUESTION [10 upvotes]: In the paper lex colimits, Garner and Lack gave a general characterization of "exactness properties" for categories (and enriched categories). A "lex-weight" is a weight for colimits whose domain category has finite limits. If $\Phi$ is a class of lex-weights, then a finitely complete category is "$\Phi$-lex-cocomplete" if it admits all $\Phi$-weighted colimits of left-exact diagrams, and "$\Phi$-exact" if moreover the functor assigning these colimits itself preserves finite limits. Intuitively, $\Phi$-exactness of $C$ means that "all exactness relations between $\Phi$-lex-colimits and finite limits which hold in Grothendieck topoi hold in $C$", which is justified by a theorem that a small $\Phi$-lex-cocomplete category is $\Phi$-exact just when it embeds in a Grothendieck topos by a functor preserving finite limits and $\Phi$-lex-colimits. -They also gave many examples of "exactness properties" which can be characterized as $\Phi$-exactness for some $\Phi$, such as regularity, Barr-exactness, lextensivity, coherency, adhesivity, etc. In basically all examples, the proof that the given explicitly defined property is sufficient for $\Phi$-exactness proceeds by constructing a subcanonical Grothendieck topology out of the colimits in question and showing that the colimits are preserved by the resulting embedding into the topos of sheaves for this topology. -My question is about whether there is a version of this theory for $(\infty,1)$-categories. The first part of it seems extremely formal, and I would expect probably works just the same, although obviously there are a lot of details to work out. At present I am more interested in the construction of examples. This is interesting for two reasons: - -In the $(\infty,1)$-categorical context, for any kind of colimit, there is an obvious notion of "exactness", namely "descent", as in for example section 6.5 of Rezk's note "Toposes and homotopy toposes" or Lemma 6.1.3.5 of Higher Topos Theory. -However, there is not (yet) a fully general notion of "Grothendieck topology" for $(\infty,1)$-categories which suffices to present all Grothendieck $(\infty,1)$-topoi. - -Here then, at last, is a concrete question. - -Suppose $C$ is a small $(\infty,1)$-category with finite limits, which also admits colimits of some specified sort, and that these colimits satisfy descent. Does $C$ necessarily embed in a Grothendieck $(\infty,1)$-topos by a functor preserving finite limits and the specified colimits? - -There are two obvious candidates for such an $(\infty,1)$-topos: - -The topos of sheaves on $C$ for the topology whose covering sieves consist of coprojections for the specified colimits. This almost certainly does not work. It doesn't even work in general in the 1-categorical case, although in many particular cases it does. In general, Garner and Lack showed that some additional covering families have to be added to ensure "postulatedness" of the colimits. It's unclear to me whether this will remain possible when we go "off to $\infty$". -The subcategory of presheaves on $C$ which preserve the specified colimits (i.e. take them to limits of $\infty$-groupoids). This satisfies the embedding property almost by definition, but I don't immediately see why it should be an $(\infty,1)$-topos. If this worked, it would be a generalization of Remark 6.3.5.17(a) of Higher Topos Theory, but I don't see a way to generalize that proof. - -REPLY [5 votes]: Probably not. -Claim: Let $C$ be a small $(\infty,1)$-category with finite limits and colimits which admits an embedding $V:C\to E$ into a Grothendieck $(\infty,1)$-topos preserving finite limits and colimits. Then $C$ has a countable coproduct of copies of the terminal object $1$. -Proof: Let $S^1\in C$ be the pushout of $1 \leftarrow 1+1 \to 1$, and let $Z\in C$ be the pullback of $1\to S^1 \leftarrow 1$. Since $V$ preserves these limits and colimits, $V(Z)$ is the analogous object of $E$, which is the countable coproduct of copies of $1$ (this is true in presheaf $(\infty,1)$-toposes and preserved by left exact localizations). Thus, since $V$ is fully faithful, $Z$ is also the countable coproduct of copies of $1$. $\quad\Box$ -It seems unlikely to me that assuming descent for finite colimits, which is (at least intuitively) a "finitary" property, would be sufficient to guarantee an infinitary property like the existence of some countable coproduct. It's reasonable to expect that the object $Z$ would be an "internal object of integers", but not (it seems to me) that it would be such a coproduct in an externally infinitary sense. -On the other hand, it does seem likely to me that Proposition 7.3 of the Garner-Lack paper (any small $\Phi$-exact category $C$ embeds $\Phi$-exactly in a topos) should work in the $(\infty,1)$-case, at least when the generating class of lex weights $\Phi$ is a small set. The topos in question is the category of presheaves on $C$ that preserve $\Phi^*$-lex-colimits, where $\Phi^*$ is the "lex saturation" of $\Phi$. The proof (in the case of small $\Phi$) shows this category to be the intersection of two lex-reflective subcategories of the topos of presheaves on $\Phi_l C$, the closure of the representables in the presheaves on $C$ under finite limits and $\Phi$-lex-colimits; thus it is also lex-reflective, hence a topos. -I haven't checked the details, but on a quick scan I don't see anything in this proof that would obviously fail in the $\infty$-world. One difference in the $\infty$-world is that apparently not every lex-reflective subcategory of a topos is accessible (hence also a topos), but in this case the two reflections are both abstractly equivalent to presheaves on $C$, hence accessible. (The case of proper-class $\Phi$ does seem suspicious, as in that case they use a proper-class intersection of lex-reflective subcategories, appealing to a fact that the lex-reflective subcategories of a topos form a small complete lattice, which seems less likely to be true for $\infty$-toposes. But for something like $\Phi = \{ \text{pushouts}\}$ it should be fine.) -I think the disconnect is that even in the $\infty$-world, descent for a certain class $\Phi$ of colimits is (probably) not strong enough to ensure $\Phi$-exactness, so the analogue of Prop. 7.3 can't be applied. A $\Phi$-exact category is one where the inclusion $C \to \Phi_l C$ has a left exact left adjoint, and so in particular it has a left adjoint, which means that $C$ must be not just $\Phi$-lex-cocomplete but $\Phi^*$-lex-cocomplete. The latter is a stronger condition in general even for 1-categories, e.g. there is a $\Phi_{rc}$ for which $\Phi_{rc}$-lex-cocompleteness means the existence of reflexive coequalizers, whereas $\Phi_{rc}^*$-lex-cocompleteness requires also the existence of "equivalence relations freely generated by reflexive relations". Similarly in the $\infty$-case, for $\Phi = \{ \text{pushouts}\}$, $\Phi^*$-lex-cocompleteness includes also countable coproducts of copies of $1$ (and lots of other stuff like $\Omega^k S^n$!). -Note also that even in the 1-categorical case, "equivalence relations freely generated by reflexive relations" are an infinite colimit, even though $\Phi_{rc}$ itself consists only of finite limits. In particular, therefore, there are colimits that an elementary 1-topos admits but is not necessarily exact for (in contrast to a Grothendieck 1-topos, which is exact for all lex colimits), hence cannot be preserved by any embedding into a Grothendieck 1-topos. So really nothing new is happening in the $\infty$-case: once again there may be colimits (such as pushouts) that an "elementary $(\infty,1)$-topos" admits but is not necessarily exact for, hence cannot be preserved by any embedding into a Grothendieck $(\infty,1)$-topos. The only difference is that perhaps this is more surprising in the $\infty$-world, where we might have acquired an intuition that "everything that's wrong with colimits in 1-categories has been fixed by descent". (-:<|endoftext|> -TITLE: Dimensional regularization in odd dimensions -QUESTION [12 upvotes]: I am not quite sure that my question below is appropriate for this site, probably it should be addressed to the physical commutity. But I hope that some (mathematical) physicists do attend MO. I have two questions on dimensional regularization used in the renormalization theory (they should be very basic, I am a mathematician, even not a mathematical physicist). -1) Is there a mathematically rigorous exposition of the dimensional regularization? -2) Let $d$ denote the dimension of the space time. -My impression is that the method of dimensional -regularization works better for even $d$ rather than for odd. Namely -for some integrals which are obviously divergent in odd dimensions, -the method of dimensional regularization gives convergent -expressions. Below I give a simple example of such a situation for -$d=3$. Thus my second question is how to resolve this apparent contradiction, and -whether the method can be modified to work in odd dimensions as -well, even at the physical level of rigour. -Here is the example. Consider the theory $\phi^4$ in Euclidean space-time. Consider the -Feynmann diagram with just one vertex, one self-loop, and two -exterior lines (though, I guess, one can construct many other -examples). The corresponding integral, up to some factors containing -the interaction constant, is equal to $\int \frac{1}{p^2+1}d^dp$ (we -take the mass $m=1$ for simplicity; it is physically impossible for -dimensional considerations, but does not influence the analysis of -convergence issues). We have -\begin{eqnarray*} -\int \frac{1}{p^2+1}d^dp=\frac{2\pi^{d/2}}{\Gamma(d/2)}\int_0^\infty -\frac{r^{d-1}}{r^2+1}dr=:A. -\end{eqnarray*} -The integral $A$ diverges for $d\geq 2$. After some -change of variables and standard computations it becomes -\begin{eqnarray*} -2\pi^{d/2} \Gamma(-\frac{d}{2}+1). -\end{eqnarray*} -The last expression has no poles at odd $d$, in -particular at $d=3$. This apparently contradicts the above -mentioned divergence of the integral $A$. On the other hand, -at even $d$, $\Gamma(-\frac{d}{2}+1)$ does have a pole -as expected, and the method works well. - -REPLY [3 votes]: Albeit rather late I would like to add the following reference as another answer to your question (1): dimensional regularization is also treated rigorously in -P. Etingof, Note on Dimensional Regularization in Vol 1, pp 597–607, of Quantum Fields and Strings: A Course for Mathematicians (see also here).<|endoftext|> -TITLE: Failure of Fredholm property of elliptic PDE systems -QUESTION [8 upvotes]: Roughly speaking, a PDE operator satisfies the Fredholm property if its principal symbol is elliptic and the information provided on the boundary satisfies the Shapiro-Lopatinskii condition. -What can happen when one of these conditions is not met, but the other holds: - -In case of failure of the Shapiro-Lopatinskii boundary condition (while having ellipticity). Does one automatically have an infinite dimensional kernel? -What happens when the operator is elliptic everywhere but at some point, while the Lopatinskii condition is still satisfied? Can the PDE problem still be well-posed? - -Is there a clear picture of what happens in these situations, or are there counterexamples of different type, depending on regularity of coefficients and boundary? -Thank you for any information. - -REPLY [2 votes]: Example for first question: -$\Delta^2 u=0$, say on the unit disk, with boundary condition ${\partial\over\partial n}\Delta u=0$, $\Delta u-u=0$. The boundary conditions do not satisfy the Lopatinskii condition (note that the $u$ term is lower order). Nevertheless, it follows from the first boundary condition that $\Delta u$ is constant, and then the second boundary condition implies that $u$ is constant on the boundary. So the kernel is one-dimensional. -Example for second question: -ODE problems like $(x^2y')'-y=0$ with Dirichlet conditions $y(1)=y(-1)=0$.<|endoftext|> -TITLE: How constructive is Dirichlet on primes in progressions? -QUESTION [8 upvotes]: Is there a known elementary function bound in terms of $a,b,n$ for the $n$-th prime equal to $b$ modulo $a$ (coprime to $b$)? -Bounds on Linnik's constant answer this for the first prime in each progression. Is there a known analogue for an $n$-th prime in a progression? And I found some references on an error term for the prime number theorem for arithmetic progressions. But I don't see how to turn these into a construction for arbitrary $a,b,n$. - -REPLY [5 votes]: Corollary 18.8 in Iwaniec-Kowalski's Analytic number theory shows the existence of an explicitly computable $L>3/2$ such that for $x>a^L$ and $(a,b)=1$, the number of primes less than $x$ and congruent to $b$ modulo $a$ is at least a constant times $\frac{x}{\varphi(a)\sqrt{a}\log x}$. So the $n$-th prime congruent to $b$ modulo $a$ is less than a constant times $a^L n\log (2n)$.<|endoftext|> -TITLE: Do proper Zariski closed sets of algebraic sets have measure zero -QUESTION [7 upvotes]: This is a question related to another question I asked: here. -Say we induce a probability measure that is absolutely continuous with respect to to Lebesgue measure onto an irreducible real algebraic set $V$. This is done by considering the $w^*$-limit of the sequence of measures $\lim_{\epsilon \to 0} \mu_{V,\epsilon}=\mu_V$, where -$\mu_{V,\epsilon}(A) := \frac{\mu(A\cap V_\epsilon)}{\mu(V_\epsilon)}$, and, $V_\epsilon := \{x:dist(x,V)<\epsilon\}$. -We assume that the induced measure, $\mu_V$, is well defined and is supported on $V$. -I would like to prove (or better, find reference) that any proper zariski closed set of $V$ has measure zero. i.e. for any irreducible algebraic set $U$ such that $U\cap V \ne V$ we have $\mu_V(U)=0$. - -This is what I have: If the Minkowski dimension of an algebraic set equals its krull dimension, because than $\mu(U_\epsilon)$ decays faster than $\mu(V_\epsilon)$ and then the result might follows. -If at any non-singular point, $V$ and $U$ are differential manifolds the former works at every non singular point. Still one needs to show that if $S$ are all the singular values then $\mu_V(S)=0$. - -REPLY [6 votes]: In your comment, you exhibit two important features of “having measure zero”: it is a local property and it is invariant under diffeomorphisms. This is because 1. you are considering a measure continuous with respect to the Lebesgue measure, 2. the ambient space is sigma countable and 3. locally compact -Now let us simplify the problem a bit: we only need to prove that -PROPOSITION. Any algebraic hypersurface of ${\bf C}^N$ is a null set for the Lebesgue measure. -To justify this simplification, observe that $\mu$ is continuous with -respect to the Lebesgue measure, so that it has more null sets -(i.e. sets of measure zero) and it is then enough to show that your -algebraic variety is a null set for the Lebesgue measure. An -algebraic variety is cut out by hypersurfaces, so if any (real) -hypersurface is a null set, any (real) algebraic variety must also be -a null set. The last point to check is that if the proposition holds -it implies the analog statement for real algebraic sets. I will not go in the -details, but the point is that if V is the complex zero locus of a -real equation, you can cover it by acountable set of self-conjugated -measurable compact sets whose union has an arbitrary small measure: -you conclude by observing that the Lebesgue measure of ${\bf R}^N$ can -be recovered from the Lebesgue measure of ${\bf C}^N$ with an -$\epsilon$-based construction similar to yours. (To deal with this -gently, use the cahracterisation of the Lebesgue measure as the -Haar-mesure for the topological group ${\bf R}^N$.) -It is not a very clean exposition, but I nevertheless assume I -convinced you that the PROPOSITION implies that algebraic subvarieties -of ${\bf R}^N$ are null sets for the Lebesgue measure. -Proof of the PROPOSITION. Let $f$ be a complex polynomial on ${\bf -C}^N$ vanishing at some point $p$. Thanks to the Weierstrass -preparation theorem, we can assume that in a coordinate system -centered on $p$, the function $f$ has the form -$$ -f(z, \zeta) = z^k + z^{k-1} g_1(\zeta) + \cdots + g_k(\zeta) -$$ -where $(z,\eta)$ belongs to some neighbourhood of the origin in ${\bf -C} \times {\bf C}^{N+1}$. Because the (graded) algebra of symmetric -polynomials in $k$ variables spanned by the $k$ elementary symmetric functions -$\sigma_j$ is isomorphic to the (graded) algebra of polynomials in $k$ -variables via the map -$$ -\Phi: \phi(X_1,\ldots,X_k) \mapsto \phi(\sigma_1, \ldots, \sigma_k) -$$ -there is polynomials $\gamma_1, \ldots, \gamma_k$ such that -$$ -f(z, \zeta) = \prod_{j = 1}^k (z - \gamma_j(\zeta)) -. -$$ -In other words we look at the morphism from ${\bf C}^k$ to the set $M$ -of monic polynomials of degree $k$ given by -$$ -\omega \mapsto \prod_{j = 1}^k (z - \omega_k), -$$ -it induces an isomorphism of algebraic varieties ${\bf C}^k / -{\mathfrak{S}_k} \simeq M$ which is precisely $\Phi$. Looking at $f$ -as a function of $\zeta$ with values in $M$, we define the $\gamma_j$ -(up to permutation!) by composing $f$ with the map $M \to {\bf C}^k / -{\mathfrak{S}_k}$. -This is very good, because we now see that the zero locus of $f$ is -near $p$ the union of the graphs $z = \gamma_j(\eta)$ of $k$ -functions. But a graph is diffeomorphic to a hyperplane and is thus a -null set.<|endoftext|> -TITLE: sequences of real numbers -QUESTION [14 upvotes]: Let $\lbrace x_i\rbrace_{i=1}^\infty$ be a sequence of distinct numbers in $(0,1)$. For any $n$ after deleting $x_1,...,x_n$ from $[0,1]$ we get $n+1$ subintervals. Let $a_n$ be the maximum length of these subintervals. Is there any sharp lower bound for $\limsup_n na_n$ ? - -REPLY [11 votes]: Here is how to achieve $1/\ln 2$. It is a modification of Harald's construction. We take the sequence $1/2$, $1/4$, $3/4$, etc. and apply the transformation $x \to \ln (1+x)/\ln 2$. -So the sequence is $\ln(3/2)/\ln(2), \ln(5/4)/\ln (2),\ln(7/4)/\ln(2),\dots$ -The general terms is $x_n$ for $n=2^a+b$ with $b<2^a$ is -$$\ln \left( \frac{2^{a+1} + 2 b+1}{2^{a+1}}\right)$$ -or, equivalently -$$\frac{\ln( 2^{a+1}+2 b + 1)}{\ln(2)} - a-1$$ -Then after $n = 2^a + b$ steps for $b<2^a$, we will have just added the cut $(2 b+1 ) /(2^{a+1})$, the largest interval will be -$$\left[\frac{\ln( 2^{a}+2 b + 1)}{\ln(2)} - a, \frac{\ln( 2^{a}+2 b + 2)}{\ln(2)} - a\right] $$ -whose length is: -$$\frac{ \ln \left(1 + \frac{1}{2^{a}+b+1}\right)}{\ln 2}$$ -If we multiply by $2^a+b+1$, the $\lim\sup$ is $1/\ln 2$. -To show this is optimal, form a binary tree where the vertices are intervals that appear at any point in a sequence. Each interval at some point splits into two intervals, which gives the tree structure. Label each vertex with the step at which it splits. For any $\beta>\lim\sup n a_n$, for all but finitely many vertices, the length of the interval labeled $n$ is at most $\beta/n$. The sum of the lengths of the intervals on each row is $1$, so the sum of the first $k$ rows is $k$. -Thus $k$ is at most the sum over $2^k-1$ distinct numbers $n$ of $\beta/n$ plus a constant coming from the finitely many vertices where the length is not at most $\beta/n$. So $k$ is at most the sum over the first $2^k-1$ numbers of $\beta/n$ plus a constant, which is at most $\beta k \ln 2$ plus a constant. So $\beta\geq 1/\ln 2$. -So $\lim\sup n a_n \geq 1/\ln 2$.<|endoftext|> -TITLE: An algebraic approach to the thermodynamic limit $N\rightarrow\infty$? -QUESTION [16 upvotes]: In physics one studies quite often the thermodynamic limit or what we call the $N\rightarrow \infty$ behavior of a system of $N\rightarrow\infty$ particles. This is of particular relevance in the study of many-body systems, namely when one looks at understanding physical properties of materials: in this case $N$ represents the number of particles, molecules, etc. of your system; the limit behavior $N\rightarrow\infty$ is of interest when this number is vast. -My question is very related to this concept. Say I am a bit of a theoretical physicist and I would like to ``formalize" the following procedure. Imagine you have a finite one-dimensional lattice modelled by the group $\mathbb{Z}_N$. Then, it is common (in physics) to invoke the trick $N\rightarrow\infty$ and voilà the lattice becomes infinite. Of course, this trick might be the right thing to do in certain problems, but I am presently trying to learn more about some aspects of lattice theory that depend on the (mathematical) group that describes your lattice (such as Wigner functions). For these reasons, I would like to understand whether and how the group of integers can be obtained via limit procedure from modular groups. In particular I would like to understand if I can construct the integers in such way using some algebraic transformations (such isomorphisms, homomorphisms, but not only) so that you I can make statements about group theoretical properties of the integers from what I know about modular groups using my (perhaps convoluted) limit procedure/map. - -A more concrete idea. -I have been reading about the inverse limit procedure used in category theory and profinite completions, (I myself do not know category theory). I am aware that you can construct a profinite completion of the integers $\widehat{\mathbb{Z}}$ that is obtained via an inverse limit over modular groups and naturally contains $\mathbb{Z}$ (via an injective group homomorphism). This looks somewhat like what I was looking for. Still, $\widehat{\mathbb{Z}}$ is stricly larger than $\mathbb{Z}$. Still, these topics are quite alien to me and I do not know whether this construction might be something interesting to look at. I have some concrete questions about this inverse limit construction that I would like to ask here: - -Might the inverse limit procedure be a rigorous possible approach to obtain $\mathbb{Z}$ from modular groups in the lattice example I mention? -How can I characterize which elements of $\widehat{\mathbb{Z}}$ are in $\mathbb{Z}$ (since the former is larger)? Is there a simple algorithm to do this? -Can you use the inverse limit construction to make meaningful statements about concepts like the topology of the integers, based on your knowledge of modular groups? For instance, is it possible to induce on the integers its usual LCA topology using this limit? -(I am particularly interested in this one) Can you derive the representation theory (irreps, regular representation, etc.) of the integers (or its Pontryagin-dual group, the torus), and its Fourier analysis using the inverse limit procedure? (Again, from what you know about modular groups.) For instance, can you characterize the irreps of $\mathbb{Z}$ in this way. - - -Both answers and references are greatly welcomed! - -REPLY [5 votes]: I doubt the following has anything to do with the thermodynamic limit. However, it has been pointed out that neither the inverse nor the projective limit of a sequence of $\mathbb{Z}/N$'s gives you $\mathbb{Z}$, so I thought I'd mention a different way of taking limits of groups which does work. -A marked group is a pair $(\Gamma,S)$ where $\Gamma$ is a group and $S$ is a finite generating set for $\Gamma$; alternatively it's an epimorphism $F_r\to\Gamma$, where $F_r$ is the free group of rank $r=|S|$. There's an obvious notion of isomorphism, and an isomorphism class can be identified with the kernel of the map $F_r\to\Gamma$. Thus, we can identify the isomorphism class of $(\Gamma,S)$ with an element of the power set $2^{F_r}$. The set of all normal subgroups of $2^{F_r}$ is called the space of marked groups of rank $r$, was introduced by Grigorchuk, and is denoted by $\mathcal{G}_r$. -We endow $2^{F_r}$ with the product topology, and $\mathcal{G}_r$ with the subspace topology. The power set $2^{F_r}$ is compact, by Tychonoff's theorem, and totally disconnected. It's easy to check that being a normal subgroup is a closed condition, and so $\mathcal{G}_r$ is also compact and totally disconnected. -It's worth spending a little while working out what the open sets look like. A system of neighbourhoods for $(\Gamma,S)$ is given by the set $U_n$ of marked groups with the same (marked) ball of radius $n$ about the identity in the Cayley graph. So convergence here is very like the intuitive picture presented in the question. -With this in hand, it's easy to see that in $\mathcal{G}_1$, if we identify $F_1=\mathbb{Z}$ and consider the obvious epimorphisms $\mathbb{Z}\to\mathbb{Z}/N$, then we obtain a sequence that converges to the identity $\mathbb{Z}\to\mathbb{Z}$. And of course one has the same phenomenon in any $\mathcal{G}_r$, for suitable choices of generating set. -A google search for 'space of marked groups' will give you lots of references, some written by Yves Cornulier, who is active on MO.<|endoftext|> -TITLE: Is there an operator algebraic reformulation of the invariant subspace problem? -QUESTION [5 upvotes]: Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. -Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace? -Remark: This problem is known for the Banach spaces in general, but is still open for an Hilbert space. -The ISP is an operator theoretic problem, and I ask here about an operator algebraic reformulation. -A counter-example $T \in B(H)$ of the ISP is a fortiori an irreducible operator, i.e. $W^{*}(T) = B(H)$. But of course the converse is false : there are many irreducible operators which are not counter-examples of the ISP (for example the unilateral shift). -Conclusion : the von Neumann algebras don't see the ISP, because the property to be a counter-exemple of the ISP can't be encode in $W^{*}(T)$. -Is there a property $P$ of $C^{*}$-algebras verifying : $T$ is a counter-example of the ISP iff $C^{*}(T)$ is $P$ ? -Perhaps the $C^{*}$-algebras are also not relevant for the ISP, I don't know... - In this case, is there a class of operator $*$-algebras which is relevant for the ISP ? -Perhaps the operator $*$-algebras are also not relevant for the ISP... -Is there a class of operator algebras (non necessarily self-adjoint), which is relevant for the ISP ? - -Is there an operator algebraic reformulation of the invariant subspace - problem ? - -To be more precise, is there a class $\mathcal{C}$ of operator algebras and a property $\mathcal{P}$, such that the algebra $\mathcal{C}(T)$ of class $\mathcal{C}$ generated by $T \in B(H)$, is $\mathcal{P}$ iff $T$ is a counter-example of the ISP ? - -REPLY [4 votes]: To the weakest form of your question ("Is there a class of operator algebras relevant to the ISP?"), there is the "reductive algebra problem," which is not quite of the form you're asking for but is, I think, in the same spirit. An algebra of bounded operators on Hilbert space is called reductive if it is WOT-closed, contains 1, and every invariant subspace for the algebra is reducing for the algebra. The reductive algebra problem asks if every reductive algebra is self-adjoint. It is known that a "yes" answer to RAP implies a "yes" answer to ISP. (In fact, if I am not mistaken, a "yes" to RAP actually implies a "yes" to the hyperinvariant subspace problem.) I am not sure about the status of (ISP implies RAP). If you haven't already, you might want to look at the book "Invariant Subspaces" by Radjavi and Rosenthal.<|endoftext|> -TITLE: Reference for embeddings of reflection groups (related to folding ADE Coxeter graphs)? -QUESTION [9 upvotes]: There are a couple of indirect methods, using Lie theory or Springer's general theory of regular elements in (real, complex) reflection groups, to construct natural embeddings among certain Weyl groups. But it should be possible to find such explicit embeddings in more elementary sources. One starts with the familiar finite real reflection groups of types $A_n \:(n \geq 2), D_n \: (n \geq 4), E_6$ and their Coxeter graphs (which are also Dynkin diagrams). Each graph has an obvious "folding" which gives respectively reflection groups of types: $BC_\ell$ (with $\ell = n/2$ if $n$ is even or $(n+1)/2$ if $n$ is odd); $BC_{n-1}$; and $F_4$. Here $BC_\ell$ is the Weyl group of Lie type $B_\ell$ or $C_\ell$ and has a normal subgroup $(\mathbb{Z}/2\mathbb{Z})^\ell$ acted on by $S_\ell$. - -Is there an elementary construction in the literature of such embeddings of finite reflection groups? - -For example, the reflection group of type $E_6$ has order $2^7 \; 3^4 \; 5$ and is realized (in Bourbaki or the Atlas of finite Groups) in various ways, for instance as the automorphism group of the 27 lines on a cubic surface having a simple subgroup of index 2. On the other hand, the group of type $F_4$ has order $2^7 \; 3^2$ and also has some standard realizations. - -REPLY [3 votes]: One reference for these embeddings is the first section of -[Stein] Robert Steinberg, Endomorphisms of linear algebraic groups. -     Memoirs of the American Mathematical Society, No. 80 -This does not quite answer Jim Humphreys' question, as Steinberg makes use of Lie theory. I mention it mainly because of the above comments poining to: -[Stem] John Stembridge, Folding by Automorphisms. -     http://www.math.lsa.umich.edu/~jrs/papers/folding.pdf -Unlike the results in [Stem], the results in [Stein] also cover the folding $A_{2n} \leadsto BC_n$. -Brief summary: Consider an automorphism $\sigma$ of a Dynkin diagram $A$, the induced folding $A \leadsto A^{\text{folded}}$ and the corresponding embedding of groups $W^{\text{folded}} \hookrightarrow W$. The nodes of $A^{\text{folded}}$ correspond to the orbits of $\sigma$ in $A$. Denote the generator of $W$ corresponding to a node $\alpha\in A$ as $s_\alpha$, and the generator of $W^{\text{folded}}$ corresponding to a $\sigma$-orbit $B$ as $s_B$. Claim 3 in [Stem] asserts that the embedding $W^{\text{folded}} \hookrightarrow W$ has the form -$$ -s_B \mapsto \prod_{\alpha\in B}s_\alpha -$$ -provided that each $\sigma$-orbit is an edge-free set in $A$. -This covers the cases $E_6\leadsto F_4$, $D_n \leadsto BC_{n-1}$ and $A_{2n-1}\leadsto BC_n$. Proposition 1.30 in [Stein] shows more generally that this embedding has the form -$$ -s_B \mapsto w_0^B, -$$ -where $w_0^B$ denotes the longest word in the parabolic subgroup corresponding to $B\subset A$. Specifically, according to this result the embedding for $A_{2n}\leadsto BC_n$ has the form -$$ -s_{\{i,2n-i\}} \mapsto -\begin{cases} -s_is_{2n-i} & \text{ for } i = 1,\dots,n-1 \\ -s_n s_{n+1} s_n & \text{ for } i = n-1 -\end{cases} -$$<|endoftext|> -TITLE: automorphisms of C*-algebras and partial isometries -QUESTION [15 upvotes]: Let $A$ be a $C^*$-algebra, let $p$ and $q$ be Murray-von Neumann equivalent projections in $A$, i.e. there is a partial isometry $v$ such that $v^*v = p$ and -$vv^* = q$. Suppose $\alpha \in Aut(A)$ is an automorphism of $A$, such that $\alpha(p) = p$ and $\alpha(q) = q$. - -Under what condition is it true that $\alpha$ fixes $v$ as well? - -For example, if we assume that $\lVert p - q \rVert < 1$, then $p$ and $q$ are Murray-von Neumann equivalent via a partial isometry explicitly constructed from $p$ and $q$ only - using continuous functional calculus. Thus, $\alpha$ fixes $v$ as well in this case. -If it helps, you may assume $A$ to be a unital Kirchberg algebra. - -REPLY [3 votes]: I don't think you'll be able to prove that $v$ itself is fixed in general, even if $p$ and $q$ are MvN equivalent in $A^\alpha$: you could perturb a fixed partial isometry to get another partial isometry from $p$ to $q$ that is no longer fixed. -It seems like you are asking for sufficient (and necessary?) conditions for the inclusion $A^\alpha\hookrightarrow A$ to induce an injective map $K_0(A^\alpha)\to K_0(A)$. The naive thing to do is to simply average the partial isometry $v$ to get something that is fixed and almost a partial isometry with the right source and range projections, and then use functional calculus in the fixed point algebra. However, this does not in general work since you don't in general get an approximate partial isometry (even if the group is compact, in which case you can take the conditional expectation). -In some cases you will get this conclusion for trivial reasons. For example, if $\alpha$ is a pointwise outer automorphism of $\mathcal{O}_2$, then $\mathcal{O}_2 \rtimes_\alpha \mathbb{Z}$ has trivial $K$-theory by the Pimsner-Voiculescu exact sequence, and this crossed product is moreover Morita equivalent to the fixed point algebra. In this case you get that $K_0(A^\alpha)\to K_0(A)$ is injective (and even an isomorphism!) because both $K$-groups are trivial. If you look at non-pointwise outer automorphisms of $\mathcal{O}_2$ (specifically, of finite order), than the examples that Gabor mentioned show that this can fail for non-pointwise outer automorphisms. -If you're working (or willing to work) with automorphisms on Kirchberg algebras of finite order (equivalently, actions of $\mathbb{Z}_n$), then pointwise outerness will of course not be enough, and the condition you need is probably the Rokhlin property (though this may be overkill). If $\alpha\colon \mathbb{Z}_n\to \mbox{Aut}(A)$ is an action with the Rokhlin property on any unital $C^*$-algebra (Kirchberg or not), then $A^\alpha\hookrightarrow A$ induces an injective map $K_\ast(A^\alpha)\to K_\ast(A)$ (you also get this for $K_1$). -There is also a definition of the Rokhlin property for automorphisms ($\mathbb{Z}$-actions), but I don't know if you get the same conclusion in this case. In the case of Kirchberg algebras, the Rokhlin property for an automorphism is equivalent to it being pointwise outer.<|endoftext|> -TITLE: Künneth theorem for fibred products -QUESTION [7 upvotes]: Given the fibred product of two manifolds over a base space $X\times_Y Z$ , is there an analogue of Künneth theorem that allows one to compute the cohomology of the fibred product? - -REPLY [5 votes]: The Eilenberg-Moore spectral sequence can be understood as a parametrized K\"unneth theorem. There is a monograph (also a paper) by L. Smith giving this approach (Lectures on the EMSS, Springer LNM 134, 1970). Section 22.7 of http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf gives a new conceptual construction and a long list of relevant references.<|endoftext|> -TITLE: spherical buildings for non-split groups -QUESTION [8 upvotes]: I am looking for references to explicit descriptions of Tits buildings for semisimple (classical) Lie groups via language of incidence geometry. Such descriptions are well-documented in the case of split groups in many sources (e.g. in Garrett's book), but there is hardly anything available in the non-split case. I know how to do so in the pseudo-orthogonal case (groups O(p,q)). In principle, I could run similar arguments for other types (unitary groups over complex numbers, quaternions, etc.), but it is a bit tedious, so I am looking for a reference. - -REPLY [12 votes]: Misha, Tits' Lecture Note "Buildings of spherical type and finite BN pairs" -gives a fairly explicit description of the buildings associated to the -classical groups (not just the split ones). I also wrote a survey called -"Buildings and classical groups" (arXiv:math/0307117) -which appeared in print some years ago. -These sources deal with general fields and division rings. Of course, -some details become easier over the reals, the complex numbers and the -quaternions, partly because one does not have to worry about characteristic 2. -The buildings associated to hermitian or symplectic forms (and hence to -orthogonal, unitary or symplectic groups) are also called polar spaces. -Under this name, one can find lots of references. -Here are some more details, based on Section 2.7 in my article arXiv:math/0307117. -Since we are interested in Lie groups, we are dealing with finite dimensional -modules over the reals, the complex numbers or the real quaternions. -We are given a nondegenerate $(\sigma,\epsilon)$-hermitian form $h$ on -a finite dimensional right module $V$ over -$\mathbb K=\mathbb R,\mathbb C,\mathbb H$. -This means that $\epsilon=\pm 1$, that $\sigma$ is an involution -(an anti-automorphism whose square is the identity) on $\mathbb K$ and -that $h$ is a $\sigma$-sesqulinear form on $V$, with -$$h(u,va)=h(u,v)a=\epsilon h(va,u)^\sigma\qquad\forall u,v\in V,\ a\in\mathbb K$$ -and with $V^\perp=0$. A submodule $W$ is called totally isotropic if -$W\subseteq W^\perp$. The building is the simplicial complex consisting -of all (partial) flags consisting of totally isotropic subspaces. -If we rescale the hermitian form $h$ by multiplying it by a (suitable) -nonzero scalar, the building is obviously not changed. -Up to this rescaling process, there are the following cases. -1) Symplectic groups over $\mathbb R,\mathbb C$, where -$\sigma=id_{\mathbb K}$ and $\epsilon=-1$. -2) Orthogonal groups over $\mathbb R,\mathbb C$, where -$\sigma=id_{\mathbb K}$ and $\epsilon=1$. -3) Unitary groups over $\mathbb C,\mathbb H$, where -$a^\sigma=\bar a$ is the standard conjugation and $\epsilon=1$. -4) Unitary groups over $\mathbb H$, where -$a^\sigma=-i\bar a i $ and $\epsilon=1$. -In case 1) and 2), the building has simplicial dimension $m-1$, were -$\dim(V)=2m$. -In case 3) the building has simplicial dimension -$m-1$, and the form $h$ has signature $(m,m+k)$ for $k\geq 0$. -In case 4), the building has simplicial dimension -$m-1$, and $\dim V=2m$ or $\dim V=2m+1$. -In case 2), if $k=0$, then -the building is not thick. If one want a thick building, one has -to consider instead the "oriflamme geometry" which is explained -for example in Tits' Lecture Note.<|endoftext|> -TITLE: Pontrjagin ring structure on homology of Eilenberg-Mac Lane spaces -QUESTION [5 upvotes]: Is there any good reference for the Pontrjagin ring structure on -$$ -H_\ast(K(\mathbb{Z}/2,k);\mathbb{Z}/2)\cong H_\ast(\Omega K(\mathbb{Z}/2,k+1);\mathbb{Z}/2)? -$$ -I am familiar with Serre's theorem describing the mod 2 cohomology ring structure. I'm also aware that the action of the Dyer--Lashof operations on this infinite loop space is trivial. - -REPLY [7 votes]: By naturality and the external Cartan formula, the standard polynomial generators of $H^*(K(\mathbf{Z}/2,k);\mathbf{Z}/2)$ given by iterated Steenrod operations on the fundamental class are primitive. Therefore the homology is a divided power algebra on the dual elements.<|endoftext|> -TITLE: Curves which do not dominate other curves -QUESTION [14 upvotes]: Let $g>1$ be an integer. Does there exist a (smooth projective) genus g curve $X$ which doesn`t dominate a curve of positive genus and genus smaller than $g$? -Surely such curves exist. Just take a curve with simple jacobian. -I`m wondering whether there are any other interesting examples, or is the above property equivalent to having simple jacobian? - -REPLY [9 votes]: I elaborate a bit on Roy Smith's comment. -One can show that the general curve of genus $g>1$ does not map onto a curve of genus $h>0$ by a dimension count, at least for complex curves. -Let $f\colon C\to D$ be a map of degree $d$ from a curve of genus $g$ to a curve of genus $h>0$ and let $B$ the branch divisor of $f$ (a point $P\in D$ appears in $B$ with multiplicitiy equal to $\sum_{Q\mapsto P}(m_Q-1)$, where $m_Q$ is the order of ramification of $f$ at $Q$. -The Hurwitz formula gives: -$$2g-2=d(2h-2)+\deg B.$$ -Let $S$ be the support of $B$ (i.e., $S$ is the set of critical values of $f$) and consider the restricted cover $f_0\colon C\setminus f^{-1}(S)\to D\setminus S$: this is a topological cover of degree $d$ and it determines $f$. There are finitely many such covers, hence the maps $f\colon C\to D$ as above depend on $3h-3+s$ parameters, where $s$ is the cardinality of $s$. -Since $s\le \deg B$, the Hurwitz formula gives: -$$(3g-3)-(3h-3+s)\ge 3(d-1)(h-1)+s/2>0.$$ -So the general curve of genus $g>1$ does not have a map of degree $d$ onto a curve of genus $h>0$. Now it is enough to observe that, again by the Hurwitz formula, there are finitely many possibilities for $h$ and $d$. -I don't think that a curve without maps onto curves of positive genus must have a simple Jacobian: if one takes a curve $C$ inside an irregular surface $S$ such that $C$ is an ample divisor of $S$, then there is an injection $Pic^0(S)\to J(C)$, and I do not see why in general $C$ should have a map onto a curve of positive genus. To get an actual counterexample, one could try to look at an abelian surface $S$ with a polarization $L$ of type $(1,3)$ and a curve $C\in |L|$, but I don't know how to show that a general such $C$ does not map onto a curve of positive genus.<|endoftext|> -TITLE: Are surjectivity and injectivity of polynomial functions from $\mathbb{Q}^n$ to $\mathbb{Q}$ algorithmically decidable? -QUESTION [33 upvotes]: Is there an algorithm which, given a polynomial $f \in \mathbb{Q}[x_1, \dots, x_n]$, -decides whether the mapping $f: \mathbb{Q}^n \rightarrow \mathbb{Q}$ is surjective, -respectively, injective? -- -And what is the answer if $\mathbb{Q}$ is replaced by $\mathbb{Z}$? -The motivation for this question is Jonas Meyer's comment on the question -Polynomial bijection from $\mathbb{Q} \times \mathbb{Q}$ to $\mathbb{Q}$ -which says that the explicit determination of an injective polynomial mapping -$f: \mathbb{Q}^2 \rightarrow \mathbb{Q}$ is already difficult, and that -checking whether the polynomial $x^7+3y^7$ is an example is also. -Added on Aug 8, 2013: SJR's nice answer still leaves the following 3 problems open: - -Is there at all an injective polynomial mapping from $\mathbb{Q}^2$ to $\mathbb{Q}$? -Would a positive answer to Hilbert's Tenth Problem over $\mathbb{Q}$ imply that -surjectivity of polynomial functions $f: \mathbb{Q}^n \rightarrow \mathbb{Q}$ is -algorithmically decidable? -Hilbert's Tenth Problem over $\mathbb{Q}$. - -REPLY [28 votes]: We treat all four problems in turn. In all that follows $n>1$. -Surjectivity over $\mathbb{Q}$: -If there is an algorithm to test whether an arbitrary polynomial with rational coefficients is surjective as a map from $\mathbb{Q}^n$ into $\mathbb{Q}$ then Hilbert's Tenth Problem for $\mathbb{Q}$ is effectively decidable. -Proof: Let $g(x_1,\ldots,x_n)$ be any nonconstant polynomial with rational coefficients. We want to construct a polynomial $H$ that is surjective if and only if $g$ has a rational zero. -First we define an auxillary polynomial $h$ as follows; -$$h(y_1,\ldots,y_6):=y_1^2+(1-y_1y_2)^2+y_3^2+y_4^2+y_5^2+y_6^2.$$ -The point of the definition is that $h(\mathbb{Q}^6)$ is precisely the set of positive rationals. This follows from Lagrange's four-square theorem and from the fact that $y_1^2+(1-y_1y_2)^2$ is never 0 but takes on arbitrarily small positive values at rational arguments. -Next, let $a$ be any positive rational such that $a$ is not the square of a rational, and such that for some tuple $b\in \mathbb{Q}^n$, it holds that $g(\bar{b})^2 -TITLE: Coequalizers in an Eilenberg-Moore category -QUESTION [5 upvotes]: Last month I proved that some category $\mathbf C$ that I happen to care about is isomorphic to the Eilenberg-Moore category for a monad on the category of bounded posets $\mathbf{BPos}$. -I know from other results that $\mathbf C$ is cocomplete. Coproducts in $\mathbf C$ are easy to describe, but I am struggling to find an explicit description of coequalizers in $\mathbf C$. I need this to prove a hypothesis that a certain complicated explicit construction in $\mathbf C$ is, in fact, just a colimit. -Is it possible to give an explicit description of coequalizers in an Eilenberg-Moore category over a concrete category, such as $\mathbf{BPos}$? -Any pointers to relevant papers are appreciated. - -REPLY [5 votes]: I'd like to add more information that is in line with Zhen's answer, but with slightly different hypotheses. -Proposition: If $C$ is cocomplete and a monad $T$ on $C$ preserves reflexive coequalizers, then the category of algebras $C^T$ is cocomplete. -Indeed, the forgetful functor $U: C^T \to C$ preserves and reflects any class of colimits that $T$ preserves, so that if $T$ preserves reflexive coequalizers and $C$ has them, then so will $C^T$. As Zhen said, we can get general coequalizers in $C^T$ if $C^T$ has binary coproducts and reflexive coequalizers, but it turns out that this follows from $C^T$ having reflexive coequalizers and $C$ having binary coproducts. See the arguments presented here for details. See particularly theorem 1, and the second corollary below it. -On the off-chance that your monad is finitary (preserves filtered colimits), this might come in handy: -Proposition: If $C$ is complete and cocomplete and $T$ is a finitary monad, then $C^T$ has coequalizers (and therefore is also complete and cocomplete). -See Barr and Wells, Toposes, Theories, and Triples, p. 267 (theorem 3.9) for a somewhat sharper statement. -For example, if products in $C$ distribute over colimits (as they do in the category of bounded posets), and your monad came from a finitary Lawvere theory $T$, this proposition would apply. See also this MO answer and this page from the ncatlab written in support of that answer.<|endoftext|> -TITLE: On the wikipedia entry for Borel-Moore homology -QUESTION [16 upvotes]: The wikipedia page on Borel-Moore homology claims to give several definitions of it, -all of which are supposed to coincide for those spaces $X$ which are homotopy equivalent to a finite CW complex and which admit a sufficiently nice embedding into a smooth manifold. -My question concerns one of the definitions on that page: it is asserted that if $X$ is a space and $\bar X$ is some compactification, then there is a definition of Borel-Moore homology given by -$$ -H_*^{\text{BM}}(X) := H_*(\bar X,\bar X \setminus X) . -$$ -The entry additionally requires that the pair $(\bar X,X)$ should be a CW pair for this to work. -It then goes on to assert that $\bar X = X^+$, the one-point compactification, will suffice for this. -Frankly, I don't see how $(X^+,X)$ will form a CW pair in most instances. For example suppose $X = \Bbb Z$ is the set of integers which is considered as a discrete space. Then in this instance $(X^+,X)$ certainly fails to be a CW pair. -Another example: $X = S^1 \setminus \ast$ with one point compactification $S^1$. Then -$(S^1,S^1 \setminus \ast)$ is not a CW pair either. -My Questions: -(1) For what class of spaces $X$ does -$H_*(X^+,+)$ coincide with definition of Borel-Moore homology given by locally finite chains? -(2) Does the wikipedia entry contain a mistaken assumption? Perhaps we do not need to assume that $(X^+,X)$ can be given the structure of a CW pair? -Note: if $X = \Bbb Z$ and if we use the above as a definition of Borel-Moore homology then -$ -H_0^{\text{BM}}(X) -$ -is a free abelian group whose generators are given by the underlying set of $\pi_0(X) = \Bbb Z$. This is clearly the wrong answer: it should be the countably infinite cartesian product of copies of the integers indexed over $X$ instead (using, say, the definition of Borel-Moore homology given by locally finite chains). -Another Question: -(3) Is there a definition the above kind (using compactifications) -which will work (i.e., coincide with the locally finite chain definition) - for $X = \Bbb Z$? -(I suspect not, since ordinary singular homology in degree zero is always free abelian.) -Incidentally, later in the page it lists the main variance property: Borel-Moore homology is supposed to be covariant with respect to proper maps. The page gives a -proof using the above definition. But since the above definition doesn't work in general, I don't see how this is really supposed to be a proof. - -REPLY [8 votes]: In Chriss & Ginzburg, it's asserted that the Borel-Moore homology is given by $H_*(\bar X, \bar X\setminus X)$ if $\bar X$ is the one-point compactification or any compactification such that the inclusion is cellular. They very pointedly do not assume that the inclusion of $X$ into $X^+$ is cellular. They reference Bredon but don't cite a particular chapter.<|endoftext|> -TITLE: The cell structure of Thom spectra -QUESTION [19 upvotes]: I would like to understand the cell structure of integrally oriented Thom spectra. A Thom spectrum over a space $X$ is something you can build from a stable spherical bundle, which is classified by a map $\varphi: X \to BGL_1 \mathbb{S}$. When $R$ is an $A_\infty$ ring spectrum, it's possible to deloop $GL_1 R$ once to get $BGL_1 R$, and the spherical bundle $\varphi$ is said to be $R$-orientable when the map $$X \xrightarrow{\varphi} BGL_1 \mathbb{S} \xrightarrow{BGL_1 \eta} BGL_1 R$$ is null. The point of an orientation (i.e., a choice of lift of $\varphi$ to the fiber of $BGL_1 \eta$) is that it buys you a Thom isomorphism $$R \wedge T(\varphi) = R \wedge \Sigma^\infty_+ X.$$ -$\newcommand{\Z}{\mathbb{Z}}\renewcommand{\S}{\mathbb{S}}\renewcommand{\phi}{\varphi}\newcommand{\sm}{\wedge}\newcommand{\Susp}{\Sigma}\newcommand{\Loops}{\Omega}$I would like to specify that my bundle is $H\Z$-oriented --- I think that'll be important for a couple reasons below. What it means homotopically for $\phi$ to be $H\Z$-orientable is evident by looking at the homotopy groups of $BGL_1 R$, which are given by the following formula: $$\pi_n GL_1 R = \begin{cases}(\pi_0 R)^\times & \hbox{when $n = 0$,} \\ \pi_n R & \hbox{otherwise.}\end{cases}$$ In the case of $BGL_1 \S$, these groups are complicated because $\pi_n \S$ is complicated, but at least we know $\pi_1 BGL_1 \S = \Z/2$. The spectrum $H\Z$ is easier, where we have the complete calculation $BGL_1 H\Z = \Susp H\Z/2$, and the map $BGL_1 \eta$ is an isomorphism in degree $1$. Altogether this means... - -... that an $H\Z$-oriented spherical fibration has a unique lift to the fiber, and -... that the fiber has known homotopy type as well. It is given as a connected cover $$\operatorname{fib}(BGL_1 \S \to BGL_1 H\Z) = \Loops^\infty \S^1[2, \infty).$$ - -Now with all the pieces in place, let me ask the baby version of my question first: suppose that $X$ takes the form $X = S^n$, $n \ge 2$. Then the spherical bundle $\phi$ is actually selecting an element $\omega \in \pi_{n-1} \S$ in the stable homotopy groups of spheres. - -Can the Thom spectrum $T(\phi)$ be identified with $M_0(\omega)$, the cone on $\omega$ with bottom cell in dimension $0$? - -This is true in the paltry number of cases that I know classically: the projective spaces $\mathbb{R}\mathrm{P}^1$ and $\mathbb{C}\mathrm{P}^1$ can be identified with the spheres $S^1$ and $S^2$ respectively, and they carry reduced tautological line bundles whose Thom spectra can be identified with $\Susp^{-1} \Susp^\infty \mathbb{R}\mathrm{P}^2$ and $\Susp^{-2} \Susp^\infty \mathbb{C}\mathrm{P}^2$, which themselves are the Moore spectra $M_0(2\iota)$ and $M_0(\eta)$. (Of course, $\mathbb{R}\mathrm{P}^1$ doesn't carry an $H\mathbb{Z}$-orientable bundle, but that's not super important to me that this example is captured by an answer --- it's just motivation.) The proofs of these that I know (from Atiyah's Thom Complexes) don't go in a way that lend themselves to generalization along the lines of the highlighted question, though. -Now the more serious question: I'm much more interested to know what happens in the case where $X$ is a cell complex with more than one cell. Toward that end: - -What information about the attaching maps of the cells of $\Susp^\infty X$ and about the action of the map $\phi$ is necessary to describe the attaching maps of the cells of $T(\phi)$? - -Here's a guess at a way of phrasing a response to this question: one possible way of encoding the attaching maps of $\Susp^\infty_+ X$ is to claim perfect knowledge of the differentials in the Atiyah-Hirzebruch spectral sequence $$H_*(X; \pi_* \S) \Rightarrow \pi_* \Susp^\infty_+ X.$$ So my question could be interpreted as: what do I need to know about the map $\phi$ to use it to twist the differentials in that Atiyah-Hirzebruch spectral sequence to study the spectral sequence $$H_*(X; \pi_* \S) \cong H_*(T(\phi); \pi_* \S) \Rightarrow \pi_* T(\phi)$$ instead? (Note: the isomorphism in this last line is where I finally use the $H\Z$-orientable hypothesis. I think it's required to begin comparing cell structures using these methods.) - -REPLY [15 votes]: EDIT: Let's try and turn this disaster into something with a moderate amount of workability by making some simplifying assumptions. I wouldn't recommend reading the previous version. -First, let's assume $X$ is a finite cell complex, coming from a filtration by subcomplexes $X^{(d)}$. (We'll allow ourselves the possibility of $X^{(d)}$ being formed from $X^{(d-1)}$ by attaching multiple cells of any necessary dimension.) The Thom spectrum functor of an infinite complex is (at the least, equivalent to) the colimit of the Thom spectra on its finite subcomplexes, and since our questions are about cellular filtrations we're not losing important details by restricting to finite objects. -Let's also assume for simplification that we have a topological group $G$ that acts by based maps on $S^n$, and a map $\alpha: X \to BG$ classifying a principal $G$-bundle $P \to X$. From this, we get a Thom space $X^\alpha = S^n \wedge_G P_+$. Let's assume for simplicity that we're talking about the associated virtual bundle of degree zero so that the Thom spectrum is the suspension spectrum $\Sigma^{-n} \Sigma^\infty X^\alpha$. -For each cell $D^k \to X$, choose a lift $D^k \to P$, which extends uniquely to a $G$-map $G \times D^k \to P$. These give $P$ a $G$-equivariant cell structure. If $P^{(d)}$ is the preimage of $X^{(d)}$, then $P^{(d)}/P^{(d-1)}$ is a wedge of copies of $G_+ \wedge S^{j_i}$. Every new cell of $P^{(d+1)}$ is attached via an equivariant boundary map $G \times S^{k-1} \to P^{(d)}$. On Thom spaces we get a filtration by $S^n \wedge_G P^{(d)}_+$, with the quotients in the filtration being wedges of $S^{n+k}$. In particular, the attaching maps between different layers are determined by $G$-equivariant maps $G_+ \wedge S^{k-1} \to \bigvee G_+ \wedge S^{j_i}$; taking the Thom space turns this into maps $S^{n+k-1} \to \bigvee S^{n+j_i}$. -Now take this whole picture and take suspension spectra. -We have a suspension spectrum $Y = \Sigma^\infty P_+$, which has an action of the group algebra $\mathbb{S}[G] = \Sigma^\infty G_+$. Taking suspension spectra preserves colimits, so that we can translate space-level identifications. We find that the suspension spectrum of $X$ is $\mathbb{S} \wedge_{\mathbb{S}[G]} Y$ where $G$ acts trivially on $\mathbb{S}$, while the Thom spectrum is $S^0 \wedge_{\mathbb{S}[G]} Y$. Here we view $S^0 = \Sigma^{-n} \Sigma^\infty S^n$ as being equivalent to $\mathbb{S}$, but inheriting its appropriate $G$-action. -We also have a filtration of $Y$ by submodules $Y^d = \Sigma^\infty_+ P^{(d)}_+$. The layers $Y^d / Y^{d-1}$ are wedges $\bigvee \mathbb{S}[G] \wedge S^{j_i}$. Let's use this filtration to build the "resolution" -$$ -Y^0 \leftarrow \Sigma^{-1} Y^1/Y^0 \leftarrow \Sigma^{-2} Y^2/Y^0 \leftarrow \cdots. -$$ -On homotopy groups, this gives a chain complex $C$ with $C_d = \pi_* (\Sigma^{-d} Y^d / Y^{d-1})$ of (graded) free modules over $\pi_* \mathbb{S}[G]$. It has the following properties. - -It gives a spectral sequence with $E_2$-term $H_{**}(C)$ converging to $\pi_* Y$. -Smashing over $\mathbb{S}[G]$ with $\mathbb{S}$ tensors this chain complex over $\pi_* \mathbb{S}[G]$ with $\pi_* \mathbb{S}$. This chain complex comes from the cellular filtration of $\Sigma^\infty X_+$, and its homology is the $E_2$-term of the cellular spectral sequence for the homotopy of $\Sigma^\infty X_+$. (Standard techniques allow us to say there's a spectral sequence starting with $Tor^{\pi_* \mathbb{S}[G]}_{***}(\pi_* \mathbb{S}, H_{**}(C))$ and converging to the $E_2$-term.) -Smashing over $\mathbb{S}[G]$ with $S^0$ tensors this chain complex over $\pi_* \mathbb{S}[G]$ with $\pi_* S^0$ (in this case, $G$ acts nontrivially!). This chain complex comes from the cellular filtration of the Thom spectrum, and its homology is the $E_2$-term of the cellular spectral sequence for the homotopy of $\Sigma^\infty X_+$. (Again, there's a spectral sequence starting with $Tor^{\pi_* \mathbb{S}[G]}_{***}(\pi_* S^0, H_{**}(C))$ and converging to the $E_2$-term.) - -The upshot of this is that, in terms of attaching maps, both the suspension spectrum of $X$ and the Thom spectrum of this degree-zero virtual bundle have cellular filtrations with the same cells. The relation between the adjacent attaching maps is this: they have common lifts from $\pi_* \mathbb{S}$ to $\pi_* \mathbb{S}[G]$, but one specialization has $G$ act trivially and one has $G$ act via its action on spheres. -Examples help. -If $G = O(1)$, then $\pi_* \mathbb{S}[G]$ is literally the group algebra over $\pi_* \mathbb{S}$ on $\mathbb{Z}/2$. The principal bundle over your space is a chain complex of free modules over this group algebra (and if $X = \mathbb{RP}^\infty$, it is a resolution of $\pi_* \mathbb{S}$). The two specializations make the generator ${-1}$ acts trivially and nontrivially respectively. -If $G = U(1)$, then $\pi_* \mathbb{S}[G]$ is even more interesting: it is a graded-commutative ring $\pi_* \mathbb{S}[d] / (d^2 = \eta d)$, where $|d| = 1$. The element $d$ comes from the canonical element in $\pi_1 U(1)$. The principal bundle again gives us a chain complex of free modules over this (and if $X = \mathbb{CP}^\infty$, it is the resolution of $\pi_* \mathbb{S}$ by free modules of rank $1$ given by alternately multiplying by $d$ and $d + \eta$). The two specializations make the generator act by $0$ and by $\eta$ respectively. -If $X = S^k$, then we can get into gory detail. We can use the cell decomposition with one zero-cell and one $k$-cell; the principal bundle is then formed by attaching $G \times D^k$ to $G$ along an equivariant map $G \times S^{k-1} \to G$ (so we need to be careful when $k=1$); without loss of generality we can make a choice sending the basepoint to the identity. This is literally an element in $\alpha \in \pi_{k-1}(G) = \pi_k BG$. Take suspension spectra and unwind: our chain complex is a two-term complex $\pi_* \mathbb{S}[G] \leftarrow \Sigma^{k-1} \pi_* \mathbb{S}[G]$. The map if $k \gt 1$ is given by (the image of) our element $\alpha$, while if $k = 1$ it is multiplication by the difference $1 - \alpha$. The two specializations make $\alpha$ go to zero (recovering the suspension of $S^k$) and to the appropriate image of $\pi_{k-1} G \to \pi_{k-1} Map(S^n,S^n) \to \pi_{k-1} \Omega^n S^n$ (recovering the cone on $\alpha$ or $1-\alpha$). -Secondary maps, like the $d_2$ in the spectral sequences I mentioned, are harder; suddenly you are trying to relate attaching maps for cells of two different objects. There may be some last bit of voodoo possible by taking these, which are basically Toda brackets, and trying to lift these to something equivariant. Seems like an interesting problem.<|endoftext|> -TITLE: Measure concentration for weakly dependent random variables -QUESTION [5 upvotes]: For an application quite alien to probability theory, I'd like to have a kind of measure concentration estimate, in the following spirit. Suppose that to every $1\le i,j\le n$ there corresponds a zero-mean random variable $X_{ij}$, all of them being identically distributed and taking values in $[-1,1]$. Let $S$ denote the normalized sum of all these $n^2$ random variables: $S:=(X_{11}+\dotsb+X_{nn})/n$. I want to conclude that under some week dependence assumption (to be stated immediately), one has - $$ {\mathsf P}(|S|>\tau) 0$. (I originally hoped to prove that this probability is $O(e^{-\gamma\tau^2})$, but Carl noticed that this is too optimistic.) Indeed, I may be happy with some slightly weaker estimate, or even with an estimate like ${\mathsf E} |S|^p\le(Mp)^p$ with an absolute constant $M$.) -Now, the weak dependence assumption just mentioned is that, viewing the $X_{ij}$ as the entries of a matrix, if several of them are dependent, then one can create a closed aligned loop in the matrix with the corresponding entries being the vertices of the loop. In particular, - -the variables are independent pairwise and in triples; -any four of them are independent, unless their indices form a rectangle in the matrix; -any system of the variables no two of which are in the same column or in the same row is independent. - -In general, I wonder whether this weak dependence assumption has ever been studied, and what conclusions can be drawn from it. Thanks in advance for any suggestions or pointers! - -REPLY [3 votes]: How about this example? All $X_{ij}$ are $\pm 1$, conditional on an even number of $+1$ at the corners of each rectangle. All entries are thus determined by the first row and column, and if these are all $+1$, the entire matrix is $+1$ and so has $S=n$ with probability $2^{-2n+1}$. Hence any bound must have $e^{-\gamma \tau}$ rather than $e^{-\gamma \tau^2}$.<|endoftext|> -TITLE: Arithmetic product of symmetric functions: why is it integral? -QUESTION [9 upvotes]: For every commutative ring $A$, let $\mathbf{Symm}_A$ be the ring of symmetric functions over $A$. Let $\mathbf{Symm}$ without a subscript denote $\mathbf{Symm}_{\mathbb{Z}}$. -We can define a bilinear map $\boxdot : \mathbf{Symm}_{\mathbb{Q}} \times \mathbf{Symm}_{\mathbb{Q}} \to \mathbf{Symm}_{\mathbb{Q}}$ by setting -$p_{\lambda} \boxdot p_{\mu} = \prod\limits_{i\geq 1,\ j\geq 1} p_{\operatorname*{lcm}\left(\lambda_i,\mu_j\right)}^{\gcd\left(\lambda_i,\mu_j\right)}$ -for any two partitions $\lambda = \left(\lambda_1,\lambda_2,\lambda_3,...\right)$ and $\mu = \left(\mu_1,\mu_2,\mu_3,...\right)$. Here, we are writing $\boxdot$ as an infix operator (that is, $a\boxdot b$ means $\boxdot\left(a,b\right)$), and $p_\nu$ means the $\nu$-power sum symmetric function. -This bilinear map $\boxdot$ is associative. I call it the "arithmetic product", as it boils down to the arithmetic product of species viewed through the cycle index series. -Now, species theory can be used to show that $\boxdot$ restricts to a well-defined map $ \mathbf{Symm} \times \mathbf{Symm} \to \mathbf{Symm}$ (that is, the restriction of $\boxdot$ to $\mathbf{Symm} \times \mathbf{Symm}$ has its image in $\mathbf{Symm}$). My question is: Can this be proven more elementarily? Is there a good way to describe this map on an actual basis of $ \mathbf{Symm}$ rather than on the power-sum symmetric functions? Is there a more direct combinatorial or even representation-theoretical significance of this map? -(This is somewhat similar to MO question #120924, where another operation on $\mathbf{Symm}$ is constructed on the power sums first and then happens to be integral for weird reasons.) - -REPLY [8 votes]: About the question - "Is there a more direct combinatorial or even representation-theoretical significance of this map?" -Probably you know that the arithmetic product is an operation between representations of the symmetric group, because the species is a theory of representations of the symmetric group that gives concrete set theoretical constructions of operations, without the use of induced representations. The ordinary species correspond to permutation-representations and the tensor species to vector representations (representations on the general linear group). -In the classical language, the arithmetic product of two represntations $\rho$ and $\tau$ of $S_m$ and $S_n$ respectively is given as the induced representation: -$$\rho\boxdot\tau=\mathrm{Ind}_{S_m\times S_n}^{S_{m.n}}\rho\otimes\tau.$$ -But this seems to me less concrete than the direct recipe, given in the language of species, of the vector space where the symmetric group $S_U$ acts naturally, -$$(R\boxdot T)[U]=\bigoplus_{(\pi,\sigma)}R[\pi]\otimes T[\sigma]$$ -The arithmetic product of two homogeneous symmetric functions in terms of the monomial s.f. is as follows -$h_m\boxdot h_n=\sum M_{\lambda} m_{\lambda}$ -where $M_{\lambda}$ is the number of $m\times n$ matrices with $\lambda_i$ i-es (up to row and column permutations). This is because: -$$Ch(E_m\boxdot E_m)(x)=\sum_{\lambda}|(E_m\boxdot E_n)[m.n]/S_{\lambda_1}\times S_{\lambda_2}\times\dots| \, m_{\lambda}(x)$$ -$Ch$ being the Frobenius character. An analogous result is obtained for the arithmetic product of elementary symmetric functions, taking the arithmetic product of the sign representations $\Lambda_m$ and $\Lambda_n$. Maybe this can be helpful for expressing their arithmetic product in terms of themselves.<|endoftext|> -TITLE: Curves on varieties, and a criterion for nef divisor -QUESTION [7 upvotes]: Let $X$ be a projective variety over $k$, and $\dim X \geq 2$. By a curve $C$ on $X$, I mean a proper,reduced subscheme of $X$ of dimension $1$. -(1)If $C$ is an irreducible curve on $X$, then is $C$ numerically equivalent to some $\sum n_i C_i$ with $n_i > 0$, and $C_i$ smooth curves ? -(2) If $D$ is a Cartier divisor on $X$, and $D\cdot C \geq 0$ for any smooth curve $C$ on $X$, then is $D$ necessarily to be a nef divisor? -Certainly, if (1) holds, then (2) holds. -(I ask this question because I curious why people usually call a nonconstant morphism $C \to X$ with $C$ being a smooth curve to be a curve on $X$, rather than a smooth curve $C$ exactly sitting inside $X$.) - -REPLY [2 votes]: Since Francesco answered your two questions very nicely, I thought I'd indicate some possible reasons why people call a nonconstant morphism from a smooth curve $C$ to $X$ a curve as opposed to one actually sitting on $X$. -One big reason is that one can reduce many statements to discussions about flat families over smooth curves. A well-known fact is that connected algebraic varieties (say projective for simplicity) are "path-connected" by smooth curves. Here the meaning is that any two points $p$ and $q$ can be connected by the $\textit{image}$ of a smooth curve $C$. For smooth varieties this can probably be made to work with $C$ actually sitting on $X$, but the proof of the above fact is not difficult (it's come up a few times on MO, once even by me :)). This fact is very useful because pulling back the question, whatever it is, to one over $C$ then allows us to use the result that a morphism $\pi:X\rightarrow C$ to a smooth curve $C$ is flat iff every associated point of $X$ gets sent to the generic point of $C$ (Hartshorne III.9.7). This allows us to use flatness in situations where a more general morphism might not be flat. -For example, this is used in Hartshorne III.9.13 to show that the Hilbert polynomial of an "algebraic family" of normal varieties are constant. Here the family wasn't originally flat but becomes so upon base change. -Uses like this also appear in Hilbert scheme-type arguments, for example to show that the Hilbert scheme is proper, since one often just restricts to ALL curves passing through some point to show the result in general. -I hope this indicates the uses of such terminology.<|endoftext|> -TITLE: Non-trivial convergent sequence in Stone-Čech compactification of $\mathbb{N}$ -QUESTION [8 upvotes]: Why are there only trivial convergent sequences in the Stone-Čech compactification of $\mathbb{N}$? - -REPLY [8 votes]: Let me give a fairly direct proof. Assume $R=(\mathcal{U}_{n})_{n}$ is a sequence of distinct ultrafilters on some set $X$. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence $(\mathcal{V}_{n})_{n}$ of $(\mathcal{U}_{n})_{n}$ such that $\{\mathcal{V}_{n}|n\in\mathbb{N}\}$ is a discrete subspace of $\beta X$. In particular, there is a sequence $(A_{n})_{n}$ of sets with $A_{n}\in\mathcal{V}_{m}$ if and only if $m=n$. If we let $B_{n}=A_{n}\setminus(A_{0}\cup...\cup A_{n-1})$, then the sequence $(B_{n})_{n}$ is pairwise disjoint and $B_{n}\in\mathcal{V}_{m}$ iff $m=n$. If we let $B=\bigcup_{n}B_{2n}$, then $B\in\mathcal{V}_{m}$ if and only if $m$ is even. In other words, if $\mathcal{B}=\{\mathcal{V}\in\beta X|B\in\mathcal{V}\}$, then $\mathcal{B}$ is a clopen set with $\mathcal{V}_{m}\in\mathcal{B}$ iff $m$ is even. Therefore, we conclude that the sequence $(\mathcal{V}_{m})_{m}$ cannot converge to any point, so the sequence $(\mathcal{U}_{n})_{n}$ cannot converge to any point either.<|endoftext|> -TITLE: Embeddings of $\overline{\mathbf{Q}}$ into $\mathbf{C}$ -QUESTION [5 upvotes]: Keenan Kidwell's answer to Place stabilizers for the absolute Galois Group mentions that "choosing a complex conjugation" in $G_{\mathbf{Q}}$ means choosing an embedding $\overline{\mathbf{Q}}\rightarrow\mathbf{C}$ so the consequent injection $\mathrm{Gal}(\mathbf{C}/\mathbf{R})\hookrightarrow G_{\mathbf{Q}}$ takes ordinary complex conjugation to an element of $G_{\mathbf{Q}}$. Obviously such an element has order 2. Is there some natural group theoretic characterization of exactly which elements of $G_{\mathbf{Q}}$ these can be? - -REPLY [10 votes]: $\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}$A community wiki answer to record the proof sketched above. I had misread the question earlier; the comments of S.Carnahan and user36938 are correct. -Let $\sigma$ be an element of order $2$ in $Gal(\bar{\QQ}/\QQ)$ and let $R$ be the fixed field of $\sigma$. By the Artin-Schrier theorem, $R$ is real closed. In particular, it comes with a natural order $\leq_R$ defined by $a \leq_R b$ if $\sqrt{b-a} \in R$. -Lemma The order $\leq_R$ is archimedean. -Proof Suppose, for the sake of contradiction, that there is some $t \in R$ with $t >_R a$ for all $a \in \QQ$. Let $t^n + a_{n-1} t^{n-1} \cdots + a_0$ be the minimal polynomial of $t$ over $\QQ$. But then -$$|a_{n-1} t^{n-1} \cdots + a_0| \leq_R t^{n-1} \sum | a_i | <_R t^n,$$ -a contradiction. Here, for $x \in R$, the notation $|x|$ means whichever of $x$ and $-x$ is nonnegative in the order $\leq_R$. $\square$ -So, as an ordered field, $R$ embeds in $\RR$. Let $\phi: R \to \RR$ be this embedding. Let $\RR^{alg}$ be the field of algebraic elements in $\RR$. -Since $R$ is algebraic over $\QQ$, we have $\phi(R) \subseteq \RR^{alg}$. Since no nontrivial finite extension of $R$ can be ordered (property 6 on Wikipedia's list of properties of real closed fields), we actually have $\phi(R) = \RR^{alg}$. We have $\bar{\QQ} = R(\sqrt{-1}) = \RR^{alg}(\sqrt{-1})$. So we can extend $\phi$ to an automorphism of $\bar{\QQ}$ by declaring it to fix $\sqrt{-1}$. Then $\phi$ conjugates $\sigma$ to complex conjugation.<|endoftext|> -TITLE: Is there a cheap proof of power savings for exponential sums over finite fields? -QUESTION [28 upvotes]: Let $p$ be a large prime, and let $f(x) = P(x)/Q(x)$ be a non-constant rational function over ${\Bbb F}_p$ of bounded degree. From the Weil conjectures for curves, we have a bound of the form -$$ |\sum_{x \in {\Bbb F}_p}^* e_p( f(x) )| \ll p^{1/2}$$ -where the asterisk denotes that the summation excludes those $x$ for which $Q(x)$ vanishes, and $e_p(x) := e^{2\pi i x/p}$ is the standard additive character. (As far as I can tell, this particular bound first appeared explicitly in this paper of Perelmuter; an elementary proof based on Stepanov's method, though still using the rationality of the zeta function, also appears in this paper of Cochrane and Pinner.) -My question is whether there is a "cheap" argument, avoiding the Weil conjectures (i.e. not using either the Riemann hypothesis or rationality of the zeta function) that gives a weaker power savings bound -$$ |\sum_{x \in {\Bbb F}_p}^* e_p( f(x) )| \ll p^{1-c}$$ -over the trivial bound for some $c>0$ (which is now allowed to depend on the degree of the polynomials $P,Q$)? I know of two special cases in which this is possible: - -If $f$ is a polynomial, then one can use the Weyl differencing method to obtain a bound of this form (with $c$ decaying exponentially in the degree of $f$). -In the case of Kloosterman sums $f(x) = ax + \frac{b}{x}$, Kloosterman obtained a power savings of $c=1/4$ by computing (or more precisely, upper bounding) the fourth moment -$$ \sum_{a,b \in {\Bbb F}_p} |\sum_{x \in {\Bbb F}_p}^* e_p( ax + \frac{b}{x} )|^4$$ -and exploiting the dilation symmetry -$$ \sum_{x \in {\Bbb F}_p}^* e_p( ax + \frac{b}{x} ) = \sum_{x \in {\Bbb F}_p}^* e_p( cax + \frac{b/c}{x} )$$ -for any $c \in {\Bbb F}_p^\times$. (EDIT: as pointed out by Felipe, Mordell extended this argument to the case when $f$ is a linear combination of monomials $x^n$. For Mordell-type sums there are also recent papers of Bourgain and co-authors using the sum-product phenomenon to get power savings estimates in certain high-degree cases, but these techniques rely again on multiplicative structure and do not seem to be available in general.) - -However, in the positive genus case it does not appear that Weyl differencing can be used to reduce the complexity of the exponential sum even if combined with changes of variable (although I do not have a rigorous proof of this assertion), while I have been unable to extract a power saving from the Kloosterman argument in the absence of a symmetry such as dilation symmetry (although the argument does give the very weak upper bound of $(1-c)p$ for some $c>0$ in general). The only other arguments I know of go through the rationality of the zeta function, and in particular on linking the above exponential sum to the companion sums -$$ \sum_{x \in {\Bbb F}_{p^n}}^* e_p( \operatorname{Tr}(f(x)) )$$ -for large $n$, the key point being that one can now lose powers of $p$ in estimates on these sums as they can be recovered using the tensor power trick. However it does require a little bit of computation to attain this rationality (e.g. the Riemann-Roch theorem, or a bit of messy linear algebra, as done in Cochrane and Pinner), and I would be interested to know if there was a more direct proof of a power saving (or even a qualitative improvement $o(p)$ over the trivial bound of $p$) that did not require the introduction of the companion sums. -(My motivation here is to explore the minimal background needed to establish Zhang's recent theorem on bounded gaps between primes; currently, the argument requires the Weil conjectures for curves, but a cheap power savings for general rational function phases would eliminate the dependence on these conjectures.) - -REPLY [5 votes]: For your main question, see L. J. Mordell, On a sum analogous to a Gauss's sum, Quart. J. Math. Oxford Ser. 3 (1932), 161–167. It is a similar trick as in your 2. -I am sure that Hasse (in the 30's) stated the fact that RH for the curve $y^p-y = f(x)$ implies the bound with $p^{1/2}$. Weil didn't feel he needed to bother to state that as a consequence of his result. -Edit: For an explicit statement see L. Carlitz and S. Uchiyama Duke Math. J. 24, (1957), 37-41.<|endoftext|> -TITLE: Why Cohen-Macaulay rings have become important in commutative algebra? -QUESTION [35 upvotes]: I want to know the historic reasons behind singling out Cohen-Macaulay rings as interesting algebraic objects. - -I'm reviewing my previous lecture notes about Cohen-Macaulay rings because now I'm studying about Stanley-Reisner rings and I think I need to have a better general understanding about why I need to study CM rings. - -REPLY [5 votes]: Richard Stanley's description on How the Upper Bound Conjecture was Proved tells much about the history of how Cohen-Macaulay rings became important in combinatorics. Additional links can be found here.<|endoftext|> -TITLE: Certain asymptotics involving double infinite sum -QUESTION [6 upvotes]: Let $1<\alpha<\beta<3/2$. Set -$$ -S(n)= \sum_{i,j>0} [i^\alpha+j^\beta]^{-1}[(i+n)^\alpha+(j+n)^\beta]^{-1}. -$$ -One can check that $S(n)$ is finite. My question is when $n\rightarrow \infty$, how does $S(n)$ behave asymptotically, e.g., if it is asymptotically a power function? If yes, what is the exponent? -Remarks: -When $\alpha=\beta$, this problem can be resolved using an integral approximation argument (rewriting the sum as a double integral by replacing $\frac{i}{n}$ with $\frac{[nx]+1}{n}$, $\frac{j}{n}$ with $\frac{[ny]+1}{n}$ and letting $n\rightarrow\infty$ through the Dominated Convergence Theorem) which yields $S(n)\sim c n^{2-2\alpha}$ for some $c>0$. But when $\alpha<\beta$, the similar argument seems difficult to apply due to the non-homogeneity of the function $g(x,y)=(x^{\alpha} +y^{\beta})^{-1}$. -It seems that if we do have $S(n)\sim cn^{2-2\gamma}$ for some $\gamma$, then $\alpha\le \gamma\le \beta$. Furthermore, by Jensen's inequality, we have for any $0<\rho<1$, $i^\alpha+j^\beta\ge c i^{\alpha\rho}j^{\beta(1-\rho)}$ (now $g(x,y)= x^{-\rho\alpha}y^{-(1-\rho)\beta}$ is homogeneous, and an integral approximation argument applies provided $\alpha\rho\in (1/2,3/4)$, $\beta(1-\rho)\in (1/2,3/4)$), we should have -$ -\gamma\ge\rho\alpha+(1-\rho)\beta. -$ -By taking $\rho$ close to $1/(2\alpha)$, we expect that $\gamma\ge \beta+(\alpha-\beta)/(2\alpha)$. -Update: Matt shows below that $cn^{2-2\gamma}\le S(n)\le C n^{2-2\gamma}$, where $$\gamma=\beta+\frac{\alpha-\beta}{2\alpha}=\rho\alpha+(1-\rho)\beta\in (\alpha,\beta),$$ with $\rho=\frac{1}{2\alpha}$. Now the problem becomes whether one can show that $S(n)\sim cn^{2-2\gamma}$ where $\gamma$ is given as above. - -REPLY [4 votes]: I can show without too much work that there exist absolute constants $c, C > 0$ so that $$c \leq \frac{S(n)}{n^{1-2\beta + \frac{\beta}{\alpha}}} \leq C.$$ This at least shows what the exponent must be if there is an asymptotic formula for $S(n)$. When $\alpha = \beta$ this gives $S(n) \asymp n^{2-2\alpha}$, which uncovers a typo in the remarks in the question. Here and below the notation $f(x) \asymp g(x)$ for positive functions $f$ and $g$ means there exist $c, C > 0$ so that $c < \frac{f(x)}{g(x)} < C$. -Proof. First note that $$(i + n)^{\alpha} + (j + n)^{\beta} \asymp i^{\alpha} + j^{\beta} + n^{\beta}.$$ The terms with $i^{\alpha} \leq j^{\beta}$ and $j \leq n$ contribute to $S(n)$ an amount -$$\asymp \sum_{j \leq n} \sum_{i \leq j^{\beta/\alpha}} \frac{1}{i^\alpha + j^{\beta}} \frac{1}{i^{\alpha} + j^{\beta} + n^{\beta}}$$ which is -$$\asymp \sum_{j \leq n} \frac{1}{j^{\beta}} \frac{1}{j^{\beta} + n^{\beta}} \sum_{i \leq j^{\beta/\alpha}} 1 \asymp n^{-\beta} \sum_{j \leq n} \frac{1}{j^{\beta -\frac{\beta}{\alpha}}}.$$ Using $0 < \beta - \frac{\beta}{\alpha} < 1$, it is not hard to check that this is $\asymp n^{1-2\beta + \frac{\beta}{\alpha}}$. Similarly, the terms with $i^{\alpha} \leq j^{\beta}$ and $j > n$ satisfy the same type of estimates, as do the terms with $i^{\alpha} \geq j^{\beta}$ where in this case one should execute the $j$-sum first and divide the $i$ sum into two pieces depending on if $i^{\alpha} \leq n^{\beta}$ or not.<|endoftext|> -TITLE: On maximal regular polyhedra inscribed in a regular polyhedron -QUESTION [10 upvotes]: Let T, C, O, D, or I be regular tetrahedron, cube, octahedron, dodecahedron, and icosahedron, respectively. Suppose that the outer polyhedron have edge-length 1. -For example, it's easy to prove that the length of each edge of the maximal T inscribed in C is $\sqrt2$. Let's call this problem T in C. -As far as I know, H. T. Croft proved 14 cases out of 20 cases, so the following 6 cases still remain unsolved. -Two reciprocal pairs, C in I, D in O, and T in I, D in T, and (self-reciprocal) cases D in I, I in D. -Croft, HT : On maximal regular polyhedra inscribed in a regular polyhedron, Proc. London Math Soc(3), 41, 279-296, 1980. -I've tried to prove these six cases, but I'm facing difficulty. I need your help. - -REPLY [18 votes]: Edit: a preprint concerning this problem can now be found on the arXiv: http://arxiv.org/abs/1407.0683 -Let me give an exhaustive answer. Croft closes his paper with a list of the unsolved cases: - -Here $\kappa$ denotes the maximal edge length of the inner polyhedron. The ratio of the volumes can be easily computed when $\kappa$ is known. -Let's fill in the blank spaces that "denote ignorance"! -Let's say we want to find the largest polyhedron similar to a regular polyhedron $P$ that fits into a regular polyhedron $Q$ with edge length equal to $1$. We set up the following optimization problem. - -real variables: $e$, and for every vertex $v$ of $P$: $(x_v,y_v,z_v)$ -objective: maximize $e$ -linear constraints: For each vertex $v$ in $P$ and each hyperplane $h^+$ defining $Q$: $$(x_v,y_v,z_v)\in h^+.$$ -quadratic constraints: For each edge $(v,w)$ in $P$: $$e=(x_v-x_w)^2+(y_v-y_w)^2+(z_v-z_w)^2.$$ -If $Q=C$ or $Q=D$ we have to include more constraints to make sure that the coordinates $(x_v,y_v,z_v)$ really give a polyhedron similar to $P$. - -In other words: find coordinates of a polyhedron similar to $P$, which are contained in $Q$ with the square of the edge length being equal to $e$. Maximize $e$ to find the solution. -This non-linear programming problem with quadratic constraints can be effectively solved with SCIP, which uses branch-and-bound methods. (Many thanks to Ambros Gleixner for help with SCIP!). In order to run fast enough one has to set up the problem a bit more clever and remove redundant constraints. -I ran the program to solve the unsolved cases and also checked to known ones. Here is a table with the results: - -I emphasized new results (where Croft had blanks). Notice also that my values differ from Croft's in three places, which I marked with an exclamation mark. - -In the case $T$ in $D$ the number is off by a power of ten. -In the case $O$ in $D$ the rounded decimal expansion should read $1.851$, not $1.1850$, since the number really is $1.85122958682191611960\ldots$. -In the dual case $D$ in $C$ the correct decimal expansion reads $0.3942834797251374518168\ldots,$ so the rounded number should be $0.394$ not $0.395$. - -However Croft gives correct exact results ($\tau$ denotes the golden ratio) and only makes these minor mistakes in the decimal approximation. -Using similar arguments about immobility as Croft, we can calculate exact results for the open pairs. Let me give you exact values for two open pairs. -For $D$ in $I$ the maximum is $$\frac{15-\sqrt{5}}{22}\approx 0.58017872829546410470867392\ldots.$$ -For $C$ in $I$ the maximum is $$\frac{5+7\sqrt{5}}{22}\approx 0.93874890193175126703928253\ldots.$$ -In some cases it is harder to find the algebraic values, but in all cases it can be done, see the arXiv preprint above for details. -I used sage to generate a few images. Click here to view a 3d animation. - -The same optimization ansatz will also work in other dimensions and even for non-regular polyhedra.<|endoftext|> -TITLE: Easy argument for "connected simple real rank zero Lie groups are compact"? -QUESTION [9 upvotes]: Let $G$ be a connected simple Lie group. It is known that if $G$ has real rank zero, then $G$ is compact. -Background: every connected (semi)simple Lie group $G$ (with Lie algebra $\mathfrak{g}$) has a polar decomposition $G=KAK$, where $K$ arises from a Cartan decomposition $\mathfrak{g}=\mathfrak{k} + \mathfrak{p}$ (the group $K$ has Lie algebra $\mathfrak{k}$) and $A$ is an abelian Lie group such that its Lie algebra $\mathfrak{a}$ is a maximal abelian subspace of $\mathfrak{p}$. The real rank of $G$ is defined as the dimension of $\mathfrak{a}$. (Equivalently, the real rank can be defined in a similar way through the Iwasawa decomposition (see http://en.wikipedia.org/wiki/Iwasawa_decomposition)). -A possible way of obtaining the result that "connected simple real rank zero implies compact" is by looking at a list of (connected) simple Lie groups (see for example http://en.wikipedia.org/wiki/List_of_simple_Lie_groups) and check that all such Lie groups with real rank zero are compact. This, however, is not as explanatory as I would like. -Question: Is there an easy and explanatory proof of the fact mentioned above that connected simple real rank zero Lie groups are compact? -Perhaps, one might be able to use that the Lie algebra of a noncompact simple Lie group contains a copy of the real rank one Lie algebra $\mathfrak{sl}(2,\mathbb{R})$? -An easy special case: If $G$ is a connected simple real rank zero Lie group with finite center, then the implication easily follows from the $KAK$-decomposition. Indeed, if $G$ has finite center, the group $K$ in this decomposition is a (maximal) compact subgroup of $G$. This directly implies that $G$ is compact if it has real rank zero. -Using Riemannian symmetric spaces: if $X$ is a Riemannian symmetric space, and $G$ is the connected component of the isometry group of $X$, then the real rank of $G$ is the largest $n \in \mathbb{N}$ such that $X$ contains an $n$-dimensional closed, simply connected, connected, totally geodesic flat submanifold. It follows that $G$ has real rank zero if and only $X$ is compact. This can also be used to prove the fact mentioned above, but as far as I know, one still needs to consider all possible spaces $X$ that might occur. Or is there an easier way here? - -REPLY [9 votes]: (Real rank 0) $\ \Rightarrow\ (\mathfrak a = 0) \ \Rightarrow\ (\mathfrak p=0) \ \Rightarrow\ (\mathfrak g$ is a compact Lie algebra). -Therefore this comes down to (e.g. Bourbaki, Lie Groups, Chap IX, §1, no 4): - -THEOREM (H. Weyl) Let G be a connected Lie group whose Lie algebra is compact semi-simple. Then G is compact and its centre is finite. - -(Of that, I don't know an "easier" proof than what you'll find in Bourbaki or Hilgert-Neeb.)<|endoftext|> -TITLE: Rigid analytic spaces vs Berkovich spaces vs Formal schemes -QUESTION [24 upvotes]: I wonder if someone could explain briefly what is the relation between these 3 formal models, of a Berkovich space, a rigid analytic space and a formal scheme? -I have been working with formal schemes in the last few years, but I know virtually nothing about these other models. -What are the relations between these categories? is one contained in the other? -Are there any recommended references about this comparison ? -Many thanks. - -REPLY [28 votes]: There are different answers to your questions depending on what you have in mind. If you want to compare the spaces themselves, then Berkovich spaces and rigid analytic spaces are very close. I would say that it is similar to the relation between schemes of finite type over $\mathbb{C}$ and complex varieties in a naive sense, i.e. with only $\mathbb{C}$-points. -You can have a look at Berkovich's book "Spectral theory etc.", section 3.3, for precise statements. Fix a non-archimedean complete valued field $k$ with non-trivial valuation. Then, there is a functor from the category of strictly $k$-analytic spaces (Berkovich spaces) to rigid spaces with good properties: it preserves cohomology, for instance. At the level of points, the functor is easy to describe: you keep the points whose residue field is a finite extension of $k$. -As for the essential image of the functor, I am not sure what to say, but you can at least build a Berkovich space from a quasi-compact and quasi-separated rigid space in a satisfactory way (see Conrad's notes mentioned in Colin McLarty's answer). -(There are technical points that show that Berkovich's theory is actually more general. First, trivial valuation on $k$ is allowed. Second, the basic algebras of rigid geometry are the Tate algebras $k\{T_1,\dots,T_n\}$ that contain power series with radius of convergence at least 1, whereas Berkovich allows any radii.) -So the spaces are more or less the same, but I think the way you think about them is different. For example, Berkovich spaces are true topological spaces (as opposed to the Grothendieck topology of rigid spaces), which makes it easier to adapt the local methods from complex analytic geometry. In my mind, rigid analytic spaces look more algebraic. -As regards formal schemes, they are quite different objects. The relation is that the rigid spaces and Berkovich spaces are their generic fibers. So a formal scheme has more information. Raynaud's theory (see his paper in Mémoires de la S.M.F 39-40 (1974), p.319-327) actually tells you how to pass from a model to another (by blowing-ups ideals supported on the special fiber), which makes it possible to recover the category of rigid spaces purely in terms of formal schemes (by a suitable localization). This makes the analytic theory very much algebraic! A big gain is that you may now use everything that has been developed in the algebraic setting and push it on the analytic side. For example, you can show the coherence of proper direct images of coherent sheaves this way (see Lütkebohmert, Math. Ann. 286 (1990), p.341-171). -I hope this helps. If you have more specific questions, please ask!<|endoftext|> -TITLE: Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using a theorem -QUESTION [19 upvotes]: Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gel'fond-Schneider's theorem. -We know that ${\sqrt2}^{\sqrt2}$ is a transcendental number by the Gel'fond-Schneider's theorem. I've tried to prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gel'fond-Schneider's theorem, but I'm facing difficulty. I need your help. -This question has been asked previously on math.SE without receiving any answers. - -REPLY [21 votes]: You do not need to use Gelfond-Schneider theorem, you can just repeat one of the well known easy proofs of that theorem. For example, a much stronger theorem is proved in an Appendix to Lang's "Algebra", the proof is only 4 pages long. The proof uses only elementary linear algebra, some calculus and first notions of Galois theory. That proof can be significantly shortened if you want to prove irrationality only. -Edit I have written a more or less complete proof here . Galois theory is not needed there (except for the fact that the norm of an integral element is an integer), but one needs the maximal modulus theorem for analytic functions. I am sure that can be avoided also. From calculus, one needs the Taylor formula (no integration is required).<|endoftext|> -TITLE: Uncertainty principle for Mellin transform -QUESTION [22 upvotes]: Let $f:\mathbb{R}^+\to \mathbb{C}$. Let $Mf$ be its Mellin transform: $Mf(s) = \int_0^\infty f(x) x^{s-1} dx$. -(a) Some time ago, I convinced myself that -$f(t)$, $Mf(\sigma+it)$ and $Mf(\sigma-it)$ cannot all decrease faster than exponentially as $t\to +\infty$. This should be a variation on G. H. Hardy's version of the uncertainty principle. I presume this is well-known? Does anybody have a reference? Does anybody know an easy proof (one that can be sketched here)? -(b) The example $f(t) = e^{-t^r}$ shows that $f(t)$ can be made to decrease much faster than exponentially, while $Mf(\sigma+it)$ and $Mf(\sigma-it)$ still decrease exponentially (meaning $\sim e^{-(\pi/2r) |t|}$ - that is, the exponent does degrade as $r$ increases). Is this in any sense optimal? That is, can one show that, if $f(t)$ decreases faster than $e^{-t^r}$ for every $r$, then $Mf(\sigma+it)$ and $Mf(\sigma-it)$ cannot both decrease exponentially? - -REPLY [6 votes]: There is something the above posts missed (perhaps because the wording didn't make it explicit): the condition on the function $f(x)$ is one sided (i.e., it assumes something on the decay as $x\to \infty$, not as $x\to -\infty$), and thus, once we take logarithms to make the Mellin transform into a Fourier transform, we end up with a one-sided condition as well. -Now, one-sided versions of the uncertainty principle do exist - notably that of Nazarov (1993); see also Bonami and Demange (2006). In the end, it turns out to be best to rework the proofs given there to get an uncertainty principle for the Mellin transform of the shape I was asking for. And yes, one does get something essentially optimal in that way, with no spurious factors of log in the exponent.<|endoftext|> -TITLE: What prevents a cover to be Galois? -QUESTION [9 upvotes]: Let $f:X\rightarrow Y$ be a ramified cover of Riemann surfaces or algebraic curves over $\mathbb{C}$. My question is can one in terms of the ramification data of $f$, determine whether the cover is Galois or not. I am Specially interested in the following question: if $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are both ramified Galois covers of curves, when $g\circ f:X\rightarrow Z$ is (or isn't) a Galois cover? - -REPLY [4 votes]: There are many automorphisms of $Y$ over $Z$. One can pull back the cover $X$ along any of these automorphisms of $Y$, producing a new cover of $Y$. If the composition is Galois, then these new covers are in fact isomorphic to the original cover - this is because the map $Gal(X|Z) \to Gal(Y|Z)$ is surjective. Taking a lift of an element of $Gal(Y|Z)$ produces an isomorphism between the original cover and the pullback. -The converse is also true. If such an isomorphism exist, they give us enough elements of the automorphism group of $X$ over $Z$ that, combined with the automorphisms of $X$ over $Y$, the cover is Galois! -So this explains why the two preimages with different ramification data will prevent the cover from being Galois. Because then the pullback of the cover along an automorphism which permutes those two preimages will not be isomorphic to the original - it will have different ramification data.<|endoftext|> -TITLE: An Entropy Inequality -QUESTION [34 upvotes]: Let $X,Y$ be probability measures on $\{1,2,\dots,n\}$, and set $K=\sum_i\sqrt{X(i)Y(i)}$ so that $Z:=\frac{1}{K}\sqrt{XY}$ is also a probability measure on $\{1,2,\dots,n\}$. How can we prove the inequality - $$H(X)+H(Y)\geq 2K^2 H(Z),$$ where $H(X)=-\sum_{i=1}^n X(i)\log X(i)$ is the Entropy function. - -The problem originates from this math stack exchange post, and cardinal's rewording of it in the comments. Despite having being asked over two years ago, with numerous bounties posted, the problem was never solved, and for that reason I am posting it here. -I checked the inequality numerically on matlab for millions of choices of $X$ and $Y$, with $n$ up to size $100$, and it always held, which suggests that finding a counter example is unlikely. -Remark: By Cauchy Schwarz, $1\geq K^2,$ so the above inequality would be implied by $H(X)+H(Y)\geq 2H(Z).$ However it is worth noting that this inequality does not hold, so the factor of $K^2$ is important. - -REPLY [4 votes]: Here is another alternative proof of -the inequality -$$ -(1+a)\ln(1+b)+(1+b)\ln(1+a)\ge 2(1+c)\ln(1+c) \tag{1} -$$ -with -$$a,b,c>0,\ ab=c^2,\tag{2}$$ -proved in the above answer. -The left-hand side of $(1)$ goes to $\infty$ as $a$ or $b$ does so while keeping condition $(2)$. So, it remains to show that a conditional local extremum given $(2)$ can only be attained when $a=b$ and hence $a=b=c$. -Without loss of generality, $a -TITLE: A question on an ordinal for ZFC- -QUESTION [9 upvotes]: ZFC-, which is ZFC minus power set, is modelled by $ L_{\delta}$ where $\delta$ is an admissible ordinal larger than -any least $\Sigma_{n}$-admissible ordinal for n a natural number. Can some provide more information? Does it have a name? What is its most natural omega sequence? - -REPLY [13 votes]: The least such ordinal $\beta$, often written $\beta_0$, for which $L_\beta$ is a $ZF^-$ is also characterised as the "ordinal of ramified analysis". This is because the ramified analytical hierarchy, which builds up cumulative second order number theoretic structures of the form $$\underline{P_\alpha}=( P_\alpha,\mathbb{N}, +, \times, \ldots)$$ with $P_\alpha\subseteq P_{\alpha +1}\subseteq -\mathcal{P}(\mathbb{N})$ by looking at all sets of integers definable over $\underline{P_\alpha}$ by using instances of second order comprehensionto to obtain $P_{\alpha+1}$. (Unions are taken at limits). This hierarchy has height exactly $\beta_0$ (meaning $P_{\beta_0}=P_{\beta_0+1}$.) Alternatively put: $\underline{P_{\beta_0}}$ is then the least model of $Z_2$, or full second order comprehension (sometimes just abbreviated to "analysis", or ``second order number theory''). -${P_{\beta_0}}$ is then $\mathcal{P}(\mathbb{N}) \cap L_{\beta_0}$. -$\beta_0$ of course is $\Sigma_n$-admissible for all $n$. $L_{\beta_0+1}$ sees that $\beta_0$ is countable, and so I suppose one may define a particular $\omega$ sequence in $L_{\beta_0+1}$ cofinal in it by just taking the $<_L$- least such. -One can also paraphrase the assertion that $L_{\beta_0}$ is the least $\beta$ with $L_\beta$ a $ZF^-$ model, as being the least such that for no $n$ is the fine-structural projectum $\rho^n_\beta$ less than $\beta$. (But this is really only dressing up one concept in the jargon of another.) -$L_{\beta_0}$ is (I believe) the least level of the $L$ hierarchy whose reals form a model of $\Delta^0_4$-Determinacy (Martin - unpublished). -For the Ramified Analytical Hierarchy: see Boyd, Hensel, Putnam "An intrinsic characterisation of the ramified analytical hierarchy", JSL, late 60's I believe. This hierarchy is not much studied these days, but was of interest in the pre-Jensen fine-structural era of the analysis of levels of $L$ (the latter's work now surpassing it). -For connections between models of subsystems of second order comprehension and constructible sets, see also S. Simpson's book "Subsystems of second order arithmetic".<|endoftext|> -TITLE: On quadratic forms, Pontryagin Squares, and H^4(K(pi,2),U(1)) -QUESTION [7 upvotes]: I am trying to get a concrete handle on the isomorphism $H^4(K(\pi_2,2),U(1)) \simeq \{$quadratic forms $\pi_2 \to U(1) \}$. This is explained in Eilenberg and Maclane's http://www.jstor.org/stable/1969702 and its companion but I am having a hard time getting just what this 4-cocycle should assign to a 4-simplex in $K(\pi_2,2)$. I am primarily interested in understanding the map from the right to the left. -I have a guess at something which may be close, which is there is a canonical closed 2-form on $K(\pi_2,2)$ valued in $\pi_2$. Using the associated bilinear form of the given quadratic form, I can wedge this form with itself to obtain a closed 4-form valued in $U(1)$. I worry that instead of the ordinary square, I need to be doing some factoring, perhaps using the Pontryagin square instead. -Any help, especially with some intuition, would be much appreciated. - -REPLY [6 votes]: To a quadratic form $q: \pi_2 \to U(1)$, we get a corresponding Pontryagin square operation $H^2(-; \pi_2) \to H^4(-; U(1))$, and such cohomology operations are given by elements of $H^4(K(\pi_2, 2); U(1))$. Unfortunately, I don't know if it's possible to get an explicit cochain-level description of the Pontryagin square operator just from the quadratic form. In Proposition 7.3 of my paper Extensions of groups by braided 2-groups, I write down a (group) cochain-level description of the Pontryagin square corresponding to an abelian 3-cocycle, but I don't know if there's a way in general to get an abelian 3-cocycle from a quadratic form. One good reference to look at is Baues's Combinatorial Homotopy and 4-Dimensional Complexes. -One intuitive way to think of the appearance of the Pontryagin square is that we are describing braided 2-groups with homotopy groups $\pi_2$ and $U(1)$, and the Pontryagin square operation $H^2(G; \pi_2) \to H^4(G; U(1))$ on a group $G$ is the obstruction to lifting an extension of $G$ by $\pi_2$ to an extension of $G$ by the entire braided 2-group (in the sense I describe in my paper). -EDIT: I should mention that you can avoid referring to the Pontryagin square directly. Per Baues, $H_4(K(\pi_2, 2); \mathbb{Z}) = \Gamma(\pi_2)$, the receptor of the universal quadratic map from $\pi_2$, so by universal coefficients, $H^4(K(\pi_2, 2); U(1)) = \operatorname{Hom}_{\mathbb{Z}}(\Gamma(\pi_2), U(1))$, which is the same as quadratic forms on $\pi_2$ valued in $U(1)$.<|endoftext|> -TITLE: What's a good example/reference for cohomology classes on Springer fibers that aren't restricted from the flag variety -QUESTION [8 upvotes]: As usual, by Springer fiber, I mean the fixed points $X^u$ of a unipotent element $u$ of the group $G$ on the flag variety $X=G/B$. It's a lovely theorem that when $G=SL_n$, the induced map on cohomology $H^*(X)\to H^*(X^u)$ is surjective. -However, I've been told many times that for other $G$, this isn't the case. Can anyone point me to a good reference for finding these, or a handy source of examples? - -REPLY [3 votes]: I think the standard early reference is the paper by Hotta and Springer here, in which they work with $\ell$-adic cohomology (but the results seem to carry over to other settings). What they show is that the map in cohomology is surjective on the top degree in the target space when mapping into the subspace fixed by the component group, while the precise image in other degrees might be more elusive when there is a nontrivial component group for the unipotent class in question. (This is essentially why all goes well for type $A$, where component groups are trivial.) -Examples in rank 2 for types $B_2, G_2$ show already what goes wrong when the component group is nontrivial (which happens there for the subregular orbit, as Jay observes). But as far as I know, the problem remains quite difficult to study beyond what is done by Hotta-Springer. -[I gave a short survey of Springer theory in Chapter 9 of my 1995 AMS book on conjugacy classes, but unfortunately overstated the Hotta-Springer result in 9.6(4). Corrections are posted on my homepage.]<|endoftext|> -TITLE: Determinacy from $\omega_1\rightarrow(\omega_1)^{\omega_1}$ -QUESTION [8 upvotes]: Assuming the Axiom of Determinacy (abbreviated AD), Martin showed how to derive a rather strong partition on $\omega_1$, namely that $\omega_1\rightarrow(\omega_1)^{\omega_1}$. In "Infinitary Combinatorics and the Axiom of Determinateness", Kleinberg goes through Martin's proof that $\omega_1\rightarrow(\omega_1)^{\omega_1}$ under AD, and from there he goes on to prove much about the large cardinal structure of ZF+AD below $\aleph_\omega$. In fact, he does this directly from the partition relation. In particular, Kleinberg shows that $\aleph_1$ and $\aleph_2$ are measurable, for each $n>2, \aleph_n$ has cofinality $\aleph_2$ and is Jonsson, and that $\aleph_\omega$ is Rowbottom. -We see that the existence of such a partition relation is inconsistent with choice. In fact, the following is a theorem of Erdos and Rado: - -For all infinite $\tau$, there is no $\kappa$ such that $\kappa\rightarrow(\tau)^\tau$ - -However, I am curious about the following: - -How much is known about the amount of determinacy one can derive from ZF + DC + $\omega_1\rightarrow(\omega_1)^{\omega_1}$? - -In particular, any references would be greatly appreciated. -Edit: -I elaborated on the nature of the question in a comment to an answer. For sake of visibility, I'm pasting the relevant portion here: -To elaborate on the question, it's interesting that one can derive so much about the large cardinal structure of ZF + AD directly from the strong partition relation on $\omega_1$. As far as I know, "$\omega_1$ is measurable" and "$\omega_2$ is measurable" alone allow us a fragment of determinacy. I was curious as to what is known beyond what can be derived from these measures, as $\omega_1\rightarrow(\omega_1)^{\omega_1}$ gives us these measures. - -REPLY [6 votes]: There is some sort of equivalence between partition properties and determinacy. The question is treated in Kechris, Kleinberg, Moschovakis, Woodin, Determinacy, partition properties, nonsingular measures in the Cabal reprints, Volume I. More specifically the weak partition property on $\omega_1$ implies analytic determinacy (which is provable in ZF). Concerning the strong partition property I am not really sure, I think it implies $\Pi^1_1$-determinacy (remember this is equivalent to $0^{\sharp}$ and originally the strong partition on $\omega_1$ was proven using properties of indiscernibles). The KKMW article has the details<|endoftext|> -TITLE: Woodin's unpublished proof of the global failure of GCH -QUESTION [18 upvotes]: An unpublished result of Woodin says the following: -Theorem. Assuming the existence of large cardinals, it is consistent that $\forall \lambda, 2^{\lambda}=\lambda^{++}.$ -In the paper "The generalized continuum hypothesis can fail everywhere, Ann. Math. 133 (1991), 1–35" it is stated that Woodin has proved this result assuming the existence of a $P^{2}(\kappa)-$hypermeasurable (or equivalently a $\kappa+2-$strong) cardinal $\kappa$. -But using core model theory we now that more than a $P^{2}(\kappa)-$hypermeasurable cardinal $\kappa$ is required for this result (though a $P^{3}(\kappa)-$hypermeasurable cardinal $\kappa$ is sufficient). -So my question is -Question. 1-What large cardinal assumption is used in the proof of the above Theorem. -2-Does anyone know the proof? Does it use the supercompact Radin forcing as in the paper stated above, or it uses the ordinary Radin forcing (like Cummings paper "A model in which GCH holds at successors but fails at limits"). - -REPLY [6 votes]: At the present there are at least three published paper concerning the global behavior of the power function: -1-Foreman-Woodin, The generalized continuum hypothesis can fail everywhere. -2- Cummings, A model in which GCH holds at successors but fails at limits. -3-Merimovich, A power function with a fixed finite gap everywhere. -The first paper uses a supercompact cardinal with infinitely many inaccessibles above it, while the last two papers use strong cardinals. Let me mention that: -(*) In all of these models cofinalities are changed and in the last two models cardinals are also collapsed. -(**) All of these models are obtained in two steps: At the first step a reverse Easton iteration is done which blows up the power of some cardinals below $\kappa$ ($\kappa$ is the large cardinal which we are using to exist in the ground model) and in the extension some guiding generics are constructed for the use of second step. In the second step a variant of Radin forcing (usually with interleaved collapsed,...) is used. Note that changing cofinalities and collapsing cardinals are presented in this step. At the end $\kappa$ remains inaccessible and below $\kappa$ we have the behavior of the function as we requested. -I would like to mention two recent results of Sy Friedman and I: -Theorem 1. Assuming the existence of a $\kappa+3-$strong cardinal $\kappa,$ there exists a pair $(W,V)$ of models of $ZFC$ such that: -1- $W$ and $V$ ahve the same cardinals, -2-GCH holds in $W$, -3-$GCH$ fails everywhere in $V$, -4-$V=W[R]$ for some real $R$. -The above theorem says that it is possible to kill the GCH everywhere just by adding a single real. It answers an open question of Shelah-Woodin "Forcing the failure of CH by adding a real" (I like the above result a lot). -Theorem 2. Assuming the existence of a $\kappa+4-$strong cardinal $\kappa$ it is consistent to have a pair $(W,V)$ of models of $ZFC$ such that: -1-$W$ and $V$ have the same cofinalities, -2-GCH holds in $W$, -3-$V\models \forall \lambda, 2^{\lambda}=\lambda^{+3}.$ -Thus it is possible to kill the GCH everywhere by a cofinality preserving forcing. -I also should mention that Moti Gitik and Carmi Merimovich are doing a project in which they are planning to obatain the global failure of GCH from the optimal hypotheses. If I remeber their result correctly it says something like this: -Theorem. The following are equiconsistent: -1-For any $\alpha,$ there are stationary many cardinals $\kappa$ with $O(\kappa)=\kappa^{++}+\alpha,$ -2-GCH fails everywhere, -3-$\forall \lambda, 2^{\lambda}=\lambda^{++}.$ -Gitik told me that their proof uses the Radin forcing and they need to embed many models into each other. My guess is that the paper " Gitik, Moti; Merimovich, Carmi Power function on stationary classes" is related to their work.<|endoftext|> -TITLE: What to do now that Lusztig's and James' conjectures have been shown to be false? -QUESTION [50 upvotes]: Lusztig and James provided conjectures for dimensions of simple modules (or decomposition numbers) for algebraic groups and symmetric groups in characteristic $p$. These conjectures have been considered almost certainly true and have guided a lot of the research in these areas for a long time. A new preprint by Geordie Williamson, (with a classily understated title: "Schubert calculus and torsion") shows that the Lusztig conjecture is false (and therefore that James' conjecture is also false). -Amongst other things, this certainly implies that the existing cases in which the conjectures have been proved become more interesting (very large $p$ for Lusztig's, RoCK blocks and certain defects for James'). -But my question is: what next? Do we just accept that we'll only understand algebraic groups in very large characteristic? Do we direct focus on non-abelian defect blocks for symmetric groups? -Or... there's a lovely conjecture due to Doty which (though less explicit than Lusztig's conjecture) could, in principle, give a character formula for the characters of simple modules for $GL_n$ and hence symmetric groups. It states that: -The modular Kostka numbers are defined as follows: $K′= [Tr^λ(E) : L(μ)]$, for -$Tr^\lambda(E)$ the truncated symmetric power and $L(\mu)$ the irreducible polynomial $GL_n$-module -of highest weight $\mu$. Then Doty’s Conjecture states that the modular Kostka matrix $K′ = (K′_{\mu,\lambda} )$, with rows and columns indexed by the set of all partitions $\lambda$ of length $\leq n$, and bounded by $n(p−1)$ (fixed in some order), is non-singular for all $n$ and all primes $p$. -Could this conjecture be the new "big problem" for algebraic and symmetric groups in positive characteristic? Are there any other conjectures out there of a similar flavour? - -References -G. Williamson, Schubert Calculus and Torsion. -http://people.mpim-bonn.mpg.de/geordie/Torsion.pdf -Doty, S., Walker, G., Modular Symmetric Functions and Irreducible Modular Representations of General Linear Groups, Journal of Pure and Applied Mathematics, 82, (1992), 1-26. -See also page 105 of S. Martin "Schur algebras and Representation theory". - -REPLY [7 votes]: Geordie made some excellent points, beyond my earlier community-wiki comments, but maybe it's worth spilling more pixels to emphasize another important but often neglected viewpoint: Lusztig's 1980 Generic Decomposition Conjecture (GDC), closely related to what is usually called the Lusztig Conjecture (LC). Here is a brief overview of the essential literature: -In a 1977 paper (J. Algebra 49), Jantzen worked out some ideas about "generic decomposition" of Weyl modules in sufficiently large characteristic $p$. This reveals patterns of composition factors (interchangeable in a sense for different alcove types) for all highest weights in general position in the lowest $p^2$-alcove for an affine Weyl group $W_p$ relative to $p$ of Langlands dual type. (For example, in type $B_2$ he finds generically 20 composition factors with multiplicity 1, whereas for $G_2$ there are 119 composition factors with multiplicity ranging from 1 to 4.) He later proved existence more efficiently in a 1980 Crelle J. 317 paper, also in German. -Jantzen's patterns look rather unpredictable, but a "dual" version (which I observed at first indirectly via the finite groups of Lie type) shows symmetry around a special point for $W_p$. From these generic patterns Jantzen also showed how to derive all patterns for arbitrary weights in the region. But the -underlying assumption is always that $p$ is at least the Coxeter number $h$, to ensure existence of weights inside $p$-alcoves and permit use of translation functors. To get generic patterns, $p$ of course has to be even bigger in general. -In his write-up of a survey of problems at the 1979 AMS summer institute in Santa Cruz, published in PSPM (1980), Lusztig first stated LC with the hypothesis $p \geq h$ and the stipulation that the dominant weight involved should lie inside an alcove within the "Jantzen region" in the lowest $p^2$-alcove for $W_p$; then the irreducible character should be a $\mathbb{Z}$-linear combination of Weyl characters with coefficients up to sign given by specializations at 1 of Kazhdan-Lusztig polynomials for the (abstract) affine Weyl group. -Eventually Andersen-Jantzen-Soergel proved this (via the successful quantum group version) for "sufficiently large" $p$, after which Fiebig derived an explicit but huge lower bound on $p$. As Williamson has observed, there are serious problems about smaller $p \geq h$. -In 1980 a more technical paper by Lusztig (in Advances 37) laid out a generic version of Jantzen's ideas. Here one has the symmetric dual patterns of alcoves in the abstract affine Weyl group, with polynomial entries in $q$ given by inverse K-L polynomials. His Remark 1.9 is the conjecture that for $q=1$ one should recover Jantzen's patterns, for $p$ "sufficiently large". -In a 1985 paper, S.I. Kato (Advances 60) calls this the GDC and proves in Thm. 5.6 that for "sufficiently large" $p$ it is equivalent to the LC. He builds on a previously accepted but later published 1986 paper by Andersen (*Advances 60) which studies inverse K-L polynomials in more detail. -With this background, it makes sense to distinguish three ranges for $p$: always we need $p \geq h$ (which is a serious limitation for type $A_n$ and related symmetric group problems), whereas for "very large" $p$ both LC and GDC are simultaneously true. For a range of "intermediate" $p$ and for $p -TITLE: Is the sequence of partition numbers log-concave? -QUESTION [24 upvotes]: Let $p(n)$ denote the number of partitions of a positive integer $n$. It seems to me that we have for all $n>25$ -$$ -p(n)^2>p(n-1)p(n+1). -$$ -In other words, the sequence $(p(n))_{n\in \mathbb{N}}$ is log-concave, or satisfies $PF_2$, with -$$ -\det \begin{pmatrix} p(n) & p(n+1) \cr p(n-1) & p(n) \end{pmatrix}>0 -$$ -for $n>25$. Is this true ? I could not find a reference in the literature so far. On the other hand, the partition function is really studied a lot. -So it seems likely that this is known. -Similarly, property $PF_3$, with the corresponding $3\times 3$ determinant, seems to hold for all $n>221$, too, and also -$PF_4$ for all $n>657$. -The question is also motivated from the study of Betti numbers for nilpotent Lie algebras, in particular filiform nilpotent Lie algebras. - -REPLY [14 votes]: The statement referenced by Igor Rivin http://www.math.clemson.edu/~janoski/ResearchStatement.pdf uses the phrase - -Computationally looking at p(n) we see that for n ≥ 26 the partition function is log-concave [2]. - -I had seen this reference before probably about the same time this research statement was first released, and I am skeptical for two reasons. - -The phrasing "Computationally..." would seem to indicate some type of calculation. This cannot involve a computer since it would have to hold for all n larger than 26, and I am not aware of any simplification that allows one to only consider a finite number of cases. It would have been helpful to at least expound on the type of computations involved. -I checked for the promised reference, and indeed I found it on the CV of the author, http://www.math.clemson.edu/~janoski/VitaTex.pdf, but it refers to the quote below. I did a quick google search and I could find no reference or anything pointing to a publication. - -Brian Bowers, Neil Calkin, Kerry Gannon, Janine E. Janoski, Katie Joes, Anna Kirkpatrick, The Log Concavity of the Partition Function, (in preparation) - -Asymptotics will not provide the answer here, since n sufficiently large doesn't hold up unless you can provide a concrete n and test everything less than it, and I don't believe the Hardy-Ramanujan asymptotic expansion yields any guaranteed error estimates. -It may be possible to use DH Lehmer's estimates to obtain a proof. In two papers (1937 and 1939) he investigated the coefficients of both the Hardy-Ramanujan asymptotic expansion and the Hardy-Ramanujan-Rademacher expansion. He provided guaranteed error bounds on the remainder terms in the asymptotic expansions so that, for example, his Theorem 13 says that for n>600, only $2/3 \sqrt n$ terms of the Hardy-Ramanujan asymptotic series are needed to estimate p(n) to the nearest integer. - -At present, I don't believe the matter is completely settled, despite the overwhelming computational evidence. -UPDATE 11-1-13: -Igor Pak and I have just uploaded a preprint to the ArXiv: http://arxiv.org/abs/1310.7982 . In it we prove the log-concavity of the partition numbers for all $n>25$, and Section 6.3 addresses Janoski's thesis. -UPDATE 11-23-15: -Igor and I were recently informed of work by Jean-Louis Nicolas which also contains a proof of the log-concavity of the partition numbers: -Sur les entiers N pour lesquels il y a beaucoup de groupes abéliens d’ordre N, -Annales de l’institut Fourier, tome 28, no 4 (1978), p. 1-16. -http://www.numdam.org/item?id=AIF_1978__28_4_1_0<|endoftext|> -TITLE: Famous vacuously true statements -QUESTION [42 upvotes]: I am interested to know other examples vacuously true statements that are non-trivial. My starting example is Turan's result in regards to the Riemann hypothesis, which states -Suppose that for each $N \in \mathbb{N}_{>0}$ the function $\displaystyle \sum_{n=1}^N n^{-s}$ has no zeroes for $\mathfrak{R}(s) > 1$. Then the function -$$\displaystyle T(x) = \sum_{1\leq n \leq x} \frac{\lambda(n)}{n}$$ -is non-negative for $x \geq 0$. In particular, this would imply the Riemann hypothesis. -Here $\lambda(n) = \lambda(p_1^{a_1} \cdots p_r^{a_r}) = (-1)^r$ is the Liouville function. -The interesting thing about this statement is that both the hypothesis and the consequence can be proven false independently. In particular, Montgomery showed in 1983 that for all sufficiently large $N$ the above sums have zeroes with real parts larger than one, and Haselgrove showed in 1958 that $T(x)$ is negative for infinitely many values of $x$. Peter Borwein et al. found the smallest such $x$ in 2008. -I find this result fascinating because it relates to a well-known conjecture, and both the hypothesis and consequence were plausible. Are there any other mathematical facts of this nature, perhaps in other areas? - -REPLY [6 votes]: I heard there was a entire theory of finite skew fields.<|endoftext|> -TITLE: What are good ways to present proofs of theorems requiring auxiliary lemmas? -QUESTION [9 upvotes]: I am writing an academic paper for submission to a journal. One of my co-authors wrote the following: - -Theorem Statement of the theorem -Proof of theorem We first show the following result -Lemma Statement of lemma used to prove the theorem - -However, I think that it is more natural to present things in the following way: - -Theorem Statement of the theorem -We first show the following result which is used to prove the theorem: -Lemma Statement of lemma used to prove the theorem -Proof of theorem blah blah - -I understand that this is a subjective question so I am happy to mark this as community wiki if most people feel that it belongs there. - -REPLY [6 votes]: The guiding principle should be: think of your readers. Nesting results inside the proofs of other results can make things confusing. I am looking right now at a paper that has three instances of that, and one of them is truly egregious: the bracket diagram is like this: $((((()))(())))$. -Maybe one nested lemma is ok, but if I could avoid it I would just put the lemma before, and say something like: "The following lemma will be used in the proof of Theorem 1." If you really want to state the theorem before the lemma, then your second option is better. -OK, I'll stop grinding this axe now.<|endoftext|> -TITLE: Modular Functions with Rational Fourier Expansions -QUESTION [5 upvotes]: I have been reading the paper of Cox, McKay and Stevenhagen "Principal Moduli and Class Fields", http://arxiv.org/pdf/math/0311202v1.pdf, and I have a question regarding the nature of the function field for $X0(N)$: if I have a modular function for $Γ_0(N)$ that has a rational $q$-expansion at $\infty$, does it follow that $f∈\mathbf{Q}(j,j_N)$? -Allow to me to elaborate on some discoveries I've made reading the above paper, and playing with sage: -I understand that the function field over $\mathbf{C}$ for $X_0(N)$ is $\mathbf{C}(j,j_N)$, and that the $\mathbf{Q}$-function field given by $\mathbf{Q}(j,j_N)$ defines the curve $X_0(N)$ over the rationals. Thus, any modular function for $\Gamma_0(N)$ can be written as rational function in $j$ and $j_N$. -Now, for the case of level $N=1$ modular functions, it is stated in the paper that if $f$ is a modular function that has a fourier expansion at $\infty$ with rational coefficients, then $f$ in fact lies in $\mathbf{Q}(j)$, which seems intuitive. However, I have recently discovered that the analogous fact does not hold for modular functions for $\Gamma_0(N)$ that have a rational fourier expansion at $\infty$. Using a paper of Maier, I have been able to play with the hauptmoduln for $\Gamma_0(N)$ when the group has genus zero. These functions have rational $q$-expansions, but don't necessarily lie in $\mathbf{Q}(j,j_N)$. Indeed, using the function $t_2$ in his paper (a hauptmodul for $\Gamma_0(2)$ that has rational $q$-expansion), I computed that -$$t_2(1/4\sqrt{-7}−1/4)$$ -is an algebraic integer of degree $2$. In particular, it is definitely not a rational number. On the other hand, the theory of complex multiplication ensures that both $j(1/4\sqrt{-7}−1/4)$ and $j_2(1/4\sqrt{-7}−1/4)$ are integers. Therefore, $t_2$ cannot lie in $\mathbf{Q}(j,j2)$, since otherwise $t_2(1/4\sqrt{-7}−1/4)$ would have to be rational as well. -What is troubling about this fact is that, in the paper of Cox et al., it is stated that if $f$ is a modular function for $\Gamma_0(N)$ and if $\tau$ is an elliptic point of order 2 for $\Gamma_0(N)^\dagger$, but not one for $\Gamma_0(N)$, then $f(\tau)$ must lie in the maximal abelian extension of $K=\mathbf{Q}(\tau)$ and that this follows from basic complex multiplication theory. Now, if it were the case that $f\in\mathbf{Q}(j,j_N)$, then I'd be happy to believe this, but I've just seen that this is not the case. Therefore, I don't see how basic complex multiplication theory says anything about the nature of $f(\tau)$ in this case. -Lastly, as mentioned above, it is stated explicitly that for level $N=1$, we do know that such a modular function with rational $q$-expansion must lie in $\mathbf{Q}(j)$. Does this result remain true for $X_0(N)$ when it has genus zero? That is, after fixing a hauptmodul $h$ with rational fourier expansion, so that any modular function for $\Gamma_0(N)$ can be written as a rational function in $h$, is it true that those functions with rational $q$-expansions can be written as a rational function of $h$ with coefficients in $\mathbf{Q}$? -If anyone can shed some led on any part of this problem, I would very much appreciate it! - -REPLY [6 votes]: The statement about being in ${\bf Q}(j,j_N)$ if the $q$-expansion is rational is true. See Prop 12.7 (a) in Cox's book "Primes of the form $x^2 + Ny^2$". Moreover, part (b) of that proposition tells you that you can specialize when the partial derivative $\partial \Phi_N(x,y)/ \partial(x)$ does not vanish at that point. That's not the case for the CM point you've chosen. In general if you have a genus zero curve for $X_0(N)$ and $h$ a hauptmodul, then this $q$-expansion fact will work, and you can certainly specialize as long as the denominator of the rational function in $h$ doesn't vanish.<|endoftext|> -TITLE: Do mutually dual finite vector spaces have the same orbit cardinalities under a linear group action? -QUESTION [17 upvotes]: Let $G$ be a finite group acting linearly on a finite dimensional vector space $V$ over a finite field. By Burnside's lemma, -$$ -|V/G| = \frac 1{|G|} \sum_{g\in G} q^{\dim(ker(g - I))}. -$$ -Since $g-I$ and its dual map $g^*-I$ have kernels of the same dimension, it follows that $|V/G|=|V^*/G|$. -The above argument shows that a vector space and its dual have the same number of orbits under the action of linear group. Can it happen that the cardinalities of the orbits are different? - -REPLY [21 votes]: Yes. In particular, it can happen that $V$ has non-zero fixed points, but $V^*$ doesn't. -For example, let $G$ be the symmetric group of degree 3 acting in the obvious way on the set $\{e_1,e_2,e_3\}$, and let $W$ be the corresponding permutation module over the field of 3 elements. Let $V$ be the submodule spanned by $e_1-e_2$ and $e_1-e_3$. Then $V$ has orbits of lengths $6,1,1,1$, but $V^*$ has orbits of lengths $3,3,2,1$.<|endoftext|> -TITLE: How to explicitly see the ramification over infinity -QUESTION [7 upvotes]: Take the equation $y^{d}=\Pi_{1}^{n}(x-t_{i})^{m_{i}}$ over $\mathbb{C}$. This affine equation gives a cyclic cover of $\mathbb{P}^{1}$. Now it is usually said without explanation that if the sum $\sum m_{i}$ is not divisible by $d$, then the family is ramified over $\infty$. My question is how to see explicitly--in terms of equations maybe--that this is the case. - -REPLY [7 votes]: Another way to do it is to compute explicitly the curve above the point at infinity, just by looking at the equation. The argument works over any field, we can of course assume that this one is algebraically closed. -The curve $y^d=\prod_{i=1}^n (x-t_i)^{m_i}$ is affine, and covers the affine line which coordinate is $x$. If you want to see what happens at infinity, you just change coordinates on $\mathbb{P}^1$, which corresponds to take affine coordinate $X=1/x$. -Assume that each $t_i$ is not equal to zero (otherwise, make an affine change of coordinates at begin). We then have -$\prod_{i=1}^n (x-t_i)^{m_i}=\frac{\prod_{i=1}^n (1/t_i-1/x)^{m_i}}{\prod_{i=1}^n (xt_i)^{m_i}}$. -(a) If $\sum m_i$ is a multiple of $d$, you write $\sum m_i=ad$ and make a change of variables $Y=yx^a$ and get -$(\prod_{i=1}^n t_i^{m_i} )Y^d=\prod_{i=1}^n (1/t_i-X)^{m_i}$ -In this case, the curve is not ramified at infinity, which is $X=0$. -(b) If $\sum m_i$ is not a multiple of $d$, you write $\sum m_i=ad-k$ where $1\le k< d$ and make a change of variables $Y=yx^a$ and get -$(\prod_{i=1}^n t_i^{m_i} )Y^d=X^k\prod_{i=1}^n (1/t_i-X)^{m_i}$ -In this case, the curve is ramified at infinity.<|endoftext|> -TITLE: "as close to being semisimple as it can possibly be." -QUESTION [5 upvotes]: I had originally asked this question on math stack exchange but I think maybe it's more appropriate to ask it here. -In the paper of Beilinson, Ginzburg and Soergel entitled "Koszul Duality Patterns..." After stating the following proposition, they make the statement: " a Koszul ring is a positively graded ring that is "as close to being semisimple as it can possibly be."" -Proposition: Let $A = \bigoplus_{j\geq 0} A_j$ be a positively graded ring and suppose that $A_0$ is semisimple. The following are equivalent. -1) $A$ is Koszul -2) For any two pure $A$-modules $M$, $N$ of weights $m$, $n$ respectively we have $ext^i_A (M,N) = 0$ unless $i = m-n$. -3) $ext_A ^i (A_0, A_0 \langle n \rangle) = 0$ unless $i=n$. -To clarify some things: A graded module $M$ over a graded ring, is pure of weight $m$ iff $M = M_{-m}$. Also the $ext$'s we are using are in the category of graded modules. -I am very new to all this and I don't even have a vague idea of what they mean. Any thoughts at all would be super helpful. - -REPLY [4 votes]: In the semisimple case it is really easy to calculate $ext_A^i(M,N)$, $i\ge 1$, with the above assumptions. It is zero. For Koszul rings this is almost -true, i.e., $ext^i(M,N)$ is concentrated in degree $i$ (that is, if $S$ is the direct sum of all simples, then the two natural gradings on the algebra $A^{!}=ext^{\bullet}(S,S)$ coincide). It seems reasonable that the authors had this in mind as they wrote: "Morally a Koszul ring is a graded ring that is as close to semisimple as a $\mathbb{Z}$-graded ring possibly can be”. -This was already said in the comments. I can just add another reference here, a discussion on Koszul algebras and Koszul duality, which explains several things and refers also to the paper of Beilinson, Ginzburg and Soergel, here: http://sbseminar.wordpress.com/2007/11/01/koszul-algebras-and-koszul-duality/.<|endoftext|> -TITLE: Validity of functional derivative using the Dirac delta function -QUESTION [7 upvotes]: In physics, it's customary to compute the functional derivative as $$\frac{\delta F[\rho(x)]}{\delta \rho(y)}=\lim_{\varepsilon\to 0}\frac{F[\rho(x)+\varepsilon\delta(x-y)]-F[\rho(x)]}{\varepsilon}.$$ -The Wikipedia states that "This works in cases when $F[\rho(x)+\varepsilon f(x)]$ formally can be expanded as a series (or at least up to first order) in $\varepsilon$. ¿Where could I find a proof or a reference to this fact? - -REPLY [8 votes]: In many variational problems one is given an action functional $f\mapsto S[f]$, described by an integral -$$ S[f]=\int_\Omega L\bigl(\;x,f(x),D f(x),\dotsc, D^k f(x)\;\bigr) dx $$ -in which - -$\Omega$ is a region in some Euclidean space $\mathbb{R}^n$, $x\in\Omega$, -$f$ is a $k$-times differentiable function $f: \Omega\to\mathbb{R}^m$, and -the Lagrangian $L$ is a function of the appropriate number of variables. - -For example, in classical mechanics $n=1$, $\Omega\subset \mathbb{R}$ is an interval and the function -$$f:\Omega\to \mathbb{R}^m, \;\;\Omega\ni t\mapsto f(t)\in\mathbb{R}^m$$ -describes a path in $\mathbb{R}^m$. The Lagrangia has the form $L: \mathbb{R}^m\times\mathbb{R}^m\to \mathbb{R}$, $\newcommand{\bR}{\mathbb{R}}$ -$$L(y,v)=\frac{1}{2}|v|^2-U(y), \;\; (y,v)\in\bR^m\times\bR^m $$ -where the potential $U$ is a function $U:\bR^m\to\bR$. Then -$$ L(f,\dot{f})= \frac{1}{2}|\dot{f}|^2-U(f), $$ -where the dot indicates the derivative with respect to the time parameter $t$ on $\Omega\subset\mathbb{R}$. -The functional (or variational) derivative of $S$ calculated at $f_0:\Omega\to\mathbb{R}^m$ is a gadget $\delta S[f_0]$ that feeds on an infinitesimal deformation $\delta f$ of $f_0$ and returns a scalar -$$ \langle \delta S[f_0], \delta f\rangle =\lim_{h\to 0}\frac{1}{h} \bigl(S[f_0+h\delta f]-S[f_0]\;\bigr). \tag{1}$$ -The deformation $\delta f$ is also a function $\Omega\to\mathbb{R}^m$. It is often desirable to identify $\delta S[f_0]$ with a function $g:\Omega\to\mathbb{R}^m$ which, if it exists, is uniquely determined by the equality -$$ \langle \delta S[f_0], \delta f\rangle=\langle g(x), \delta f(x)\rangle =\int_\Omega \bigl( g(x), \delta f(x) \bigr) dx,\tag{2} $$ -where $(-,-)$ denotes the natural inner product on $\mathbb{R}^m$. The value of $g$ at $x_0$ can be obtained from the equality -$$ g(x_0)= \langle g(x), \delta(x-x_0)\rangle. \tag{3} $$ -This means that the value of $g$ at $x_0$ is obtained by formally replacing $\delta f$ with $\delta(y-x_0)$ in (2). -Making the same formal replacement $\delta f(x)\to\delta(x-x_0)$ in (1) one obtains the physicists' functional derivative in your question. -How does one identify $\delta S[f_0]$ with a function? In the example from classical mechanics one has -$$ S[f_0](t)=-\frac{d}{dt}\frac{\partial L}{\partial v}(f_0(t), \dot{f_0}(t))+\frac{\partial L}{\partial y}(f_0,\dot{f}_0). $$ -How can one see this? Fix a function (path) $f_0=f_0(t)$. For simplicity set $\alpha:=\delta f$ and assume $\Omega=[0,1]$. We have a Taylor approximation -$$ L(f_0+h \alpha, \dot{f}_0+h\dot{\alpha}) = L(f_0,\dot{f}_0)+ h\frac{\partial L}{\partial y}(f_0,\dot{f_0})\alpha+h\frac{\partial L}{\partial v}(f_0,\dot{f_0})\dot{\alpha} +O(h^2). $$ -Hence -$$\frac{1}{h} \bigl(\; S[f_0+h\alpha]-S[f_0]\;\bigr)=\int_0^1 \frac{\partial L}{\partial y}(f_0,\dot{f_0})\alpha+\frac{\partial L}{\partial v}(f_0,\dot{f_0})\dot \alpha dt +O(h). $$ -Letting $h\to 0$ we deduce -$$\langle \delta S[f_0],\alpha\rangle = \int_0^1\frac{\partial L}{\partial y}(f_0,\dot{f_0})\alpha+\frac{\partial L}{\partial v}(f_0,\dot{f_0})\dot{\alpha} dt. $$ -If we further assume that $\alpha(0)=\alpha(1)$ then upon integrating by parts we deduce -$$ \langle \delta S[f_0],\alpha\rangle=\int_0^1\Bigl(\frac{\partial L}{\partial y}(f_0,\dot{f_0}) -\frac{d}{dt}\frac{\partial L}{\partial v}(f_0,\dot{f_0})\Bigr) \alpha dt. $$ -A good place to look for more details is the book "Calculus of Variations" by Gelfand and Fomin, Dover 2000.<|endoftext|> -TITLE: Existence property for ordered fields -QUESTION [14 upvotes]: A theory $T$ has the existence property (EP) if the following holds: -Let $\phi(x)$ be a formula with one free variable (and no parameters) such that $T \vdash (\exists x) \phi(x)$. Then there is another formula $\psi(x)$ (again no parameters) such that $T \vdash (\exists ! x)\psi(x)$ (ie there is a unique $x$ such that $\psi(x)$) and $T \vdash(\forall x)\psi(x) \rightarrow \phi(x)$. -For example, $T := \operatorname{Th}(\langle \mathbb{C}, 0, 1, +, \times \rangle)$ does not have EP, because if one takes $\phi(x)$ to be $x^2 + 1 = 0$, then neither of the two solutions $\pm i$ is fixed by the automorphism $z \mapsto \bar{z}$. -The question I would like to ask is: Does the theory of ordered fields have the existence property? -I can see that $\operatorname{Th}(\langle \mathbb{R}, 0, 1, +, \times, < \rangle)$ does have EP. Since the theory is o-minimal, $\{x | \phi(x)\}$ is either finite, in which case it has a greatest element, or it contains an interval, in which case there is a rational, $q$ such that $\phi(q)$. Also, one can show that the theory of ordered fields with intuitionistic logic has EP using Kripke models. (Incidentally if that last remark is already known, I would be grateful if someone can provide a reference for it). However, I can't see how to adjust either proof to work with (classical) ordered fields in general. - -REPLY [8 votes]: The answer is negative. -For any model $M$, let $M_d$ denote the submodel of its parameter-free definable elements. We have the following characterization for classical theories. -Lemma: $T$ has EP iff $M_d\preceq M$ for every $M\models T$. -Proof: $\leftarrow$ is left as an exercise. $\to$: By Tarski’s test, it suffices to show that whenever $M\models\exists x\,\phi(x,a_1,\dots,a_n)$ for some $a_1,\dots,a_n\in M_d$, there is $b\in M_d$ such that $M\models\phi(b,a_1,\dots,a_n)$. Since we can plug in the definition of each $a_i$, it suffices to show this for $n=0$. We have $T\vdash\exists x\,(\exists y\,\phi(y)\to\phi(x))$, hence by EP, there exists $\psi(x)$ such that $T\vdash\exists!x\,\psi(x)$, and $T\vdash\exists y\,\phi(y)\to\forall x\,(\psi(x)\to\phi(x))$. Thus, if $b$ is the element defined by $\psi(x)$ in $M$, we have $M\models\phi(b)$. QED -So, the question is whether every ordered field is an elementary extension of its subfield of definable elements. Here is an easy counterexample. -Example: Let $F$ be the rational function field $\mathbb Q(x)$, ordered so that $x>q$ for every $q\in\mathbb Q$. - -$F_d=\mathbb Q$. -$F$ is not elementarily equivalent to $\mathbb Q$. - -Proof: - -For any $q\in\mathbb Q$, there is a unique automorphism $\sigma_q$ of $F$ such that $\sigma_q(x)=x+q$, specifically $\sigma_q(f(x)/g(x))=f(x+q)/g(x+q)$. Definable elements are fixed by every automorphism, and it is easy to check that a rational function $f(x)/g(x)$ is fixed by $\sigma_q$ for $q\ne0$ only when it is a constant. -In $\mathbb Q$, every positive element is the sum of four squares. This sentence does not hold in $F$. In particular, $x$ is not a sum of four (or more) squares, as we can make $\mathbb Q(x)$ into an ordered field in a different way so that $x$ becomes negative. - -EDIT: Since this may be lost in the argument, let me state explicitly that an example of a formula $\phi(x)$ which violates EP in the theory of ordered fields is -\begin{multline}\forall y\,(y>0\to\exists z_1,z_2,z_3,z_4\,z_1^2+z_2^2+z_3^2+z_4^2=y)\\\lor (x>0\land\forall z_1,z_2,z_3,z_4\,z_1^2+z_2^2+z_3^2+z_4^2\ne x).\end{multline}<|endoftext|> -TITLE: Idempotent ultrafilters and the Rudin-Keisler ordering -QUESTION [13 upvotes]: Short version: what can we say about the place of idempotent ultrafilters in the Rudin-Keisler ordering? -Longer version: -If $U$, $V$ are (nonprincipal) ultrafilters on $\omega$, then we write $U\ge_{RK}V$ in case there is some function $f:\omega\rightarrow\omega$ such that $$ \forall X\subseteq\omega,\quad f^{-1}(X)\in U\iff X\in V.$$ -An ultrafilter $U$ is Ramsey if given any two-coloring of pairs $c: [\omega]^2\rightarrow 2$, there is a homogeneous set for $c$ in $U$. Equivalently, $U$ is Ramsey if whenever $\lbrace C_n\rbrace_{n\in\omega}$ is a partition of $\omega$ with each $C_n\not\in U$, there is some $H\in U$ such that for all $n\in\omega$, $$\vert H\cap C_n\vert=1.$$ An ultrafilter $U$ is idempotent if $U\oplus U=U$, where $$ V\oplus W=\lbrace X: \lbrace x: \lbrace y: x+y\in X \rbrace \in W \rbrace \in V\rbrace.$$ Idempotent ultrafilters can be proved to exist in $ZFC$: this amounts to showing that $\oplus$ is left-continuous and associative on the compact space $\beta\mathbb{N}$, and then applying Ellis' theorem that every left-continuous semigroup on a compact space has an idempotent element. By contrast, Ramsey ultrafilters cannot be shown to exist in $ZFC$, although their existence is equiconsistent with $ZFC$ (in particular, if the Continuum Hypothesis - or weaker statements - holds, then there are Ramsey ultrafilters). -Now, an easy argument shows that no Ramsey ultrafilter is idempotent. On the other hand, the Ramsey ultrafilters enjoy a special property with respect to the RK-ordering: they are precisely the RK-minimal ultrafilters. So combining these facts shows that no idempotent ultrafilter can be RK-minimal. -My question is, what else can be said about the idempotent ultrafilters in terms of RK-reducibility? For example, can we have an idempotent ultrafilter U with exactly one RK-class of (necessarily, Ramsey) ultrafilters strictly RK-below U? This seems clearly impossible, but I don't see how to prove it. - -Motivation: a few weeks ago, I taught a one-week course on the proof of Hindman's theorem from additive combinatorics using idempotent ultrafilters. On the last day, I talked a bit about other types of ultrafilters, and spent a bit of time defining Ramsey ultrafilters and explaining (not proving) their place in the RK-ordering. One of my students asked whether anything similar could be said about idempotent ultrafilters; besides the obvious, I couldn't come up with anything, so I'm asking here. - -REPLY [12 votes]: The idempotent ultrafilters closest to being Ramsey are the stable ordered-union ultrafilters. These are officially defined as certain ultrafilters on the set $\mathbb F$ of finite subsets of $\omega$, but they can be transferred to $\omega$ via the "binary expansion" map $\mathbb F\to\omega: s\mapsto\sum_{n\in s}2^n$. The image on $\omega$ of a stable ordered-union ultrafilter is idempotent and has has exactly three non-isomorphic non-principal ultrafilters RK-below it. Two of these are the ones Todd Eisworth mentioned in his comment; the third is the "pairing" of these two, i.e, the image of the idempotent under the map to $\omega^2$ given by $n\mapsto($position of leftmost $1,\,$position of rightmost $1)$. For details, see my paper "Ultrafilters related to Hindman's finite-unions theorem and its extensions" ["Logic and Combinatorics", Contemporary Math 65 (1987) 89-124] also available at http://www.math.lsa.umich.edu/~ablass/uf-hindman.pdf (this is a scanned picture and therefore not searchable; the stable ordered-union stuff starts on page 113).<|endoftext|> -TITLE: A property of the unit circle -QUESTION [28 upvotes]: Let $(X,d)$ be a compact connected metric space with the property that for any distinct points $a,b$, $X\backslash \lbrace a,b\rbrace$ is disconnected. Clearly the unit circle has this property. Is there any other example (up to isomorphism) ?? - -REPLY [23 votes]: This property indeed characterizes the circle, but this is not obvious. -This was shown by R. L. Moore, according to Sam Nadler's Continuum Theory p. 156. -Added: the precise reference is [522] in this historical survey of continuum theory. - -REPLY [14 votes]: Yes and this is a nontrivial result, see e.g. J. G. Hocking, G. S. Young, "Topology", page 55, theorem 2-28. See also their theorem 2-27 for topological characterization of the interval.<|endoftext|> -TITLE: Transfinite recursion, collection and replacement in KP and KF -QUESTION [5 upvotes]: Kripke Platek set theory has collection instead of replacement, and it is a weakening of KP if one has replacement instead of collection. Call KP minus collection plus replacement KF for Kripke Fraenkel. Is KF weaker than KP in that some transfinite recursion can be done by KP which cannot be done by KF? If so, is that a difference inherited by strengthened theories as $\Sigma _{n} KP$ and $\Sigma _{n} KF$ for $n>1$? - -REPLY [8 votes]: In our paper, V. Gitman, J. D. Hamkins, T.A. Johnstone, What is the theory ZFC without powerset?, which arose also in one of your previous questions, we prove that there is a model of ZFC-powerset using only replacement and not collection (and so it satisfies your theory KF and indeed $\Sigma_n$-KF for every $n$), which has sets of reals of size $\aleph_n$ for every finite $n$, yet no set of reals of size $\aleph_\omega$, even though this model has $\aleph_\omega$ and indeed satisfies Hartog's theorem "$\forall\alpha\aleph_\alpha$ exists". Thus, in this model, which has the axiom of choice, there is no largest well-ordered set of real numbers. -One can view this as a failure of transfinite recursion. Specifically, KP proves that if -if every set of reals is well-orderable, and for every $n$ there is a set of reals of size $\aleph_n$, then there is a set of reals of size $\aleph_\omega$. One simply collects the smaller sets, takes the union, and well-orders it. But KF and indeed $\bigcup_n\Sigma_n$-KF does not prove this, because of our model. -But perhaps you had a more specific kind or instance of recursion in mind? If so, please describe in more detail what forms of recursion you are considering. - -REPLY [6 votes]: The main deficit of KF over KP is that one cannot prove that the set of formulae equivalent to a $\Sigma_1$ or $\Pi_1$ formula is closed under bounded quantification. (And this persists up through the $KF_n/KP_n$ you define.) -Therefore recursions involving schemes defined by a formula of the form $\forall u \in v \varphi$ cannot in general be effected in KF. -At the level n=1: Mathias in: ``Weak Systems of Gandy, Jensen and Devlin'', (Trends in Mathematics, Ed. Bagaria, Todorcevic, Birkhäuser Press) states that methods of Zarach should be sufficient to give an example showing that $\Delta_0$-collection cannot be substituted -by $\Delta_0$-replacement, but does not give an example (and I don't have one). -At levels > 1: See Gitman, Johnstone, Hamkins ``What is the theory ZFC without Power Set". -(Archive for Math.Logic, 2012) where they give examples (some derived from Zarach) of failures of even low levels of $\Sigma_n$-Collection even in models of full $\Sigma_\omega$-Replacement. They observe that one can inductively define in one such model a family of bijective functions $f:\omega \longrightarrow \aleph_k$ for $k\leq n$, but not altogether for all $n<\omega$. Thus strengthening the replacement scheme does not help.<|endoftext|> -TITLE: What is known about the spectrum of a Cauchy matrix? -QUESTION [5 upvotes]: Math people: -A Cauchy matrix is an $m$-by-$n$ matrix $A$ whose elements have the form -$a_{i,j} = \frac{1}{x_i-y_j}$, with $x_i \neq y_j$ for all $(i, j)$, and the $x_i$'s and $y_i$'s belong to a field (http://en.wikipedia.org/wiki/Cauchy_matrix). Also it seems to be part of the definition that the $x_i$'s and $y_j$'s are all distinct (does anyone know why?). I am only interested in the case where the field is the real numbers, and all the $x_i$'s and $y_j$'s are positive integers. My question is, what is known about the eigenvalues of a square, real Cauchy matrix? There is a formula for its determinant, which gives you their product, and the trace of the matrix, which is their sum, is easy to find. I have Googled this extensively and found almost nothing. -I originally posted this on Math Stack Exchange but I got no answers so I removed the question and I am posting it here. - -REPLY [3 votes]: Suppose $x_i > 0$ and $y_j -x_j$, then $c_{ij} = 1/(x_i+x_j)$. These matrices are infinitely divisible, i.e., $[c_{ij}^r]$ is also positive definite for all $r > 0$. -Spectral properties of Cauchy-like matrices and kernels are studied here. -For additional information and an easier read, on the general case ($1/(x_i+y_j)$), you might enjoy looking at the recent book by Pinkus.<|endoftext|> -TITLE: Idempotent polynomials -QUESTION [11 upvotes]: Let $R$ be a commutative ring with identity and let $f \in R[x]$. There are well known characterizations for $f$ to be a nilpotent element of $R[x]$ or to have a multiplicative inverse in $R[x]$. Is there any characterization for idempotent elements in $R[x]$ ? - -REPLY [4 votes]: Based on @user30230's great answer, but avoiding induction: Let $f=a+gx$, with $a\in R$ and $g\in R[x]$. Then $f=f^2$ yields $a+gx=a^2+2agx+g^2x^2$, so $a=a^2$ and $(1-2a)gx=g^2x^2$. Since $(1-2a)^2=1$, then $(1-2a)gx=\bigl[(1-2a)gx\bigr]^2$, so $(1-2a)gx=\bigl[(1-2a)g\bigr]^nx^n$ for all $n\geq1$. Thus, every power of $x$ divides $(1-2a)gx$, which forces $(1-2a)gx=0$. As $(1-2a)x$ is regular, it follows that $g=0$. Note that the argument also works for formal power series.<|endoftext|> -TITLE: Finding the vertices of a convex polyhedron from a set of planes -QUESTION [5 upvotes]: I'm new to computational geometry and advanced mathematics in general here so bear with me. I've spent a decent amount of time attempting to figure out this problem and I just can't find a solution. -My problem is to find the vertices that make up each face of a convex polyhedron. At my disposal are a set of planes, with each plane corresponding to a face. Each plane is derived from exactly three points that I know already exist on the corresponding face. -How do I solve this problem in an efficient manner? Apologies if this isn't in the right section, since it involves both math and computer science. - -REPLY [5 votes]: For the description of a pretty good algorithm see Avis and Fukuda. For an efficient implementation (in any dimension), and much additional discussion and references, see Komei Fukuda's page.<|endoftext|> -TITLE: Sets which are not fixed by any non-identity isomorphism -QUESTION [8 upvotes]: Consider a finite dimensional vector space $V$ over a field (finite or infinite but big enough). I am looking for a subset $W$ of $V$ such that for any bijective but non-identity linear map $T: V \longrightarrow V$, $T(W) \not = W$. Is it always possible to find such a set ? - -REPLY [2 votes]: Here is a method that works for all fields with more than $2$ elements. -I will denote the underlying field by $F$ and assume that $|F|>2$. -The idea is to use an induction to construct a solution -for $n$-dimensional spaces. I will consider $F^n$ as a linear subspace of -$F^{n+1}$ by using the embedding $(x_1,\dots,x_n) \mapsto (x_1,\dots,x_n,0)$. -Assume that we have a subspace $W_n$ of $F^n$ that is a solution, and -consider the subspace $W_{n+1}$ of $F^{n+1}$ defined as the union of $W_n$ -with the set of all vectors $(x_1,\dots,x_n,1)$ with $(x_1,\dots,x_n) \neq 0$. -I contend that, unless $n=1$ and $|F|=3$, the subset $W_{n+1}$ is a solution for the -space $F^{n+1}$. Indeed, let $f$ be an automorphism of $F^{n+1}$ that leaves $W_{n+1}$ -invariant. Unless $n=1$ and $|F|=3$, one checks that the affine hyperplane $\mathcal{H}$ with equation $x_n=1$ is the sole non-linear affine hyperplane of $V$ such that -$\mathcal{H} \setminus W$ contains at most one point. Thus, $\mathcal{H}$ is stable under $f$. It follows that $f$ fixes $(0,\dots,0,1)$, as it is the only point of $\mathcal{H} \setminus W$, and on the other hand the translation vector space $H$ of $\mathcal{H}$ - associated with the equation $x_n=0$ - is stable under $f$. By induction, $f$ is the identity on $H$. As $f$ is linear, one concludes that it is the identity. -If $|F|>3$, this method yields a solution in every situation as one can start from -$W_1:=\{1\}$. If $|F|=3$, there is no solution for $n=2$ as was noted by Harry Altman, -but there is one for $n=3$: one sets $\mathcal{H}:=\{(x_1,x_2,1) \mid (x_1,x_2) \in F^2\}$ and - $W_3=\{(1,0,0)\} \cup (\mathcal{H} \setminus \bigl\{(0,0,1),(1,0,1),(0,1,1)\}\bigr\}$. -Then, one can check that $W_3$ is a solution for $F^3$: -indeed, one sees that $\mathcal{H}$ -is the sole affine hyperplane of $F^3$ such that $|\mathcal{H} \setminus W_3|=3$. -Let $f$ be an automorphism of $F^3$ that leaves $W_3$ invariant. Then, $f$ -also leaves $\mathcal{H}$ invariant, and it must induce an affine automorphism of -it that permutes the three points $(0,0,1)$, $(1,0,1)$ and $(0,1,1)$, -and it must induce an automorphism of $F^2$ that fixes $(1,0,0)$ (as $F^2$ -is the translation vector space of $\mathcal{H}$). -Note that $f$ must fix $(0,0,1)+(1,0,1)+(0,1,1)=(1,1,0)$, -whence $f$ induces the identity on $F^2$. It follows that the restriction of $f$ -to $\mathcal{H}$ is a translation, and from there it is easily checked that -this translation is the identity by using the fact that $f$ leaves $\{(0,0,1),(1,0,1),(0,1,1)\}$ invariant. Thus, $f$ is the identity and $W_3$ is a solution, as claimed. -With the above method, one obtains a solution for all $n\geq 3$ if $|F|=3$.<|endoftext|> -TITLE: Generalization of analytic functors -QUESTION [6 upvotes]: It's been a long time since I posted the following question on stackexchange. -Now I think it's better to adress it to you, in the hope I will reach the right audience: Martin's comment, albeit useful, didn't help me to satisfy my curiosity. -=== -A functor $F\colon \bf Sets\to Sets$ is said to be analytic if it results from the left Kan extension of a functor $f\colon \mathbf{Bij}(\mathbb N)\to \bf Sets$ (the "species" of the functors $F$) along the natural inclusion $\mathbf{Bij}(\mathbb N)\to \bf Sets$, where $\mathbf{Bij}(\mathbb N)$ is the category having objects natural numbers and where $\mathbf{Bij}(\mathbb N)(m,n)$ are the bijective functions $\{1,\dots,m\}\to \{1,\dots,n\}$ (empty if $n\neq m$). Representing a left Kan extension as a coend it means that -$$ -F(T)\cong \int^n T^n\times f(n) -$$ -(the most of you will recognize the fact that a functor is "anaytic" if it can be written in Taylor form, and the coend is in a suitable sense exactly that Taylor series) This can be expressed replacing $\bf Sets$ with any symmetric monoidally cocomplete category: there is a functor $\mathbf{Bij}(\mathbb N)^\text{op}\times \mathcal V\to \mathcal V\colon (n,V)\mapsto V\otimes\dots\otimes V=V^{\otimes n}$, which allows to define -$$ -\int^n V^{\otimes n}\otimes f(n) -$$ -for any "species" $f\colon \mathbf{Bij}(\mathbb N)\to \mathcal V$. -In the case of $\bf Sets$, it seems to be possible to extend the definition of an analytic functor to the case of an arbitrary cardinal $\kappa$: - -the category $\mathbf{Bij}(\kappa)$ is defined to be the category having objects cardinals $<\kappa$ and $\mathbf{Bij}(\kappa)(\mu,\nu)$ bijections between those cardinals (hence empty if $\mu\neq\nu$). -I say that a functor $F\colon\bf Sets\to Sets$ is $\kappa$-analytic if it results from $\text{Lan}_jf$ in the diagram -$$ -\begin{array}{ccc} -\mathbf{Bij}(\kappa) &\to& \bf Sets \\ -\downarrow&\nearrow_F&\\ -\bf Sets & -\end{array} -$$ -$F(T)\cong \int^{\mu}T^\mu\times f(\mu)$ where $T^\mu$ is the product of $\mu$ copies of $T$, in the obvious sense. - -Now my question: can I define something analogous in the case of a generic monoidal(ly cocomplete) category, maybe imposing some additional property? In the end all boils down to the possibility of defining a $\kappa$-fold tensor product $\otimes\colon \mathcal V^\kappa\to \mathcal V$, for any cardinal $\kappa$. Modules over a ring $R$ seem to have this "extension property", together with sets. -Being these two very explicit and natural examples of a monoidal category where "tensoring has an arbitrarily large arity", I think the problem has its own interest, and I would be surprised if the topic was new. -=== -As I pointed out in the comments on math.SE, I was motivated by Martin's topic about k-ary tensors in $Mod_R$ to pose the question in the form of a "k-ary tensor product". I think there is no hope to recover "usual properties"; it's better to think about an axiomatization of the structure I need, but for the moment I don't have the slightest idea. I repeat myself: on the one side I find quite astounding the lack of a theory of those monoidal categories where tensoring has an arbitrarily large arity, and on the other, it's obviously due to the absence of "good" properties for such a structure. - -REPLY [5 votes]: Not sure this will meet your expectations, but maybe Mark Weber's work on functors with arities is relevant. See Remark 2.12 therein, in particular. -The upshot, if I'm correct, is that any analytic endofunctor $F$ obtained from $f$ as above may be recovered from a polynomial diagram along the lines of -$$1 \leftarrow D \to \int^{n} f(n) \to 1,$$ -where $1 \leftarrow D \to \int^{n} f(n)$ is the comma cone from -the Yoneda embedding $1 \to Sets$ to the functor $\int^{n} f(n) \to Sets$ mapping $(n,x)$ to the finite ordinal $n$. -The functor $Sets \to Sets$ associated to such a diagram takes any set $X$, viewed as a functor $1^{op} \to Sets$, and - - restricts it along (the opposite of) $1 \leftarrow D$, then - right extends it along (...) $D \to \int^{n} f(n)$, and finally - left extends it along (...) $\int^{n} f(n) \to 1.$ - -In our case, this should map $X$ to $\int^p X^p \times {f(p)}$, as desired. -So we obtain $F$ as a so-called parametric right adjoint functor (I think Mark now calls these local right adjoints). There is an even more general notion of functor with arities, defined in the particular case of monads in the above paper, and made more explicit in Berger, Melliès, and Weber.<|endoftext|> -TITLE: Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem? -QUESTION [12 upvotes]: This post is an appendix of this one. -Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. -Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace? -Hypothesis : The ISP admits a negative answer, i.e., there are ISP counter-examples. -Definition : A category $\mathcal{S}$ of operator algebras see the ISP if $ \forall T, T' \in B(H)$ with $\mathcal{S}(T) \simeq \mathcal{S}(T')$: -$$ -T \text{ is an ISP counter-example} \Leftrightarrow T' \text{ is an ISP counter-example } -$$ -Proposition: The category $W^{*}$ of von Neumann algebras, doesn't see the ISP. -proof: Under the previous hypothesis, let $T \in B(H)$ be an ISP counter-example. Then $T$ is irreducible, i.e., $W^{*}(T) = B(H)$. But there are many irreducible operators checking the ISP, for example, the unilateral shift $S$. So $W^{*}(T) \simeq W^{*}(S)$, $S$ checks the ISP and $T$ not. $\square$ -This post asks about an equivalent result for the category of $C^{*}$-algebras : - -Is there a proof that the category of $C^{*}$-algebras doesn't see the - ISP ? - -REPLY [17 votes]: C*-algebras don't see the ISP. The operators $T\in B(H)$ and $T\oplus T\in B(H\oplus H)$ -generate isomorphic C*-algebras, but the latter clearly has non-trivial invariant subspaces. -To have both operators in the same Hilbert space, pick isometries $v_1,v_2\in B(H)$ with orthogonal ranges that add up to $H$. Then -$$ -T\mapsto v_1Tv_1^*+v_2Tv_2^* -$$ -is an injective *-endomorphism of $B(H)$ that maps $T$ to an operator with non-trivial invariant subspaces.<|endoftext|> -TITLE: subconvexity problem for $GL(3) × GL(2)$ $L$-function without involving in symmetric lift -QUESTION [6 upvotes]: A question in study of subconvexity topic puzzles me for a long time, which mabe a stupid question for many experts. I really wish someone to help me out, and any advice will be highly appreciated. -Let g a Hecke-Maass form for SL3(Z) which do not come from a symmetric square lift, and f be a Hecke-Maass cusp form for SL2(Z) of level $q$. The Rankin-Selberg L-function is defined by $$L( s,g\times f)=\sum_{m,n\ge 1}\frac{A(m,n)a(n)}{(m^2n)^s}.$$ -My question is - how to prove subconvexity bound on level aspect - -$$L(\frac{1}{2},g\times f)\ll q^{3/4-\epsilon},\quad \text{some constant }\epsilon>0 \hskip2em ?$$ And who has studied it? please show me their works. -Remark: Suppose $f_1,f_2$ be Hecke-Maass cusp forms on SL2(Z). In many papers, the upperbound for $L(\frac{1}{2},\text{sym}^2(f_1)\times f_2)$ has greatly improved. However there was few literature involving in L-function with general GL(3) Hecke-Maass form twisted by a GL(2) cusp form, i.e. $L(\frac{1}{2},g\times f)$. -So far I know that Rizwanur Khan ( link his paper here) prove a conditional result, he proved that suppose $f$ be holomorphic, and $\sum_{nq^{\frac{1}{4}+\frac{1}{2001}}$, then $L(\frac{1}{2},g\times f)\ll q^{3/4-1/2001}.$ If there are other literature studying $L(\frac{1}{2},g\times f)$, please guide me their names or papers. -Another stupid little question is let $g$ be a a Hecke-Maass form for SL3(Z) which is self-dual, I'm not sure whether $g$ must come from a symmetric square lift or not? - -REPLY [5 votes]: I think your question is an unsolved problem (I suspect Khan's conditional result can be extended to Maass forms, and it would be interesting to do so). Even the GL(3)xGL(1) version of it (when $f$ is replaced by a Dirichlet character) is still unsolved, see the works of Munshi. -To your second question I recommend this paper by Ramakrishnan. It proves that on GL(3) any self-dual cuspidal representation $\Pi$ is the adjoint square lift of a non-dihedral cuspidal representation $\pi$ on GL(2) twisted by a quadratic character (which is the central character of $\Pi$). Moreover, the result in the paper implies that for $\Pi$ unramified everywhere, one can take $\pi$ to be unramified everywhere, while the quadratic character is clearly also unramified everywhere (i.e. it corresponds to an ideal class character). In particular, the self-dual cusp forms on SL(3,Z) are precisely the adjoint square lifts of non-dihedral cusp forms on SL(2,Z). Note that in this last case the adjoint square lift is identical to the symmetric square lift.<|endoftext|> -TITLE: (A question about)${}^3$ 3-dimensional convex bodies -QUESTION [15 upvotes]: Related to the questions mathoverflow.net question No. 137850 and mathoverflow.net question No. 39127, is there a 3-dimensional convex body other than a ball whose perpendicular projections in all directions are of the same area? - -REPLY [15 votes]: A convex body $K\subset\mathbb{R}^n$ all of whose $(n-1)$-dimensional projections have the same $n-1$-content is known as a body of constant brightness, by analogy with bodies of constant width. The theory is very similar to that of bodies of constant width. The surface area measure $dS_K(\mathbf{x})$ takes the place of the support function $h_K$. The brightness in the direction $\mathbf{u}$ is given by $V(K_\mathbf{u}) = \tfrac{1}{2}\int |\langle\mathbf{x},\mathbf{u}\rangle| dS_K(\mathbf{x})$. If we expand $S_K$ in spherical harmonics $\sum_{l,m}s_{l,m}Y_{l,m}$, we get that $V(K_\mathbf{u}) = \sum_{l,m} c_l s_{l,m} Y_{l,m}(\mathbf{u})$, and $c_l=0$ whenever $l$ is odd. Therefore, $K$ has constant brightness if and only if $\frac{S_K(U)+S_K(-U)}{|U|}$ is constant over all Borel sets $U\subset S^{n-1}$ (that is, apart from a constant term, $S_K$ is antisymmetric). From the existence theorem for the Minkowski problem, we can easily construct examples.<|endoftext|> -TITLE: Are smooth functions tame? -QUESTION [14 upvotes]: I know the article of Hamilton on the inverse function theorem of Nash and Moser (with the same title) where he proves that $C^\infty(M)$ is a tame Fréchet space, when $M$ is closed or compact with boundary. -However, as far as I see no word is lost on the case of non-compact $M$. In particular, what if $M$ is an open subset of $\mathbb{R}^n$? Of course, we need to endow this with its Fréchet topology (uniform $C^m$ convergence on compact subsets). -I assume that this may not be a tame space, but I couldn't find a reference. At least it has a sequence of seminorms of increasing strength that induce the topology, by taking an exhaustion with compact sets and simultaneously letting the order of differentiation increase. - -REPLY [8 votes]: Kofi: "... what if $M$ is an open subset of $\mathbb R^n$?" -Kofi has already accepted Peter Michor's answer, but anyway I post the following to give a different argument for seeing that $C^\infty(\Omega)$ is not tame. Tameness in the sense of Hamilton implies the interpolation inequalities $$\|x\|_{i+j}\le C_{i,j+k}\|x\|_i^{\frac k{j+k}}\|x\|_{i+j+k}^{\frac j{j+k}}$$ for a taming sequence $\nu_i:x\mapsto\|x\|_i$ of seminorms. This in turn implies existence of a continuous norm on the space. Indeed, every $\nu_i$ is necessarily a norm, since otherwise there would exist a nonzero $x$ with $\|x\|_i=0$ for all $i$, and then the space would not be Hausdorff. Consequently, for $\Omega$ a (nonempty) open set in a Euclidean space, the space $C^\infty(\Omega)$ is not tame.<|endoftext|> -TITLE: the number of Sylow subgroups -QUESTION [5 upvotes]: Let $p>3$ be a prime number and $G$ be a finite group of order $2p(p^2+1)$. -Is it true that always the Sylow $p$- subgroup of $G$ is a normal subgroup of $G$? -As I checked by Gap it seems true. -Thanks for your answer. - -REPLY [14 votes]: As $p^2+1 \equiv 2 \bmod 4$, a Sylow $2$-subgroup of $G$ has order four. As $p^2+1 \equiv 2 \bmod 3$ and $p>3$, $G$ has no element of order three. It follows now that a Sylow $2$-subgroup $X$ of $G$ is central in its normalizer in $G$. By Burnside's normal $p$-complement theorem, $G$ contains a normal complement $N$ to $X$. It suffices to show that $N$ has a normal Sylow $p$-subgroup, as such a subgroup is characteristic in $N$ and thus normal in $G$. -Let $t$ be the number of Sylow $p$-subgroups of $N$. Then $t \equiv 1 \bmod p$ and $t$ divides $\frac{p^2+1}{2}$. Assume for contradiction that $t>1$, write $t=ap+1$ with $a>0$. Then $p^2+1=2b(ap+1)$ for some integer $b$. Now $2b \equiv 1 \bmod p$. So, $2b>p$ and $2b(ap+1)>p^2+p$, giving the desired contradiction.<|endoftext|> -TITLE: What are current trends/questions in algebraic logic? -QUESTION [7 upvotes]: What are current trends/questions in algebraic logic? I mean the research developed by Paul Halmos. -Could anyone give some references for the overview of its history? Any overview of its application to computer science and computability theory is welcome. - -REPLY [7 votes]: I ran across this today, and was reminded of this question: I think the article "Algebraic logic, Where does it stand today?" by T. S. Ahmed addresses your question.<|endoftext|> -TITLE: Is the Thompson group F locally indicable? -QUESTION [12 upvotes]: A group $G$ is called locally indicable if for any finitely generated subgroup $H \subset G$, there is a non-trivial homomorphism from $H$ to the real additive group $(\mathbb{R},+)$. -Is the Thompson group $F$ locally indicable? - -REPLY [19 votes]: Yes. Let $H$ be a finitely generated subgroup of $F$. Let $[0,a]$ be the largest interval where all elements of $H$ are equal to the identity. Then consider the map that sends $h\in H$ to the $\log_2$ of the (right) slope of $h$ at $a$. This map is a non-trivial homomorphism of $H$ into $\mathbb{Z}$. -A more fancy (but essentialy the same) proof is this: by Thurston's stability theorem (W. Thurston. A generalization of the Reeb stability theorem. Topology 13 (1974), 347- -352), every group of $C^1$ diffeomorphisms of the closed interval is locally indicable, and by Ghys and Sergiescu, $F$ is a subgroup of the group of $C^\infty$ diffeomorphisms of the interval $[0,1]$ (see E. Ghys and V. Sergiescu: Sur un groupe remarquable -de diffeomorphisms du cercle, Comm. Math. Helv. 62 -(1987) 185–239, an easy proof using diagram groups can be found in V.S. Guba and M. Sapir. Diagram groups are totally orderable. J. Pure Appl. Algebra, -205(1):48–73, 2006).<|endoftext|> -TITLE: Renaud-Sarvate limitation on Frankl's Conjecture -QUESTION [9 upvotes]: I was wondering if someone could give me a specific example of union closed family with a three element set, none of whose elements are in half the members of the family. -Thanks in advance! - -REPLY [8 votes]: This example with minimal size $3$ is from D. G. Sarvate and J-C. Renaud, "Improved bounds for the union-closed sets conjecture", Ars Combinatoria 29 (1990), pp. 181-185: -$\mathcal A=\{A_1,\dots,A_{27}\},$ with -$A_1=\{1,2,3\}$ -$A_2=\{1,2,3,6,7,8,9\}$ -$A_3=\{1,2,3,4,6,7,8,9\}$ -$A_4=\{1,2,3,4,5,6,7,8,9\}$ -$A_5=\{1,2,3,4,5,8,9\}$ -$A_6=\{1,2,3,4,5,6,8,9\}$ -$A_7=\{1,2,3,4,5,6,7\}$ -$A_8=\{1,2,3,4,5,6,7,8\}$ -$A_9=\{6,7,8,9\}$ -$A_{10}=\{4,6,7,8,9\}$ -$A_{11}=\{4,5,6,7,8,9\}$ -$A_{12}=\{4,5,8,9\}$ -$A_{13}=\{4,5,6,8,9\}$ -$A_{14}=\{4,5,6,7\}$ -$A_{15}=\{4,5,6,7,8\}$ -$A_{16}=\{1,6,7,8,9\}$ -$A_{17}=\{1,4,6,7,8,9\}$ -$A_{18}=\{1,4,5,6,7,8,9\}$ -$A_{19}=\{2,4,5,8,9\}$ -$A_{20}=\{2,4,5,6,8,9\}$ -$A_{21}=\{2,4,5,6,7,8,9\}$ -$A_{22}=\{3,4,5,6,7\}$ -$A_{23}=\{3,4,5,6,7,8\}$ -$A_{24}=\{3,4,5,6,7,8,9\}$ -$A_{25}=\{1,2,4,5,6,7,8,9\}$ -$A_{26}=\{1,3,4,5,6,7,8,9\}$ -$A_{27}=\{2,3,4,5,6,7,8,9\}$ -Here $A_1=\{1,2,3\}$ is the unique set of minimal size, and each of the elements $1,2,3$ is in exactly $13$ of the $27$ sets.<|endoftext|> -TITLE: Are most curves over Q pointless? -QUESTION [26 upvotes]: Fresh out of the arXiv press is the remarkable result of Manjul Bhargava saying that most hyperelliptic curves over $\mathbf{Q}$ have no rational points. Don Zagier suggests the paraphrase : Most hyperelliptic curves are pointless. -Crucial to the precise mathematical formulation of the statement is a kind of canonical equation for hyperelliptic curves (of a fixed genus) permitting one to define the density of those which have no rational points. -What is the corresponding statement for all curves over $\mathbf{Q}$ ? -Addendum (2013/09/28) A very nice introduction to the work of Bhargava can be found in How many rational points does a random curve have? by Wei Ho. - -REPLY [9 votes]: I suppose I would say this. Let h: M_g(Q) -> R be the height function corresponding to the Hodge class. Let S(N) be the set of points of M_g(Q) of height at most N. Finally, let P be the image of the projection from M_{g,1}(Q) to M_g(Q). (i.e. "the set of pointy curves.") Then one version of the assertion you're looking for would be: -$\lim_{N \rightarrow \infty} \frac{|S(N) \cap P|}{|S(N)|} = 0$. -But as Felipe says, it's possible that asymptotically 100% of the points in $S(N)$ are hyperelliptic, which makes this not very general at all. So you could go more hardcore and try this. -For all subvarieties X of M_g with infinitely many rational points, either: - -$X(\mathbf{Q}) \subset P$ (i.e. every curve parametrized by X has a point) - -or - -$\lim_{N \rightarrow \infty} \frac{|S(N) \cap P \cap X|}{|S(N) \cap X|} = 0$ - -That is, either every curve in the family X has a point, or almost none of them do. -Of course, I have no idea whether this is true -- just trying to write something down in the vein of your question which is not "secretly just about hyperelliptic curves."<|endoftext|> -TITLE: What does the moduli stack of G-torsors over the multiplicative group look like? -QUESTION [10 upvotes]: I am an algebraic topologist and am trying to understand some computations related to p-adic complex K-theory and equivariant K-theory. However this has led me into the world of algebraic geometry over $\mathbb{Z}_p$ and I need some help with some basic computations. -I would like to understand the moduli stack of G-torsors over the scheme underlying the multiplicative group $\mathbb{G}_m$ where we are working with schemes over the p-adic integers $\mathbb{Z}_p$ and $G$ is a finite group (which we also view as a group scheme over $\mathbb{Z}_p$). I would even be happy to know the cases where $G = \mathbb{Z}/p$ and $G= \mathbb{Z}/q$ with $q \neq p$, i.e., for cyclic groups. -To clarify, for a scheme S defined over the p-adic integers $\mathbb{Z}_p$, the S-points of the stack in question form a groupoid, the groupoid whose objects are S-families of G-torsors over the p-adic multiplicative group $\mathbb{G}_m$. In other words the objects are G-torsors over $$S \times_{spec \; \mathbb{Z}_p} \mathbb{G}_m.$$ -The morphisms are the obvious ones, and -everything is defined over the base ring $\mathbb{Z}_p$, which should probably be viewed as either a topolological ring or a pro-ring/ind-scheme. I am willing to be quite flexible on this point, as well as the particular topology used (Zariski, etale, flat, etc). -Sometimes it is useful to use the heuristic that $\mathbb{G}_m$ is like the circle $S^1 = \mathbb{R}/\mathbb{Z}$. For the circle the stack of G-torsors looks like the quotient stack $[G/G]$ for the conjugation action. Perhaps the $G$-torsors on $\mathbb{G}_m$ hava a similar description? A simple description of this type would be very useful to me, i.e., as a quotient stack or really any presentation at all. I am not very picky, but I need this for explicit calculations and so certainly need the description to be as explicit as possible. -A good start would be computing the groupoid of G-torsors over $\mathbb{G}_m$ (ignoring S-families) or even the set of isomorphism classes of $G$-torsors over $\mathbb{G}_m$. I tried to compute this by finding some nice cover of $\mathbb{G}_m$, but it is getting late here and I haven't found something that works yet. -I have tried googling around and have come up short. Most of the stuff I have found is either about $\mathbb{G}_m$-torsors, or about $G$-torsors for groups related to $\mathbb{G}_m$, or work which only works over a field of some sort or another. To clarify I am interested in $\mathbb{G}_m$ over $\mathbb{Z}_p$, so not over a field, in particular not over a field which has pth roots of unity. However I would be interested in the answer over those other rings in-so-far as they help get me the answer over $\mathbb{Z}_p$. -If this question is too basic, I would be happy even for a reference with some similar examples carefully worked out so I can learn explicitly how to do this kind of calculation. - -REPLY [2 votes]: I think if you look, you can find the answer in the literature. In the case that $G$ is Abelian and has order prime to $p$, this should follow from Mumford's theory of biextensions, "Cartier duality", Exercise 7.1 of Dolgachev's "Invariant Theory", etc.<|endoftext|> -TITLE: Motives over the complex numbers versus mixed Hodge structures -QUESTION [8 upvotes]: Let $\mathsf{MM}(\mathbf C)$ be the hypothetical category of mixed motives over the complex numbers, and consider the realization functor $\Phi : \mathsf{MM}( \mathbf C) \to \mathsf{MHS}$ to integral mixed Hodge structures. -Is $\Phi$ expected to be faithful? If not, what kind of information does one expect to lose? - -REPLY [10 votes]: Yes, this functor is expected to be faithful, but this has very little to do with Hodge theory. The reason for this is that, if $\Psi$ denotes the forgetful functor from integral mixed Hodge structures to abelian groups, then the composed functor $\Psi\circ\Phi$ simply is the Betti realization functor. As $\Psi$ itself is faithful, it is thus sufficient to prove that the Betti realization functor is faithful. The reason why the Betti realization functor is expected to be faithful is the following. The abelian category $\mathsf{MM}(\mathbf{C})$ is expected to be the heart of the motivic $t$-structure on the triangulated category $\mathsf{DM}_{et}(\mathbf{C})$ of Voevodsky's constructible (or geometric) étale motives (the one obtained from étale sheaves with transfers; the triangulated category of motives obtained from Nisnevich sheaves with transfers (which is known to coincide with $\mathsf{DM}_{et}(\mathbf{C})$ with rational coefficients) is known not to have a motivic $t$-structure with integral coefficients anyway). The faithfulness of the Betti realization functor as above boils down to the following two properties of the triangulated version of the Betti realization functor -$$R:\mathsf{DM}_{et}(\mathbf{C})\to\mathsf{D}^b(\mathit{Ab})$$ -1) it must be conservative; -2) it must be $t$-exact. -Note that the functor $R$ is actually well defined and that property 1) is equivalent to its version with rational coefficients (because, for torsion coefficients, the functor $R$ is known to be an equivalence, by the rigidity theorem of Suslin and Voevodsky). Property 2) is required in the very definition of a motivic $t$-structure. This essentially means that, for this problem, the main difficulty is about rational coefficients. This is then very related with the standard conjectures, as explained in this paper of Beilinson: arXiv:1006.1116. -Finally, it might be worth to mention that there is an actual candidate for $\mathsf{MM}(\mathbf{C})$, constructed by Madhav Nori, which, by definition, comes with an exact and faithful functor to $\mathsf{MHS}$.<|endoftext|> -TITLE: A problem about Ramsey Property -QUESTION [6 upvotes]: It is known that Ramsey property is a kind of generalizition of pigeon hole principle, and some kinds of Ramsey properties have lots of equivalent forms. -We often deal with the case $a\rightarrow (b)^r_c $,when $a,b,c$ are cardinals; however, if we consider the ordertype of the homogeneous set, the question becomes much more complicated. -For instance, we can prove that any two-coloring of $\omega_1$ must have a homogeneous subset of ordertype $\omega+1$, rather than only stating that can have a countable homogeneous subset. And also it's easy to show that for any countable ordinal $\alpha$, $\alpha\rightarrow (\omega+1)^2_2$ fails. -I wonder: what is the least ordinal $c$ such that any 2-coloring of $c$ must have a homogeneous set of ordertype $\omega+2$? And can one prove that for any ordinal $b$, the least ordinal $a$ with $a\rightarrow b^2_2$ must be a cardinal? -In this case, the properties of such $a$ is quite different from Erdős cardinal, any tricks used to show an Erdős cardinal must be inaccessible fail here. - -REPLY [6 votes]: For $b=\omega+2$, see - -András Hajnal. Some results and problems on set theory, Acta Math. Acad. Sci. Hungar., 11, (1960), 277–298. MR0150044 (27 #47). - -In this paper, András shows that $\omega_1\to(\omega\cdot n,\omega\cdot 2)^2_2$ for all $n<\omega$. This is best possible, in the sense that if $\alpha$ is countable, then $\alpha\not\to(\omega,\omega+1)^2_2$. -He also shows that $\mathsf{CH}$ implies that $\omega_1\not\to(\omega_1,\omega+2)^2_2$. I seem to remember Stevo saying that the same negative relation follows from adding a Cohen real, though I do not have details of the proof. -On the other hand, Stevo showed that the following statement $(1)$ is consistent: - $$\omega_1\to(\omega_1,\alpha)^2_2\mbox{ for all countable ordinals }\alpha $$ -(in fact, he established a significant strengthening). His argument uses proper forcing, and it follows that $\mathsf{PFA}$ implies (1) (though no large cardinals are needed to establish its consistency together with $\mathsf{MA}+2^{\aleph_0}=\aleph_2$). See - -Stevo Todorcevic. Forcing positive partition relations, Trans. Amer. Math. Soc., 280 (2), (1983), 703–720. MR0716846 (85d:03102). - -As Péter Komjáth pointed out in a comment, there is a generalization of András result I should mention: The Baumgartner-Hajnal theorem states that if $\alpha<\omega_1$, then - $$ \omega_1\to(\alpha)^2_2, $$ -provably in $\mathsf{ZFC}$. This solves your problem, as long as $b$ is countable. -In fact, Baumgartner-Hajnal admits a significant generalization that was pursued by several authors, most notably Fred Galvin and Stevo: Say that a poset $\phi$ is non-special iff $\phi\to(\omega)^1_\omega$. Any such poset satisfies $\phi\to(\alpha)^2_2$ for any countable $\alpha$, and this is best possible, since $\phi\to(\omega,\omega+1)^2_2$ already implies that $\phi$ is non-special. See - -James Baumgartner, and András Hajnal. A proof (involving Martin's axiom) of a partition relation, Fund. Math., 78 (3), (1973), 193–203. MR0319768 (47 #8310), - -and - -Stevo Todorcevic. Partition relations for partially ordered sets, Acta Math., 155 (1-2),(1985), 1–25. MR0793235 (87d:03126). - -For the general case, $b$ an arbitrary ordinal, not much seems known. There is of course the Erdős-Rado theorem, that gives in particular that if $\kappa$ is regular, then - $$ (2^{<\kappa})^+\to(\kappa+1)^2_\mu \mbox{ for all }\mu<\kappa, $$ -and this is best possible in that no ordinal smaller than $(2^{<\kappa})^+$ works here, even if $\mu=2$. -This has been somewhat extender by James Baumgartner, András, and Stevo. See - -James E. Baumgartner, András Hajnal, and Stevo Todorcevic. Extensions of the Erdős-Rado theorem. In Finite and infinite combinatorics in sets and logic (Banff, AB, 1991), pp. 1–17, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 411, Kluwer Acad. Publ., Dordrecht, 1993. MR1261193 (95c:03111). - -There, they show that if $\kappa$ is regular and uncountable, then forall $k<\omega$ and $\xi<\log\kappa$ (where $\log\kappa$ is the least cardinal $\mu$ such that $2^\mu\ge\kappa$), we have - $$ (2^{<\kappa})^+\to(\kappa+\xi)^2_k. $$ -They also show that, under the same assumption on $\kappa$, $(2^{<\kappa})^+\to(\rho,(\kappa+n)_k)^2$, where $\rho=\kappa^{\omega+2}+1$ (ordinal exponentiation) and, improving results of Saharon Shelah, (the only result here where the ordinal on the left is not a cardinal) for all $n<\omega$, - $$ (2^{<\kappa})^+·\omega\to(\kappa·n)^2_2. $$ -(On the other hand, I do not see any evidence that the ordinal on the left is least possible, or that a cardinal does not work.) -A good reference for the state of the art on partition calculus is - -András Hajnal, and Jean A. Larson. Partition relations. In Handbook of set theory. Vols. 1, 2, 3, Akihiro Kanamori, and Matthew Foreman, eds., pp. 129–213, Springer, Dordrecht, 2010. - -What follows comes from this paper, to which I refer for additional references (the numbers listed in brackets are as in their bibliography). In page 152, they write: - -It was already asked in the Erdős-Hajnal problem lists [12, 13] if the partition - relations $\omega_2\to(\alpha)^2_2$ were consistent for $\alpha<\omega_2$. Though there is nothing to refute such consistency, the results going in this direction are weak and rare. - -They add that Richard Laver [$40$] showed, under appropriate assumptions of large cardinal character, that $\omega_2\to(\omega_1\cdot2+1,\alpha)^2_2$ for all $\alpha<\omega_2$. In general, Richard showed that $\kappa^+\to(\kappa\cdot2+1,\alpha)^2_2$ for all $\alpha<\kappa^+$, provided that $\kappa$ carries a Laver ideal, that is, a non-trivial, $\kappa$-complete ideal such that given $\kappa^+$ sets not in the ideal, there are $\kappa^+$ of them, the intersection of any fewer than $\kappa$ of these also not in the ideal. -For $\omega_2$, the best result I recall is due to Matthew Foreman and András Hajnal. In [$20$] they show, under appropriate large cardinal assumptions, that $\omega_2\to(\omega_1^2+1,\alpha)^2_2$ for all $\alpha<\omega_2$. They also proved that if $\kappa$ is measurable, there is an ordinal $\Omega(\kappa)$ smaller than $\kappa^+$, but closed under of ordinal addition, multiplication, exponentiation, and taking fixed points of these operations, such that - $$ \kappa^+\to(\alpha)^2_m\mbox{ for all }\alpha<\Omega(\kappa).$$ -For the precise definition of $\Omega(\kappa)$, see Section 5 of the Hajnal-Larson paper. -Finally, Shelah [$59$] showed that if $\kappa$ is strongly compact, $\lambda>\kappa$ is a cardinal, and either $\lambda$ is regular or $\mathrm{cf}(\lambda)\ge\kappa$, then - $$ (2^{<\lambda})^+\to(\lambda+\zeta)^2_\theta\mbox{ for all }\zeta,\theta<\kappa. $$<|endoftext|> -TITLE: Software computation with arithmetic schemes -QUESTION [6 upvotes]: For rings such as $\mathbb{Z}[x,y]$ is there software to compute any of: -1.) The integral closure of $\mathbb{Z}[x,y]/(f)$. de Jong has a very general algorithm that works in this context (http://arxiv.org/abs/alg-geom/9704017) -2.) The radical of an ideal $I$. -3.) Minimal primes of an ideal/the primary decomposition. -4.) The dimension of $\mathbb{Z}[x,y]/I$ -5.) Any invariants of a singularity on such a scheme. -Over fields $\mathbb{Q}, \mathbb{F}_p$ I think Sage, Macaulay 2 etc., implement 1-4. Are 1-4 difficult to compute for such rings, or are there just no implementations yet? - -REPLY [2 votes]: To my knowledge, -2-4 are also implemented in Singular (http://www.singular.uni-kl.de/), in primdecint.lib or in kernel, but the Gröbner basis computation over integers and other algrorithms in Singular seems still buggy. -We are currently trying to pinpoint the bugs and fix them in the development version: https://github.com/Singular/Sources -Their bugtracker: www.singular.uni-kl.de:8002/trac/report/1?sort=ticket&asc=0&page=1 -Radical over integers is computed in Singular via radicalZ(I), dimension via dim(std(I)) and the primary decomposition via primdecZ(I). Sagemath has an interface to Singular 3.1.6 but not to the recent version. -Jack<|endoftext|> -TITLE: Transitive geodesics on closed surfaces of genus greater than one -QUESTION [9 upvotes]: A well-known result of Hedlund and Morse states that if a Riemannian metric on a closed surface of genus $g > 1$ has no conjugate points, then it carries transitive geodesics (i.e., geodesics whose velocity vectors are dense in the unit tangent bundle). -Are there Riemannian metrics on closed surfaces of genus $g > 1$ that do not carry a transitive geodesic and if so what is the weakest condition known under which the existence of transitive geodesics has been proved? -Addendum. As Misha remarks in his answer, it is easy to construct to metrics on any closed surface that do not carry a transitive geodesic. However, the metrics I'm interested in have the additional property that the lifted metric on the universal cover has no trapped geodesics. In other words, no geodesic stays forever in a compact subset of the open unit disc. - -REPLY [4 votes]: This is both: answer on the new version of the question and comment on comment of Andrey Gogolev, who asked whether one can make the question more complicated assuming - additionally that the set of non-periodic geodesics is dense. -Hier is an example that answers both: it is not much different from the answer of Andrey from his comment (and also from katz's answer). -Take a torus with periodic coordinates $x,y$ and - a liouville metric $(X(x) -Y(y)) (dx^2 + dy^2)$ on it (whose geodesic flow is integrable). Make two small holes on the torus near the points where $X$ has minimum and $Y$ has maximum and connect by a neck. For generic $X$ and $Y$ and generic neck the set of nonperiodic geodesic is dence but still it has a 'regular' region in the tangent space where no geodesic passing through at least one point of neck can come. The closure of a typical geodesic from the regular region is a torus in the tangent space which is projected to an annulus on the torus and after passing to the universal cover this annulus becomes an infinite unboundend band. -One more remark is that -by KAM theory, this examples survives if you slightly perturbe the metric.<|endoftext|> -TITLE: Is there an algebraically normal function from $\mathbb{Z}^{n}$ to $\{ 0 , 1\}$? -QUESTION [7 upvotes]: Definition: Let $h$ be a polynomial in $n$ variables, then : -$\gamma(h,r,R):=\{ v \in \mathbb{Z}^{n} : \vert h(v) \vert \leq r, \Vert v \Vert < R \}$ -Let $\omega : \mathbb{Z}^{n} \to \{ 0 , 1\}$ be a function. -Definition : $\omega$ is algebraically normal if for all $h$ polynomial and $\forall r \geq 0$ s. t. $\vert \gamma(h,r,\infty) \vert = \infty$ : -$$ \lim\limits_{R \to \infty} -\frac{\vert \omega^{-1}(0) \cap \gamma(h,r,R) \vert}{\vert \gamma(h,r,R) \vert} = \frac{1}{2} -$$ - -Is there such an algebraically normal function $\omega$, for all $n \geq 2$ ? - -Remark on groups : This question could be generalized by replacing $\mathbb{Z}^{n}$ by a countable subset of an Euclidean space $\mathbb{R}^{n}$ or an Hyperbolic space $\mathbb{H}^{n}$ (which are diffeomorphic), or also a separable Hilbert space $H$ (by taking $h$ a polynomial in finitely many variables). As an application, a coarsely embeddable finitely generated group $\Gamma$ (with its word metric) into $\mathbb{R}^{n}$, $\mathbb{H}^{n}$ or $H$, could be called algebraically normal if there is such an embedding, admitting an algebraically normal function $\omega$. -The question becomes : Is there an algebraically normal group ? - -REPLY [2 votes]: $\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I suspect this is false! At least, I'll show that a similar statement is false for $\ZZ^4$ and I'll sketch how I think a similar construction should work for $\ZZ^2$. -Fix a coloring $\omega : \ZZ^4 \to \{ 0, 1 \}$. Define a directed graph whose vertices are quadruples $(p,q,p',q') \in \ZZ_{\geq 0}^4$ obeying $p q' - q p' = \pm 1$, $p \leq p'$, $q \leq q'$ and where there is an edge $(p,q, p', q') \to (p', q', p'', q'')$ if there is a positive integer $a$ such that $(p'', q'') = a (p', q') + (p,q)$. -Case 1 There is some vertex $(p,q,p',q')$ all of whose descendents are of the opposite color from it. -Then, for $a > 0$, the sequence of vertices $(p', q', a p' + p, a q'+q)$ is monochromatic. We can describe points of the form $(p', q', a p' + p, a q'+q)$ as -$$\left\{ (w,x,y,z) : (w-p')^2 + (x-q')^2 + \left( p' (z-q) - q' (y-p) \right)^2 < 1 \right\}.$$ -(Since $\det \left( \begin{smallmatrix} p & p' \\ q & q' \end{smallmatrix} \right) = \pm 1$, we see that $GCD(p',q')=1$ so the condition that $a$ is integral comes for free.) So, if we following this inequality in the direction of large $z$, all the points have the same color. -Case 2 There is an infinite monochromatic path starting from $(1,0,0,1)$. -Let the vertices on this path be $(p_{n-1}, q_{n-1}, p_n, q_n)$ with -$$(p_{n+1}, q_{n+1}) = a_n (p_n, q_n) + (p_{n-1}, q_{n-1}).$$ -Let $\alpha$ be the value of the continued fraction $a_0+1/(a_1+1/(a_2+1/(\cdots)))$. Then $p_i/q_i$ are the convergents of $\alpha$. (A good reference for all the facts I am using about continued fractions is Chapter X in Hardy and Wright, Introduction to the Theory of Numbers.) -We now quote Theorem 172 from Hardy and Wright: - -Suppose that $\zeta$ is a real number $>1$, that $P$, $Q$, $R$, $S$ are integers with $Q>S>0$ and $PS-QR = \pm 1$, and $\alpha = (P \zeta+R)/(Q \zeta+S)$. Then $P/Q$ and $R/S$ are consecutive convergents of $\alpha$. Conversely, if $P/Q$ and $R/S$ are consecutive convergents in lowest terms, then there is a $\zeta$ such that these conditions hold. - -Although Hardy and Wright don't point it out, the condition $PS-QR = \pm 1$ forces $GCD(P,Q) = GCD(R,S)=1$, so we get to conclude not only that $P/Q = p_n/q_n$ and $R/S = p_{n+1}/q_{n+1}$ as fractions, but actually that $(P,Q,R,S) = (p_n, q_n, p_{n+1}, q_{n+1})$. Also, they don't state the converse, but they are proving it. -We restate the hypotheses of their theorem: - -The quadruple $(P,Q,R,S)$ is of the form $(p_n, q_n, p_{n+1}, q_{n+1})$ if and only if $Q>S>0$, $PS-QR = \pm 1$ and $(R-\alpha S)/(Q\alpha - P) > 1$. - -We can rewrite $u>1$ as $|1-(u-1)/(1+(u-1)^2)| < 1$ so - -The quadruple $(P,Q,R,S)$ is of the form $(p_n, q_n, p_{n+1}, q_{n+1})$ if and only if $Q>S>0$ and - $$ ((PS-QR)^2-1)^2 + (1-(u-1)/(1+(u-1)^2))^2 <1$$ - where $u = (R-\alpha S)/(Q\alpha - P)$. - -This is an inequality of rational functions, but $F/G < 1$ is equivalent to $FG < G^2$. So we have a polynomial inequality with real coefficients which, together with $Q>S>0$, encodes that $(P,Q,R,S)$ is of the form $(p_n, q_n, p_{n+1}, q_{n+1})$. By construction, this means that as we go to $\infty$ along points satisfying this polynomial inequality and lying in the cone $Q>S>0$, we will see only one color for $\omega$. - -If we wanted to work the same trick for $\ZZ^2$, we would need an inequality for two variables which forces $(p,q)$ to be of the form $p_n/q_n$. Theorem 184 in Hardy and Wright looks good: - -If $\left| \alpha - \frac{p}{q} \right| < \frac{1}{2 q^2}$, then $p/q$ is a convergent. - -But it's a trap! This forces $p/q = p_n/q_n$, but it doesn't force $p/q$ to be in lowest terms. I suspect sufficient cleverness could route around this, but I didn't see how. -UPDATE This is also false for $\ZZ^2$. I don't have the energy to write down a complete proof, but here is the sketch. -Case 1 There are slopes $0 < m_1 < m_2$ and a radius $R$ such that the region -$$\left\{ (x,y) : m_1 < \frac{y}{x} < m_2, \ x^2+y^2 > R^2, \ GCD(x,y)=1 \right\}$$ -is monochromatic. Then we can find a fraction $p/q$ with $m_1 < p/q < m_2$. All points on the line $py-qx=1$ for $y>>0$ will lie in the above region, so we have a monochromatic ray. -Case 2 Every region as above is bi-colored. -Mimicing fedja's argument here, recursively build an $a$ so that all sufficiently large solutions to $|x^5 - a^5 y^5| <1$ have the same color. - -REPLY [2 votes]: It is possible that the OP intended $h$ to have integer coefficients. If so, the answer is yes. More generally, the answer is "yes" if we restrict $h$ to any countable list of polynomials. -Let $p: \mathbb{Z}^2 \to \mathbb{Z}_{\geq 0}$ be an injective polynomial; such polynomials are constructed here. I will show that, if we restrict $h$ to a countable list of polynomials, then $(1/2) + (1/2) (-1)^{\lfloor \phi p(x,y) \rfloor}$ works for almost all real numbers $\phi$. -We have the following Theorem of Bernstein: - -If $b_n$ is a sequence of distinct integers then, for almost all $\theta$, the sequence $\theta b_n \bmod 1$ is equidistributed. - -(Bernstein's paper is in German, but the statement is repeated on the Wikipedia page for equidistributed.) Since a union of countably many sets of measure zero is of measure zero, this shows: - -Let $\mathcal{B}$ be a countable collection of sequences of distinct integers. (I.e. each sequence in $\mathcal{B}$ is a sequence of distinct integers.) Then, for almost all $\theta$, for all $(b_1, b_2, \ldots)$ in $\mathcal{B}$, the sequence $\theta b_n \bmod 1$ is equidistributed. - -For every polynomial $h$ with integer coefficients, and every integer $r$, such that there are infinitely many solutions to $|h(x,y)| \leq r$, order those solutions in order of increasing $x^2+y^2$, breaking ties in some manner. Applying $p$ then turns this into a sequence of distinct integers. Find $\theta$ as above and put $\phi = 2\theta$. Then the integer $\lfloor \phi p(x,y) \rfloor$ is equally often even and odd in each of these sequences.<|endoftext|> -TITLE: Weil pairing, fixed field of a $p$-adic Galois representation -QUESTION [5 upvotes]: Let $A$ be an abelian variety over a $p$-adic field $K$. If $K(A_{p^\infty})$ is the field extension of $K$ obtained by adjoining the coordinates of all $p$-power division points of $A$. By the Weil pairing, it is known that $K(A_{p^\infty})$ contains the field $K(\mu_{p^\infty})$, the field obtained by adjoining to $K$ all the $p$-power roots of unity. -Now, let $X$ be a proper smooth variety over $K$ and consider the $p$-adic Galois representation $$ \rho : G_K \rightarrow GL(V),$$ -where $G_K:=\text{Gal}(\bar{K}/K)$ and $V = H^i_{et} (X_{\bar{K}}, \mathbb{Q}_p)$. -Let $F := {\bar{K}}^{\text{ker}\rho}$ be the fixed subfield of $\bar{K}$ by the kernel of $\rho$. -Question 1: Does $F$ contain $K(\mu_{p^\infty})$ in this general case? -By comparing with the abelian variety case, I think that for the answer to Question 1 to be yes, one needs to generalize the Weil pairing. -Question 2: Is there a "generalized Weil pairing" for proper smooth varieties $X$ given above? - -REPLY [10 votes]: The answer to question 1 as states is no. For example if $i=0$, $V$ will be the trivial Galois representation (assuming your variety to be geometrically connected). -But the answer to question 2 is yes (which implies that some corrected version of question 1 holds as well): a "generalized Weil pairing" is given by the theory of Poincaré duality in étale cohomology: there is a Galois-equivariant pairing $H^i_{et}(X_{\bar K},\mathbb Q_p)\times H^{2n-i}_{et}(X_{\overline K},\mathbb Q_p) \rightarrow \mathbb Q_p(1)$, where $n$ is the dimension of your variety. -When $X$ is an elliptic curve, $n=1$, and taking $i=1$ gives you exactly the Weil pairing. -There is more to say about your question 2: there is a Lefschetz isomorphism $H^i_{et} \rightarrow H^{2n-i}_{et}$ which is a also Galois equivariant up to a suitable Tate twist. -Hence you get a perfect equivariant pairing $H^i_{et} \times H^i_{et} \rightarrow \mathbb Q_p(i)$. Hence back to your question 1, you see that $\ker \rho$ is contained in the kernel of the power $i$ of the cyclotomic character, hence that -your field $F$ contains if $i>0$ a subfield $K'$ of $K(\mu_{p^\infty})$ such that -$K(\mu_{p^\infty})$ is finite degree over $K'$. Moreover, if $i$ is relatively prime to $2(p-1)$, the answer to your question 1 is yes, as the $i$-th power of the cyclotomic character has then same kernel as the cyclotomic character itself.<|endoftext|> -TITLE: Twisted random walks -QUESTION [9 upvotes]: Suppose the points of two random walks in $\mathbb{R}^2$ are given the -step number (or time) as a third coordinate, so that they become paths in $\mathbb{R}^3$. -Here are several pairs of walks of $n=100$ steps, both starting at the origin, -with steps normally distributed with $\sigma=1$: -      -I would like to know the expected number of times that one path winds about the other, -as a function of $n$, the number of steps. -I believe for the three pairs illustrated, there is zero winding by $n=100$. -Experiments indicate winding becomes less likely as $n$ grows. -This is a bit counterintuitive to me. -The winding number of path $b(t)$ about path $a(t)$ up to $t=T$ could be defined by counting -the number of times (the normalization of) the vector $b(t)-a(t)$ turns around the origin for $t\in[0,T]$. - -REPLY [4 votes]: Please find here three papers giving at least partial answers to your question: - -M.A. Berger, "The random walk winding number problem: convergence to a diffusion process with excluded area." 1987: -http://iopscience.iop.org/0305-4470/20/17/028 -Z. Shi, "Windings of Brownian motion and random walks in the plane." 1998: -http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aop/1022855413 -Y. Sun, "On the Expected Winding Number of a Random Walk on the Unit Lattice." 2005: -http://www.artofproblemsolving.com/Resources/Papers/YiSunIntel.pdf - -The upshot is that the root mean square winding number grows logarithmically with the number of steps N for $N \to \infty$.<|endoftext|> -TITLE: Conditions for the contractibility of subvarieties -QUESTION [11 upvotes]: One often finds statements of the sort "and one can contract this subvariety $E\subset X$ to a point in the projective variety $W$," without any explanation of the reasons such a contraction exists, let alone why the variety $W$ is projective. I was wondering if there are any known criteria (obviously necessary and sufficient criteria would be the best) for when one can do this. Staying algebraic/projective is my main issue here since I'm aware that Grauert has many theorems allowing this in the analytic category. -I'm aware of Ishii's paper, "Some Projective Contraction Theorems" which pretty much sums up the story very nicely in the case of divisors on projective varieties, and everything there remains in the projective category. -I was wondering about contractions of higher codimension subvarieties. Here are two examples from the literature of contractions which are claimed to exist, but no reason is given. Is there a well-known theorem (just unknown to me) lurking in the background here? -1) In Namikawa's paper "Mukai flops and derived categories", he takes a subvariety $Y\cong \mathbb P^n$ on a smooth $2n$-dimensional projective variety $X$ with normal bundle $N_{Y/X}\cong \Omega^1_{\mathbb P^n}$. Then he blows $X$ up along $Y$ and blows down the exceptional divisor in the other direction to obtain another subvariety $Y^+$ in $X^+$ satisfying the same conditions as the "non-plus" versions. Then he claims that $Y$ and $Y^+$ can be contracted down to the same point on a projective variety $\overline{X}$. -2) In Kawamata's paper "D-equivalence and K-equivalence", he has something probably easier to explain: the smooth $2m+1$-dimensional projective variety $X$ contains a subvariety $E\cong \mathbb P^m$ with normal bundle $N_{E/X}\cong \mathcal O_{\mathbb P^m}(-1)^{m+1}$. Again he blows up $X$ along $E$ and then blows down the exceptional divisor in the other direction to obtain $F\subset Y$ which again satisfy the same conditions as $E\subset X$. Again there is a claim that both $E$ and $F$ can be contracted in $X$ and $Y$, respectively, to points in the same projective variety $W$. -These are the kind of contractions I'm wondering about. Why do these two contractions exist (and specifically why in the projective/algebraic category), and why do both subvarieties in each case get contracted to the same point in the same projective variety? - -REPLY [5 votes]: My understanding is that there are essentially no results of the form you are looking for, i.e. there are no local criteria for contractibility in the category of projective varieties. -The problem is: assume such a contraction exists; it would be induced by a nef divisor L with L.C = 0 for any curve C in Y. But then the map induced by L would also contract any curve numerically equivalent to C. This is an inherently global phenomenon. -In fact, regarding your examples 1): Namikawa assumes to start with that the variety $X^+$ is also projective. There are counter-examples to such a Mukai flop being projective, see e.g. section 4.4 in Yoshioka, arXiv:math/0009001: there is a projective flop of $X$, given by simultaneously flopping a number of projective subspaces in $X$; the lines in the different projective spaces all have the same numerical class $l$ on $X$. Thus, in the projective category, no single $\mathbb P^r$ can be flopped individually (then $l$ and $-l$ would be effective curve classes, contradicting the existence of ample divisors), and no single $\mathbb P^r$ can be contracted individually (by the argument in the previous paragraphs). Other examples could be constructed using section 14 of my own paper with E. Macrì, arXiv:1301.6968, where we give examples of flopping contraction whose exceptional locus has arbitrary many connected components; I would be surprised if the formal local structure of these components is distinguishable from cases where the flopping contraction has just one irreducible component in the exceptional locus.<|endoftext|> -TITLE: Are formal completions along a subvariety only dependent on the normal bundle? -QUESTION [5 upvotes]: In his paper "Mukai flops and derived categories", Namikawa reduces a general Mukai flop of a smooth projective $2n$-dimensional variety $Z$ along a subvariety $W\cong \mathbb P^n$ with $N_{W/Z}\cong \Omega^1_{\mathbb P^n}$ to another smooth projective variety $Z^+$ with subvariety $W^+$, satisfying the above properties and obtained by blowing up $Z$ along $W$ and then blowing down the exceptional divisor in the other direction, to the simpler case where $X,X^+$ are $\mathbb P^n$ bundles over $\mathbb P^n$ (given by $\mathbb P(\Omega^1_{\mathbb P^n}\oplus \mathcal O_{\mathbb P^n})$) with the projective space subvarieties $Y,Y^+$ coming from the sections of these bundles corresponding to the quotient line bundle $\mathcal O_{\mathbb P^n}$ where the "plus" versions are obtained again by blowing up and blowing down in the other direction. -He proves the simpler case directly, and then reduces the above case to this one by noting that the formal completion of $X$ along $Y$, $X_Y$, is naturally isomorphic to $Z_W$, the formal completion of $Z$ along $W$, and likewise for the "plus" versions. -I was wondering if these identifications only relied on the fact that the normal bundles and underlying variety completed along were the same in each case. If so, can someone point me to a general reference for this technique? If not, what is the proof of these identifications, since no mention of a proof is given there? - -REPLY [6 votes]: Even the first-order infinitesimal neighborhood of a subvariety typically depends on more than the normal bundle, cf. Exercise III.4.10, p. 225 of Hartshorne's "Algebraic Geometry". However, for some special choices of subvariety and normal bundle, the formal neighborhood is automatically formally isomorphic to the formal neighborhood of the zero section in the normal bundle.<|endoftext|> -TITLE: A (very naive) question about Homotopy Type Theory -QUESTION [9 upvotes]: In homotopy type theory, homotopy types can be viewed as logical types and it is possible to prove some theorems about them without using any underlying space (no simplicial set, no topological space). It is a kind of synthetic algebraic topology. Is it just a coincidence that the word "type" may have these two meanings ? What I mean is: what had in mind the one who invented this terminology of "homotopy types" ? And do we even know where that terminology of "homotopy types" comes from. I guess that the answer will be that it is a coincidence but since it is written nowhere in the book, I ask the question in order to be sure. - -REPLY [9 votes]: There seem to be at least two distinct traditions in mathematics for the use of 'type'. -The 'Homotopy types' of algebraic topology seem to have been defined in analogy to 'order types'. Two ordered sets have the same order type when they are isomorphic. Cantor defined the ordinal numbers to be the order types of well-ordered sets. I have not checked whether he used the German equivalent of the word 'type'. -The 'intensional types' of HoTT originate with Bertrand Russell's approach to resolving his paradox by requiring each mathematical object to be of some type, with constraints on which objects are allowed to belong to any given type. Any formal quantifier could only be a quantifier over some type. This was in contrast to Frege who allowed all of his objects to belong to a single realm which could be quantified over. -No doubt both traditions have a source in everyday usage; e.g. from the free dictionary -Type: A number of people or things having in common traits or characteristics that distinguish them as a group or class.<|endoftext|> -TITLE: Is Prikry forcing minimal? -QUESTION [14 upvotes]: Let $M$ be a model of $\sf ZFC$ in which $\kappa$ is a measurable cardinal, and $\cal U$ is a normal measure on $\kappa$. We can define the Prikry forcing (the most simple one) as the poset: $$\Bbb P=\left\{(p,A)\mid p\in[\kappa]^{<\omega}, A\in\mathcal U, \max p<\min A\right\}.$$ -We also define the order, $(q,B)$ is stronger than $(p,A)$ if $q$ is an end-extension of $p$, $B\subseteq A$ and $q\setminus p\subseteq A$. The generic $G$ adds an $\omega$-sequence, $x_G=\bigcup\{q\mid\exists A:(q,A)\in G\}$ which we call a Prikry sequence. We can show that $x\subseteq^* A$, for all $A\in\cal U$, that is $x\setminus A$ is finite. -In the other direction, if $M$ is a model of $\sf ZFC$ in which $\kappa$ is measurable, and $M\subseteq V$, such that in $V$ we have some $x\in[\kappa]^\omega$ such that for some $\cal U$ in $M$ which is a normal measure on $\kappa$, $x\subseteq^*A$ for all $A\in\cal U$, then $x=x_G$ for some generic filter $G$ over the Prikry forcing defined from $\cal U$. -One conclusion from this last statement is that if $x$ is a Prirky sequence for $\kappa$, and $y\subseteq x$ is infinite then $y$ is also a Prikry sequence. This raises the following question. - -Suppose that $x$ is a Prikry sequence for $\kappa$, and $y\subseteq x$ is infinite. Is $x$ generic over $y$? More generally, if $y,w\subseteq x$ are disjoint infinite subsets, are they pairwise generic? What about the weaker condition $y\cap w$ being finite? - -REPLY [6 votes]: It is possible to define a version of Prikry forcing which gives a minimal extension of the universe. See the paper -"A minimal Prikry-type forcing for singularizing a measurable cardinal, J. Symbolic Logic Volume 78, Issue 1 (2013), 85-100." -by Peter Koepke, Karen Räsch, and Philipp Schlicht.<|endoftext|> -TITLE: KK-theory as a stable infinity-category and KU Mod -QUESTION [23 upvotes]: The category KK of bivariant operator K-theory (or possibly its E-theory variant) ought to be the homotopy category of something at least close to a stable infinity-category; notably in that it carries a by now well-known triangulated category structure. -What seems like a step in the direction of establishing such a stable $\infty$-category structure is in the note - -Michael Joachim, Stephan Stolz, An enrichment of KK-theory over the category of symmetric spectra Münster J. of Math. 2 (2009), 143–182 (pdf) - -which produces - -an enrichment $\mathbb{KK}$ of $KK$ in symmetric spectra, in fact in KU-module spectra; -a symmetric monoidal enriched functor $\mathbb{KK} \to \mathrm{KU} Mod$. - -(A partial equivariant generalization of this is given by Mitchener in arXiv:0711.2152.) -This prompts some evident questions: - -Does this enrichment exhibit a presentation of a stable $\infty$-category structure (or close)? -How far is that functor from being homotopy full and faithful? - -Has anyone thought about this? What can one say? -(I see that Mahanta has a note arXiv:1211.6576 along these lines, but not sure yet if it helps with KK.) - -REPLY [2 votes]: Answering your first question in more generality: an $\infty$-category that is enriched over spectra is a stable $\infty$-category if it is closed under finite limits.<|endoftext|> -TITLE: Conceptual explanation for the relationship between Clifford algebras and KO -QUESTION [29 upvotes]: Recall the following table of Clifford algebras: -$$\begin{array}{ccc} -n & Cl_n & M_n/i^{*}M_{n+1}\\ - 1 & \mathbb{C} & \mathbb{Z}/2\mathbb{Z} \\ - 2 & \mathbb{H} & \mathbb{Z}/2\mathbb{Z} \\ - 3 & \mathbb{H}+\mathbb{H} & 0 \\ - 4 & \mathbb{R}(4) & \mathbb{Z} \\ - 5 & \mathbb{C}(4) & 0 \\ - 6 & \mathbb{R}(8) & 0 \\ - 7 & \mathbb{R}(8)+\mathbb{R}(8) & 0 \\ - 8 & \mathbb{R}(16) & \mathbb{Z} \\ -\end{array}$$ -Here $Cl_n=T(\mathbb{R}^{n})/(v\otimes v-q(v))$ where $q$ is the quadratic function associated to the standard inner product on $\mathbb{R}^{n}$. $M_k$ denotes the free abelian group on irreducible $\mathbb{Z}/2\mathbb{Z}$ graded modules (these are easy to work out as this group is isomorphic to the free abelian group of ungraded irreducible modules on the algebra of one dimension lower and these are all matrix algebras). Finally $i:Cl_n\rightarrow Cl_{n+1}$ is the inclusion. For example (using the above isomorphism) the first two groups correspond to $\mathbb{C}$ splitting as two copies of $\mathbb{R}$ as a real representation and $\mathbb{H}$ splitting as two copies of $\mathbb{C}$ as a complex representation. From here the table is periodic mod 8 (up to graded Morita equivalence). -The last column certainly looks eerily familiar, it is the "Bott song" i.e. the coefficients of $\mathrm{KO}$. In their classic paper "Clifford Modules", Atiyah, Bott and Shaprio construct a class in $KO(B^n,S^{n-1})$ from a graded irreducible $Cl_n$ module using the difference construction where the trivialisation on the boundary is given by Clifford multiplication. They use this as well as knowledge of $KO_*$ to show that the aforementioned map gives a ring isomorphism: -$$ \bigoplus_{k\geq 0} M_k/M_{k+1}\rightarrow KO_{*} $$ -Here the ring structure on the left is given by the super tensor product of modules. This is all rather spectacular. It seems reasonable from the perspective of the construction of the ABS map (i.e. by taking the difference bundle of a $\mathbb{Z}/2\mathbb{Z}$ graded vector bundle) that some aspect of the representation theory of superalgebras over $\mathbb{R}$ (in particular the Brauer--Wall group) should be closely conceptually related to real K-theory. Another basic observation is that, via the clutching construction, one identifies the coefficient ring of real K-theory with the (shifted) homotopy groups of the stable orthogonal group. The group $Pin$ may be constructed as the subgroup generated by the unit sphere in $V$ inside the group of units of $Cl_n(V)$ and this is the universal cover of the orthogonal group. It is conceivable that these groups and their representation theory are of more fundamental importance in understanding the rôle Clifford algebras play. -In Karoubi's book "K-theory: An Introduction" he provides an explicit construction of real and complex K-theory in terms of the K-theory of the Banach category of vector bundles with a $Cl_n$ action (this K-theory group is very similar in spirit to the above quotients $M_k/M_{k+1}$) and uses this to prove real Bott periodicity. Others have followed the same path and constructed K-theory classes in terms of something like bundles of $\mathbb{Z}/2\mathbb{Z}$ graded Hilbert spaces with a $Cl_{n}$ action and a self adjoint operator switching degrees. Unfortunately I find these accounts rather technical and don't really intuit why such constructions are reasonable to expect. -Question: Is there a conceptual explanation for the tight relationship (or some aspect thereof) between $KO$ and the representation theory of Clifford algebras? -Finally the work of Douglas and Henriques aims to replicate the above for $TMF$. Here they replace Clifford algebras with conformal nets, the spin groups with the string groups etc. If there is any general philosophical perspective on what drives such a generalisation and in particular on which aspects are important in the classical picture, I would also be extremely interested. - -REPLY [21 votes]: Here is one conceptual description of the relationship. (I wouldn't call it an explanation; I don't really know why it's true, except that it's because Bott periodicity is true.) -KO-theory is the first Weiss-derivative of the $K$-theory of Clifford algebras. -More precisely: given a real inner product space $V$, we get a category $M(V)=\mathrm{Mod}(Cl(V))$ of finite dimentional, $\mathbb{Z}/2$-graded modules over the real Clifford algebra $Cl(V)$. We can consider the $K$-theory space $K(M(V))$ of the topological category $M(V)$; take this to mean something like group completion with respect to direct sum. -Thus, -$$ M_n = \pi_0 K(M(\mathbb{R}^n)),$$ -using the notation in your question. -We thus have a functor -$$ -V\mapsto K(M(V))\colon \mathcal{L} \to \mathrm{Top}_* -$$ -from the category $\mathcal{L}$ of finite dimentional inner product spaces and isometries, to pointed spaces. -Michael Weiss came up with an "orthogonal calculus", which produces a tower of functors approximating a functor $F\colon \mathcal{L}\to \mathrm{Top}_*$, whose $n$th layer takes the form -$$V\mapsto \Omega^\infty( ((D_nF) \wedge S^{V\otimes \mathbb{R}^n})_{hO(n)}),$$ -where $D_nF$ is some spectrum equipped with an action by the orthogonal group $O(n)$. -If we take $F(V)= K(M(V))$, then it turns out that - -$D_nF\approx *$ for $n>1$, and -$D_1F\approx KO$. - -So $KO$ is the first Weiss derivative of $F$. -Why is this true? Weiss gives an easy formula for $D_1F$. Define -$$F^1(V) = \mathrm{ho.fib.}\bigl( F(V) \to F(V\oplus \mathbb{R})).$$ -This turns out to come with a tautological map -$$a_V\colon F^1(V) \to \Omega (F^1(V\oplus \mathbb{R})).$$ -It turns out that $D_1F$ is exactly the spectrum associated to the prespectrum $\{F^1(V), a_V\}$. -In our case, the spaces $F^1(V)$ are the spaces of the $KO$-spectrum (as in Karoubi, I guess), and the maps $a_V$ turn out to be the Bott maps. Which are equivalences by Bott's theorem. -The higher derivatives $D_nF$ are computed using the fibers of $a_V$, but as these are already contractible, $D_2(V),D_3(V),\dots$ must vanish. -Replace $Cl(V)$ with $Cl(V)\otimes \mathbb{C}$ to get $KU$. -This seems like a neat fact, though I've never been able to figure out what it's good for.<|endoftext|> -TITLE: What kind of category is a cyclically ordered set? -QUESTION [9 upvotes]: Background: A preorder is a binary relation $\leq$ which is reflexive and transitive. We can write the transitive property as ${\leq}(a,b)\wedge{\leq}(b,c)\to{\leq}(a,c)$. There are additional axioms that give us partial orders etc., so plenty of everyday "order" concepts can be modeled with such a binary relation. -Similarly, a category has a composition of morphisms $\circ$, which has identities and sends $C(a,b)\times C(b,c)\to C(a,c)$. So we can also model an preordered set as a category in which every hom-set has at most one morphism. I haven't really studied this perspective, but I gather that it's useful. -Now, a cyclic ordering is more naturally a ternary relation! Its version of transitivity states that $(a,b,c)\wedge(a,c,d)\to(a,b,d)$. A cyclic ordering is also cyclic: $(a,b,c)\to(b,c,a)$. If you hunt around Wikipedia you can find a few different kinds of cyclic orders with more or less restrictive axioms. For this question, let's understand "cyclic order" broadly, so it might be strict or not, and it might be total or not: whatever is convenient! -Question: Would it be useful to model cyclically ordered sets as categories? How would you do it? -I'm guessing that higher categories might be useful, so that you can replace the ternary relation $(a,b,c)$ with a 2-morphism like $(a\rightarrow b)\Rightarrow(a\rightarrow c)$, and there could be a composition of 2-morphisms like -$$\left[(a\rightarrow b)\Rightarrow(a\rightarrow c)\right]\times\left[(a\rightarrow c)\Rightarrow(a\rightarrow d)\right]\mapsto\left[(a\rightarrow b)\Rightarrow(a\rightarrow d)\right].$$ -But I'm getting way out of my depth here! I've skimmed over some higher categories on nLab, and I don't see a kind of 2-morphism that does quite what I want. Simplicial 2-morphisms looked promising at first, but they don't seem to compose in the right way. Is that idea a dead end? -Note that I'm not asking about any category of monotone functions between cyclically ordered sets. At least, I don't think that's what I'm asking. -This question was previously asked on Math.SE, but it hasn't prompted an answer. I'm hoping someone here might be able to help, even though it's not a research question for me. I'm just curious. Thanks! - -REPLY [8 votes]: It seems like cyclically ordered sets ought to be regarded as cousins of categories rather than as categories themselves. More precisely, a cyclically ordered set $C$ has a nerve $N(C)$ analogous to the nerve of a (higher) category or (higher) groupoid. The nerve is first of all a sequence $N(C)_n$ of sets, namely the set of $n$-tuples $(a_1, ... a_n)$ such that the relation $(a_i, a_j, a_k)$ holds for all $1 \le i < j < k \le n$ (or two of $a_i, a_j, a_k$ are equal). There are various natural maps between the $N(C)_n$ which make $N(C)$ a cyclic set in the sense of Connes. -Cyclic sets lie intermediate between simplicial sets (which is where the nerve construction takes its values for (higher) categories) and symmetric sets (which is where the nerve construction takes its values for (higher) groupoids). They are in particular simplicial sets, so you can think of cyclically ordered sets as modeled by simplicial sets with extra data. This construction is a beefed-up version of the construction of the order complex of a poset (which is the nerve (regarded as an abstract simplicial complex) of the poset (regarded as a category)).<|endoftext|> -TITLE: Tetrahedra passing through a hole -QUESTION [20 upvotes]: Assume a plane $P\subset\mathbb R^3$ has a hole $H$, and that the hole is topologically a compact disc. Being so, $P\setminus H$ does not separate the space. A regular tetrahedron $\sigma^3$ (of edge-length 1, say) wants to pass through the hole. -As far as I know, there are some papers about this. - -H. Maehara, N. Tokushige, A regular tetrahedron passes through a hole smaller than its face, preprint. -J. Itoh, Y. Tanoue, T. Zamfirescu, Tetrahedra passing through a circular or square hole, Rendiconti del Circolo Matematico di Palermo, Suppl. 77 (2006), 349-354. -J. Itoh, T. Zamfirescu, Simplicies passing through a hole, J. of Geometry, 83 (2005), 65-70. - -The paper 1 shows that the minimum side-length of holes in the shape of a regular triangle is $\frac{1+\sqrt2}{\sqrt6}\approx0.985599$. -The paper 2 shows that the minimum diameter of circular holes is $\frac{t^2-t+1}{\sqrt{\frac{3}{4}t^2-t+1}}\approx0.8957$ $\left(3t=2+\sqrt[3]{\sqrt{43}-4}-\sqrt[3]{\sqrt{43}+4}\right)$ and that the minimum diagonal-length of holes in the shape of a square is 1. -The paper 3 shows that there exists a convex hole $H\subset P$ of diameter $\frac{\sqrt3}{2}$ and width $\frac{\sqrt2}{2}$ such that the regular tetrahedron $\sigma^3$ moving in $\mathbb R^3$ can pass through $H$. In the first paragraph of the proof, they say -"Take a square $Q\subset P$ of edge-length $\frac12$, with vertices $q_{\pm,+}=\left(\pm{\frac14}, \frac12\right)$ and $q_{\pm,-}=(\pm{\frac14}, 0)$. Denote the point $\left(0, \frac{\sqrt11}{4}\right)\in P$ by $v$. Take a disc $D$ of center $\left(0, \frac{\sqrt2}{4}\right)$ and radius $\frac{\sqrt2}{4}$. Define the hole $H$ as the convex hull of $D\cup T$, where $T$ is the triangle $vq_{+-}q_{--}$." -Then, here is my question. -Question: What is the shape of the holes which have the minimum area? -As far as I know, this question still remains unsolved. I have tried to solve this question, but I don't have any good idea. I suspect the paper 3 would be a key. I need your help. - -REPLY [25 votes]: Did you ever find any answer to this? -I find it intriguing that figuring out which shapes of holes a given solid object can pass through is widely considered to be a suitable puzzle for 2 year olds, yet we still don't know the smallest possible hole even for a regular simplex. -edit 29/04/21 — as pointed out by @mjqxxxx in the comments, there's a smaller area hole possible than what I described below by simply projecting onto a plane along the direction of one of the tetrahedron edges, giving a triangle with area $1/\sqrt{8}$ ≈ 0.3536. -I'll leave below my earlier answer, even though it looks rather silly in retrospect :) - -Just playing I found a hole shape which I believe brings the upper bound down from anything I've seen described so far. It is not a convex shape and the tetrahedron does not pass through with a simple translation, but has to also perform multiple rotations. -Say instead of starting with the tetrahedron above the plane and figuring out how to get it below it, we start in the middle with the tetrahedron trapped in a square hole of side 0.5 — we can free one of its vertices by rotating it about one edge of this square, and the opposite edges sweep out a pair of hyperbolas, cutting away a slightly curved triangle. Once a vertex is free, the tetrahedron can slide out along the direction of one of the connected edges. -We can apply this sequence in reverse to get from above the plane to the middle, and then forwards about one of the other sides of the square to pass all the way through, cutting away a second curved triangle similar to the first. - -This gives a shape with a total area of ≈ 0.4044 -For comparison, the minimal equilateral triangle has an area ≈ 0.4206, -and the minimal circle has area ≈ 0.6301. -I don't have any proof that this is the actual minimal-area hole, and indeed suspect that even smaller might be possible — for instance perhaps there is some way to twist the tetrahedron after it is half way through, so that the top half gets through using some of the same curved triangle we cut away for the lower half.<|endoftext|> -TITLE: Can we see invisible sets? -QUESTION [12 upvotes]: What are invisible sets? In order to illustrate what we mean here let me to explain some examples:‎ -‎ -‎ -Consider you want to find the biggest treasure of the world and you have found a magic map of some hidden treasures. If you look at it in "sun light", you will see some objects, signs and sentences which guide you to a treasure. But if you look at it in "moon light", theses signs will vanish and you can see another objects, signs and sentences which guide you to a bigger treasure. But you know these treasures are not the "biggest" one which you want to discover. So what will you do? Probably you ask: "Is there any other light which can uncover guide signs of the biggest treasure of the world? What is the correct light? Candle or firefly?" ‎ -‎ -‎ -As a more real example note that a same situation takes place in astronomy. Many astronomical phenomenas are not detectable in the visible light spectrum. But if you look at them by other electromagnetic frequences such as infra - red, ultra - violet, X - ray or gamma spectrum you can see them but at the same time, most of visible objects will be vanished or deformed.‎ -‎ -‎ -The above examples show that in many cases the objects which thought to have ‎"no existence"‎ are just ‎"invisible"��� under our prevailing paradigm and we can uncover them easily by changing our glasses.‎ -‎ -‎ -Now back to set theory. Redundancy of independence results in set theory says the current mathematics and its foundation (‎$‎‎ZFC^{-}$‎) are very weak. In the other words actually our mathematical game is too superficial and boring and we must richen it. But what is a "rich" mathematics? Intuitionally richness ‎of a‎ ‎game such as mathematics has a direct relevance with t‎he ‎situations which ‎can ‎take ‎place during its playing and this depends ‎on‎ the number of playable objects. So the richest mathematics must have the most number of mathematical objects (sets). As same as above examples, the world of "all" mathematical objects is such a magic map or cosmos and every axiomatization for set theory is a particular light. If you are looking on the map of mathematical objects using ‎$ZFC‎$‎-‎‎light ‎you ‎can ‎see ‎‎$‎‎‎\emptyset‎$ , ‎‎‎‎$‎‎‎\omega‎$, ‎choice ‎functions,... ‎but ‎the ‎"non well founded sets" and "set ‎of ‎all ‎sets" ‎are ‎invisible. Indeed under ‎$‎‎NF$-‎‎light ‎"‎set ‎of ‎all ‎sets" is a visible object and with the light of ‎$‎‎ZFC^{-}+AFA$ ‎some ‎non ‎well ‎founded ‎sets will ‎appear ‎in ‎the ‎world. But beside the number of objects in theory, "consistency" is an important problem too. Although by an inconsistent light we can see whole of the secret objects and sentences of the magic map but this is completely useless and by these signs you can go nowhere because the treasure island is everywhere! ‎Now ‎the ‎main question ‎is: ‎‎ -‎ -‎‎ -Main Question: Which one of the axiomatizations of set theory are the best and give us the richest (relative consistent) mathematics? In other words: Which one of theses axiomatic systems have the most number of mathematical objects? -‎ -‎‎ -In order to clarify our main question we need to investigate the structure of producing objects in axiomatizations of set theory. In this direction we can consider any axiomatic foundation of mathematics as a "set producing factory". They have two main parts. First "atomic existencial axioms" which produce new sets with no need to any former existent object such as axioms of "empty set" and "infinity" in ‎$‎ZFC‎$ which provide some "initial inputs" for this "factory"‎. Second are "relative existential axioms" which produce new sets using former objects as same as a "machine" which takes some "inputs" and give us a "production". In ‎$‎‎ZFC$, ‎axioms ‎of ‎"union", "pairing", ‎"separation", ‎"replacement", ‎"choice", ‎... ‎are ‎relative ‎existential‎. Now intuitionally we can consider a "rich" mathematics as a "factory" (theory) which has the most number of "initial inputs" and "machines" for a huge production of mathematical objects. Now we can explain an exact formalization for these notions: ‎ -‎ -‎ -‎ -Definition (1) : Let ‎$‎‎‎\mathcal{L}‎=‎\lbrace ‎\in ‎\rbrace‎$‎ ‎and ‎‎$‎T‎$ ‎be ‎an ‎‎$‎‎‎\mathcal{L}‎$ - ‎‎theory. ‎Define:‎ -‎ -‎ -$‎‎O_{0}(T):=‎\lbrace\varphi‎(x)\in ‎\mathcal{L}-Form |~T\vDash ‎\exists !‎ x \varphi ‎(x)‎ ‎\rbrace‎‎$‎‎‎‎ -‎ -‎ -‎$\forall n>0~;~~O_{n}(T):=‎\lbrace‎‎‎\varphi‎(x,y_{1},...,y_{n})\in ‎\mathcal{L}-Form |$ -$T\vDash ‎‎\forall ‎y_{1},...,y_{n} \exists ! ‎x‎\varphi ‎(x, y_{1},...,y_{n})‎ ‎\rbrace‎‎$‎‎‎ -‎ -‎‎ -Informally any formula in ‎$‎‎O_{0}(T)$ ‎describes a‎n "‎atomic ‎set" ‎and ‎any ‎formula ‎in ‎‎$‎‎O_{n}(T)$ ‎(for some ‎$‎‎n>0$‎) ‎describes ‎an ‎‎$‎‎n$ -‎ ‎ary "‎set ‎producing machine".‎ ‎For ‎example consider:‎ -‎ -‎ -‎$‎‎emp(x):~~‎\forall ‎y~\neg (y\in x)‎$‎(which says ‎$"‎‎x=‎\emptyset"‎$‎‎) -‎ -‎‎ -$‎‎uni(x,y):~~‎\forall ‎z~(z\in x ‎\longleftrightarrow ‎‎\exists ‎t~(z\in t \wedge t\in y)‎‎)‎$‎‎‎‎ ‎(which says ‎$"‎‎x=‎\cup ‎y‎"‎$‎‎) -‎ -‎ -$int(x,y,z):~~‎\forall ‎t~(t\in x ‎\longleftrightarrow ‎t\in y \wedge t\in z‎)‎$‎ ‎‎(which says ‎$"‎‎x=y ‎\cap ‎z‎"‎$‎‎)‎ -‎ -‎And we have:‎ -‎ -$‎‎emp(x)\in O_{0}(ZFC)‎$‎ -‎ -‎ -‎$‎uni(x,y)\in O_{1}(ZFC)‎$‎‎ -‎ -‎ -‎$‎int(x,y,z)\in O_{2}(ZFC)$‎‎ -‎ -Definition (2) : Let ‎$‎‎‎\mathcal{L}‎=‎\lbrace ‎\in ‎\rbrace‎$‎, define a partial order (i.e. reflexive and transitive) on ‎$‎‎‎\mathcal{L}‎$ ‎-theories ‎as ‎follows:‎ -‎ -$‎‎T\sqsubseteq T' ‎\Longleftrightarrow ‎\forall ‎n\in ‎\omega‎~~O_{n}(T)\subseteq O_{n}(T')‎‎$‎‎ -‎ -‎ -‎‎Definition (3) : Let ‎$‎‎‎\mathcal{L}‎=‎\lbrace ‎\in ‎\rbrace‎$‎, an ‎$‎‎‎\mathcal{L}‎$ ‎-theory ‎$‎T‎$‎ called a "foundation of mathematics" iff ‎$T\vDash ZFC^{-}‎$‎.‎ -‎ -‎‎‎ -Definition (4) : Let ‎$‎‎‎\mathcal{L}‎=‎\lbrace ‎\in ‎\rbrace‎$‎, an ‎$‎‎‎\mathcal{L}‎$ ‎-theory ‎$‎T‎$‎ called "almost consistent" iff ‎$‎‎Con(ZF+ "some~large~cardinal~axiom")‎\longrightarrow Con(T)‎$‎.‎ -‎ -‎‎Definition (5) : Let ‎$‎‎‎\mathcal{L}‎=‎\lbrace ‎\in ‎\rbrace‎$, then ‎define $‎\mathcal{F}‎$ to be the ‎set ‎of ‎all almost ‎‎consistent ‎foundations ‎of ‎mathematics. -‎ -‎ -Question (1) : Is ‎there ‎any ‎maximal ‎element ‎in ‎‎$‎‎‎\langle ‎\mathcal{F},\sqsubseteq ‎\rangle‎‎$‎? -‎ -‎ -‎‎Question (2) : Are there any better definitions for the "object oriented" notion of richness on ‎$‎‎‎\mathcal{F}‎$‎‎?‎ -‎ - -REPLY [3 votes]: If you restrict attention to recursively axiomatized theories, then the answer to your Question 1 is "no." (I'm replacing "consistent relative to a large cardinal axiom" with "consistent," here, for simplicity; it doesn't make much difference.) -EDIT: This argument is much more involved than it needs to be; see Noel's answer. -Suppose $T$ were a maximal element of $\mathcal{F}$. Then we must have that for every formula $\varphi(x; \overline{y}),$ $$ \text{Either }T\vdash\forall\overline{y}\exists!x\varphi(x; \overline{y})\text{ or $T\vdash\neg(\forall\overline{y}\exists!x\varphi(x; \overline{y})).$}$$ Otherwise, we could adjoin $\forall \overline{y}\exists!x\varphi(x;\overline{y})$ to $T$ to get a still consistent theory $> T$ in $\mathcal{F}$. So $T$ has a weak form of completeness. I'll show that - under the assumption that $T$ contains $ZF^-$ (actually, much less is needed) - this contradicts Goedel's Theorem. (Actually, the $\overline{y}$s are completely unnecessary here - this argument still applies if we just pay attention to how big $O_0(T)$ is.) -The key is our assumption that $T$ is recursively axiomatizable. This, together with the assumption that $T$ is a "foundation of mathematics," lets us perform Goedel's arguments formalizing $T$-provability inside $T$. Now consider the formula $$ \beta(x)\equiv\text{$x=0$ OR (For every $T$-proof of $\beta(x)$ there is a shorter $T$-proof of $\neg\beta(x)$)}.$$ Let $(*)$ be the statement "$\exists!x\beta(x)$." By our assumption on $T$, we have $$ T\vdash (*)\quad OR \quad T\vdash\neg(*).$$ But for the former to be true, $T$ would have to disprove its own Rosser sentence; and for the latter to be true, $T$ would have to prove its own Rosser sentence. Either way, this contradicts the assumption that $T$ is consistent. $\Box$ -If we drop the restriction that $T$ be recursively axiomatizable, this of course fails completely; but then again, the theories also become much less interesting (at least from a foundations point of view). In this case, the answer to your Question 1 becomes "yes," for trivial reasons: -Let $\mathbb{P}$ be the poset of all sequences of sets of formulas $(F_i)_{i\in\omega}$ such that for some consistent foundation of mathematics $T$, we have $O_i(T)=F_i$; with elements of $\mathbb{P}$ ordered by inclusion. Then since first-order logic is compact, chains in $\mathbb{P}$ have upper bounds; we could use Zorn's Lemma here for overkill, but actually we don't need any choice in this context, and can now directly prove the existence of a maximal element, since we can well-order the set of formulas. $\Box$ - -As for your Question 2, one partial ordering that is frequently studied is the interpretability ordering on theories (not just "foundations of mathematics," or even recursively axiomatized theories) in a finite signature (interpretability gets weird if the signature is infinite). One theory $T$ interprets a theory $T'$ if there is a (finite collection of) formulas $\Phi$ which, in any model of $T$, define a model of $T'$. So, for example, the theory $ZFC^-$ interprets $PA$, by identifying hereditarily finite sets with natural numbers in definable fashion. -Motivated by this partial ordering, set theories which "interpret more" are more desirable; and axioms such as $V=L$, for example, are undesirable since they limit the achievable interpretive strength of the theory. Some philosophers of mathematics have argued - most notably, Penelope Maddy - that this "maximality" principle should be taken as a key criterion for evaluating the philosophical suitability of set theories (see http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.lnl/1235415905). However, there are intuitively "bad" theories which have extremely high interpretability strength (I believe this is detailed in Benedikt Loewe's "A second glance at non-restrictiveness," but it's behind a paywall so I can't check my memory), and so obviously this picture needs more details. -CAVEAT: this is all half-remembered, so I might have made some very silly mistakes in the previous couple paragraphs.<|endoftext|> -TITLE: A naive question on eigensheaves for group actions on derived categories -QUESTION [5 upvotes]: In this Mathoverflow question, Examples of Eigensheaves outside of langlands, David Ben-Zvi says -" Given a G -space X you can recover quasicoherent sheaves on X from sheaves on X/G (ie equivariant sheaves) as eigenobjects for the natural action of Rep(G)= QC(BG) on QC(X/G)." -Here QC seems to denote say the dg enhancement of the derived category of quasicoherent sheaves. I understand that there is a natural map from -$X/G \to BG$ -and the monoidal action is by pulling back along this map. What I don't see is how to formulate the notion of eigensheaf in this setting. Can someone give some more detail on this statement, e.g. what is the definition of eigensheaf in this statement, what is the precise way to recover the derived category of quasi-coherent sheaves on $X$, etc.? I get the feeling that at least in spirit this is some really concrete thing, so a down-to-earth explanation would be much preferable then a reference to the Geometric Laglands literature. - -REPLY [2 votes]: Given a group $G$ acting on a category $QC(X)$, you get a sheaf of quasicoherent categories over $BG$, whose fiber at $pt\rightarrow BG$ is $QC(X)$. Global sections of this sheaf are exactly invariants $QC(X/G)$ with an action of "global functions" $QC(BG)$. See http://arxiv.org/abs/1306.4304 for a reference on quasicoherent sheaves of categories. -For nice groups $G$ (affine algebraic groups of finite type), $BG$ is 1-affine. In other words, the functor of taking global sections is an equivalence between sheaves of categories over $BG$ and $QC(BG)$-modules. -So, to recover $QC(X)$, you have to localize $QC(X/G)$ over $BG$ and take the fiber over $pt\rightarrow BG$. This is simply -$$QC(X) = Vect\otimes_{QC(BG)}QC(X/G),$$ -where $QC(BG)\rightarrow Vect$ is the pullback to the basepoint $pt\rightarrow BG$. It acquires an action of $Vect\otimes_{QC(BG)} Vect\cong QC(\Omega BG)\cong QC(G)$. Of course, one can just write $X = pt\times_{BG} X/G$ and use the fact that all stacks are perfect, so $QC(pt\times_{BG} X/G) \cong QC(pt) \otimes_{QC(BG)} QC(X/G)$. -$QC(BG)$ is rigid (since $BG$ is a perfect stack), so instead of looking at an action of $QC(BG)$, by adjunction one can look at a coaction of $QC(BG)^\vee\cong QC(BG)$ and take the "cotensor" product. This is just a totalization of the cobar resolution $QC(X/G)\rightrightarrows QC(BG)\otimes QC(X/G)\rightrightarrows...$ -To conclude, an object in $QC(X)$ is an object $\mathcal{F}\in QC(X/G)$ with an identification $\mathcal{O}_{triv}\boxtimes\mathcal{F} \cong (f\times id)_* \mathcal{F}$ and higher homotopies expressing associativity. Here $\mathcal{O}_{triv}$ is the skyscraper sheaf at the trivial bundle (under the identification $QC(BG)\cong\mathcal{O}(G)-comod$ it is $\mathcal{O}(G)$ as a comodule over itself) and $f\times id:X\rightarrow BG\times X$. In other words, $\mathcal{F}$ is an eigensheaf with eigenvalue $\mathcal{O}_{triv}$. -Here is one way to think of the "cobar" presentation. Quasicoherent sheaves on $X$ are the same as sheaves on $X/G$ with an action of $p_*\mathcal{O}_X$ for $p:X\rightarrow X/G$. By base change, $p_*\mathcal{O}_X\cong f^*\mathcal{O}_{triv}$. Finally, $f^*\mathcal{O}_{triv}\otimes\mathcal{F}\rightarrow\mathcal{F}$ is equivalent to a map $\mathcal{O}_{triv}\boxtimes\mathcal{F}\rightarrow (f\times id)_*\mathcal{F}$ on $BG\times X$. Its restriction to $pt\times X\rightarrow BG\times X$ is an isomorphism, hence the map itself is an isomorphism (the forgetful functor $Rep(G)\rightarrow Vect$ reflects isomorphisms). -ADDED (in response to a question in the comments): -An $H$-equivariant sheaf (or a $D$-module, or a local system) on $G/N$ is a sheaf $\mathcal{F}$ on $G/N$ together with an isomorphism $p_1^*\mathcal{F}\cong a^*\mathcal{F}$ for $a: G/N\times H\rightarrow G/N$ and $p_1$ the projection to the first factor. One can also write this as $a^*\mathcal{F}\cong \mathcal{F}\boxtimes \mathcal{O}_H$, i.e. $\mathcal{F}$ is an eigensheaf with eigenvalue $\mathcal{O}_H$. Similarly, given any multiplicative sheaf $\Lambda$ on $H$, you can consider $\Lambda$-twisted $H$-equivariant sheaves on $G/N$: these are sheaves $\mathcal{F}$ together with an isomorphism $a^*\mathcal{F}\cong \mathcal{F}\boxtimes \Lambda$. (A multiplicative sheaf is a sheaf $\Lambda$ on $H$ together with an isomorphism $m^*\Lambda\cong \Lambda\boxtimes \Lambda$ for $m:H\times H\rightarrow H$ the multiplication map.) Multiplicativity is needed to make sense of the associativity conditions on the action. Remark: just as above, you can write this as a cotensor product: $QC(G/N)\otimes^{QC(H)} Vect$ for $Vect\rightarrow QC(H)$ the inclusion of $\Lambda$.<|endoftext|> -TITLE: Does $P(X_1>X_2)$ and $P(X_1=X_2)$, where $X_1$ and $X_2$ are independent and Poisson distributed, uniquely determine the parameters? -QUESTION [13 upvotes]: Let $X_1$ and $X_2$ be independent Poisson distributed random variables with parameters $\lambda_1$ and $\lambda_2$, respectively. -Let $a = P(X_1 > X_2)$ and $b = P(X_1 = X_2)$. -Question: regarding $(a,b)$ as data, does it uniquely determine $\lambda_1$ and $\lambda_2$? -Idea 1: we have the expressions $a = a(\lambda_1,\lambda_2) = e^{-\lambda_1-\lambda_2} \sum_{k=0}^\infty \sum_{j > k}(\lambda_1^j\lambda_2^k)/(j!k!)$ and $b = b(\lambda_1,\lambda_2) = e^{-\lambda_1-\lambda_2} \sum_{j = 0}^\infty (\lambda_1\lambda_2)^j/(j!)^2$, so perhaps one can use this to directly show that $a(\lambda_1,\lambda_2) = a(\mu_1,\mu_2)$ and $b(\lambda_1,\lambda_2) = b(\mu_1,\mu_2)$ implies $(\lambda_1,\lambda_2) = (\mu_1,\mu_2)$. -Idea 2: the random variable $Z = X_1 - X_2$ is Skellam distributed with parameters $(\lambda_1,\lambda_2)$. The conditions amount to specifying the probability mass to the right of $0$ (this is $a$), and the probability mass at $0$ (this is $b$). Perhaps there is a clever way to see that this determines the distribution of $Z$. - -REPLY [7 votes]: Consider the function $F: (\lambda_1,\lambda_2) \rightarrow (a,c)$, where $a = P(X_1 > X_2)$ and $c = 1 - a - b = P(X_1 < X_2)$. We claim that $F$ is invertible, considered as a map from $(0,\infty)^2$ to the open triangular set $T = \{(x,y) \in \mathbb{R}^2; x>0, y>0, x + y < 1\}$. -$F$ is injective -First, we compute the Jacobian. -\begin{eqnarray} -\frac{da}{d\lambda_1} &=&-a + e^{-\lambda_1-\lambda_2}\sum_{k = 0}^\infty\sum_{j > k}\frac{\lambda_1^{j-1}\lambda_2^k}{(j-1)!k!}\\&=&-a + e^{-\lambda_1-\lambda_2}\sum_{k = 0}^\infty\sum_{j \geq k}\frac{\lambda_1^j\lambda_2^k}{j!k!}\\ -&=&-P(X_1>X_2) + P(X_1\geq X_2)\\ -&=&P(X_1 = X_2). -\end{eqnarray} -Similar calculations for the other partial derivatives gives: -$$ -J = \left(\begin{array}{cc}P(Z = 0) &-P(Z = 1)\\-P(Z = -1) &P(Z=0)\end{array}\right), -$$ -where $Z = X_1 - X_2$ is the Skellam distributed random variable. The probability mass function for $Z$ is given by $P(Z=k) = e^{-\lambda_1 -\lambda_2}(\lambda_1/\lambda_2)^{k/2}I_{|k|}(2\sqrt{\lambda_1\lambda_2})$, where $I_{k}$ is the modified Bessel function of the first kind. -Now, $J$ is a P-matrix, i.e. all its principal minors are positive. Indeed, $P(Z=0) > 0$, and the determinant $P(Z=0)^2 - P(Z=1)P(Z=-1)$ equals $e^{-2\lambda_1 -2\lambda_2}(I_0(2\sqrt{\lambda_1\lambda_2})^2-I_1(2\sqrt{\lambda_1\lambda_2})^2)$, which is positive since $I_0 > I_1$. -But now we have everything we need to apply the "Fundamental Global Univalence Theorem (Gale-Nikaido-Inada)" (see On Global Univalence Theorems, Lecture Notes in Mathematics Volume 977, 1983, pp 17-27), which gives us injectivity of $F$. -$F$ is surjective -It is clear by construction that $F = (a,c)$ does not take values outside the closure of $T$: $a$ and $c$ are probabilities of disjoint events, so they are non-negative and sum to at most $1$. Moreover, the range of $F$ is simply connected, since $F$ is continuous and its domain is simply connected. Consider the three line segments that constitute the boundary of $T$: $R_1 = \{(x,0); 0 < x < 1\}$, $R_2 = \{(0,y); 0 < y < 1\}$, and $R_3 = \{(x,y); x>0,y>0,x + y = 1\}$. We are done if we can show that $F$ attains values arbitrarily close to each point in $R_1\cup R_2\cup R_3$, but never takes values in that set. (One should also consider the three corners, $(0,0)$, $(1,0)$ and $(1,0)$, but the arguments are similar for these cases.) -Consider a point $(x,0) \in R_1$. It is clear that this value is never attained by $F$, since $P(X_2 > X_1) = 0$ would imply either vanishing $\lambda_2$, or infinite $\lambda_1$. However, values arbitrarily close to $(x,0)$ are attained: choose $\lambda_1$ such that $P(X_1 > 0) = x$. Then $F(\lambda_1,\lambda_2) \rightarrow (x,0)$ as $\lambda_2 \rightarrow 0$. The set $R_2$ is treated similarly. -Finally, consider a point $(x,y) \in R_3$. By symmetry, we can assume that $x > y$. This point can not be in the range of $F$, since that would imply $P(X_1 = X_2) = 0$, which is impossible for finite $\lambda_1$ and $\lambda_2$. Now, $1 - a - c \rightarrow 0$ when $\lambda_2 \leq \lambda_1$ and $\lambda_2\rightarrow \infty$, so we can choose $\lambda_2$ sufficiently large and send $\lambda_1$ from $\lambda_1 = \lambda_2$ towards infinity: $F(\lambda_1,\lambda_2)$ will describe a curve from a point in the epsilon ball $B_\varepsilon(0.5,0.5)$ to a point in $B_\varepsilon(1,0)$, always staying at most $\varepsilon$ away from $R_3$, hence passing within $\varepsilon$ of $(x,y)$.<|endoftext|> -TITLE: How should you respond to a student who asks whether a very nice physical example constitutes a proof? -QUESTION [7 upvotes]: "Is this really a proof?" is the exact question e-mailed to me today from an undergraduate mathematics student whom I know as a highly competent student. The one sentence question was accompanied with the following demo: - -I am looking for a down-to-earth, non-authoritative answer who one may give to such a student. What would be your answer if you were faced with such a question? -Update after closure. Reading the comments you may realized that most of them answer the original title of the post: "Is this really a proof?" Of course, the answer to such a question is as clear as the daylight for MO users. And, such a question should be closed asap. But, the actual question was (is) in the body of the post, and it was (is) what your constructive answer would be to such a student if you were faced with such a question. Now, with the change of the title, the actual question is much more clearer, and I hope, worthy of MO attention. - -REPLY [10 votes]: Approximate equality in one or more examples is all that can be demonstrated by physical measurements and pictures. This certainly is a useful first step in proof of "exact" (ideal) equality, or in suggesting that there might be an underlying causal mechanism. But there are many geometric dissection problems/puzzles (which I cannot quickly locate, unfortunately) which seem, pictorially, to decompose a figure into pieces whose total area does not add up to the true total. Of course, the "catch" is a tiny imprecision in drawing. -Nevertheless, if a physical demonstration is not deliberately "rigged" to give a deceitful result, an accidental measured-equality is on the whole very convincing, in the same way that so-called Monte-Carlo testing is fairly convincing. -Methodologically, too, I suppose one routinely checks the plausibility of an assertion before allocating much effort to proving it, and physical demonstrations can be quick and effective. (Maybe construction of such an elaborate model as in the demo above wouldn't be usual!) -For that matter, the Euclidean picture-drawing rules-of-proof are themselves a fairly stylized game, as corroborated by Hilbert-et-al's eventual observation that there were some implicit assumptions. Not that the two-thousand-year-old conclusions were wrong, but only that some visual/physical assumptions were being used, in addition to a supposed axiomatic set-up. -Pictures and physical demos certainly capture a diffident audience's attention better than narrative.<|endoftext|> -TITLE: Ubiquity/scarcity of non-analytically continuable functions -QUESTION [5 upvotes]: Suppose f(z) is a power series with positive integer coefficients centered at zero and positive radius of convergence. What is the likelihood that f has a dense set of singularities on its circle of convergence (or any larger circle for that matter). -For instance, a classical theorem shown in Polya & Szego assumes f has integer coefficients and radius of convergence one. In this case, f is either rational or has a dense set of singularities on the unit circle. There are only countably many of the former and uncountably many of the latter, so the latter case is ubiquitous. (The supplied proof depends heavily on integer coefficients and radius 1.) -What can be said for more general radii? This question arises from combinatorial investigations: such f are generating functions for counting problems and several that I have run into seem to have dense sets of singularities on a circle (they also have a dominant real singularity, per Pringsheim's theorem, with slightly smaller modulus). - -REPLY [3 votes]: An old an famous result of Borel states that this happens most of the time. Here is the precise statement $\newcommand{\bsP}{\boldsymbol{P}}$ -Suppose that $(X_n)_{n\geq 0}$ are independent complex valued random variables on the same probability space $(\Omega, \mathscr{A},\bsP)$. Assume additionally that each $X_n$ is symmetric, i.e., $X_n$ and $-X_n$ have identical distributions. Form the random Taylor series -$$f_\omega(z)= \sum_{n\geq 0} X_n(\omega) z^n. $$ -Then the probability that $F_\omega$ has a dense set of singularities on its circle of convergence is $1$. -For details see Chapter 4 of the book Some random series of functions by Jean Pierre Kahane. -Here is a simple application of this result. -Suppose that -$$ f(z)=\sum_{n\geq 0} a_nz^n $$ -is a series with positive radius of convergence. Consider its modifications $\newcommand{\ve}{\varepsilon}$ -$$f_{\ve}(z):= \sum_{n\geq 0} \ve_n a_n z^n, $$ -where $\ve_n=\pm 1$. If the sign changes $\ve_n$ are chosen randomly, independently, each sign with probability $1/2$, then with probability $1$ the series $f_\ve$ will have a dense set of singularities on its circle of convergence.<|endoftext|> -TITLE: On vanishing signed measure -QUESTION [5 upvotes]: It is quite easy to show that if $\mu$ is positive finite Borel measure on, say $[0,1]$, and for all $n \in \mathbb{N}$ -$$\int_{[0,1]} e^{-nx}\mu(dx)=0$$ -holds true, then $\mu=0$. Does this still hold if $\mu$ is signed finite Borel measure? -For positive measure, by applying Holder's inequality it can be shown that $\int_{[0,1]} e^{-rx} \mu(dx)=0$ holds for all positive rational $r$, hence by DCT holds for all $r\geq 0$. By taking differentiation under the integral sign and set $t=0$ we find $$\int_{[0,1]}f(x)\mu(dx)=0$$ holds for all polynomial $f(\cdot)$ and hence for continuous $f(\cdot)$ by Weistrass's approximation theorem. Now Lusin's theorem for Radon measure implies the above equality also holds for measurable $f(\cdot)$, especially for indicating functions so we reach the conclusion. -But for signed measure, it seems Holder's inequality does not work out so I tried to show -$\int_{[0,1]}e^{-nx}\mu_+(dx)=0$ and $\int_{[0,1]}e^{-nx}\mu_{-}(dx)=0$ where $\mu=\mu_+-\mu_-$ is the Hahn decomposition. But it did not work out so well. -[Possible solution]Below is one possible solution I gave. -Let $T:\mathbb{R}\supset A \to B\subset \mathbb{R}$ be diffeomorphism,($\mathcal{C}^1$-diffeomorphism seems to be sufficient but smoothness isn't really what matters here), then -$$\int_A f(x) \mu(dx)=\int_B f\circ T^{-1} (y) (\mu\circ T^{-1})(dy)$$ -holds for all $f$ that is $\mu$-measurable and $\mu\circ T^{-1}$ is the image measure induced naturally by $T$. Now let $T={x \mapsto e^{-nx}}$, and $A=[0,1]$, then $B=[e^{-1},1]$, and -$$\int_{[e^{-1},1]}x^n \tilde{\mu}(dx)=0$$ -where $\tilde{\mu}=\mu\circ T^{-1}$. Now the conclusion follows immediately by similar argument as stated above. - -REPLY [5 votes]: This is an old Theorem of M. Lerch, 1903. For more info see Theorem 6.2, Chap.II, $\S 6$ of - -D.V. Widder: The Laplace Transform, Princeton University Press 1941 (or Dover 2010).<|endoftext|> -TITLE: Computing multiplication in the Ext (for a simple example) -QUESTION [5 upvotes]: This is a follow-up to Computing Ext in Exterior algebra (related to Koszul duality) . -Let $V = \mathbb{C} x$, $A = \Lambda^{\bullet}(V) = A_0 \oplus A_1$ is graded (with $A_0 = -\mathbb{C}, A_1 = V$). Consider $A_0$ as a left $A$-module, how do we compute the multiplication in the algebra $\text{Ext}^{\bullet}_A(A_0, A_0)$? I've heard the word "Yoneda product" mentioned, but I don't know how it works. -We can compute $\text{Ext}^{k}_A(A_0, A_0)$ as follows (see the answer in my above post for more details). We have a Koszul complex $\cdots \rightarrow S^2 V \otimes \Lambda^{\bullet}V[-2] \rightarrow V \otimes \Lambda^{\bullet}V[-1] \rightarrow \Lambda^{\bullet}V \rightarrow A_0$; and we have to compute the cohomology of the complex $\text{Hom}_A(A_0, A_0) \rightarrow \text{Hom}_A(A, A_0) \rightarrow \text{Hom}_A(V \otimes A[-1], A_0) \rightarrow \cdots$. It's easy to check that $\text{Hom}_A(S^k V \otimes A[-k], A_0)=(S^k V)^*[-k]$, and that all maps in this chain complex are $0$ (except for the first); this means $\text{Ext}^k_A(A_0, A_0)=(S^k V)^*[-k]$. -I'm also interested in multiplication for the general case (where $\text{dim } V>1$), but I'm guessing it follows from using the exact same method. - -REPLY [15 votes]: This is a standard computation, if I understand the question -correctly. You are interested in computing $Ext_A(k,k)$ as an -algebra, starting -with an exterior algebra $A$ on $n$ generators $x_1,\cdots, x_n$ -over a field $k$ (no reason to restrict to $\mathbb C$); $A$ is -graded with $k$ in degree $0$ and the $x_i$ in degree $1$. It -is a Hopf algebra, with the $x_i$ primitive. Let $\Gamma$ denote -the divided polynomial Hopf algebra on generators $y_i$, $1\leq i\leq n$; -it is bigraded with the $y_i$ in homological degree $1$ and also internal -degree $1$. We are mainly interested in the coproduct on $\Gamma$, and that is -defined so that its vector space dual is the polynomial algebra on $n$ -generators $y_i^*$. The basis elements dual to $(y_i^*)^r$ are denoted -$\gamma_r(y_i)$. Write $\otimes = \otimes_k$ and form $K = \Gamma \otimes A$. It is a differential graded $A$-algebra with differential defined on the -$\gamma_r(y_i)\otimes 1$ by -$$ d(\gamma_r(y_i)\otimes 1) = \gamma_{r-1}(y_i)\otimes x_i. $$ -With the natural augmentation to $k$, $K$ is a free $A$-resolution of $k$. -Therefore -$$ Hom_A(K,k)\cong Hom_k(\Gamma,k) \cong P[y_1^*,\cdots,y_n^*] $$ is a -cochain complex suitable for computing $Ext_A(k,k)$, and its differential is -zero. We are entitled to conclude that -$$ Ext_A(k,k) = P[y_1^*,\cdots,y_n^*] $$ -as an algebra (indeed as a Hopf algebra, with the $y_i^*$ primitive). -Indeed, via the coproduct on $A$, $K\otimes K$ is a chain complex of $A$-modules, -and it is clearly a free $A$-resolution of $k\cong k\otimes k$. The coproduct $K\longrightarrow K\otimes K$ is a map of chain complexes over the -identification $k\longrightarrow k\otimes k$. We have -$$Ext_A(k,k) \otimes Ext_A(k,k) \cong Ext_{A\otimes A}(k\otimes k,k) $$ -The product on $A$ induces a product -$$ Ext_{A\otimes A}(k\otimes k,k)\longrightarrow Ext_{A}(k,k).$$ -The composite is the product in question, and it is computed using -$$ Hom_A(K,k) \otimes Hom_A(K,k) \cong Hom_{A\otimes A}(K\otimes K,k) \longrightarrow Hom_A(K,k). $$ -This is nothing but the multiplication on the polynomial algebra $P[y_1^*,\cdots,y_n^*]$.<|endoftext|> -TITLE: Is there any forcing free proof for hard independence results? -QUESTION [9 upvotes]: We are forced to use forcing for almost all "hard" independence results such as: $Con(ZFC)\longrightarrow Con (ZFC+\neg CH) $. The question simply is: -Primary Question: Is there any "forcing free" proof for $Con(ZFC)\longrightarrow Con (ZFC+\neg CH) $ or $Con(ZF)\longrightarrow Con (ZF+\neg AC) $ or any other "hard" independence results? -Secondary Question: Please list all "non simple" consistency results which have two proofs one by forcing and another without using it. - -REPLY [6 votes]: Krivine realizability can be used to obtain independence results over $\mathsf{ZF}$. For instance, in Krivine's paper Realizability algebras II : new models of ZF + DC, Logical Methods in Computer Science 8 (1:10) p. 1-28 (2012), he constructs a realizability model in which $\mathsf{AC}$ fails in a pretty strong way: there is an infinite sequence of infinite subsets of $\mathbb{R}$ strictly decreasing in cardinality.<|endoftext|> -TITLE: A question about conjugacy in Higman's group -QUESTION [10 upvotes]: In his paper A finitely generated infinite simple group (J. London Math. Soc., 1951), Higman introduced the following finitely presented group: -$$ -H = \langle x,y,z,w \mid [x,y]=y, \, [y,z]=z, \, [z,w]=w, \, [w,x]=x \rangle . -$$ -This group has many remarkable properties, including being acyclic. -Let $H_{x,y} \le H$ be the subgroup generated by $x$ and $y$, and let $H_{z,w}\le H$ be the subgroup generated by $z$ and $w$. Both $H_{x,y}$ and $H_{z,w}$ are isomorphic to the Baumslag--Solitar group $BS(1,2)=\langle a,b \mid aba^{-1}=b^2\rangle$ (although this is not particularly relevant to my question). - -Does there exist a non-trivial element of $H_{x,y}$ which is conjugate in $H$ to an element of $H_{z,w}$? - -It would seem to me that the answer is no, but I'm having difficulty proving or disproving it. Probably I am missing some well-known technique for deciding this type of question, which I would be very happy to hear about. -Motivation: My co-authors and I are writing about the topological complexity (in the sense of Farber) of aspherical spaces, and the above fact (if it were true) would give a nice example illustrating our techniques. - -REPLY [7 votes]: Below is a geometric argument based on the action of Higman's group on its natural CAT(0) square complex. (Of course, Yves' argument is more elementary, but I find this alternative viewpoint enjoyable.) -The key observation is that $H$ can be described as the fundamental group of a square of groups: -$$\begin{array}{ccc} \langle x,y \rangle & \leftarrow \langle y \rangle \rightarrow & \langle y,z \rangle \\ \uparrow & \uparrow & \uparrow \\ \langle x \rangle & \leftarrow \{1\} \rightarrow & \langle z \rangle \\ \downarrow & \downarrow & \downarrow \\ \langle w,x \rangle & \leftarrow \langle w \rangle \rightarrow & \langle z,w \rangle \end{array}$$ -Now, this square of groups is nonpositively curved (as defined by Gersten and Stallings), so it comes from an action of $H$ on some CAT(0) square complex $X$. (Following the construction of the Bass-Serre tree associated to a graph of groups, $X$ can be described directly in terms of cosets of the vertex-, edge-, square-groups of the square of groups.) The action $H \curvearrowright X$ is studied in details in Alexandre Martin's paper On the cubical geometry of Higman's group, and there it is proved the following weak acylindricity (see Corollary 2.2, whose proof uses only elementary arguments): - -Weak acylindricity: For any two vertices $x,y \in X$ at distance at least three in the $1$-skeleton $X^{(1)}$, the intersection $\mathrm{stab}(x) \cap \mathrm{stab}(y)$ is trivial. - -Now, let $\alpha$ be an element of $H$ which belongs to $\langle x,y \rangle$ and which is conjugate to an element of $\langle z,w \rangle$, say $g^{-1}\alpha g \in \langle z,w \rangle$ for some $g \in H$. Consequently, if $A$ (resp. $B$) denotes the vertex of $X$ corresponding to the coset $\langle x,y \rangle$ (resp. $\langle z,w \rangle$), then $\alpha$ belongs to $\mathrm{stab}(A) \cap \mathrm{stab}(gB)$. Two cases may happen: - -Either $d(A,gB) \geq 3$, so that $\mathrm{stab}(A) \cap \mathrm{stab}(gB)=\{1\}$ as a consequence of the weak acylindricity. -Or $d(A,gB) \leq 2$. In this case, it is not difficult to show that the only possible configuration is that $A$ and $gB$ are two diametrically opposed vertices of some square $C$, hence $$\mathrm{stab}(A) \cap \mathrm{stab}(gB) \subset \mathrm{stab}(C) = \{1\},$$ since square-stabilisers are trivial (this follows from the fact that the square-group of the square of groups defining $H$ is itself trivial). - -The conclusion is that $\alpha$ is necessarily trivial.<|endoftext|> -TITLE: Two cubes in unit cube -QUESTION [8 upvotes]: A cube of side one contains two cubes of sides a and b having non-overlapping interiors. How to prove the inequality $a+b \le 1$? The same question in higher dimensions. It was asked, but not answered in SE. - -REPLY [6 votes]: A better bound, namely -$\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ -is obtained as follows. -The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respectively. Now, forget these two cubes for a while and maximize the sum $a+b$ of the balls' diameters, assuming that the (variable) balls are contained in the unit cube and their interiors are disjoint. Obviously, the maximum possible value of $a+b$ will be an upper bound for the sum of the edges of the cubes. Easy to see, in the optimal configuration of the balls -(1) the two balls must be tangent to each other, -(2) the balls' centers must lie on the main diagonal of the unit cube, -and -(3) each of the two balls must be tangent to the unit cube's boundary. -A brief explanation for (2): the center of the first ball lies somewhere in a cube of edge length $1-a$, concentric with, and parallel to the unit cube. Likewise, the other ball's center is confined to a cube of edge length $1-b$, concentric with, and parallel to the unit cube. The maximum distance between the balls' centers is reached when the centers lie in the "opposite" corners of their confining cubes. -In this position, by (1), the distance between the balls' centers must be $a+b$. Easy to calculate, $a+b=\frac{2\sqrt n}{\sqrt n+1}$. -Somewhat surprisingly, there is a continuum of optimal configurations, where -$\ \frac{\sqrt n-1}{\sqrt n+1}\le a\le1$, $\ b=\frac{2\sqrt n}{\sqrt n+1}-a$. -Neither this, nor the bound given by asatzhh (above) proves the conjectured inequality for cubes, since each of them gives values greater than $1$ for every $n\ge2$, and they both approach $2$ monotonically from below as $n\to\infty$. However, -$$\frac{2\sqrt n}{\sqrt n+1}<2^{\frac{n-1}{n}}\ \ {\rm for\ every}\ \ n\ge2.$$ -For $n=3,\ $ $\frac{2\sqrt n}{\sqrt n+1}\approx 1.26,\ $ while -$\ 2^{\frac{n-1}{n}}\approx 1.58$, -and for $n=9,\ $ $\frac{2\sqrt n}{\sqrt n+1}=1.5,\ $ while $\ 2^{\frac{n-1}{n}}\approx 1.85$.<|endoftext|> -TITLE: Yau's conjecture for positive Chern class -QUESTION [7 upvotes]: I heard in a conference that Yau's conjecture is open for positive Chern class. I read in an article that talked about some stability conditions necessary in this case. So I want to know if this stability condition is well-determined or still conjectural. More precisely, is there any precise statement of this conjecture or is it open ended? - -REPLY [7 votes]: A precise statement and proof of the relationship between stability and the existence of Calabi-Yau metrics is in: - -arXiv:1302.0282, Xiuxiong Chen, Simon Donaldson, Song Sun, -Kahler-Einstein metrics on Fano manifolds, III: limits as cone angle approaches 2π and completion of the main proof -arXiv:1212.4714, Xiuxiong Chen, Simon Donaldson, Song Sun, -Kahler-Einstein metrics on Fano manifolds, II: limits with cone angle less than 2 π -arXiv:1211.4566, Xiu-Xiong Chen, Simon Donaldson, Song Sun, -Kahler-Einstein metrics on Fano manifolds, I: approximation of metrics with cone singularities -arXiv:1210.7494, Xiu-Xiong Chen, Simon Donaldson, Song Sun -Kahler-Einstein metrics and stability -arXiv:1211.4669, Gang Tian, K-stability and Kähler-Einstein metrics - -(so I understand; I haven't read all of this)<|endoftext|> -TITLE: Siegel's theorem with real coefficients -QUESTION [7 upvotes]: Let $h(x,y)$ be a polynomial with real coefficients. Suppose there are infinitely many integer solutions to $|h(x,y)|<1$. What can I say about $h$? -When $h$ itself has integer coefficients, a famous theorem of Siegel tells me that the curve $h(x,y)$ has geometric genus zero and either $1$ or $2$ points at infinity. The main reason I can make no progress on this question is that I know no analogous result for $h \in \mathbb{R}[x,y]$. -All I need is something very crude. Basically, if $(x_n,y_n)$ is the sequence of solutions ordered by $x_n^2+y_n^2$, and you can give me any reasonable upper bound on the growth rate of $x_n^2+y_n^2$, that is good enough to solve the linked problem. (When $h$ has integer coefficients, it follows from Siegel's theorem that $(x_n, y_n)$ are more or less the images of $(a_n, b_n)$ under a polynomial map, where $(a_n, b_n)$ are the solutions to a Pell equation, or else are of the form $(f(n), g(n))$ for some polynomials $f$ and $g$. So $x_n^2+y_n^2$ can grow at worst exponentially.) - -Adding more details here: If $h$ has integer coefficients, then $|h(x,y)| < 1$ is the same as $h(x,y)=0$. Curves of genus $\geq 2$ have only finitely many rational points (Faltings). Affine curves of genus $1$ have only finitely many integer points (Siegel). $\mathbb{P}^1 \setminus \{ 0,1, \infty \}$ has only finitely many $\mathcal{O}_{K,S}$ points for any number field $K$ and any finite $S$ (this is the $S$-unit equation, I think finiteness was also proved by Siegel.) If the normalization of our curve is isomorphic over $\bar{\mathbb{Q}}$ to $\mathbb{A}^1 \setminus \{ z_1, z_2, \ldots, z_s \}$ for some $s \geq 2$ and some $z_i \in \bar{\mathbb{Q}}$ then, after extending the ground field and inverting finitely many primes, we can apply a linear change of variables making $z_1=0$ and $z_2=1$. So we can embed integer solutions into the $S$-unit equation for some $(K,S)$. -Thus, the only remaining options are a genus $0$ curve with one puncture or two punctures. -A genus zero curve with one puncture is rational over $\mathbb{Q}$, since it has a point (the puncture). So there is a parametrization $(f(t), g(t))$ of $h(x,y)=0$ where $f$ and $g$ are polynomials with rational coefficients. I haven't quite been careful with the details here, but the integer points should wind up being the image of some finite collection of arithmetic progressions under $(f(t), g(t))$. -A genus zero curve with two punctures is either $uv=1$ or $u^2-D v^2 = C$ for some nonsquare $D$. Once again, we get a parametrization $(u,v) \to (f(u,v), g(u,v))$ for some polynomials $(f,g)$ with rational coefficients. Again, we need to be careful with denominators from the coefficients of $(f,g)$, but we should get more or less the image of a pell sequence under a polynomial map in the second case, and only finitely many integer points in the first case. -I'm sure that I have seen (Faltings)+(Siegel)+($S$-unit) all stated together as "only finitely many integer points on a curve with $3g+n \geq 3$" and treated as Siegel's result; but I couldn't quickly find a reference that puts it that way. - -REPLY [7 votes]: Nice question! And it is also not hard at all if the degree is not too small. The only downside is that the answer is negative: the growth can be as fast as one wishes. -For a counterexample, we'll just use $P(x,y)=a^px^p-y^p$ with $a\in(0.4,0,6)$ and $p$ to be chosen later. Note that, given an integer $x>0$, $P(x,y)$ is less than $1$ in absolute value for some $y>0$ if and only if $a\in E_x=(\cup_ y I_{x,y})$ where $I_{x,y}$ is an interval almost centered at $\frac yx$ of length $\ell_{x,y}\approx x^{-p}$. -Now we just make a sequence of simple claims: -Claim 1: Assume that $x,y\in\mathbb N$ and $x$ is prime and not ridiculously small. Then $|I_{x,y}\setminus(\cup_{X>x}E_X)|>\frac 18\ell_{x,y}$. -Indeed, if $X>x$ and $I_{X,Y}\cap I_{x,y}\ne\varnothing$, then either $x\vert X$ and $\frac YX=\frac yx$, so $I_{X,Y}$ is contained in the middle half of $I_{x,y}$, or $\frac 1{xX}\le |\frac YX-\frac yx|\le \ell_{x,y}+\ell_{X,y}\le Cx^{-p}$, so $X\ge cx^{p-1}$, in which case the total measure of $E_X$ is about $X^{-(p-1)}$, and the whole union of such pieces can occupy only the length $\sum_{X> cx^{p-1}}X^{-(p-1)}\approx Cx^{-(p-1)(p-2)}\ll x^{-p}\approx \ell_{x,y}$, provided that $p\ge 5$. A lot is, obviously, left. -Claim 2: Under the same assumptions, there exists an arbitrarily large prime $X>x$ and $Y\in\mathbb N$ such that $\operatorname{Clos}I_{X,Y}\subset I_{x,y}$ and $I_{X,Y}$ is disjoint with $E_z$ for $xx$ has measure about $\ell_{x,y}$. Choose large $N$ and notice that the difference $I_{x,y}\setminus(\cup_{x\frac 1{Xz}>\frac {\ell_{x,y}}{z^3}\gg z^{-p}$ if $N$ is large enough and $p\ge 4$, so they just stand outside as poor beggars with hands too short to reach and steal a point from $I_{X,Y}$. -Claim 3: The rest is obvious (the straightforward inductive choice with as fast growth as you wish and the nested interval lemma).<|endoftext|> -TITLE: Rings with a finite maximal ideal -QUESTION [5 upvotes]: A field has a finite maximal ideal. Of course the converse is not true. So is there a characterization for those rings having a finite maximal ideal ?? -EDIT: The characterization below is very nice. I would like to know What happens in non-commutative case ? I mean, Rings with a finite maximal left ideal. Of course, it can be shown easily that $_RR$ has finite composition length. - -REPLY [7 votes]: EDIT : The first version of my answer was unnecessarily complicated. The following avoids lifting of idempotents and structure theory of artinian rings and answers the original question: -So let $R$ be a ring with $1$ and finite maximal ideal $M$. (Note that if $R$ contains a finite maximal left ideal, then it contains a finite maximal ideal.) Let -$$ I = \{ r\in R \mid rM = 0 \}. $$ -I claim that if $R/M$ is infinite, then $R = I \oplus M \cong (R/M)\times (R/I)$. Thus the answer in the noncommutative case is: finite ring or direct product of simple ring and finite ring. -Proof of $R=I+M$: Note that $I$ is the kernel of the map -$$ R \to \bigoplus_{m\in M} (Rm)\subseteq M^M,\quad r\mapsto (rm)_{m\in M}. $$ -As $M$ is finite, $R/I$ is finite. Since we assume $R/M$ infinite, it follows that $I+M> M$ and so $I+M=R$ by maximality. -Proof of $I\cap M=0$: Assume $I\cap M\neq 0$. Since $M$ is finite, $(I\cap M)_R$ contains a simple right $R$-module $S_R$. As $IM=0$, we may view $S$ as a module over $R/M$. Let $D= \text{End}(S_R)$. Since $|S|$ is finite, it follows $S\cong D^n$ for some $n$ and $D$ is finite. By the density theorem of Jacobson-Chevalley then $R/M$ is isomorphic to an $n\times n$-matrix ring over $D$ and thus finite, contradiction. Thus $I\cap M=0$ as claimed. -In the special case where $R$ has a finite maximal left ideal, an occuring infinite simple ring factor is necessarily a division ring.<|endoftext|> -TITLE: Piecewise Smooth Knot Theory -QUESTION [8 upvotes]: In introductory knot theory books, authors usually make a choice of smooth knots or piecewise-linear knots. I often find myself wanting to work in the larger setting of piecewise-smooth knots which subsumes both smooth and PL knots. To do this I would need to prove a "piecewise smooth isotopy-extension theorem" in order to show that knots which are equivalent in the piecewise-smooth sense are equivalent in the usual sense that there exists a continuous ambient isotopy or an ambient homemorphism throwing one onto the other: -Suppose $H: S^1 \times I \rightarrow S^3$ is a continuous isotopy and there exist $$0=t_0 < t_1 < t_2 < ... < t_k = 2\pi$$ such that $H|_{[t_i, t_{i+1}] \times I}$ is $C^\infty$ for all $i$. Then show that $H$ extends to a continuous isotopy of $S^3$, or at the very least there exists a homeomorphism of $S^3$ carrying $H_0$ to $H_1$. Feel free to add hypotheses as needed. -One approach to a proof would replace the piecewise smooth knots $H_0$ and $H_1$ with smooth knots using continuous ambient isotopies of $S^3$ (I can do this). Then if I knew that every piecewise smooth isotopy (defined as above) of smooth knots can be replaced by a smooth isotopy, I could use the smooth isotopy extension theorem to get a continuous ambient isotopy from $H_0$ to $H_1$. -Any thoughts? - -REPLY [5 votes]: Everything that you wish for is true. -You can approximate "ambient" (in the sense explained below) -isotopies by smooth isotopies and keep the ends (the knots) fixed or not. This is spelled out in great detail in: - -MR0674117, Bröcker, Theodor; Jänich, Klaus: Introduction to differential topology. Translated from the German by C. B. Thomas and M. J. Thomas. Cambridge University Press, Cambridge-New York, 1982. vii+160 pp. - -Also the book [Hirsch: Differential topology] has many details. -Edit: -Sorry, I looked up ambient isotopy in Wikipedia just now, and it means something different than what I remember. I meant an isotopy which can be extended to an isotopy of tubular neighborhoods. The Wikipedia notion was called diffeotopy, or something like that, was it not? -Of course it gives a finer equivalence relation on closed curves (embedded with a tubular neighborhood).<|endoftext|> -TITLE: Why do stacked quantifiers in PA correspond to ordinals up to $\epsilon_0$? -QUESTION [36 upvotes]: I am trying to understand why induction up to exactly $\epsilon_0$ is necessary to prove the cut-elimination theorem for first-order Peano Arithmetic; or, as I understand, equivalently, why the length of a PA-proof with all cuts eliminated grows (in the worst case) as fast as $f_{\epsilon_0}$ in the fast-growing hierarchy. -I can understand why use of an induction axiom corresponds to ordinal multiplication by $\omega$. As induction axioms are written in the sequent calculus, as a premise we have a proof of $\phi(x)\vdash\phi(Sx)$ for a free variable $x$, and as a conclusion we get $\phi(0)\vdash\phi(y)$ for any term $y$. -Then if the proof of $\phi(x)\vdash\phi(Sx)$ has the ordinal $\omega^\alpha$, the proof with the conclusion $\phi(0)\vdash\phi(y)$ is assigned the ordinal $\omega^{\alpha+1}$. E.g. if the proof of the induction step had ordinal $\omega^0 = 1$, then the conclusion $\phi(y)$ has the ordinal $\omega^1 = \omega$. -This part makes perfect sense to me, because if I wanted to eliminate the use of an induction rule (CJ sequent) I might just start with $\phi(0)$ and repeatedly go through the part of the proof that I used to prove $\phi(x)\vdash\phi(Sx)$. Say, I'd repeat the sub-proof of the induction step 17 times, e.g. with $\phi(14)\vdash\phi(15)$ and so on, until I got to $\phi(17)$, or whichever number I wanted to prove $\phi$ about. -I don't quite understand how this corresponds to eliminating cuts, per se (it seems to create a host of new cuts, in fact). But it is still very intuitive to me that Peano Arithmetic's power to invoke the induction axiom would correspond to multiplication by $\omega$. If we take a proof in PA that ends with a single use of the induction axiom to get $\phi(y)$, and we want to translate it into a proof in an arithmetic theory that doesn't have induction, then we might have to multiply the PA-proof-length by any finite number (e.g. multiply it by 17) to get the proof length in the inductionless theory. -Repeated multiplication by $\omega$ only takes us up to $\omega^\omega$ as a proof-theoretic ordinal, though. -According to Gentzen, when we use cuts on formulas involving quantifiers, e.g. $\forall p:\exists q:\psi(p,q)$, this corresponds to ordinal exponentiation. If the proof above the cut has ordinal $\alpha$ and we cut a formula with one quantifier, then the proof below the cut has ordinal $\omega^\alpha$. If we cut on two quantifiers, the proof below the cut has ordinal $\omega^{\omega^\alpha}$. This fits with other things I've heard about how PA using formulas with only N quantifiers can be proven consistent by PA using only N+1 quantifiers. (As a side issue I'd be interested in knowing how you use N quantifiers to prove wellfoundedness of an ordinal notation for a stack of $\omega$s N layers high.) -What I don't understand is why the ability to use quantifiers corresponds to ordinal exponentiation. I can guess in a vague sense that if we have a proof using $\forall p:\phi(p)\vdash\phi(17)$, and we want to eliminate the use of $\forall p:\phi(p)$, then we need to repeat that part of the proof each time we want to prove $\phi(17)$, $\phi(q)$, and so on. But this would again just imply that we needed to repeat that part of a proof a finite number of times, and iterating this takes us up to only $\omega^\omega$, not the desired $\epsilon_0$, so I must be missing something. -I asked a friend to help with this and she's read through Gentzen's relevant papers, and she's shown me the relevant parts, and I'd previously checked several standard texts on proof theory and Googled around, and she's also shopped the question around the logic department of a major university, and we still don't know any answer to this except "because Gentzen says to use ordinal exponentiation". -We also can't find any examples of cut-elimination being carried out on a proof with cut on a quantified formula, which shows how the size of the resulting cut-free proof could grow faster than $f_{\omega^\omega}$. An example like this for some particular proof would be very helpful, even if, of necessity, the repeated steps for eliminating the cut are sketched more than shown. I can understand why the length of a Kirby-Paris hydra game grows at the same rate as the Goodstein sequence, and visualize both processes insofar as a human being possibly can. I cannot visualize why the length of a PA-proof heading for cut-freeness would grow at that same rate as cuts were repeatedly eliminated. (Mapping the process onto a Kirby-Paris hydra game of starting height at least 3 would answer the question!) - -REPLY [21 votes]: 8/14: Substantially edited in response to comments: added to 1st part, added new 2nd and 4th parts -There's also some discussion underneath, and a link to a partial write-up of a case of cut-elimination involving the Ackermann function -If you haven't found the Gentzen-style proof illustrative, I recommend trying the infinitary proof. (Not everyone likes it, but it's at least an alternate perspective on it.) Pohlers "Proof Theory: An Introduction" gives a nice presentation of the proof, and it includes the proof in PA that ordinals below $\epsilon_0$ are well-founded. (This includes that fact that you use $N$-quantifier induction to proof well-foundedness up to $N+1$ exponents, and provides some insight into way additional quantifiers make it have the effect they do.) -One nice feature of the infinitary proof is that the ordinals are outright bounds on the heights of proofs, so if you're comfortable visualizing infinitary proofs, the source of the increase in bounds is quite explicit. -It's worth noting that "infinitary" is a bit of a misnomer. The infinitary proof takes the perspective that a proof of $\forall x\phi(x)$ should be a computable function $f$ so that, for each $n$, $f(n)$ is a proof of $\phi(n)$. Since the functions are all computable, there's nothing genuinely infinitary about it. People usually ignore the "computable" part for expository purposes, since it makes sense without that requirement, at which point it does look infinitary, but this is just because many of the ideas are clearer without constantly rechecking that the operations we're describing really are computable. -$ $ -A follow-up point about what ordinals mean in proof theory. As pointed out in the comments, $\omega$ doesn't really mean "infinite" it means "an unspecified fixed integer". Similarly, $\omega+3$ really means "not only is this integer unspecified, but it will take three steps to figure out which integer it is". So in the infinitary proof of cut-elimination, when we say that a proof has height $\omega$, we mean that it is a proof of, say, $\forall x\phi(x)$, where the height of proof of $\phi(n)$ is $g(n)$ with $g(n)\rightarrow\infty$. -A proof of height $\omega+3$ is a proof with three additional steps below a proof of height $\omega$. More interestingly, a proof of height $\omega+\omega$ might be a proof of $\forall x\phi(x)\vee\forall y\psi(y)$ where for each $n$, we have a proof of $\phi(n)\vee\forall y\psi(y)$ of height $\omega+n$. To find a quantifier-free instantiation of this proof, we'd have to first plug in an $n$ for $x$ and then, at the appropriate level, an $m$ for $y$ to get a proof of $\phi(n)\vee\psi(m)$ of height $g(n)+h(n,m)$. -$ $ -Speaking very loosely, the intuitions you describe above are perfectly sensible, but they basically only correspond to what's happening with fairly simple (mostly one quantifier) formulas. Cut-elimination corresponds to extracting computable information, and when you cut formulas with multiple quantifiers together, information has to flow back and forth between the two sides. -Informally, I think of it like this: suppose I want to cut a proof of -$$(\forall x\exists y\phi)\vee\psi$$ -with a proof of -$$(\exists x\forall y\neg\phi)\vee\psi'$$ -(cut-elimination is usually easier to think of in a one-sided form). The second proof produces a first proposal for a value of $x$ witnessing $\forall y\neg\phi$. The first proof might refute it by exhibiting a $y$. The second proof can now use that value of $y$ to compute a second guess at what $x$ is, the first proof refutes it again, and so on back and forth. This could go back and forth a large number of times (on the order of the ordinal of the second proof---that is, not just a fixed number of times, but, if the ordinal is something like $\omega+1$, a number of times based on the value of the first $y$, or if it's $\omega+\omega+1$, the first $y$ tells us how many steps we go before a value of $y$ which tells us how many steps we go, etc.) -This description is literally true when viewed through the lens of the functional interpretation. When viewed in terms of infinitary cut-elimination, these "back and forths" correspond to interleaved cuts: we place many copies of the first proof (more precisely, proofs of $\exists y\phi(n,y)\vee\psi$ for various values of $n$) throughout the second proof. Each of these creates new cuts, which correspond to taking sections of the second proof and placing them inside the many copies of the first proof which we've just created. -$ $ -I think your indexing on the relationship between the fast growing hierarchy and proof-theoretic ordinals is off by an exponential. The proof theoretic ordinal of $I\Sigma_1$ is $\omega^\omega$, but it only proves the primitive recursive functions total (i.e. $f_n$ with $n<\omega$ in the fast-growing hierarchy). I'm not finding a reference on the exact relationship, but I would guess that the gap is consistent---that in the way two quantifier induction is like $\omega^{\omega^n}$, the Conway chain notation is like $\omega^{\omega^2}$. So, while two quantifier induction suffices, dealing with Conway chain notation will look like a substantial use of two-quantifier induction (indeed, I think it involves three levels of nested induction, two of which are over two-quantifier formulas, which would line up well with an ordinal of $\omega^{\omega^2}$). -Having said that, your question is still perfectly sensible: how does the back and forth I described above get things that are very fast growing. Suppose the first proof, of $(\forall x\exists y\phi)\vee\psi$, has a sensible rate of growth, say, exponential. The second proof is iterating the rate of growth of the first proof, and can iterate it an ordinal number of times based on the height of the second proof. Since the second proof could easily (given perhaps some additional cuts over smaller formulas) have, say, height $\omega^2$. That is, if we take $f_0(n)=n^n$, $f_{\alpha+1}(n)=f_\alpha(f_\alpha(n))$, and $f_\lambda(n)=f_{\lambda[n]}(n)$, we should hit around level $f_{\omega^2}(n)$ in this hierarchy.<|endoftext|> -TITLE: Which integers can be expressed as a sum of three cubes in infinitely many ways? -QUESTION [42 upvotes]: For fixed $n \in \mathbb{N}$ consider integer solutions to -$$x^3+y^3+z^3=n \qquad (1) $$ -If $n$ is a cube or twice a cube, identities exist. -Elkies suggests no other polynomial identities are known. - -For which $n$ (1) has infinitely many integer solutions? - -Added - -Is there $n$, not a cube or twice a cube, which allows infinitely - many solutions? - -Added 2019-09-23: -The number of solutions can be unbounded. -For integers $n_0,A,B$ set $z=Ax+By$ and consider -$x^3+y^3+(Ax+By)^3=n_0$. This is elliptic curve -and it may have infinitely many rational points -coming from the group law. Take $k$ rational points -$(X_i/Z_i,Y_i/Z_i)$. Set $Z=\rm{lcm}\{Z_i\}$. -Then $n_0 Z^3$ has the $k$ integer solutions $(Z X_i/Z_i,Z Y_i/Z_i)$. - -REPLY [27 votes]: As stated by Dietrich Burde, it is known that there is no solution for $n\equiv \pm 4 \pmod{9}$, and conjectured that there are infinitely many solutions otherwise. -A cryptic aspect is that it is not even known that there exists one solution for all $n \not\equiv \pm 4 \pmod{9}$. -Today the smallest number for which the problem is open is $n=114$. -Here is a (non-exhaustive) history of the latest solutions found for $n \le 100$ (see here and there): -(1960s) - -$87 = 4271^3 – 4126^3 – 1972^3$ -$96 = 13139^3 -15250^3 + 10853^3$ -$91 = 83538^3 – 67134^3 – 65453^3$ -$80 = 103532^3 -112969^3 + 69241^3$ - -(1990s) - -$39 = 134476^3 - 159380^3 + 117367^3$ -$75 = 435203083^3 – 435203231^3 + 4381159^3$ -$84 = 41639611^3 – 41531726^3 – 8241191^3$ - -(2000s) - -$30 = 2220422932^3 – 2218888517^3 – 283059965^3$ -$52 = 23961292454^3 - 61922712865^3 + 60702901317^3$ -$74 = 66229832190556^3 − 284650292555885^3 + 283450105697727^3$ - -(2019) - -$33 = 8866128975287528^3 - 8778405442862239^3 -2736111468807040^3$ -$42 = 80435758145817515^3 - 80538738812075974^3 + 12602123297335631^3$ -$3 = 569936821221962380720^3 - 569936821113563493509^3 - 472715493453327032^3$ -$906 = 72054089679353378^3 -74924259395610397^3 + 35961979615356503^3$ -$165 = 383344975542639445^3 -385495523231271884^3 + 98422560467622814^3$ - -Remark: for $n \le 1000$, the problem is still open only for $114$, $390$, $579$, $627$, $633$, $732$, $921$, and $975$ (see this paper and this paper, and also this). -Numberphile's videos: - -The Uncracked Problem with 33 -42 is the new 33 -NEWS: The Mystery of 42 is Solved<|endoftext|> -TITLE: Convergence and Closed Form of an Integral Involving Bell Numbers -QUESTION [7 upvotes]: 1. Does the following integral converge ? -$$\int_0^\infty \frac{b(x)}{B(x)} dx$$ -where -$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$ -2. Does it possess a closed form, or some other alternative expression ? - -REPLY [4 votes]: The question on convergence is certainly not research level. -I'm too lazy to give a detailed answer, but here is a direction in which I believe one can obtain a proof with a little effort. -Obviously, all we need is to estimate the asymptotics of $B(x)$ and $b(x)$ for $x \to \infty$. Using a variant of Laplace's method one can show that for large $x$ the value of the sums $b(x)$ and $B(x)$ is asymptotic to the maximal term in these sums, up to a polynomial factor. The maximal term is somewhere around $n(x) \sim \frac{x}{\log x}$, so $b(x) / B(x) \sim {n(x)!} / {n(x)^{n(x)}} \sim e^{-n(x)}$, up to a polynomial factor again. This $e^{-n(x)}$ is approximately $e^{-x / \log x}$, so it decays exponentially at infinity, therefore the integral converges.<|endoftext|> -TITLE: Is it possible to have the set $f^{-1}(\lbrace x \rbrace)$ perfect for every $x$? -QUESTION [12 upvotes]: There are examples of functions $f \colon [0,1] \longrightarrow [0,1]$ such that -for any $\alpha $, $f^{-1}(\lbrace \alpha \rbrace)$ is uncountable. My favorite example is $$f(r) = \limsup_n \frac{a_1 + a_2 +\cdots + a_n}{n}$$ where $0.a_1a_2\cdots$ is the (non-terminated) binary expansion of $r$. -Is there a continuous function $f \colon [0,1] \longrightarrow \Bbb{R}$ such that for any $\alpha \in \Bbb{R}$, $f^{-1}(\lbrace \alpha \rbrace)$ is (non-empty) perfect ? - -REPLY [6 votes]: Existence of such a function goes back to 1939: -J. Gillies, Note on a conjecture of Erdos, Quart. J. Math. Oxford 10, 1939, 151-154 -Also, it can be shown that there is a residual set (a set whose complement is of first category) of continuous functions on $[0,1]$ such that for any $f$ in that set $f^{-1}(\lbrace \alpha \rbrace)$ is perfect except for countably many $\alpha$ and for each of the exceptional $\alpha 's$, $f^{-1}(\lbrace \alpha \rbrace)$ has the form $P\cup\lbrace t\rbrace$ where $P$ is perfect and $t$ is an isolated point of $f^{-1}(\lbrace \alpha \rbrace)$.<|endoftext|> -TITLE: Geometric realization of simplicial spaces and finite limits -QUESTION [7 upvotes]: Let $X_{\bullet}$ be a simplicial space and denote the (non-fat!) geometric realization by $\lvert X_{\bullet} \rvert$. - -Does this geometric realization of simplicial spaces preserve finite limits? - -This is well-known to be true for simplicial sets instead of simplicial spaces. Moreover, the fat geometric realization preserves finite limits up to homotopy and the fat-free geometric realization preserves pullbacks and products on the nose (see Proposition 8 and 9 here for references to proofs). So there might be some hope. - -REPLY [3 votes]: To avoid leaving this question open: -Assuming we work in the category of compactly generated spaces, geometric realization commutes with pullbacks.(It's crucial that we use the compactly generated product.) The proof is basically the same as for simplicial sets. A reference in the space-case is Corollary 11.6 of Peter May's book 'The Geometry of Iterated Loop Space'. The terminal object is preserved as well, so the claim follows by abstract nonsense, since a functor preserves all finite limits iff it preserves pullbacks and the terminal object.<|endoftext|> -TITLE: Contemporary mathematical themes -QUESTION [22 upvotes]: The presence of fruitful mathematical themes suggests the unity of mathematics. What I mean by a mathematical theme here is a basic idea or guiding principle that motivates or directs the central questions of a subject. A few classic examples are "representation", "classification" and perhaps "duality". Another example that seems (to me) currently more active is "rigidity", by which I mean the exploration of conditions under which weak equivalence of a pair of objects implies stronger equivalence. -In the interest of seeing the general direction of contemporary mathematics as the resultant of such themes, I ask: - -Question: What are the major mathematical themes driving mathematical exploration now? - -A good answer ought to not only include the theme in question, but at least two specific areas of mathematics in which strong currents of research are driven by the theme. For example, the "rigidity" theme above is strongly driving the theory of finite von Neumann algebras (as can be seen for example in Popa's deformation/rigidity theory), but appears also in ergodic theory, geometric group theory and differential geometry. Of course, rigidity questions make sense in any area in which there is a heierarchy of equivalences of various strengths, but certain areas (due to the suitability of available techniques) are more strongly driven by efforts to address such questions than other areas are. It would be nice to have a sense of which areas are driven by which themes and (perhaps) why. Arguably, it is the state of the art of techniques in an area that drive the themes, but also those techniques were probably developed because their associated theme was natural. - -REPLY [7 votes]: The dichotomy between structure and randomness is one such theme. Tao's paper focuses on additive number theory, where the idea is that almost all sets are either highly structured (e.g., contain arithmetic progressions) or similar to a random set. But similar themes appear in computational complexity theory; low computational complexity is associated with structure, and high computational complexity is associated with randomness. Razborov and Rudich's result on natural proofs can loosely be thought of as an argument that certain kinds of simple proofs of P≠NP are highly unlikely because they would imply the existence of a lot more structure in randomness than most people expect there to be.<|endoftext|> -TITLE: Vanishing of integral on hemispheres implies vanishing of function? -QUESTION [6 upvotes]: Consider a function $F$ on the half space $\{(x,y,z)|z>0\}$. If $F$ is analytic, it is straightforward to show that -A) The integral of $F$ over the hemisphere $(x-x_0)^2 + (y-y_0)^2 + z^2 = R^2$ vanishes for all $x_0$, $y_0$, and $R$. -implies -B) $F = 0$ everywhere -My question is whether this is known to be true for any continuous function (or under some less restrictive conditions). - -REPLY [3 votes]: As @alvarezpaiva notes, this is a question about hyperbolic hyperplane (Radon) transform. This has been studied. See, e.g., Kurusa, the Radon transform in hyperbolic space [Geometriae Dedicata, 1991] (available for free, thanks to Springer). and references therein (Kurusa inverts it on fairly natural subspaces of $L^2,$ I am not sure if his results are optimal.<|endoftext|> -TITLE: Convergent subsequence of $\sin n$ -QUESTION [9 upvotes]: It is well known (not to me -- ed.) that for every real number $\theta \in [0, 1]$ there exists a sequence $(k_i)$ such that $\lim\sin k_i = \theta,$ but there appear to be no explicit such (infinite) sequences, even for $\theta=0.$ Does anyone know of such? - -REPLY [5 votes]: Since $\pi$ is irrational, the sequence $n\bmod 2\pi$ is dense in $[0,2\pi]$. Now use the fact that $\sin$ is continuous.<|endoftext|> -TITLE: Uniqueness up to isometric isomorphism of predual of $(\sum_{\lambda\in\Lambda} H_\lambda)_{l_\infty}$ where $H_\lambda$ are Hilbert spaces -QUESTION [8 upvotes]: This fact is an easy consequence of results of the paper by Leon Brown and Takashi Ito, but it looks like an overkill. Does anyone know a simpler proof? - -REPLY [6 votes]: In this particular case, you can proceed similarly as in Example 2.1 here. For more general results please consult a fantastic survey on unique preduals by G. Godefroy: - -G. Godefroy. Existence and uniqueness of isometric preduals: a survey. In: BorLuh Lin, editor, Banach Space Theory. Proc. of the Iowa Workshop on Banach Space Theory 1987, 131–193. Contemp. Math. 85, 1989.<|endoftext|> -TITLE: Infinitesimal equivalence of admissible representations -QUESTION [6 upvotes]: Let $G_0$ be the $\mathbb{R}$-points of a real reductive group with complexified Lie algebra $\mathfrak{g}$ and maximal compact subgroup $K$. What is the precise relation between the category of admissible representations of $G_0$ and the category of admissible $(\mathfrak{g},K)$-modules? If we restrict to unitary (admissible) representations and $(\mathfrak{g},K)$-modules, they are supposed to be equivalent, but what about non-unitary representations? Are there counterexamples for e.g. $\text{SL}_2(\mathbb{R})$? - -REPLY [6 votes]: Probably the relevant work is that of Casselman and Wallach (independently) on (in effect) adjoints (right? left?) to the forgetful functor taking Lie group repns to Lie algebra repns. The keyword is "globalization". It turns out that a right adjoint is not the left adjoint (which could be anticipated by observing that many different Lie group repns have the same "smooth vectors", already for the Lie group $SO(2,\mathbb R)$. -So, within mild constraints, it really is true that there exists (more than one) "globalization" of a $(\mathfrak g,K)$ repn, because the worst obstruction is just the topology of $K$. -The papers of Casselman, and of Wallach, should be easily visible on MathSciNet. If it seems otherwise, I can add information here.<|endoftext|> -TITLE: Minimum dilatation pseudo-anosovs on non-orientable surfaces -QUESTION [6 upvotes]: Is anything known about minimum dilatation pseudo-anosovs on non-orientable surfaces? -More specifically it is known whether the asymptotic behavior for log(minimal dilatation) is 1/genus? (The lower bound follows from the lower bound for pseudo-anosovs on orientable surfaces, but I do not know of any construction that realizes it.) - -REPLY [3 votes]: One may construct upper bounds in the non-orientable case the same way as McMullen does in his paper (see p. 523 for a short description of his "renormalization" procedure, or Section 10).<|endoftext|> -TITLE: Homology of localisations of spectra -QUESTION [10 upvotes]: Let $H^*$ and $K^*$ be two cohomology theories, and $X$ a reasonable spectrum. Here, I'm thinking that $H^*$ is singular cohomology (and for my purposes, rational cohomology will suffice), and $K$ is a Morava K-theory. -Suppose that I understand $H^*(X)$ and $K^*(X)$. Is there any way to use this (or other!) information to get a handle on $H^*(L_{K} X)$?, the $H$-cohomology of the Bousfield localisation of $X$ with respect to $K$? -I know that $L_K X$ may sometimes be defined as the totalisation of a cosimplicial spectrum $K^{\bullet +1} \wedge X$. Does the associated Eilenberg-Moore spectral sequence have good convergence properties? This seems doubtful in the case where $H$ is rational cohomology and $K$ Morava K-theory; I think that the $E_1$ term of the spectral sequence is $0$, but it's certainly the case that some $K$-localisations have torsion free homotopy. - -REPLY [8 votes]: I'm going to phrase in terms of $H$-homology instead of cohomology. - -If $H$ is a finite spectrum, smashing with it always commutes across the limit. More generally, if you express $H$ as a (homotopy) colimit of finite spectra -- eg, by taking a cellular filtration -- then you get a directed system of spectral sequences, and the groups you're looking for are the direct limits of the abutments. However, in most circumstances you can't compute this via the direct limit of the spectral sequences and need to be sensitive to interchanging the direct and inverse limits. This might work if you can get a very tight computational handle on things. -If $H$ is a spectrum of finite type (connective and with finitely generated homology groups, or equivalently having a CW model with only finitely many cells in each dimension), then smashing with $H$ commutes across inverse limits of towers where there is some $r$ so that all the terms are $r$-connected. This doesn't apply in your circumstance, but would often apply when $K$ is connective. -Let's take an opposing case where $H = H\mathbb{Q}$ and $K = H\mathbb{F}_p$. Then the spectral sequence you're describing is the Adams spectral sequence and you certainly can't pass rationalization in through the spectral sequence. However, you can often get at the rationalization of the target because you're localizing with respect to an operation (multiplication by p) which lifts to an operation in the spectral sequence that shifts filtration (sometimes called multiplication by $h_0$). A paper of Miller's ("On relations between Adams spectral sequences, with an application to the stable homotopy of a Moore space") introduces these kinds of techniques, and they also showed up in Mahowald-Ravenel-Shick's paper on the telescope conjecture. Unfortunately integral homology can't be expressed as one of these, - - -Now let's get to some details about the more specific version that you asked. -Since $K(n)$-localizations are always $E(n)$-local, and $E(n)$-localization is smashing, $H \wedge L_{K(n)} X \simeq L_{E(n)} H \wedge L_{K(n)} X$. -Further, the map $E(n)_* H\mathbb{Z} \to E(n)_* H\mathbb{Q}$ is an isomorphism: the former is the universal ring where formal group law of $E(n)_*$ obtains a logarithm, and that ring is rational. (Otherwise the ideal $(p)$ would give a nontrivial quotient, but that quotient is $E(n)_* H\mathbb{F}_p = 0$.) Moreover, $H\mathbb{Q}$ is $E(n)$-local, and so $L_{E(n)} H\mathbb{Z} \simeq H\mathbb{Q}$. -EDIT (some elaboration): The spectrum $E(n) \wedge H\mathbb{F}_p$ has a complex orientation inherited from $E(n)$, and one from $H\mathbb{F}_p$. This gives the coefficient ring $E(n)_* H\mathbb{F}_p$ two formal group laws which must be isomorphic. The first has $p$-series satisfying $[p](x) = v_k x^{p^k}$ mod $(v_0,\cdots,v_{k-1})$; $v_0$ is $p$ by convention and $v_n$ is a unit. The second has $p$-series $[p]'(x) = 0$. A strict isomorphism $f$ implies $[p](x) = f^{-1} [p]' f(x) = 0$. This inductively implies that $v_k$ maps to zero for all $k$, and thus the unit $v_n$ is zero; the entire ring $E(n)_* H\mathbb{F}_p$ must be zero. -EDIT (correcting accidental switch to cohomology): Therefore, the integral homology of the spectrum you're asking about is actually $H\mathbb{Q}_* L_{K(n)} X$, and the universal coefficient theorem implies that the cohomology is given by applying $Ext(-,\mathbb{Z})$ to this (with a shift). This reduces the problem to the chromatic splitting conjecture that Eric described (i.e. you may want to stop working on it).<|endoftext|> -TITLE: Uniform hyperbolicity decay estimate -QUESTION [5 upvotes]: I have been trying to understand the proof of the following result, which is considered well-known. -Theorem: Fix a compact metric space $X$, a homeomorphism $T:X \to X$, and a continuous map $ A : X \to \mathrm{SL}_2(\mathbb{R}) $. Define the skew-product -$$ - (T,A): X \times \mathbb{R}^2 \to X \times \mathbb{R}^2, - \quad - (x,v) \mapsto (Tx, A(x)v), -$$ -and set $ A_n(x) = A(T^{n-1}x) \cdots A(x) $ for $x \in X, n > 0$ and similarly for $n \leq 0 $ so that $ (T,A)^n = (T^n,A_n) $. If there are uniform constants $ C > 0, \lambda > 1 $ such that -$$ -\| A_n(x) \| \geq C \lambda^{|n|} -$$ -for all $n,x$, then the cocycle $(T,A)$ is uniformly hyperbolic. More precisely, there are continuous maps $ \Lambda^s,\Lambda^u: X \to \mathbb{RP}^1 $ and constants $ c >0 , L>1$ such that -$$ -A(x) \Lambda^{\bullet}(x) = \Lambda^{\bullet}(Tx), \quad \bullet \in \{ s,u \} -$$ -and -$$ - \| A_n(x) v_s \| \leq cL^{-n} \|v_s\|, \quad \| A_{-n}(x) v_u \| \leq cL^{-n} \| v_u \| -$$ - for all $n \geq 0, x \in X, v_s \in \Lambda^s(x), v_u \in \Lambda^u(x)$. -The consruction of $\Lambda^s$ goes like this: Given $ x \in X, n >0 $, let $ \Lambda_n^s(x) $ be the most contracted subspace of $ A_n(x) $ (i.e. the eigenspace of $ A_n(x)^* A_n(x) $ corresponding to the eigenvalue $ \|A_n(x) \|^{-2} $). Of course, one needs $ \| A_n(x) \| > 1 $ for this subspace to be one-dimensional, but the growth condition assures us that this happens for sufficiently large $n$. -One then proves readily that there are constants $C_0,C_1$ independent of $x$ and $n$ such that the angles between these singular subspaces obey -$$ -\angle \left( \Lambda_n^s(x), \Lambda_{n+1}^s(x) \right) \leq C_0 \| A_n(x) \|^{-2} \leq C_1 \lambda^{-2n}. -$$ -In particular, $ \Lambda_n^s(\cdot) $ converges (uniformly) to a limiting map $ \Lambda^s(\cdot) $. Continuity of $\Lambda^s$ is immediate, and the $A$-invariance condition follows from a straightforward calculation. -Now, here is the part of the proof with which I am having difficulties - the exponential decay estimates. Pick a unit vector $v_s \in \Lambda^s(x)$, and let $ \theta_n = \theta_n(x) $ denote the (smallest nonnegative) angle between $\Lambda_n^s(x)$ and $\Lambda^s(x)$. One can check that -$$ -\| A_n(x) v_s \|^2 - = -\| A_n(x) \|^{-2} \cos^2(\theta_n) + \| A_n(x) \|^{2} \sin^2(\theta_n) -$$ -(simply decompose $v_s$ in an orthonormal basis consisting of a unit vector from $\Lambda_n^s(x)$ and a unit vector from $\Lambda_n^u(x)$). We want to see that this decays exponentially. That the first term on the RHS decays exponentially is obvious, but the second term is bothersome. The factor $ \| A_n(x) \|^2 $ grows exponentially, and the factor $ \sin^2(\theta_n) $ decays exponentially, but it is not clear to me why the exponential decay of $\sin^2(\theta_n)$ should necessarily ``win'' and produce a net result of exponential decay. -I have a nice reference for this result, namely Yoccoz' article ``Some questions and remarks about $ \mathrm{SL}_2(\mathbb{R}) $ cocycles.'' Unfortunately for me, the decay estimate I want to understand is referred to as something easily checked, so I am likely missing something very obvious. I would be very grateful for any helpful remarks. -EDIT: Due to lack of interest, I have cross-posted this question on math stack exchange: -https://math.stackexchange.com/questions/467478/uniform-hyperbolicity-decay-estimate -EDIT (Inspired by A. Blumenthal's comment on Math Stack Exchange): The bound on the angle between $ \Lambda_n $ and $\Lambda_{n+1}$ implies that -$$ -\theta_n(x) \leq C_0 \sum_{m=n}^\infty \| A_m(x) \|^{-2}, -$$ -so it would be enough to prove that -$$ -\| A_n(x) \|^2 \left( \sum_{m=n}^\infty \| A_m(x) \|^{-2} \right)^2, -$$ -decays exponentially. If we define $ a_n(x) = \| A_n(x) \| $ and $ B = \sup_{x\in X} \|A(x) \| $, then we have a sequence of functions $ a_n:X \to \mathbb{R}_{\geq 0} $ with the following properties: -$\bullet$ $ C \lambda^n \leq a_n(x) \leq B^n $ -$\bullet$ $ B^{-1} \cdot a_n(x) \leq a_{n+1}(x) \leq B\cdot a_n(x) $ -$\bullet$ $ B^{-2} \cdot a_n(x) \leq a_n(Tx) \leq B^2 \cdot a_n(x) $ -for all $x \in X$, $n \geq 0$. -We would then like to show that -$$ -a_n(x) \sum_{m=n}^\infty a_m(x)^{-2} -$$ -decays exponentially (uniformly in $x$). This looks more promising, but the desired estimate remains elusive. -In particular, if the sequence $a_n(x)^{-2}$ decreases monotonically, or if $ B<\lambda^2 $, the desired estimate is obvious. However, if $B $ is much larger than $\lambda^2$ and the sequence is wildly non-monotonic, then the waters remain murky to me. - -REPLY [2 votes]: I nearly asked this exact question earlier this year, after coming to the same series and being confounded by the problems which arise when the decay of $\|A_m(x)\|^{-2}$ is too irregular. I discussed this result this summer with an expert in the field and we agreed that behind Yoccoz's "easily checked" lurks perhaps half of the entire proof. -You can find a complete proof of this result in Bochi and Gourmelon's article "Some characterizations of domination" where it arises as a special case of a more general theorem: if a cocycle taking values in $GL_d(\mathbb{R})$ has the property that the $k^{\mathrm{th}}$ singular value of $A_m(x)$ grows faster than the $(k+1)^{\mathrm{st}}$ singular value at an exponential rate which is uniform in $x$ and $m$, then the cocycle has a continuous splitting into a $k$-dimensional invariant bundle and a $(d-k)$-dimensional invariant bundle, and the growth rate in the top bundle always dominates the growth in the bottom bundle along the same orbit by a uniform exponential factor. To overcome the problem of the series they use ergodic theory. Once an ergodic measure is chosen, the series is well-behaved because there is a well-defined exponential growth rate at almost every $x$, and we obtain the desired growth rate almost everywhere. This can be done for every ergodic measure, and we obtain an upper bound on the asymptotic exponential growth rate of $\|A_m(x)s(x)\|$ which is uniform across all measures. By appealing to abstract results such as Schreiber's theorem ("On growth rates of subadditive functions for semiflows", Journal of Differential Equations 1998) one can obtain an exponential growth estimate which is uniform across all $x$. -I believe that it is possible to prove the result without ergodic theory, but this gets quite involved.<|endoftext|> -TITLE: Is the localisation of a product of categories the product of the localisation? -QUESTION [8 upvotes]: Let $\cal C, \cal D$ be model categories. Hovey says in his monograph "Model Categories" that the homotopy category $\operatorname{Ho}(\cal C \times D)$ is isomorphic to $\operatorname{Ho}(\cal C) \times \operatorname{Ho} (\cal D)$, and that this is true for any (finite I assume) number of model categories. -Is this true in general? Let $\{ \mathcal C_i \}_{i \in I}$ be a family of categories, let $W_i \subseteq \operatorname{Mor} \cal C_i$ be families of morphisms and let $\mathcal C = \prod_{i \in I} \cal C_i$. Are the localisations $$\mathcal C\left[ \prod_{i \in I} W_i ^{-1}\right] \cong \prod_{i \in I} \mathcal C_i\left[W_i^{-1}\right]?$$ -I can find a functor from the left to the right in general. -Is it at least true for infinite families of model categories? - -REPLY [12 votes]: It is true for arbitrary products of model categories (or just cofibration categories) as proven in Theorem 7.1.1 of http://arxiv.org/abs/math/0610009v4. -It is also true for finite products of arbitrary relative categories as discussed in this answser: Localizing an arbitrary additive category. (For more details on this and related arguments see http://nforum.mathforge.org/discussion/4769/connected-components-preserve-finite-products/.) -Addendum: This doesn't hold for infinite products of arbitrary relative categories. Consider a sequence of relative categories $\mathcal{C}_0, \mathcal{C}_1, \mathcal{C}_2, \ldots$ such that $\mathcal{C}_i$ is saturated and has objects $X_i$ and $Y_i$ that are weakly equivalent, but the shortest zig-zag of weak equivalences witnessing this is at least $i$ arrows long. Then $(X_0, X_1, \ldots)$ and $(Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(\mathcal{C}_i)$ but not as objects of $\mathrm{Ho}(\prod_i \mathcal{C}_i)$ since this would imply that the length of the shortest zig-zag connecting $X_i$ to $Y_i$ is bounded.<|endoftext|> -TITLE: Good reference for studying operads? -QUESTION [20 upvotes]: Can you, please, recommend a good text about algebraic operads? -I know the main one, namely, Loday, Vallette "Algebraic operads". But it is very big and there is no way you can read it fast. Also there are notes by Vallette "Algebra+Homotopy=Operad", but they don't have much information and are too combinatorial. So what I am looking for is a pretty concise introduction to the theory of algebraic operads, that will be more algebraic then combinatorial, and that will give enough information to actually start working with operads. -Thank you very much for your help! -Edit: I have also found this interesting paper Modules and Morita Theorem for Operads by Kapranov--Manin. Maybe it's a bit too concise for the first time reading about operads, but it has a lot of really nice examples and theorems. -There are also notes by Vatne (only in PostScript). - -REPLY [2 votes]: The book by M. Bremner and V. Dotsenko titled Algebraic Operads: an algorithmic companion (published in 2016) is (in my perhaps biased opinion) a must-have for those wishing to complement their reading of Loday--Vallette. As the authors explain : - -It is fairly accurate to say that the aim of this book is to create an -accessible companion book to [180] which would, in the spirit of [64] contain enough hands-on methods for working with specific operads: making experiments, formulating conjectures and, hopefully, proving theorems, as well as, in the spirit of [252], include enough interesting examples to stimulate the reader toward those experiments, conjectures and theorems. - -As the back-matter explains, it contains a systematic treatment of Groebner bases in several contexts, starting with non-commutative polynomials, and then moving to richer structure like twisted and shuffle algebras, and operads (ns, shuffle, symmetric), the main topic of the book. Like the book of Loday--Vallette, many instances of the book record relatively recent results concerning operads and related structures, and at the same time provides the reader with many challenging exercises (sometimes prompting them to use a CAS, if necessary) that provide invaluable insight for those aiming to make concrete computations using rewriting systems and their kin to study and prove results about operads. -[64] is Ideals, Varieties and Algorithms by Cox, Little and O'Shea, -[180] is Algebraic Operads by Loday and Vallette and -[252] is Combinatorial and Asymptotic Methods in Algebra by Ufnarovski.<|endoftext|> -TITLE: Probability that a random edge coloring of the complete graph is proper -QUESTION [6 upvotes]: This is a repost of this math.se question that I am posting here since it received no attention there. - -What is the probability that a random edge coloring of $K_n$ with - $m \geq n$ colors results in a proper coloring ? - -By random edge coloring I mean that every edge is assigned a color from $\{1,\ldots,m\}$ uniformly at random. -If we define the event $E_i$ to mean that edge $e_i$ is not incident with an edge of the same color class of the event $V_i$ that vertex $v_i$ is incident with edges of distinct colors, then it is easy to get a bound for the above probabiity in terms of $Pr[E_i]$ or $Pr[V_i]$ by using the union bound. -The problem is that it appears that such approach only gives bounds that make sense only when $m = \mathcal(n^2)$ which is not interesting. -Hence I am wondering if there is a more sensitive way to bound/compute the above probability? - -REPLY [3 votes]: Let $p(n)$ be this probability. Now consider you coloured $K_{n-1}$ successfully and add one new vertex. As the edges from the new vertex is coloured, there are simple bounds on how many colours are available. I get that the $i$-th new edge ($i$ starting at 0) has between $m-n+2-i$ and $m-n+2$ colours available. Therefore -$$ \frac {(m-n+2)!}{m^{n-1} (m-2n+3)!} \le \frac{p(n)}{p(n-1)} \le \frac{(m-n+2)^{n-1}}{m^{n-1}}.$$ -Multiply these ratios together to bound $p(n)$. The left bound starts to bite when $m\ge 2n-3$, the right bound earlier. -I'm sure this can be improved.<|endoftext|> -TITLE: Fixed points on boundary of hyperbolic group -QUESTION [5 upvotes]: Let G be a word-hyperbolic group with torsion and let ∂G be its boundary. Do there exist criteria that imply that all non-trivial finite order elements of G act fixed-point freely on ∂G? - -REPLY [6 votes]: Let $G$ be a hyperbolic group with the Cayley graph $X$. -Let $F -TITLE: Definitions of ordinal besides von Neumann & Frege-Russel? -QUESTION [8 upvotes]: So my Google-fu didn't show any references on this. I'm studying an obscure set theory (ML, a variation on NF with proper classes) and it seems to not deal well with the standard definitions of ordinal number. I'm wondering if anyone knows of any references (if they exist) to approaches besides the von Neumann or Frege-Russel ones. -(The specific problems are that the von Neumann ordinals are unstratified, so even ones that exist can't seem to measure wellorders [the usual recursion fails]. The Frege-Russel version works, but the numbers can't generally be guaranteed to be sets since quantifiers in set abstraction must be restricted to sets, and some sets have subclasses which are proper classes [so the usual meaning of "wellorder", among other things, would need to be altered].) - -REPLY [8 votes]: If you really are studying Quine's ML you could read the book by Quine whose title gave the name `mathematical logic' to the system you are looking at. IN ML ordinals are isomorphism classes of wellorderings, as they are in NF and in Principia mathematica. Ordinals in this style are discussed throughout the NF literature. (Look at the New Foundations Home page maintained by Randall Holmes); make me a rich man by buying a copy of Set theory with a universal set! -There are also Scott's trick ordinals, but they are no use in ML. Another thing one can do is define ordinals as isomorphism classes locally (living inside some set). Then one has to prove that all the local systems of ordinals that one obtains cohere. -How on earth did you get into ML?<|endoftext|> -TITLE: "Ultracomposite" numbers -QUESTION [6 upvotes]: If $d(n)$ denotes the number of divisors of $n \in \mathbb{N}$, we may define the function $$C(n) = \frac{\log(d(n)) \cdot \log(\log n)}{\log 2 \log n}.$$ -According to Wikipedia, the Swedish mathematician Carl Severin Wigert proved that $\displaystyle \limsup C(n) = 1.$ $C$ can therefore reasonably be used as a measure of degree of compositeness. However, the natural log base is somewhat arbitrary, and arguably a better choice would be to use logs base 2, $\log_2 n$, and set -$$\displaystyle C_b(n) = \frac{\log_2(d(n)) \cdot \log_2(\log_2 n)}{\log_2 n}.$$ -This also has $\displaystyle \limsup$ equal to 1, and has the further nice property that $C_b(2) = 0$, so that the minimal degree of compositeness is zero. -Suppose we define an "ultracomposite number" to be an $n \in \mathbb{N}$ such that $C_b(n) > 1$, and for every $m < n$, we have $C_b(m) < C_b(n).$ -It follows that the set of ultracomposite numbers is finite. Two questions now arise: -(1) What is the largest ultracomposite number, or failing that, what is a bound or estimate? -(2) How large is the set of ultracomposite numbers, or at least what is a bound or estimate for the size? -(3) More specifically, I conjecture there are seventeen ultracomposite numbers, of which 55440 is the largest. -It should be noted that the interest of this question depends on the acceptance of log base 2 as the privileged choice for this problem--natural logs, for example, lead to far more ultracomposites. - -REPLY [7 votes]: EDIT, 11:36 pm. The conjecture is correct. From the first result of Nicolas and Robin, -$$ C_2(n) \leq 1.53793986 + \frac{0.56367483}{\log \log n}. $$ -This bound decreases with $n.$ Furthermore this gives -$ 1.743557976 $ when $n = 5433960.$ This says that we need only confirm the value of the ratio $C_2(n)$ for $n \leq 5433960,$ by any means that brings joy, because $$ C_2(55440) = 1.743557976.$$ -ORIGINAL: Did it myself. For logs base 2 as described, 55440 now seems a good bet. For natural logarithms, Nicolas and Robin's number 6,983,776,800. Note where the "ratio" 1.537939861 appears in the first Nicolas-Robin result in my earlier answer. That is, it is a theorem that the relevant ratio attains its maximum at 6,983,776,800 and no other place. It may well be that the apparent list for the logs base 2 come as a corollary of one of the Nicolas-Robin results. - logs base 2: - - n 2 divisors 2 ratio 0 2 bump 2^1 - n 6 divisors 4 ratio 1.060087604 6 bump 3^1 - n 12 divisors 6 ratio 1.32815683 12 bump 2^2 - n 60 divisors 12 ratio 1.555150513 60 bump 5^1 - n 120 divisors 16 ratio 1.614640529 120 bump 2^3 - n 360 divisors 24 ratio 1.666250961 360 bump 3^2 - n 2520 divisors 48 ratio 1.729062351 2520 bump 7^1 - n 5040 divisors 60 ratio 1.73879969 5040 bump 2^4 - n 55440 divisors 120 ratio 1.743557976 55440 bump 11^1 - -natural logs: - - n 2 divisors 2 ratio -0.5287663729 2 bump 2^1 - n 6 divisors 4 ratio 0.6509780925 6 bump 3^1 - n 12 divisors 6 ratio 0.9468861067 12 bump 2^2 - n 60 divisors 12 ratio 1.234235881 60 bump 5^1 - n 120 divisors 16 ratio 1.308415107 120 bump 2^3 - n 360 divisors 24 ratio 1.380756857 360 bump 3^2 - n 2520 divisors 48 ratio 1.467704178 2520 bump 7^1 - n 5040 divisors 60 ratio 1.484851215 5040 bump 2^4 - n 55440 divisors 120 ratio 1.51180374 55440 bump 11^1 - n 720720 divisors 240 ratio 1.525218379 720720 bump 13^1 - n 1441440 divisors 288 ratio 1.527797982 1441440 bump 2^5 - n 4324320 divisors 384 ratio 1.531905251 4324320 bump 3^3 - n 21621600 divisors 576 ratio 1.534731579 21621600 bump 5^2 - n 367567200 divisors 1152 ratio 1.537551622 367567200 bump 17^1 - n 6983776800 divisors 2304 ratio 1.537939861 6983776800 bump 19^1<|endoftext|> -TITLE: a question of ranks of matrices -QUESTION [6 upvotes]: Let $S=\{A_{1}, ..., A_{m}\}$ be a set of $n \times n$ symmetric matrices and $m>n$, the rank $r(A_{i})=1$ for each $i$. Suppose that for any $m-1$ matrices $\{A_{i_{1}},...,A_{i_{m-1}}\}$ in $S$, we have the rank of matrix $r(\Sigma_{j}A_{i_{j}}) \leq t$. -My question is: -$r(\Sigma_{i} A_{i}) \leq t$ ? -I think this is true for $n \leq 2$, but I have no idea to prove the general case. - -REPLY [4 votes]: If the characteristic of your base field is 0 the answer is yes. If the characteristic is positive, the answer is no in general. First the proof for characteristic 0, since it will also show how to get counterexamples in positive characteristic. -Assume that the sum of all $A_i$ (denote it by $B$) has rank $t+1$. By general theory on symmetric matrices resp. quadratic forms there is an invertible matrix $S$ such that $SBS^t$ is of diagonal form and the first $t+1$ diagonal entries are nonzero whereas the others are $0$. By also transforming the $A_i$ we therefore can assume that already $B$ has this diagonal form. -By assumption on the sum of $m-1$ matrices we have that $B-A_i$ has rank $t$ for all $i$. Since $A_i$ has rank $1$, its columns are multiples of some column vector $w$, and since it is also symmetric, it is even of the form $a_w\cdot w\cdot w^t$ for some number $a_w$. The columns of $B-A_i$ are therefore of the form $d_j \cdot \underline{e}_j-a_w w_j\cdot w$ ($\underline{e}_j$ the $j$-th standard basis vector, $d_j$ the $j$-th diagonal entry of $B$.), and it is not hard to see that the rank is even $t+2$ if $w$ is not in the span of $\underline{e}_1,\dots,\underline{e}_{t+1}$. -Therefore, everything takes place in the $(t+1)\times (t+1)$-upper left submatrix. By ignoring the rest of the matrix, we are therefore reduced to the case $t+1=n$. -Now back to the condition that $B-A_i$ has rank $t=n-1$. Then there is a vector $v\ne 0$ such that $(B-A_i)v=0$, i.e. -$$\begin{pmatrix} d_1v_1\\ \vdots \\ d_nv_n\end{pmatrix}=a_w (\sum_{i=1}^n v_iw_i) \begin{pmatrix} w_1\\ \vdots \\ w_n\end{pmatrix} $$ -(where we use the $d_j$, $a_w$ and $w$ as above). -Since all the $d_j$ are nonzero, we obtain -$v_j=\frac{a_ws}{d_j}w_j$ where -$0\ne s:=\sum_{i=1}^n v_iw_i$. This leads to -$$s=\sum_{j=1}^n v_jw_j =a_ws \sum_{j=1}^n \frac{w_j^2}{d_j}$$ -resp. $$a_w \sum_{j=1}^n \frac{w_j^2}{d_j}=1.\tag{1}$$ -After all this preparatory work, we consider the diagonal entries of $B$ which are on one hand $d_j$ and on the other $\sum_{w} a_w w_j^2$. Hence, -$$\sum_{w} a_w \frac{w_j^2}{d_j}=1 $$ -We finally get: -$$n=\sum_{j=1}^n \left(\sum_{w} a_w \frac{w_j^2}{d_j}\right)=\sum_{w} \left( \sum_{j=1}^n a_w \frac{w_j^2}{d_j}\right)=\text{number of }w\text{'s}=m$$ -Remark:This even shows more than the question, namely if the rank of the sum of $m$ matrices is $t+1$ and the rank of the sum of arbitrary $m-1$ of those is $t$, then $m=t+1$. -Why does it not work in characteristic $p>0$? Because the last equation is an equation in the field, i.e. we only obtain $n\equiv m \mod{p}$. Indeed, since the condition (1) is sufficient for the $w$ (resp. the matrices $A$) one can construct examples for which $m=n+p$, even for $n=2$.<|endoftext|> -TITLE: Beyond Collatz: A $5n+1$ conjecture? -QUESTION [8 upvotes]: Let -$$x_{n+1} = \begin{cases} x_n/2 &;\text{if } x_n \equiv 0 \pmod{2}\\ k\,x_n+1 &; \text{if } x_n\equiv 1 \pmod{2} \end{cases}$$ -and $k=3$ and $x_n\in\Bbb N$. Collatz conjectured for this recurrence system that starting with any $x_n^{(start)}$ the system converges to a limit cycle (an atractor, orbit) of period $3$: -$$\dots\rightarrow16\rightarrow8\rightarrow4\rightarrow2\rightarrow1\to4\to2\to1$$ -Independent of the above allow me conjecture the following: - -Let -$$x_{n+1} = \begin{cases} x_n/2 &;\text{if } x_n \equiv 0 \pmod{2}\\ -k\,x_n+1 &; \text{if } x_n\equiv 1 \pmod{2} \end{cases}$$ -and $k=5$ and $x_n\in\Bbb N$. -(Part 1) For this recurrence system starting with - any $x_n^{(start)}$ the system either - -converges (stable) to the following limit cycle (an attractor of a repeating sequence, orbit) of period $7$: $$\dots\to16\to8\to4\to2\to1\to 6\to3\to16\to8\to4\to2\to1$$ - examples for this are when we start the recurrence with one of the following integers $3$, $15$, $19$, $51$, $65$, $97$, $137$, $155$, $163$, $175$ -or converges (stable) to a limit cycle (an attractor of a repeating sequence, orbit) of period $10$. Example: $$\dots\to13\to 66\to 33\to 166 \to 83 \to 416\to 208\to 104\to 52\to26\to 13\to \dots$$ - examples for this are when we start the recurrence with one of the following integers $5$, $13$, $17$, $27$, $33$, $43$, $83$ -or diverges (intsable, ecape to infinity) - -(Part 2) Hence if the Collatz conjecture would be true then the transition from $k=3$ to $k=5$ would represent a bifurcation from one single limit cycle of period $3$ to one specific limit cycle of periods $7$ (see above) and some other limit cycles of each period $10$. - -Question (1): Could you contradict the above conjecture with a counter example? -Question (2): Is this conjecture genuine or has it been stated exactly like this in literature earlier? - -citation Vaseghi 2013 - -At reqeust for a Heuristic below a Mathematica program that we applied at TrueNorth Research. - -ClearAll[collatz]; -collatz1 = 1; -collatz[n_ /; EvenQ[n]] := (Sow[n];collatz[n/2]) -collatz[n_ /; OddQ[n]] := (Sow[n]; collatz[5 n + 1]) -runcoll[n_] := Last@Last@Reap[collatz[n]] -runcoll[13] - -you can change $13$ with any other integer. - -REPLY [13 votes]: The reason why there should be values that escape to infinity in this case is that if one considers a so to say "random model" for the $k=3$ variant one has roughly speaking a change of the size by roughly $3/2$ half the time and a change of $1/2$ half-the time, and the product being less than $1$ one would expect a long-term decay. -By contrast, for the current variant, one has an increase by about $5/2$ half the time and a decrease of $1/2$ half the time. The product being larger than one one would expect a long term increase. -Thus, one would expect there are some values where the function will escape to infinity (so option three should occur sometimes). -Also it seems there is $17 \to 86 \to 43 \to 216 \to 108 \to 54 \to 27 \to 136 \to 68 \to 34 \to 17$. -Added: again I misunderstood the question. As it leaves opent the existence of other such cycles, and only fixes their lengths. The one in OP and the one I recall are it seems known to be the only ones of this length. -It seems feasible there are actually no others if one does not find any somewhat soon. The existence of certain cycles is excluded in a paper mentioned below. -To consider this variant is not original, for example it is mentioned in passing in a blog post by Tao The Collatz conjecture, Littlewood-Offord theory, and powers of 2 and 3 with a more sophisticated form of the argument above (for escaping values). -Also an older math.SE questions (I think not the one mentioned in comments; added: I meant the one by OP, the one by David Speyer mentioning it appeared while I edited) discusses this precise problem giving some additional information and references https://math.stackexchange.com/questions/14569/the-5n1-problem specfically to some paper of Metzger that determines which scyles of certain lengths can exist (also mentioning the one reproduced above, in addition to the one in OP).<|endoftext|> -TITLE: Does forcing generally go one way? -QUESTION [7 upvotes]: Question Is there any forcing free proof for hard independence results? talks about the use of forcing for independence results such as: $Con(ZFC)\longrightarrow Con (ZFC+\neg CH) $. For that case (and its companions $V=L$ and $AC$) constructibility is used for the other half of the independence: $Con(ZFC)\longrightarrow Con (ZFC+ CH) $. -Is this typical? Do later independence results still tend to use forcing one way, and $V=L$ the other? Are there independence results where forcing is used in both directions? -Are there statements $S$ where forcing is used for both $Con(ZF+S)$ and $Con(ZF+\neg S)$? -Question Is every class that does not add sets necessarily added by forcing? mentions use of forcing over a model of $NBG$ to add a universal choice function. But I do not know for whether you would also use forcing to eliminate that version of global choice in that context. - -REPLY [14 votes]: There is a wide variety of statements $S$ for which forcing can be used to prove the (relative) consistency of both $S$ and $\lnot S$ with $\mathsf{ZFC}$. -For example, Cohen developed forcing to prove the consistency of $\lnot\mathsf{CH}$, but we can also prove $\mathsf{CH}$ consistent this way: We can add to the universe a well-ordering of the reals of type $\omega_1$ without adding any reals. This, by the way, shows that any projective statement provable from $\mathsf{CH}$ is provable without it. -There are some restrictions: $V=L$ cannot be forced. This is because $L$ is forcing invariant (we can also say generically absoute): If $W$ is a forcing extension of $V$, then $L$ in the sense of $W$ is the same as $L$ in the sense of $V$. This is true of larger core models, and is one of their key features. It allows us to prove results about the universe by arguing about forcing extensions, and is particularly useful when proving consistency strength lower bounds (and in the development of the general theory of core models). -Similarly, if $\alpha$ is a countable ordinal, then we cannot use forcing to make $\alpha$ uncountable (the idea being that if we already see a surjection from $\omega$ onto $\alpha$, adding more sets is not going to make this go away). -Other examples of statements $S$ that can be forced either way are: $2^{\aleph_1}=\aleph_{17}$, $\mathfrak b=\omega_1<\mathfrak c$, any algebraic homomorphism between Banach algebras is continuous, etc. The list can go on forever. If we assume the presence of large cardinals in the universe, then even more examples can be given. -Joel Hamkins and his collaborators (notably, Benedikt Loewe) have worked on this, making explicit interesting connections with modal logic (think of $\diamondsuit\phi$ as "it is forceable that $\phi$" and of $\square\phi$ as "$\phi$ holds in all extensions"). See here for a starting point. - -Now, an interesting fact about forcing is how ubiquitous it is. This distinguishes it from other techniques. Woodin's $\Omega$-conjecture can be informally described as stating that any $\Pi_2$ statement we can verify consistent (with large cardinals) in fact can be forced (over arbitrary models with large cardinals). (This is an informal description. A more technical presentation would perhaps obscure the point here. See this survey by Bagaria, Castells, and Larson, for details.) -One of the reasons why the $\Omega$-conjecture is of interest is that it holds in all known "$L$-like" models. It is also forcing absolute, so its consistency cannot be verified by forcing (unless we already know it is true). Another reason is that there are a few examples of $\Pi_2$ statements we do not know how to force over arbitrary models, though we can prove consistent with appropriate large cardinals (by forcing over special ground models). One is: "There is a $\mathbf{\Sigma}^2_1$-well-ordering of the reals, and the continuum is a real-valued measurable cardinal", see here. -In his book The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal, Hugh suggests a few statements of large cardinal character (stronger than "there is a limit $\lambda$ and a nontrivial $j:V_\lambda\to V_\lambda$") that appear to be fragile, meaning that it is not clear that they are preserved by small forcing. Whether these statements can be proved consistent with $\mathsf{CH}$ or with $\lnot\mathsf{CH}$ does not seem to be a matter of forcing. Something different would be needed here. - -REPLY [8 votes]: There are plenty of results where one uses forcing to make both parts of the independence. Even the case of CH can be done this way, since Solovay proved shortly after forcing was invented that one can force CH as well as $\neg$CH. Thus, CH is an example of what Benedikt Löwe and I call a switch, a statement $\varphi$ such that $\varphi$ and $\neg\varphi$ are each forceable over any forcing extension, so that you can turn it on and off again at will. A button, in contrast, is a statement that you can force in such a way that it remains true in all further extensions. These concepts were important in our work on The modal logic of forcing. -Meanwhile, there are several examples where the only known proofs use forcing on both sides, particularly in the context of very large cardinals. For example: - -The existence of a supercompact cardinal $\kappa$ is relatively consistent with $2^\kappa=\kappa^+$ and with $2^\kappa\gt\kappa^+$. Both sides of these proofs use forcing. -Similar consistency results of the very largest large cardinals with CH and GCH and their negations are only known by forcing. -Many other consistency results concerning the very largest large cardinals are only known by forcing, because there is no inner model theory analogue of $L$ for those cardinals.<|endoftext|> -TITLE: Lower bound of integral involving Laguerre polynomials -QUESTION [5 upvotes]: I want to lower bound the expected value of the square root of a randomly chosen eigenvalue of a Wishart matrix. -To get the bound I want I need a lower bound on -$$T_n = \int_0^\infty\sqrt{x}e^{-x}L_n(x)^2dx,$$ -for $n>1$ where $L_n(x)$ is the $n$-th Laguerre polynomial. -Mathematica convinced me that $T_n > \sqrt{n+1}$ but I wasn't able to find a proof for this. -Does anyone have an idea of how to estimate this integral? -Note: if the $\sqrt{x}$ is replaced by $x$ then Wikipedia has a closed form solution for it. -Thanks! - -REPLY [6 votes]: Here is a proof. -First evaluate the integral. I use the generating function for squares of Laguerre polynomials ($I_0$ the modified Bessel function): -$$ -\sum_{n=0}^{\infty}L_{n}(x)^2 z^n = \frac{1}{1-z} \exp\left(-\frac{2 x z}{1-z}\right) I_{0}\left(\frac{2 x \sqrt{z}}{1-z}\right) -$$ -for $|z|<1$ (All the formulas I use, one can find it in e.g. Gradshteyn and Ryzhik.) -This gives for the integral -$$ -T_{n} = \int_{0}^{\infty} dx \sqrt{x} e^{-x} L_{n}(x)^2 = [z^n] \frac{1}{1-z} \int_{0}^{\infty} dx \sqrt{x} \exp\left(-x \frac{1+z}{1-z}\right) I_{0}\left(x \frac{2 \sqrt{z}}{1-z}\right) -$$ -$[z^n]$ means, as usual, "take the coefficient of $z^n$ in the (formal) power expansion in $z$ of the following expression". -The integral can be evaluated (it is in Gradshteyn and Ryzhik) and the result is -$$ -T_{n} = [z^n] \frac{\sqrt{\pi}}{2} (1-z)^{-1} P_{1/2}\left(\frac{1+z}{1-z}\right) -$$ -with $P_{1/2}(x)$ the Legendre Function. -Expanding the Legendre function and the prefactor in powers of $z$ gives for the coefficient of $[z^n]$ -$$ -T_{n} = \frac{1}{8\sqrt{\pi}}\sum_{m=0}^{n}\frac{\Gamma\left(m-\frac{1}{2}\right) (2(n-m)+1)!}{4^{n-m} m!^2 (n-m)!^2} -$$ -Now I was lazy and asked Mathematica to simplify, which gives -$$ -T_{n}= \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} {_3}F_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right) -$$ -with the generalized hypergeometric function ${_3}F_{2}$ evaluated at $1$. -With the exact solution at hands we are almost done. We observe that the sum defining the hypergeometric function -$$ -{_3}F_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right) -=\sum_{m=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{m}^{2}(-n)_m}{m!^2 \left(-n-\frac{1}{2}\right)_{m} } -$$ -has only positive terms (the minus signs in the Pochhammer symbols $(-n)_m$ and $(-n-\frac{1}{2})_m$ cancel.). We get a lower bound by taking only the first two expansion terms of ${_3}F_{2}$ and the well known estimate for the Gamma functions in front -$$ -\frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} \geq n^{1/2} -$$ -and find -$$ -T_{n} \geq n^{1/2} \left(1+\frac{n}{2(n+1)}\right) -$$ -which is larger than $\sqrt{n+1}$ for $n\geq 2$. -Btw, the asymptotic ($n\rightarrow \infty$) value of $T_n$ is -$$ -T_{n}\sim \frac{4}{\pi} n^{1/2} -$$ -I am pretty sure that the calculation can be done more elegantly, but had not the time to dig deeper. -Edit: -With the same trick one finds more generally: -$$ - \int_{0}^{\infty} dx \, x^{\nu} e^{-x} L_{n}(x)^2 = \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {_3}F_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {_2}F_{1}(-\nu,-\nu;1;1)) (1+ O(n^{-1})) -$$ -The last term simplifies to: -$$ -\frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} \frac{\Gamma(2 \nu +1)}{\Gamma(\nu+1)^2} (1+ O(n^{-1})) -$$ -for $\nu > 0$. -Edit: -After some massaging with Mathematica I could determine the coefficient of $n^{-1}$ in the above expansion: -$$ - \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {_3}F_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {_2}F_{1}(-\nu,-\nu;1;1)) (1 + \frac{\nu}{2 n} + O(n^{-1-2 \nu})) -$$ -The determination of the next order coefficient as well as the exact exponent is rather cumbersome. One could do better than $O(n^{-1-2 \nu})$.<|endoftext|> -TITLE: Can a tangle of arcs interlock? -QUESTION [14 upvotes]: Can a (finite) collection of disjoint circle arcs in $\mathbb{R}^3$ be interlocked in the sense in that they cannot be separated, i.e. each moved arbitrarily far from one another while remaining disjoint (or at least never crossing) throughout? -(Imagine the arcs are made of rigid steel; but infinitely thin.) -The arcs may have different radii; each spans strictly less than $2 \pi$ in angle, so each has a positive "gap" through which arcs may pass: -      -Of course, if one could prove that in any such collection, one arc can be removed to infinity, the result would follow by induction. -But an impediment to that approach is that sometimes there is no arc than can be removed while all the others remain fixed. -Another approach would be to reduce the piercing number of the configuration: -the number of intersections of an arc with the disks on whose boundary the arcs lie. If the piercing number could always be reduced in any configuration, then it would "only" remain to prove that if there are no disk-arc piercings at all, the configuration can be separated. -Intuitively it seems that no such collection can interlock, but I am not seeing a proof. -I'd appreciate any proof ideas—or interlocked configurations! - -REPLY [2 votes]: @Cristi: Perhaps your first example needs an argument to exclude the following motion, -which might only be possible with my generous arc-gap?<|endoftext|> -TITLE: What does the Jacobian of a curve tell us about the curve? -QUESTION [17 upvotes]: A natural object in the study of curves is the Jacobian of a curve. What are some natural geometric properties of the curve that the Jacobian encapsulates? In other words, what can the Jacobian tell us about the curve that we didn't know already? -Note, I am asking for concrete examples, statements like "The Jacobian having property blah implies the curve has property blah." -Ideally these will be statements that are easier to prove using the Jacobian (whose construction is not so easy!) rather than directly from the curve. -(Also, if this question is more appropriate for math.se, I'd be happy to delete it.) - -REPLY [12 votes]: I think at least historically the Jacobian is related to the function theory over a curve which was one the main areas of research back in the 19th century. In that time given a compact Riemann surface $X$ over $\mathbb{C}$, the question was to understand the behavior of holomorphis and mermorphic functions on this curve. If we have two effective divisors $D$ and $E$ on $X$, when is $D-E$ the divisors of zeros and poles of a meromorphic function $f$ on $X$? Let the genus of $X$ be $g$. Then there are $g$ basis elements of the vector space of differential forms on $X$. The clever solution that Abel proposed for this question was this: let $\omega_{1}$,...,$\omega_{g}$ be the generators of $\Omega(X)$, the space of holomorphic differentials of $X$. Given a path $\gamma$ in $X$, the set $L=\{(\int_{\gamma}\omega_{1},...,\int_{\gamma}\omega_{g})\}$ is additive in $\mathbb{C}^{g}\cong\Omega(X)$ because of additivity property of integrals and in fact is a lattice. Therefore we can quotient out and get a group $\mathbb{C}^{g}/L$. We also get a map $ A:X\rightarrow J(X)$ by choosing a base point and $p_{0}$ and sending each point $p\in X$ to $(\int_{p_{0}}^{p}\omega_{1},...,\int_{p_{0}}^{p}\omega_{g})$ mod $L$. Abel realized that two divisors $D$ and $E$ (viewed as a collection of points on $X$) are linearly equivalent if and only if they have the same image under the map $A$. Note that the map $A:X\rightarrow J(X)$ in itself is a very interesting map: we have constructed an almost natural holomorphic map form $X$ to a variety that has a structure of a group. In the first glance it is not at all clear that we can have such a map. The second funny property is that this map is not injective if and only if $X\cong \mathbb{P}^{1}$.<|endoftext|> -TITLE: Can a tangle of arcs of ellipses interlock -QUESTION [12 upvotes]: This is a variation on an earlier question resolved by user35353: Can a tangle of arcs interlock? In that question, the arcs were restricted to circular arcs, and user35353's proof that one arc can be removed without disturbing the others relies on the circularity of the arcs. That proof fails with elliptical arcs. -The length of the major and minor axes are arbitrary, and, just as in the previous question, the arcs must leave a positive gap, i.e., they cannot be complete ellipses. - -REPLY [13 votes]: Three tangled ellipses that can't be unlocked. The major radii of the smaller ellipses have to be a bit smaller than the minor radius of the large ellipse. The minor radii of the small ellipses have to allow them to be linked, but of course smaller than the large radius, so that they can't rotate. - -Added -Let's consider an ellipse with radii $A,B$ $A\gt B$, and two identical ellipses with radii $a,b$, $a\gt b$. Assume that $a$ is very close to $B$, but still $a\lt B$. -Because $a\lt B$, there is a minimal distance $d$, so that, if we introduce the large ellipse through a small one, the distance between their centers cannot be smaller than $d$. -Because $a\gt b$, there is a maximal distance $D$, so that, if we introduce the large ellipse through a small one, and the distance between their centers is smaller than $D$, the small ellipse can't be rotated around the large one. -To find a configuration as in the figure which is locked, we look to satisfy the following conditions. - -$b\gt d$, because we want the two small ellipses to be tangled. -$b\lt D$, because we want to prevent the rotation of the small ellipses around the large one. - -To find the values satisfying these conditions, start from a configuration with $b\lt a\lt B\lt A$. Then, if the condition is not satisfied, replace $a\mapsto (a+B)/2$. Then, $d$ decreases, but $D$ remains unchanged, since it depends only on $b$. If we repeat this, $d\to 0$, so at some point $d\lt D$ and $d\lt b$. Then, $b$ can be replaced by a smaller value, which is larger than $d$, but smaller than $D$.<|endoftext|> -TITLE: Serre functor of orthogonal subcategory of derived category of cubic 4-fold -QUESTION [5 upvotes]: Let $Y$ be a smooth cubic 4-fold in $\mathbf{P}^5$. The derived category of $Y$ admits a semiorthogonal decomposition -$$D^b(Y) = \langle \mathcal{A}_Y, \mathcal{O}_Y, \mathcal{O}_{Y}(1), \mathcal{O}_{Y}(2) \rangle,$$ -where $\mathcal{A}_Y$ is the right-orthogonal to the triangulated subcategory generated by $\mathcal{O}_Y, \mathcal{O}_{Y}(1), \mathcal{O}_{Y}(2)$. Namely, $\mathcal{A}_Y$ is the full subcategory with objects those $F \in D^b(Y)$ such that $Hom_{D^b(Y)}(\mathcal{O}_Y(j), F[i]) = 0$ for $j=0,1,2$ and all $i$. -Kuznetsov studies $\mathcal{A}_Y$ in his paper "Derived categories of cubic fourfolds" http://arxiv.org/abs/0808.3351. He parenthetically remarks that the Serre functor of $\mathcal A_Y$ is the shift by two functor $[2]$. -The question is: How do you show this? -The only reference I could find for this result is Corollary 4.4 of the paper http://arxiv.org/abs/math/0303037. However, I don't understand the proof given there. Specifically, I don't know how to justify the following points in Lemma 4.2 preceding the Corollary (notation as in the paper): -1) How do you get the expression Kuznetsov gives for the kernel of $O^d$ from the kernel of $O$? -2) Why is the restriction to $Y \times Y$ of the Beilinson resolution of the diagonal in $\mathbf{P}^{n+1} \times \mathbf{P}^{n+1}$ quasi-isomorphic to ${\Delta_{Y}}_*(\mathcal O_{Y})$? It seems the paper is being sloppy here with what degrees the complexes live in: Let's call $A$ the restriction of the Beilinson resolution (including the final term ${\Delta_{Y}}_*(\mathcal O_{Y}(d))$ as in the paper). Let's place A in nonpositive degrees so that ${\Delta_{Y}}_*(\mathcal O_{Y}(d))$ is in degree $0$ (I assume this is what is intended in the paper). Then I think the precise statement should be that $A$ is quasi-isomorphic to ${\Delta_{Y}}_*(\mathcal O_{Y})[2]$. Moreover, since $A$ is the restriction of a resolution of $\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))$ by locally free sheaves, this can be rephrased as the following computation of left-derived functors: if $i : Y \times Y \hookrightarrow \mathbf{P}^{n+1} \times \mathbf{P}^{n+1}$ is the inclusion, then -$$L_1 i^*(\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))) = {\Delta_{Y}}_*(\mathcal O_{Y}) $$ -and -$$L_k i^*(\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))) = 0$$ -for $k > 1$. -How do you show this? - -REPLY [6 votes]: The first is a straightforward computation. For example, to compute $O^2$ you note that the convolution of kernels preserves exactness of the triangles, hence one has distinguished triangles -$$ -\Delta_{Y*}O_Y(1) \circ K_1 \to O_Y(2)\boxtimes O_Y \to \Delta_*O_Y(2), -$$ -$$ -(O_Y(1)\boxtimes O_Y)\circ K_1 \to V^*\otimes O_Y(1)\boxtimes O \to O_Y(1)\boxtimes O_Y(1), -$$ -(where $V^* = H^0(O_Y(1))$), and -$$ -K_1\circ K_1 \to (O_Y(1)\boxtimes O_Y)\circ K_1 \to \Delta_{Y*}O_Y(1) \circ K_1. -$$ -The second triangle implies that $(O_Y(1)\boxtimes O_Y)\circ K_1 \cong O_Y(1)\boxtimes \Omega(1)_{|Y}$, so substituting this and the first triangle into the third triangle you get an expression for $O^2$. Then you repeat this computation. -For the second you need to compute $L(\alpha\times\alpha)^*\Delta_*O$ (in the notation of the paper). Let us first compute $(\alpha\times\alpha)_*L(\alpha\times\alpha)^*\Delta_*O$ instead. By projection formula it is isomorphic to -$$ -(\alpha\times\alpha)_*L(\alpha\times\alpha)^*\Delta_*O \cong (\alpha\times\alpha)_*O_{Y\times Y} \otimes^L \Delta_*O_P. -$$ -To compute this we use the resolution -$$ -0 \to O(-d)\boxtimes O(-d) \to O(-d)\boxtimes O \oplus O \boxtimes O(-d) \to O \boxtimes O -$$ -of $(\alpha\times\alpha)_*O_{Y\times Y}$. Tensoring it with $\Delta_*O_P$ one gets the complex -$$ -0 \to \Delta_*O_P(-2d) \to \Delta_*O_P(-d) \oplus \Delta_*O_P(-d) \to \Delta_*O_P -$$ -which is a direct sum od the resolution of $\Delta_*O_Y$ and of $\Delta_*O_Y(-d)[1]$. Thus -$$ -(\alpha\times\alpha)_*L(\alpha\times\alpha)^*\Delta_*O \cong \Delta_*O_Y \oplus \Delta_*O_Y(-d)[1]. -$$ -Finally, since $\alpha\times\alpha$ is a closed embedding, the functor $(\alpha\times\alpha)_*$ is exact and conservative, hence the claim.<|endoftext|> -TITLE: Are there infinitely many integer-valued polynomials dominated by $1.9^n$ on all of $\mathbb{N}$? -QUESTION [63 upvotes]: The original post is below. Question 1 was solved in the negative by David Speyer, and the title has now been changed to reflect Question 2, which turned out to be the more difficult one. A bounty of 100 is offered for a complete solution. -Original post. It follows from the prime number theorem and the periodicity properties $f(n+p) \equiv f(n) \mod{p}$ that for each $A < e$ there are only finitely many integer polynomials $f \in \mathbb{Z}[x]$ such that $|f(n)| < A^n$ for all $n \in \mathbb{N}$. On the other hand, for each $k \in \mathbb{N}$ the binomial coefficient $\binom{n}{k}$ is an integer-valued polynomial in $n$ bounded by $2^n$. -Question 1. Are there infinitely many integer polynomials with $|f(n)| < e^n$ for all $n \in \mathbb{N}$? -Question 2. Given $A < 2$, are there only finitely many integer-valued polynomials $f \in \mathbb{Q}[x]$ with $|f(n)| < A^n$ for all $n \in \mathbb{N}$? - -REPLY [44 votes]: The optimal growth rate is $\tau:= (1+\sqrt{5})/2$. Specifically, for any $\epsilon>0$, there are infinitely many integer valued polynomials bounded by $(\tau+\epsilon)^n$, but only finitely many below $(\tau-\epsilon)^n$. The first part of this answer (written first) proves the finiteness; the second uses Noam Elkies' idea combined with a theorem of Fekete to prove the infinitude. - -Fix $\epsilon>0$. I will show that there are only finitely many integer values polynomial $f(z)$ with $|f(n)| < (\tau-\epsilon)^n$. -Let $f$ be such a polynomial of degree $d$. Set -$$\frac{p(z)}{(1-z)^{d+1}} = \phi(z) = \sum_{n=0}^{\infty} f(n) z^n$$ -Then $p(z)$ has integer coefficients, $p(1) \neq 0$, and we can uniquely recover $f$ from $p$. Moreover, there is some $M$ and some $\delta_1>0$ (dependent on $\epsilon$) so that $|\phi(z)| < M$ on $|z|=\tau^{-1}+\delta_1$. -We make the change of variables $u = 1/(1-z)$, so $z=1-1/u$. We have $\phi(1-1/u) = p(1-1/u) u^{d+1}$. Set $q(u) = p(1-1/u) u^{d+1}$. From the properties of $p$ above, $q$ is a polynomial with integer coefficients of degree $d+1$, and $|q(1/(1-z))| < M$ when $|z|=\tau^{-1}+\delta_1$. The map $z \mapsto 1/(1-z)$ sends $|z|=\tau^{-1}+\delta_1$ to a circle which contains the circle of radius $1+\delta_2$ around $\tau$ (for some $\delta_2>0$). So, using the maximum modulus principle, $|q(u)| T$, we have -$$r^N + (1/2) r^{N-1} + (1/2) r^{N-2} + \cdots + (1/2) r^{T+1} + (1/2) r^T < 1/3.$$ -Take $N$ larger than $T$. Define $q^N_N(u) = (u-\tau)^N$. Define $q^N_i(u)$ to be the unique polynomial of the form -$$q^N_i(u) = q^N_{i+1}(u) + \theta_i \cdot (u-\tau)^{i}$$ -so that $|\theta_i| \leq 1/2$ and the coefficient of $u^{i}$ in $q^N_i$ is an integer. Set $q^N(u) = q^N_T(u)$. So the coefficient of $u^k$ in $q^N(u)$ is an integer for $T \leq k \leq N$. -For $u$ on the circle $|u-\tau|=r$, we get -$$|q^N_T(u)| \leq r^N + (1/2) r^{N-1} + \cdots + (1/2) r^T < 1/3.$$ -Let $(c^N_{T-1}, C^N_{T-2}, \ldots, c^T_0)$ be the last $T$, noninteger, coefficients of $q^N$. By the Pigeonhole principle, we can find $q^M$ and $q^N$ so that -$$\sum_i |\{ c^N_i - c^M_i \}| r^i < 1/3$$ -where $\{ x \}$ is the distance from $x$ to the nearest integer. We define $q(u)$ to be the result of taking $q^N(u) - q^M(u)$ and rounding the last $T$ coefficients to the nearest integer. We have now constructed $q$. -We now undo the above argument. Since $|q(u)|<1$ for $|u-\tau| -TITLE: An intuition for three different types of subgradients (proximal, regular, limiting) -QUESTION [13 upvotes]: I'm having a bit of difficulty getting my head around the different types of subgradients we're currently covering in a nonsmooth optimisation class I'm taking. -These subgradients are (assume $x \in$ dom$f$): - -Proximal subgradient: -$\partial_p(f(x)) = \{v\ |\ \exists \delta>0,\rho > 0\ s.t.\ f(y) \geq f(x) + \langle v, y-x\rangle - \frac{1}{2}\rho ||y - x||^2\ \forall\ y \in B_\delta(x) \}$ -regular subgradient: $\{v\ |\ \exists\ \delta > 0\ s.t. f(y) \geq f(x) + \langle v, - y-x\rangle + o(||y-x||)\ \forall\ y\in B_\delta(x) \}$ -limiting subgradient, defined by $\partial^\infty(f(x)) = \{v\ |\ (v, 0) \in N_{epi f}(x, f(x))\}$, where $N_C(x)$ is the limiting normal cone $\limsup_{x'\rightarrow _C x}\hat{N}_C(x)$ and $\hat{N}_C(x)$ is the normal cone to $C$ at $x$, with $v \in \hat{N}_C(x)$ iff $\langle v, y - x\rangle \leq o(||y-x||)\ \forall\ y\in C$ - -However, I'm having trouble conceptualising these; for instance, playing with the relatively straightforward function -$f(x) = -|x|$ -it seems like all three of these subgradients coincide ($x = 0$ has no subgradient, everything else has $\partial f(x) = \nabla f(x)$ since the limiting normal cone coincides with the normal cone and there exists no $\delta$ for which $B_\delta(0)$ does not intersect $f(x)$), but playing with the less straightfoward -$f(x) = \begin{cases}x^2 \sin(1/x) &\text{if }x \not= 0\\0 &\text{if } x = 0\end{cases}$ -just leaves me completely utterly lost; I can't see how the difference between a limiting normal cone and a normal cone makes a difference, I'm struggling to derive the proximal subgradient, etc. -If anyone could provide some intuition for these three concepts, preferably while working through the two functions I've been playing with, that would be fantastic! - -REPLY [6 votes]: This might not be a satisfying answer, but this is how I personally deal with this issue (at least in the topic of optimization). -I think these concepts are not made for actually computing them but rather for proofs and theoretical statements. -Let's say you want to study a nonsmooth, nonconvex optimization problem. Usually, you cannot hope to find the global minimum, but only apply some kind of iterative method. But what is this iterative method going to compute? A local minimum? Even in the smooth, nonconvex case this might fail and you can only hope for stationary points (which might turn out to be saddle points). -So what kind of object is your iterative method even supposed to compute? Evidently we need to generalize the concept of first order stationarity. What is the "right" generalization, you might ask. Well this is where the different subdifferentials come into play. Depending on the properties of your objective function (e.g. locally lipschitz or not) and the type of iterative scheme you are using the "right" notion might vary. - -Do you need good calculus? -Do you need the subdifferential to be convex? -Do you need to graph of the subdifferential map to be closed? (e.g. Murdukhovich, aka limiting differential) - -To be at least a bit more concrete: Consider an iterative scheme to solve a given nonsmooth, nonconvex optimization problem -$$ -\min_x f(x) -$$ -that produces a sequence ${(x_n)}$ with $x_n \to \bar{x}$. Let's assume that you can prove that at each iteration there is an object $w_n \in \partial f(x_n)$ that fulfills $ w_n \to 0$. You clearly would like to deduce that $0 \in \partial f(\bar{x})$. For this you would need the closedness of the graph of the map $ x \mapsto \partial f(x)$. Last but not least it would be nice to have a theorem saying that if $x$ is a local minimizer, then it stationary wrt your given subdifferential. -If you can prove all these statements, you know that the iterates of your optimization method converge to "good" points. (And as you can see - we used the concept without actually ever computing it.) -Here a very concrete example showing how you can work with different subdifferential without ever computing them: -Assume the problem at hand is -$$ -\min_x h(x) + g(x) -$$ -such that $h$ is smooth, but nonconvex and $g$ is nonsmooth, but convex. A possible algorithm, utilizing the proximal operator, would be -$$ -x_{k+1} = \text{prox}_g (x_k - \nabla h (x_k) ) -$$ -which is equivalent to -$$ -x_{k+1} = \text{argmin}_y \left\{ g(y) + \langle \nabla h (x_k), y - x_k \rangle + \frac12 || y - x_k ||^2 \right\}. -$$ -The function to be minimized in the last expression is a convex function so we can use first order optimality and some calculus of the convex subdifferential, to deduce -$$ -0 \in \partial g(x_{k+1}) + \nabla h (x_k) + x_{k+1} - x_k. -$$ -Thus -$$ -\nabla h (x_{k+1}) - \nabla h (x_k) + x_k - x_{k+1} \in \partial g(x_{k+1}) + \nabla h (x_{k+1}) . -$$ -Here is where the magic happens! If you have good calculus for your subdifferential, e.g. the Frechet subdifferential $\partial^F$, then -$$ -\partial g(x_{k+1}) + \nabla h (x_{k+1}) = \partial^F (g + h) (x_{k+1}). -$$ -Now you have an expression, namely $\nabla h (x_{k+1}) - \nabla h (x_k) + x_k - x_{k+1}$, for which you can usually deduce convergence to zero, inside your subdifferential. The only things you ever had to compute where regular gradients and and proximal operators. Note however, that proximal operators can usually only be analytically computed for "simple" functions (similar to the subdifferential). In nonsmooth optimization it is therefore assumed that the nonsmoothness comes from a "simple" term (like the absolute value) and one tries to put all the "complicated" parts into the smooth function.<|endoftext|> -TITLE: When did the meaning of the term "metabelian" change? -QUESTION [6 upvotes]: I just realised that the meaning of the term "metabelian", when applied to groups, or Lie algebras, seems to have changed over years. (These days, it means that $[[G,G],[G,G]]$ is trivial, while in the past it was occasionally used to indicate that $[[G,G],G]$ is trivial. The difference here is that between solvability and nilpotence, that is.) -This wiki says "The concept and term metabelian group was introduced by Furtwangler in 1930. -The term metabelian was earlier used for groups of nilpotence class 2, but is no longer used in that sense." (I don't understand "earlier" here. Can that sentence be parsed uniquely? Earlier than Furtwangler introduced the term? Earlier than the wiki article was written?) -I know at least one reference from mid 1960s (a PhD thesis from the US) where that old-fashioned usage is present, and I'd like to understand the history better in this instance. Thanks for help. -EDIT: it appears that the situation with usage and its history may be even different for groups and Lie algebras; I only dealt with literature on metabelian Lie algebras, and from answers so far I gather that there may be difference. - -REPLY [3 votes]: The 1967 book Varieties of Groups by Hanna Neumann has the following footnote (p. 21): The term metabelian will always mean solvable of length two in agreement with current English usage; note however that in much of the Russian literature the term is used in the sense of `nilpotent of class two'.<|endoftext|> -TITLE: Are non-isomorphic covers of riemann surfaces also generally nonisomorphic as riemann surfaces? -QUESTION [7 upvotes]: Suppose you've got a Riemann surface $E$, and two topological covers $X,Y\rightarrow E$. Suppose $X,Y$ are nonisomorphic topological covers of $E$, then would you expect $X,Y$ as Riemann surfaces (with their unique complex structure coming from the covering map) to be nonisomorphic? -I know there are some counterexamples. I was just wondering if these counterexamples are rare, and in what sense (and why) they're rare. - -REPLY [6 votes]: There's a remarkable theorem of Margulis that pertains to your question. Let $G$ be a semisimple Lie group (in your case, $PSL_2(\mathbb{R})$), and let $\Gamma$ be an irreducible lattice in $G$. The commensurator of $\Gamma$ is defined as $$Comm(\Gamma)=\{ g\in G\ |\ [\Gamma: g\Gamma g^{-1}\cap \Gamma]<\infty\},$$ the subset of $G$ which "almost" normalizes $\Gamma$. For example, if $\Gamma <\Gamma'$ a larger lattice, then $\Gamma' \leq Comm(\Gamma)$ (notice that if $\Gamma$ is not normal in $\Gamma'$, then $g\Gamma g^{-1}\neq \Gamma$ for some $g\in \Gamma'$). -Margulis' theorem implies that either $Comm(\Gamma)$ is discrete, or $\Gamma$ is arithmetic. If $\Gamma$ is non-uniform (and $G=PSL_2(\mathbb{R})$), so $\mathbb{H}^2/\Gamma$ is finite volume but non-compact, then $\Gamma$ is commensurable with $PSL_2(\mathbb{Z})$, in other words, there is $g\in PSL_2(\mathbb{R})$ so that $[\Gamma: \Gamma \cap g PSL_2(\mathbb{Z}) g^{-1}] <\infty$. If $\Gamma$ is uniform, so $\mathbb{H}^2/\Gamma$ is a compact Riemann surface, then the description is a bit more complicated, but basically $\Gamma$ is commensurable with a discrete group defined by integral automorphisms of a ternary quadratic form of signature $(2,1)$ defined over a totally real number field, and which is definite at all other infinite places, using the isogeny $PGL_2(\mathbb{R})\cong O(2,1;\mathbb{R})$ (I think in the number theory lingo these are known as Shimura curves). -Now, suppose in your question that $X$ and $Y$ are finite area (in the unique complete hyperbolic metric given by the uniformization theorem; I can't say much of anything about the infinite area case), so define finite-sheeted covers of $E$. We may associate a torsion-free lattice $\Gamma < PSL_2(\mathbb{R})$ so that $E=\mathbb{H}^2/\Gamma$, and finite-index subgroups $\chi, \gamma < \Gamma$ so that $\mathbb{H}^2/\chi = X, \mathbb{H}^2/\gamma=Y$. Suppose $X$ and $Y$ are isomorphic Riemann surfaces, then there is $g\in PSL_2(\mathbb{R})$ so that $\chi = g\gamma g^{-1}$. Therefore $\Gamma > \chi = g \gamma g^{-1} < g \Gamma g^{-1}$, so $[\Gamma : \Gamma \cap g\Gamma g^{-1}] <\infty$, so $g\in Comm(\Gamma)$. -Thus, by Margulis' theorem, either $\Gamma$ is arithmetic, or $\Gamma$ is non-arithmetic, and therefore $\Lambda=Comm(\Gamma)$ is discrete. Then $\mathbb{H}^2/\Lambda$ is a finite-area hyperbolic orbifold $\mathcal{O}_\Lambda$, and one has that $g\Lambda g^{-1} =\Lambda$. This implies that $E\to \mathcal{O}_\Lambda$ is a finite-sheeted orbifold cover, and the two covers $X,Y\to E\to \mathcal{O}_\Lambda$ are equivalent (irregular) covers of $\mathcal{O}_\Lambda$. -Notice that if $\mathcal{O}_\Lambda$ has non-trivial modulus (so it's not a turnover), then this gives rise to parameter space of such covers of Riemann surfaces of the same genus as $E$. -In the case $\Gamma$ is arithmetic, $Comm(\Gamma)$ is much larger. For example, $Comm(PSL_2(\mathbb{Z}))\cong PGL_2(\mathbb{Q})$. Such pairs of covers then are an important area of study in the theory of autormorphic forms, since they give rise to Hecke operators. However, for a fixed genus, there are only finitely many arithmetic Riemann surfaces of that genus (see e.g. Long-Reid for a more general result for orbifolds), so in particular there are only finitely many such examples for which $X$ and $Y$ have bounded area. There is (in principle) a complete arithmetic prescription for how such covers may occur in this case as well which is determined by the description of the commensurator.<|endoftext|> -TITLE: growth rate of $\mathbb{Z}^2\rtimes_{\sigma} \mathbb{Z}$? -QUESTION [6 upvotes]: I am interested in the growth rate of this type of group: $G=\mathbb{Z}^2\rtimes_{\sigma} \mathbb{Z}$, where $\sigma(a)=\begin{pmatrix}x&y\\z&w\end{pmatrix}\in SL_2(\mathbb{Z})$, where $a$ is the generator on the right copy of $\mathbb{Z}$ and the action is just by matrix multiplication. -Here are two examples: -For $\sigma(a)=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, this gives us the discrete Heisenberg group $H_3$, which is nilpotent, and hence by Gromov's theorem, it has polynomial growth rate(see here). -When $\sigma(a)=\begin{pmatrix}2&1\\1&1\end{pmatrix}$, it was mentioned here this group has exponential growth rate. -So my first question is: -1, Could anyone give me a reference to show the link between whether the above group $G$ has polynomial growth rate or not and the property, say eignvalue, of the matrix $\sigma(a)$? -Note that the above $G$ is a polycyclic-by-finite group, my question is: -2, Could anyone give me a polycyclic-by-finite group not of the type of $G$ with exponential growth rate? - -REPLY [4 votes]: To answer your 1st question see the paper of Milnor, "Growth of finitely generated solvable groups." -To answer your 2nd question, simply generalize your second example to higher dimensions, e.g. take $\mathbb{Z}^3\rtimes_{\sigma} \mathbb{Z}$ where $\sigma \in SL_3(\mathbb{Z})$ has an eigenvalue not on the unit circle.<|endoftext|> -TITLE: Counterexamples to analogue of Cannon conjecture in higher dimensions -QUESTION [8 upvotes]: It is known that a group $G$ acts geometrically on $\mathbb{H}^2$ if and only if $G$ is word-hyperbolic and its boundary $\partial G$ is homeomorphic to $S^1$. -The analogous statement for $\mathbb{H}^3$ and $S^2$ is open and is a conjecture of Cannon. -I read somewhere that this fails in higher dimensions, but I can't find an explicit counterexample. Could somebody provide one (in dimension 4)? -Thanks. - -REPLY [10 votes]: There are various compact manifolds of negative curvature which are not homnotopy-equivalent to closed hyperbolic manifolds: Locally symmetric ones (complex hyperbolic, etc) as well as Gromov-Thurston and Mostow-Siu examples. Their $\pi_1$'s are Gromov-hyperbolic, boundary is a topological sphere. The examples exist in all dimensions $\ge 4$ (Gromov-Thurston).<|endoftext|> -TITLE: Smallest regular simplex containing the unit cube in $R^n$ -QUESTION [15 upvotes]: What is the length $e_n$ of the edge of the smallest $n$-dimensional regular simplex $S_n$ containing the $n$-dimensional unit cube $Q_n$? -In particular, is there $n$ such that $e_n<\sqrt{2}(n+1-\sqrt{n+1})$? - -The question is motivated by the question Two cubes in unit cube. A positive answer to the second part would (almost) imply that two cubes with side of length greater than $1/2$ would fit inside the unit cube without overlap. Where "almost" means "if the centers of the cube and the simplex are not too far away": the plan would be to cut the unit cube by a hyperplane orthogonal to a main diagonal and fit the small cube inside each of the two halves. -Since $Q_n$ contains a ball $B_n$ of diameter $1$, the simplex $S_n$ also contains $B_n$. Therefore, the altitude of $S_n$ is at least $(n+1)/2$ and so $e_n \ge \frac{1}{\sqrt 2} \sqrt{n(n+1)}$. -On the other hand, a "greedy" recursive construction (where a facet of $Q_n$ is contained in a facet of $S_n$) gives the upper bound $e_n \le \sqrt{2}\left(\sqrt{\frac{1}{2}} + \sqrt{\frac{2}{3}} + \cdots +\sqrt{\frac{n}{n+1}}\right) = \sqrt{2}n - \Theta(\log n)$. -The three-dimensional case hase been completely solved: Ogilvy and Robbins, 1976, Croft, 1980 -It has been shown that the greedy construction is indeed optimal in $\mathbb{R}^3$; however, two other locally minimal regular tetrahedra are larger just by a few percent. - -REPLY [7 votes]: This paper by M. Nevskii is relevant. It is shown there that a simplex contains a translate of $[0,1]^n$ iff $\sum_i\frac1{d_i}\leq 1$, where $d_i$ is the length of the longest segment inside the simplex which is parallel to the $i$th axis. In particular, this clearly implies $e_n\geq n$ (and this is far from being sharp).<|endoftext|> -TITLE: What is a homotopy between $L_\infty$-algebra morphisms -QUESTION [18 upvotes]: A $L_\infty$-algebra can be defined in many different ways. One common way, that -gives the 'right' kind of morphisms, is that a $L_\infty$-algebra is a graded cocommutative and coassociative coalgebra, cofree in the category of locally nilpotent differential graded coalgebras and their morphisms are coalgebra -morphisms that commute with the codifferential. -Breaking this compact definition down into something more concrete, the category of $L_\infty$-algebras can equally be defined in the following way: -A $L_\infty$-algebra is a $\mathbb{Z}$-graded vector space $V$, together with a sequence of graded anti-symmetric, $k$-linear maps -$D_k:V \times \cdots \times V \to V$, -homogeneous of degree $-1$,such that the 'weak' Jacobi identity -$ -\sum_{p+q=n+1}\sum_{\sigma \in Sh(q,n-q)}\epsilon(\sigma;x_1,\ldots,x_n) -D_p(D_q(x_{\sigma(1)},\ldots,x_{\sigma(q)}),x_{\sigma(q+1)},\ldots,x_{\sigma(n)})=0 -$ -is satisfied, for any $n\in\mathbb{N}$. Where $\epsilon$ is the Koszul sign and $Sh(p,q)$ is the set of suffle permutations. -A morphism of $L_\infty$-algebras $(V,D_{k\in\mathbb{N}})$ -and $(W,l_{k\in\mathbb{N}})$ is a sequence $f_{k\in\mathbb{N}}$ of -graded-antisymmetric, $k$-linear maps -$ -f_k : V\times \cdots \times V \to W -$ -homogeneous of degree zero, such that the equation -$ -\sum_{p+q=n+1}\sum_{\sigma \in Sh(q,n-q)}\epsilon(\sigma;x_1,\ldots,x_n) -f_p(D_q(x_{\sigma(1)},\ldots,x_{\sigma(q)}),x_{\sigma(q+1)},\ldots,x_{\sigma(n)})=\\ -\sum_{k_1+\cdots+k_j=n}^{k_i\geq 1}\sum_{\sigma \in Sh(k_1,\ldots,k_j)} -\epsilon(\sigma;x_1,\ldots,x_n) -l_j(f_{k_1}(x_{\sigma(1)},\ldots,x_{\sigma(k_1)}),\ldots, -f_{k_j}(x_{\sigma(n-k_j+1)},\ldots,x_{\sigma(n)})) -$ -is satisfied, for any $n\in\mathbb{N}$. -This defines the category of $L_\infty$-algebras, sometimes called the category of $L_\infty$-algebras with weak morphisms. -Now after that long and tedious definition, the question is: -What is a reasonable definition of a homotopy between two (weak) morphisms -$f:V\to W$ and $g:V\to W$ of $L_\infty$-algebras? (And why?) -Edit: A lot of information pointing towards a definition of such a homotopy -(or 2-morphism in $(\infty,1)$-categorical language) is spread out in the net. -Much on the $n$-category cafe, like in -https://golem.ph.utexas.edu/category/2007/02/higher_morphisms_of_lie_nalgeb.html -and in the nLab. However it looks like an explicit equation still isn't available. -I would do the tedious calculations myself, since I can get a lot of joy out of -such huge and delicate computations, but I'm unable to finde a calculable way to -achive that goal. (Such a way should have the potential to apply to the higher homotopies too, hopefully leading towards an explicit description of the hom-space in this category) -P.S.: The tags are not very well suited, feel free to change them - -REPLY [2 votes]: There is a plethora of model structures for L-infinity algebras (going back to Quillen of course, but notably described and related in the great article by Jonathan Pridham arXiv:0705.0344). Also structures of categories of fibrant objects. Each of these induces a model for homotopies of 1-morphisms of $L_\infty$-algebras, for instance a right homotopy given by mapping into a path space object. What these are can be worked out for each of these model category/category of fibrant object structures (and all these notions will be suitably equivalent). -An explicit model of such path space objects for $L_\infty$-algebras is discussed by Dolgushev in section 5 of his article arXiv:0703113. -See on the nLab at model structure for L-infinity algebra -- Homotopies and derived hom spaces.<|endoftext|> -TITLE: How hard is Heyting satisfiability, i.e. the constructive version of SAT? In particular, is 2-HSAT NL-complete or is it harder? -QUESTION [10 upvotes]: First of all, is it clear what I mean by $k$-HSAT? -I'm assuming that for $k>2$, $k$-HSAT is NP-complete, but the details of the reductions between $k$-HSAT and $k$-SAT aren't obvious to me. -I'm more curious about whether $2$-HSAT is as easy as $2$-SAT, since as far as I know the latter is easy because, in a Boolean algebra, $a\vee b$ is the same as $\neg a\implies b$, so that a conjunction of clauses of the form $a\vee b$ is just a bunch of implications, and the algorithms I've seen for efficiently solving $2$-SAT use the implication graph that this defines. - -REPLY [6 votes]: First of all, we need to be clear about how we are defining HSAT. Over classical logic we could consider the following two definitions: - -A formula $\phi(x_1,\ldots,x_n)$ with proposition variables amongst $x_1,\ldots,x_n$ is satisfiable if there are $b_i \in \{\top, \bot\}$ such that $\phi(b_1,\ldots,b_n)$ is true. I'll refer to the corresponding decision problem as SAT. -A formula $\phi(x_1,\ldots,x_n)$ with proposition variables amongst $x_1,\ldots,x_n$ is falsifiable (I'm not sure what the standard name for this is) if there are $b_i \in \{\top, \bot\}$ such that $\phi(b_1,\ldots,b_n)$ is not true. I'll refer to the corresponding decision problem as FALSE. - -Over classical logic, this isn't such a useful distinction to make, because $\phi$ is satisfiable if and only if $\neg \phi$ is falsifiable (and vice versa). In particular, solving FALSE is equivalent to solving SAT. -By analogy with satisfiability, we could give the following definitions for a propositional formula $\phi(x_1,\ldots,x_n)$: - -there is a non-trivial Heyting algebra $\mathcal{H}$ and $h_1,\ldots,h_n \in \mathcal{H}$ such that $\phi(h_1,\ldots,h_n) = \top$ -there is a Kripke model, $\langle W, \leq, \Vdash \rangle$ with root node $0 \in W$, such that $0 \Vdash \phi$ -$\neg \phi$ is not provable in intuitionistic logic - -By applying some soundness and completeness theorems, the three definitions above are equivalent. If they hold for $\phi(x_1,\ldots,x_n)$, say that $\phi$ is H-satisfiable. I'll refer to the corresponding decision problem as HSAT. -By analogy with falsifiability, we could give the following definitions for a propositional formula $\phi(x_1,\ldots,x_n)$: - -there is a Heyting algebra $\mathcal{H}$ and $h_1,\ldots,h_n \in \mathcal{H}$ such that $\phi(h_1,\ldots,h_n) \neq \top$ -there is a Kripke model, $\langle W, \leq, \Vdash \rangle$ with root node $0 \in W$, such that $0 \nVdash \phi$ -$\phi$ is not provable in intuitionistic logic - -By applying some soundness and completeness theorems, the three definitions above are equivalent. If they hold for $\phi(x_1,\ldots,x_n)$, say that $\phi$ is H-falsifiable. I'll refer to the corresponding decision problem as HFALSE. -H-Satisfiability -Note that the correspondence we had before in classical logic no longer holds. In fact, for satisfiability we can say the following. By applying Glivenko's theorem, we can show that $\phi$ is H-satisfiable if and only if it is satisfiable. In particular HSAT is in fact exactly the same problem as SAT and $k$-HSAT is the same as $k$-SAT. -H-Falsifiability -As Yoav Kallus pointed out in the comment, this is covered in the paper R. Statman, Intuitionistic propositional logic is polynomial-space complete Theoretical Comput. Sci., 9:67{72, 1979. What I called HFALSE is PSPACE-complete. -This only leaves the case of $k$-HFALSE, which unfortunately turns out to not be that interesting. $k$-SAT is the decision problem for formulas in conjunctive normal form whose clauses have at most $k$ variables. Note that if $\phi$ is such a formula, then we can easily put $\neg \phi$ into disjunctive normal form where each conjunctive clause has at most $k$ variables, and vice versa. Hence it makes sense to define $k$-FALSE and $k$-HFALSE as the decision problems for formulas in disjunctive normal form with at most $k$ variables in each clause. In particular, this makes $k$-FALSE equivalent to $k$-SAT. -However, it turns out that for any $\phi$ in disjunctive normal form, $\phi$ is H-falsifiable (by the way, this is also true for conjunctive normal form). You can show this by constructing a Kripke model with two nodes. At the root node set every proposition variable to false and at the other node set every proposition variable to true. For any variable, $x$, neither $x$ nor $\neg x$ holds at the root node, so neither does any formula built from them with only conjunction and disjunction. -So, $k$-HFALSE is trivial.<|endoftext|> -TITLE: Shimura surfaces that do not contain a Shimura curve -QUESTION [10 upvotes]: Let $S$ be a Shimura surface i.e. a Shimura variety with $dimS=2$. Does $S$ necessarily contain a Shimura curve? I know that probably the answer is No, but do not have an explicit example. What is the best example of such a surface? - -REPLY [4 votes]: A trick: for a totally real abelian cubic number field $k$, let $B$ be a quaternion algebra over $k$ that is split at exactly two of the real places of $k$. The associated Shimura variety is two-dimensional, but does not contain any natural Shimura curve. (But I do not know how to exclude the possibility of "un-natural" imbeddings...)<|endoftext|> -TITLE: Limits in span categories -QUESTION [5 upvotes]: What are the limits in the span categories? and what is known about them in the literature? - -REPLY [3 votes]: I can think of two interesting classes of limits in bicategories of spans (which, as Urs mentioned, are necessarily also colimits, since $\mathrm{Span}(C)$ is equivalent to its opposite). -The first is a generalization of Urs's comment: any van Kampen colimit in $C$ is a (co)limit in $\mathrm{Span}(C)$ — and conversely, this property characterizes van Kampen colimits. This is in - -Pawel Sobocinski and Tobias Heindel, Being Van Kampen is a universal property, arXiv:1101.4594 - -The second is that $\mathrm{Span}(C)$ always has Eilenberg-Moore objects for comonads (a sort of 2-categorical limit) — and this figures in a characterization of bicategories of the form $\mathrm{Span}(C)$. This is in - -Stephen Lack, R.F.C. Walters, and R.J. Wood, Bicategories of spans as cartesian bicategories, TAC 24 (1).<|endoftext|> -TITLE: Theorems on sets provable in KM but not in ZFC set theory -QUESTION [8 upvotes]: While the language of ZFC set theory does not admit classes, this is not the case for NBG set theory. But the language of NBG does not allow quantification on symbols for classes, and it is known that NBG is conservative on ZFC for sets, meaning that every theorem concerning only sets provable in NBG is also provable in ZFC. But Kelley-Morse (KM) set theory allows quantification on symbols for classes, and is known not to be conservative for sets on ZFC. Indeed, the language of KM admits more formulas for classes, so probably more classes, and among them more sets. My question is: "What are examples of theorems on sets provable in KM but not in ZFC set theory?" -Gérard Lang - -REPLY [10 votes]: KM proves that for any class $X$, there are class club many cardinals $\delta$ that are fully $X$-correct, meaning that $\langle V_\delta,{\in},X\cap V_\delta\rangle\prec\langle V,{\in},X\rangle$ (and furthermore, this elementarity is expressible in KM, unlike GBC or ZFC). The reason is that KM proves that there is a satisfaction class $S$ for $\langle V,{\in},X\rangle$, a truth predicate indicating which formulas are true at which points in this (limited) structure, and we may then apply the reflection theorem to $\langle V,{\in},S\rangle$, to find a club of $\delta$ which are $X$-correct. -This way of thinking produces many theorems purely about sets that are provable in KM, such as: -Theorems. (KM) - -Con(ZFC). Indeed, there is a transitive model of ZFC. -Moreover, every set is in some $V_\delta$ that is a model of ZFC. In other words, there is a proper class of worldly cardinals. This is a weak formulation of the Grothendieck axiom of universes, using worldly cardinals in place of inaccessible cardinals. -Even stronger, the worldly cardinals form a stationary proper class. And this is true even when one adds any given class predicate $X$, since any fully $X$-correct cardinal is $X$-worldly. So there are $\alpha$-worldly cardinals of any degree $\alpha$. -If there are unboundedly many inaccessible cardinals (or measurable cardinals or supercompat cardinals, as you like), then there is a worldly limit of such cardinals. - -Nevertheless, one should not think of the large cardinal strength of KM as being very great, since if $\kappa$ is an inaccessible cardinal, then $\langle V_\kappa,{\in},V_{\kappa+1}\rangle$ satisfies KM, and so we have a comparatively low upper bound on consistency strength. -Meanwhile, I don't quite agree with your remarks about the syntactic difference between GBC and KM. One can view them both as formalized in the two-sorted language, with variable symbols for sets and variable symbols for classes, and there are exactly the same formulas in the two theories. It is just that GBC does not allow the replacement and separation axioms for formulas using class quantifiers, while KM does allow them. But meanwhile, it is perfectly sensible to speak of GBC proving or refuting formulas that happen to have class quantifiers, or to speak of such formulas being independent of GBC.<|endoftext|> -TITLE: Iwasawa and $KAK$ Decomposition for Diff$(S^1)$ -QUESTION [10 upvotes]: It is well known -$\newcommand{\Diff}{\operatorname{Diff}}$ - that the group $\Diff(S^1)$ of smooth diffeomorphisms of the circle behaves in many ways like $SL(2,\mathbb R)$. For example, if $S^1\subseteq \Diff(S^1)$ is the group of rotations then $\Diff(S^1)/S^1$ is an infinite dimensional complex domain (apparently this is due to Kirillov and Yuriev). This is analogous to the symmetric space $SL(2,\mathbb R)/S^1$ which is the complex upper half plane. -Now $SL(2,\mathbb R)$ has some very useful decompositions: the Iwasawa decomposition (or the $KAN$ decomposition) and the $KAK$ decomposition. -Does $\Diff(S^1)$ also have $KAN$ and $KAK$ decompositions? - -REPLY [7 votes]: At least for the $KAN$ decomposition, there is a candidate. -Namely take $K$ the rotation group, $A$ the group of hyperbolic elements of $PSL(2,\mathbb{R})$ fixing $0$ and $\infty$, and $N$ the group of diffeos fixing $0$ and tangent to the identity there. -Then $G=KAN$ is the group $Diff_+(\mathbb{S}^1)$, and the decomposition of elements is unique, as is easily checked. -No idea yet for the $KAK$ decomposition.<|endoftext|> -TITLE: Hn(X, OX) = 0 for X birational to a regular affine variety? -QUESTION [7 upvotes]: It is a basic fact that $H^n(X, F) = 0$ if $X$ is noetherian affine, $n > 0$, and $F$ a quasi-coherent sheaf. -If $Y \to X$ is a blow-up of a smooth variety in a smooth center, then then exceptional divisor is a projective bundle over the center, and so $H^n(Y, \mathcal{O}_Y) = H^n(X, \mathcal{O}_X)$. (right?) -I have not seen any examples of blow-ups of smooth varieties (with arbitrary center) whose fibres are not unions of projective spaces (sub-question: do such blow-ups exist?). -So I am lead to wonder: is it always true that $H^n(Y, \mathcal{O}_Y) = 0$ if $f: Y \to X$ is birational, $Y$ is integral, and $X$ is smooth and affine? -Edit: I mean't to include the hypothesis that $f$ is proper, and that $n$ is any integer $> 0$. - -REPLY [4 votes]: As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper. -For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599. -For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here). -The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities -Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in. -Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.<|endoftext|> -TITLE: Perturbation of unbounded self-adjoint operators -QUESTION [16 upvotes]: In the paper "A CRITERION FOR THE NORMALITY OF UNBOUNDED OPERATORS AND APPLICATIONS TO SELF-ADJOINTNESS" by M. H MORTAD (http://arxiv.org/pdf/1301.0241.pdf), the author states the following theorem -Theorem 1 -Let $A$ and $B$ be two operators with domains $D(A)$ and $D(B)$ respectively, such that $A+B$ is densely defined. Then $A+B$ is - -self-adjoint on $D(A)$ if $A$ and $B$ are selfadjoint and if B is bounded. -self-adjoint on $D(A) \cap D(B)$ whenever $A$ and $B$ are commuting selfadjoint and positive operators. -self-adjoint on $D(A) \cap D(B)$ whenever $A$ and $B$ are anti-commuting selfadjoint operators. -self-adjoint on $D(A)$ if $B$ is symmteric and $A$-bounded with relative bounded $a<1$, and $A$ is selfadjoint (Kato-Rellich). - -My question concerns point 2. In this point the author refers the reader to C. Putnam, Commutation Properties of Hilbert Space Operators, Springer, 1967. I didn't find this theorem in that book, so if somebody knew that it is there for sure I would be really grateful for giving me more precise information about the location of that theorem. I'm curious why the positivity assumption is important there, in 3. for anti-commuting operators no positivity assumption was made. - -REPLY [18 votes]: I'm not familiar with Putnam's book, but part (2) of this theorem should be available in any of the standard references, e.g., Conway's Course in Functional Analysis or Reed and Simon, Functional Analysis vol. 1. -To understand why he requires $A$ and $B$ to be positive, think of them as multiplication operators on some measure space. If they are commuting self-adjoint operators this is legitimate. So if $A$ is multiplication by $f$ on $L^2(X,\mu)$ and $B$ is multiplication by $g$ on $L^2(X,\mu)$, $A + B$ should be multiplication by $f + g$ --- but what is the domain of this operator? If $f$ and $g$ are positive then it is just the common domain of $A$ and $B$, the $L^2$ functions which when multiplied by either $f$ or $g$ remain $L^2$. But if we don't assume $f$ and $g$ are positive, there can be cancellation which could enlarge the domain. The obvious example is $g = -f$. -If $A$ and $B$ anticommute the picture is totally different. The simplest example of this is $A = \left[\begin{matrix}1&0\cr 0&-1\end{matrix}\right]$ and $B = \left[\begin{matrix}0&1\cr 1&0\end{matrix}\right]$. I guess the intuition could be that $A$ and $B$ are $45^\circ$ from each other. But formally the point is that when you're checking convergence, anticommutation yields $$\|(A + B)v\|^2 = \|Av\|^2 + \|Bv\|^2,$$ so if $(v_n)$ converges and $(A + B)v_n$ converges, then both $(Av_n)$ and $(Bv_n)$ converge. So $A+B$ is already closed on $D(A) \cap D(B)$. - -REPLY [6 votes]: Your point 2. is a special case of Lemma 4.16.1 from Putnam's book.<|endoftext|> -TITLE: Local boundedness of weak solutions of heat equations...? -QUESTION [5 upvotes]: (Please, what is going on?) The following claim is from a paper which was apparently reviewed by László Erdös, Zhongwei Shen, and Bernard Heffler. Someone tell me it's true. Surely it's true. The entire paper depends on it being true. -I am looking at a 2000 article by Kazuhiro Kurata in the J. London Math. Soc.: An Estimate on the Heat Kernel of Magnetic Schrödinger Operators and Uniformly Elliptic Operators with Nonnegative Potentials. The claim is Lemma 1 on page 893. (In the preprint, available online here, the claim appears more casually, inside a proof on the top of page 12.) -Kurata is studying the heat kernels of various Schrödinger operators, the simplest example being $H = -\Delta + V(x)$. His key condition is that $V(x)$ belongs to a reverse Hölder class $RH_{q}$ with $q > n/2$. -Now suppose $u(x,t)$ is a weak solution of $(\partial_{t} + H)u = 0$ in some cylinder -$$Q_{2r}(x_{0}, t_{0}) = B(x_{0}, 2r) \times (t_{0}-(2r)^{2}, t_{0})$$ -According to Kurata, under just these hypotheses, there is a ``standard subsolution estimate'' -$$\sup_{(x,t) \in Q_{r/2}(x_{0}, t_{0})} |u| \leq \biggl(\frac{C}{r^{n+2}} -\iint_{Q_{2r/3}(x_{0},t_{0})}|u|^2\,dx\,dt\biggr)^{1/2}$$ -with $C$ independent of $V$, $t_{0}$, and $r$. He cites Aronson and Serrin's foundational '67 paper ``Local behavior of solutions of quasilinear parabolic equations'', but gives no specifics. -In every local boundedness estimate I've seen for parabolic equations (including, incidentally, Theorem 2 in Aronson and Serrin '67), the constants seem to depend on $V$, $t_{0}$, and $r$...!?! Is there just something magical about the reverse Hölder hypothesis? -Life, I guess. - -REPLY [4 votes]: I think by "standard subsolution estimate", Kurata means the following: -Let $u\ge0$ be a subsolution in $Q_2:=B_2(0)\times(-4,0)$, i.e. $\partial_t u - \nabla\cdot(A\nabla u)\le 0$. Then $u$ satisfies -$$sup_{Q_{1/2}} u \le C\Big( \int_{Q_1} |u|^2 \;dxdt \Big)^{\frac{1}{2}}$$ -for some constant depending only on the ellipticity contrast $\lambda$ and the dimension $d$. -Then the result of Kurata with a constant independent of $r$ follows by scaling space with $r$ and time with $r^2$. Likewise the constant may be chosen independently of $t_0$ and $x_0$ by translation invariance of the problem. -I am guessing that Kurata assumes $u\ge 0$ and $V\ge 0$, so that a solution to his problem is indeed a subsolution: $\partial_t u - \nabla\cdot(A\nabla u)\le 0$. -The above "standard subsolution estimate" may be obtained by parabolic Moser iteration, cf. -J. Moser, A harnack inequality for parabolic differential equations, Comm. Pure Appl. Math., link. -Moser iteration corresponds to testing the equation with clever test functions. The simplest example would be the following: consider an elliptic problem (e.g. $u$ above does not depend on time) $-\nabla\cdot A \nabla u \le 0$ in $B_2$ and test with $\eta^2 u$, where $\eta$ is a test function which is one in $B_{1/2}$ and zero outside $B_1$. Then after integration by parts -$$0\ge-\int \eta^2 u \nabla\cdot A \nabla u \;dx = \int \Big(\eta^2 \nabla u \cdot A \nabla u + 2\eta u\nabla\eta \cdot A \nabla u \Big) \;dx.$$ -Ellipticity and Young's inequality yield -$$\int \eta^2 |\nabla u|^2 \;dx\le C(\lambda) \int |\nabla\eta|^2 u^2 \;dx.$$ -This is usually called Caccioppoli's inequality. By the product rule, it can be upgraded to -$$\int |\nabla (\eta u)|^2 \;dx\le C(\lambda) \int |\nabla\eta|^2 u^2 \;dx.$$ -Now we use the Sobolev embedding $\|v\|_{L^p(R^d)}\le C(d) \|\nabla v\|_{L^2(R^d)}$ with $p=2d/(d-2)$ (let's assume $d>2$) to obtain -$$\Big(\int |\eta|^p |u|^p \;dx\Big)^{\frac{2}{p}}\le C(d,\lambda) \int |\nabla\eta|^2 u^2 \;dx.$$ -Finally, by the choice of test function, e.g. $|\nabla \eta| \le C(d)$ and we obtain -$$\Big(\int_{B_{1/2}} |u|^p \;dx\Big)^{\frac{2}{p}}\le C(d,\lambda) \int_{B_1} u^2 \;dx.$$ -This was the first step to give you some gain of integrability (giving up a little bit on the domain of integration on the right hand side). For the next step, you test with something like $\eta^2u^{p-1}$... -I hid a lot of technical details and the argument gets more complicated when you iterate higher and if you do parabolic estimates, but the idea is there. Anyway, I've rambled on much longer than I intended to... -PS: The reason why the constant in Aronson and Serrin seems to depend on $V$ and $r$ is that they consider a (much) more general problem which in particular is not scaling invariant in the same way that $\partial_t u - \nabla \cdot A \nabla u \le 0$ is.<|endoftext|> -TITLE: Maximal chains in a quasi-order of linear order types -QUESTION [10 upvotes]: Let $\mathcal{T}_\kappa$ be the set of all linear order types of cardinality $\kappa$. Let $\prec$ denote a binary relation on $\mathcal{T}_\kappa$ representing embeddability of order types (note that $\prec$ is a quasi-order). - -What is the cardinality of a subset of $\mathcal{T}_\kappa$ of the largest size that can be equipped with a linear order $<$ consistent with $\prec$ (such that $\tau<\mu$ implies $\tau\prec\mu$)? -What is the cardinality of a subset of $\mathcal{T}_\kappa$ of the largest size such that all its elements are mutually embeddable? - -I am particularly interested in cases where $\kappa$ is $\aleph_0$, $\aleph_1$ and $\mathfrak{c}$. - -REPLY [2 votes]: (Really a comment, but too long:) -If you strengthen your first question to demand that the subset actually be linearly ordered by $\prec$ (that is, strict chains) - which rules out, for example, Goldstern's construction - then the answer is different. -In the case $\kappa=\aleph_0$, we can divide the countable ordertypes into the scattered (does not embed $\eta$) and unscattered cases. Since unscattered linear orders all embed $\eta$, any countable linear order embeds into any countable unscattered order, so in any strict chain $C$, at most one unscattered order occurs, and if it occurs it occurs as the maximal element. So the question reduces to how long a strict chain of scattered orders can be. -Clearly there are strict chains of cardinality $\aleph_1$ (e.g., the countable ordinals). This answers the question fully, in case $2^{\aleph_0}=\aleph_1$, since there are exactly continuum many isomorphism types of countable linear orders. Now, I believe that this can be extended to the case where $CH$ fails - that is, no matter what the value of the continuum, strict chains of ordertypes under embeddability have cardinality $\le\aleph_1$ - as follows: there is a way of assigning a countable ordinal to a countable scattered linear order, called the Hausdorff rank of the linear order, and it can be shown (I believe) by induction on rank that any strict chain of (countable scattered) linear orders of bounded rank is countable. It then follows that the maximum cardinality of a strict chain of scattered countable linear orders (and hence countable linear orders) is $\aleph_1$. What this means, concretely, is that we can have universes of ZFC in which there are no strict chains of countable linear orders of length continuum. -Even better, this actually shows that for any strict chain $\mathcal{C}$, we have $$ \omega_1+2\not\prec \mathcal{C},$$ which is better than just a cardinality bound. -I have no idea what happens in the case $\kappa>\aleph_0$, if we look at strict chains. - -CAVEAT: this is all just a sketch, and it may well be wrong; it's been a while since I've looked at this stuff.<|endoftext|> -TITLE: Is there a probability theory developed in intuitionistic logic? -QUESTION [23 upvotes]: Since Boole it is known that probability theory is closely related to logic. -According to the axioms of Kolmogorov, probability theory is formulated with a (normalized) -probability measure $\mbox{Pr}\colon \Sigma \to [0,1]$ on a Boolean -$\sigma$-algebra $\Sigma$ (of events). -Realizing these data by a set $X$ (sample space of elementary events) and a corresponding $\sigma$-algebra $\Sigma(X)\subseteq P(X)$ of subsets of $X$, one obtains a probability space $(X,\Sigma(X),\mbox{Pr})$. -The $\sigma$-homomorphisms $f \colon {\cal B}({\mathbb R})\to \Sigma$ (real $\Sigma$-valued measures) are defined on the Borel $\sigma$-algebra -${\cal B}({\mathbb R})$ of the real Borel sets. They can be realized by real-valued measurable functions $F\colon X\to {\mathbb R}$ (random variables). -I wonder how this theory extends from the classical to the intutionistic logic i.e. from the Boolean to the Heyting ($\sigma$-)algebras and what the major differences between the two theories are. -Where can I find precise descriptions of the following topics: - -Definition and properties of probability measures on a Heyting algebra ${\cal H}$. -Definition and properties of real ${\cal H}$-valued measures $f \colon {\cal B}({\mathbb R})\to {\cal H}$. - -(Already the discrete case would be of interest.) -(BTW: Boole 1815–1864; Heyting 1898–1980; Kolmogorov 1903–1987) - -REPLY [2 votes]: There is a probability theory developed in Łukasiewicz logic. -If you understand the french (and if you are still interested): -http://www.brunodefinetti.it/Opere/logiquedelaProbabilite.pdf<|endoftext|> -TITLE: How few terms may appear in a polynomial with given (cyclotomic) roots and nonnegative coefficients? -QUESTION [7 upvotes]: Given $W \subset \mathbb C$, let $S_W$ be the set of polynomials in $\mathbb R[x]$ that vanish on $W$ and have only nonnegative coefficients. - -Warm-up question: It's clear that if $W$ contains a positive real number, then $S_W = \{0\}$. Is the converse true? - -I'm pretty sure the answer is "yes" but I don't know if I'd quite call what I have a proof. -More generally, how do the combinatorics of $W$ affect what you can say about $S_W$? In particular, I'd like a handle on the following problem. - -Problem: Find (bounds on) the smallest number of nonzero terms in a polynomial belonging to $S_W$. - -What conditions on $W$ would allow us to do this? As I alluded to above, I'm especially interested in combinatorial conditions, e.g., on where the numbers in $W$ lie on the complex plane. -Maybe the following special case is more tractable. Actually, it's the only one that really matters to me anyway. - -Special case: What if $W$ is a set of complex $n$th roots of unity? - -In this case, we'll want to assume $1 \not\in W$. Then the polynomial $x^{n-1}+x^{n-2}+\cdots+x+1$ gives an upper bound of $n$. If every number in $W$ lies in the left half-plane and $W$ is a self-conjugate set, then an upper bound of $|W|+1$ can be obtained by taking the product of the corresponding quadratic factors. -Meanwhile, I'm not sure I know how to derive a lower bound better than $2$ under any conditions! -Any thoughts or theory out there that might help shed light on these questions? - -REPLY [8 votes]: $\def\ZZ{\mathbb{Z}}$ $\def\RR{\mathbb{R}}$ $\def\QQ{\mathbb{Q}}$ $\def\Re{\mathrm{Re}}$ $\def\Im{\mathrm{Im}}$Let $|W|=n$. I will show that $2n+1$ monomials are always sufficient, and are generically necessary. (Noam Elkies has already shown that $2n+1$ is generically sufficient.) If $W$ is closed under complex conjugation and has no real points, then this argument shows that $n+1$ is enough, since we could replace $W$ by $W' := \{ \omega \in W : \Im(\omega) > 0 \}$ and any real polynomial which vanishes on $W'$ will also vanish on $W$. -Let the elements of $W$ be $\omega_j = x_j + i y_j = r_j e^{i \theta_j}$. For any nonnegative integer $m$, let $v_m$ be the vector -$$v_m := (\Re(\omega_1^m), \Im(\omega_1^m), \ldots, \Re(\omega_n^{m}), \Im(\omega_n^{m}))$$ -So the $v_i$ are vectors in $\RR^{2n}$, and our goal is to find a positive linear relation between them. -Generic necessity: Suppose that we had a linear relation $\sum a_i v_{m_i}=0$ using only $2n$ terms. Then the vectors $v_{m_1}$, ..., $v_{m_{2n}}$ would be linearly dependent, so the $2n \times 2n$ matrix they form would have determinant zero. This is a nontrivial polynomial relation between the $x_j$ and $y_j$ with integer coefficients. If the $\omega_j$ are chosen generically then the $x_j$ and $y_j$ will be algebraically independent over $\QQ$, and no such relation will exist. -Sufficiency: This is essentially Noam's proof when the $\theta_j$'s are linearly independent over $\mathbb{Q}$, but with a lot more checking of degenerate cases. -Suppose, for the sake of contradiction, that we cannot write $0$ as $\sum_{i=1}^{2n+1} a_i v_{m_i}$ with the $a_i \geq 0$ and not all $0$. -Equivalently, assume that, for any $(2n+1)$-tuple of vectors of the form $v_m$, the origin is not in the convex hull of the tuple. -By the contrapositive of Carathéodory's theorem, we conclude that $0$ is not in the convex hull of the vectors $v_m$. Let $K$ be the closure of the convex hull of the $v$'s. -We conclude that $0$ is not in the interior of $K$. -By Farkas's lemma, there is a linear function $\lambda : \RR^{2n} \to \RR$ with is $\geq 0$ on $K$ and not identically $0$ on $K$. -Equivalently, $\lambda(v_m)$ is nonnegative for all $m$ and is positive for some $m$. -We can write $\lambda$ in the form -$$(f_1, g_1, f_2, g_2, \ldots, f_n, g_n) \mapsto \Re \left( \sum (a_j+i b_j) (f_j+i g_j) \right)$$ -for some $a_j$ and $b_j$. Set $\zeta_j = a_j+i b_j$. Our hypothesis now is that -$$\phi(m) : = \Re \left( \sum_j \zeta_j \omega_j^m \right)$$ -is nonnegative for all $m$ and positive for some $m$; our goal is to deduce a contradiction. -Let $R$ be the set of distinct values of $|\omega_j|$, and let the elements of $R$ be $r_1 > r_2 > \cdots > r_p$. -Let the elements of $R$ with norm $r_j$ be $r_j \exp(i \theta^j_1)$, $r_j \exp(i \theta^j_2)$, ..., $r_j \exp(i \theta^j_{k(j)})$. -Reindex the $\zeta$'s accordingly as $\zeta^1_1$, $\zeta^1_2$, ...., $\zeta^1_{k(1)}$, .... $\zeta^p_1$, ...., $\zeta^p_{k(p)}$. -Put -$$\phi_j(m) := \Re \left( \sum_{\ell=1}^{k(j)} \zeta^j_{\ell} \exp(i m \theta^j_{\ell}) \right)$$ -so -$$\phi(m) = \sum_j r_j^m \phi_j(m). \quad (\ast)$$ -Since $\phi(m)$ is not identically zero, not all of the $\phi_j(m)$ are identically zero. Let $j_0$ be minimal such that $\phi_{j_0}(m)$ takes nonzero values. -Lemma There is $c>0$ so that $\phi_{j_0}(m)$ is infinitely often less than $-c$. -Proof By assumption, $\phi_{j_0}$ is not identically zero. Since none of the $\theta$'s are $0 \bmod 2 \pi \ZZ$, the Cesaro limit $\lim_{M \to \infty} \frac{1}{M} \sum_{1 \leq m \leq M} \phi_{j_0}(m)$ is $0$. (This is where we use that the $\omega$'s are not positive reals. Note that this is true even if some of the $\theta$'s are in $2 \pi \QQ$.) So $\phi_{j_0}$ is negative for some $m$, say $m_0$. Let $\phi_{j_0}(m_0)=-2c$. -Consider any $\delta>0$. By a basic pigeonhole argument, we can find infinitely many $N$ such that $| N \theta^{j_0}_{\ell} \bmod 2 \pi \ZZ| < \delta$. (Note that this is true even if the $\theta$'s are not linearly independent over $\QQ$.) Choosing $\delta$ small enough, for such $N$'s, we have $\phi_{j_0}(m_0+N) < -c$. $\square$. -Therefore, there are infinitely many $m$ for which the leading term of $(\ast)$ is as negative as $-c r_{j_0}^m$, and all the other terms are exponentially less. Thus, we have found an $m$ for which $\phi(m)<0$. This is a contradiction and the theorem follows. -Remark: Carathéodory + Farkas + (something clever) is a general proof technique. -Chapter 2 of Barvinok's A Course in Convexity has many nontrivial exercises of this form.<|endoftext|> -TITLE: How to make countably closed forcing "nice" without choice -QUESTION [10 upvotes]: When working over a model $V$ of $ZFC$, countably closed forcings are extremely nice: - -If $\mathbb{P}$ is countably closed, then $V[G]$ has no new $\omega$-sequences of elements of $V$. In particular, countably closed forcing adds no new reals. - -This can fail miserably if $V\models ZF+\neg AC$. In particular, it is consistent with $ZF$ that there is an infinite, Dedekind-finite set $X$ of reals. Letting $\mathbb{P}$ be the poset of finite partial injective maps $\subseteq\omega\rightarrow X$, we have the peculiar result that $\mathbb{P}$ is countably closed inside $V$, since there are no countably infinite subsets of $X$! -The general question is: without $AC$, what are good ways to tell that forcing with a given poset adds no new reals? -A particular sub-question, that I am separately interested in: if $W\models ZF$, $W$ is an inner model of $V\models ZFC$ with $\mathbb{R}^V=\mathbb{R}^W$, and $\mathbb{P}\in W$ is countably closed in $V$, then forcing with $\mathbb{P}$ over $W$ adds no new reals. However, I don't actually know any ways of building pairs $(V, W)$ with these properties. So, my subquestion is: -How does one show (without extra consistency strength, so not $L(\mathbb{R})$) that there are models $V$ and $W$ of $ZFC$ and $ZF$ respectively, such that $W\models$ "The reals are not well-ordered" and $\mathbb{R}^V=\mathbb{R}^W$? - -REPLY [10 votes]: This doesn't answer the full version of your question, but it seems pertinent. -Theorem. (Gunter Fuchs) The following are equivalent over ZF: - -DC -Countably-closed forcing does not add new $\omega$-sequences. - -Proof. The forward direction is the usual argument. For the converse, suppose DC fails, so there is a relation $R$ on a set $X$ with no terminal nodes and no $\omega$-sequence through it. Let $\mathbb{P}$ be the forcing notion to add a threading $\omega$-sequence through $R$, consisting of finite sequences $\langle a_0,\ldots,a_n\rangle$, with $a_i\mathrel{R} a_{i+1}$, ordered by extension. The assumption ensures that this forcing is countably closed, for reasons similar to the ones you mention in your question. But the generic filter will create a new threading $\omega$-sequence, since every finite sequence can be extended one more node. QED - -REPLY [10 votes]: This is actually a question I am quite concern with at the moment. In its generality, let me give a slight re-hash of what Joel said, with perhaps a slight generalization.$\newcommand{\forces}{\Vdash}$ - -We say that $\Bbb P$ is $\kappa$-closed if every decreasing sequence of length $\kappa$ (or shorter) has a lower bound. We say that $\Bbb P$ is $\kappa$-Baire (or $\kappa$-distributive) if the intersection of $\kappa$ many dense open sets is dense (and open, trivially). -In $\sf ZF$ we can prove these two theorems: - -Theorem I. If $\Bbb P$ is a $\kappa$-Baire forcing then $\Bbb P$ does not add new subsets of size $\leq\kappa$ to the universe. - -Proof. Suppose that $p\forces\dot f\colon\check\kappa\to\check V$. By the definition of the forcing relation, for every $\alpha<\kappa$ we have a dense set $$D_\alpha=\{q\leq p\mid\exists x\in V, q\forces\dot f(\check\alpha)=\check x\}.$$ -Let $D$ be the intersection of these dense open sets, then by the $\kappa$-Baire property $D$ is dense, and therefore non-empty. Let $q\in D$ then for every $\alpha<\kappa$ there is some $x\in V$ such that $q\forces\dot f(\check\alpha)=\check x$. However $q$ forces that $\dot f$ is a function, therefore this $x$ must be unique. If so the function $g(\alpha)=x$ for which $q\forces\dot f(\check\alpha)=\check x$ is such for which $q\forces\check g=\dot f$, as wanted. $\square$ - -Theorem II. $\sf DC_\kappa$ holds if and only if every $\kappa$-closed forcing is $\kappa$-Baire. - -Proof. Suppose that $\sf DC_\kappa$ holds, the proof goes the same as in $\sf ZFC$. In the other direction, recall that $\sf DC_\kappa$ is equivalent to the following statement: - -Every tree $T$ of height $\kappa$, where every sequence of $<\kappa$ has a proper end extension has a branch of length $\kappa$. - -Suppose now that $\sf DC_\kappa$ fails, then there is a tree $T$ of height $\kappa$ in which every sequence can be extended, but there is no branch of length $\kappa$. Define $P$ to be the reverse tree order on $T$. This forcing is trivially $\kappa$-closed, because there are no decreasing sequences of length $\kappa$. On the other hand, consider the dense sets $\{q\in T\mid\operatorname{lvl}_T(q)>\alpha\}$. Each is dense open, but their intersection is empty. $\square$ - -Finally, for real numbers. Or more specifically, sets of ordinals. If a forcing is $\kappa$-Baire then it doesn't add any new sets of size $\kappa$, but we are only interested in sets of ordinals. -I haven't managed to figure out the exact property, and working with $\kappa$-Baire forcings seems reasonable enough (and controllable enough). -To your subquestion, without verifying the details (I might do so tomorrow, but feel free to let me know if you do it yourself and find a mistake). -Consider the forcing which adds $\aleph_1$ Cohen generics, i.e. $p\in\Bbb P$ is a finite function from $\omega_1\times\omega$ to $2$, with the usual order. Now apply permutations (perhaps with a finite or countable support) of $\omega_1$ onto the left coordinate of the conditions. -Consider the names which are fixed by a countable subset of $\omega_1$. That is to say, there exists $\alpha<\omega_1$ such that whenever $\pi$ fixes pointwise $\alpha$ (as a set), it fixes the name. -I suspect that you can prove that all the real numbers of $V[G]$ enter this symmetric extension (perhaps by some argument that every real number is decided by some countable subset of the forcing, and therefore we can find it a support). -It is easy to show by the usual arguments, though, that there is no well-ordering of $\Bbb R$ in the symmetric extension.<|endoftext|> -TITLE: Grothendieck's letter to Faltings: On the yoga of motives and the degeneration of Leray spectral sequence -QUESTION [35 upvotes]: In Grothendieck's letter to Faltings, he writes - -There exists at this time a kind of “yoga des motifs”, which is familiar to a handful of initiates, and in some situations provides a firm support for guessing precise relations, which can then sometimes be actually proved in one way or - another (somewhat as, in your last work, the statement on the Galois action - on the Tate module of abelian varieties). It has the status, it seems to me, - of some sort of secret science – Deligne seems to me to be the person who - is most fluent in it. His first [published] work, about the degeneration of - the Leray spectral sequence for a smooth proper map between algebraic varieties over $\mathbf C$, sprang from a simple reflection on “weights” of cohomology groups, which at that time was purely heuristic, but now (since the proof of the Weil conjectures) can be realised over an arbitrary base scheme. - -Just what is this "simple reflection on 'weights' of cohomology groups" that he's referring to?? The proof in Deligne's paper doesn't use weights at all, but the relative version of the hard Lefschetz theorem. -Often when people say that a certain spectral sequence degenerates because of a weight argument, they mean that all differentials past a certain page go between cohomology groups of different weights. But this is not the case here: we can certainly today put a mixed Hodge structure and a weight filtration on all the terms in the sequence -$$ H^p(Y,R^qf_\ast \mathbf Q) \implies H^{p+q}(X,\mathbf Q)$$ -(using Saito's theory of mixed Hodge modules), and if $f$ is smooth and proper then $R^qf_\ast \mathbf Q$ is a VHS pure of weight $q$. Even then, this does not imply anything about the weight filtration on $H^p(Y,R^qf_\ast \mathbf Q)$ unless we put more assumptions on $Y$ (and in Deligne's result $Y$ is arbitrary). If we in addition assume $Y$ to be smooth and proper then $E_2^{pq}$ is pure of weight $p+q$ and the spectral sequence will indeed degenerate for formal weight reasons, but surely this is not what Grothendieck is referring to? - -REPLY [2 votes]: Well, everything follows from the fact that the total direct image $Rf_*\mathbb{Q}$ splits as the direct sum of its (shifted, co)homology sheaves (so, it is "formal"). Now, this statement can be proved using weights of mixed Hodge modules; it is a consequence of the triviality of $DHM(Y)$-1-extensions between (pure) Hodge modules of the same weight (here I am using the convention in which $[q]$ increases weights by $q$).<|endoftext|> -TITLE: Online high quality colloquium talks -QUESTION [21 upvotes]: In my department we're thinking about showing online lectures one day per week at lunch, as sort of a virtual colloquium appropriate to mathematics undergraduates as well as faculty. To start with we'd probably not want to launch into a lecture series on a single topic, but instead show high quality colloquium talks. I'm looking for links to things like Voevodsky's lecture "An Intuitive Introduction to Motivic Homotopy Theory", for example. (We give only the title here because links to YouTube aren't allowed on MO. If you intend to answer this question with a YouTube link, instead please type the title of the talk so it can be googled.) -We intend to start with MSRI talks and ICM lectures, but would like to know which are the best quality talks in these collections! -So by now you've probably figured out that the purpose of the current question is to collect other high-quality online colloquium talks: - -Question: What is a link to your favorite online talk suitable for (reasonably) general audiences, that is not part of a lecture series? - -It would even be useful for us if you provided a recommendation for a particularly high quality talk found in the MSRI or ICM archives, as all of these talks are not created equal. -Thanks in advance for allowing us to benefit from your experience! -To clarify how this question is different from, for example this one, I'm looking for specific lecture recommendations and not only general collections of lectures. The hope is that the community can share its good taste in order to benefit our virtual colloquium (and help other departments who might want to try this). - -REPLY [2 votes]: For specific topics, I found very high quality and worth watching videos at the Arizona Winter School. Those are yearly 5 days schools around arithmetic geometry, providing 4 lectures with one daily session, videos and notes, so it is a bit of a lecture series, however they are short, willing to be pedagogical and introductive, hence the first talk of a topic is probably suitable for a lunch colloquium.<|endoftext|> -TITLE: Is there an "exponential law" for differentiable maps between smooth manifolds? -QUESTION [10 upvotes]: Although it seems like a textbook question, I was not able to find a textbook or even a research article answering the following question: -Let $M$, $N$ and $P$ be finite-dimensional smooth manifolds and let $f \in C^r(M \times N,P)$ for a given $r \in \mathbb{N}$. -Let $\hat{f}: M \to C^0(N,P)$ denote the adjoint map, given by $\hat{f}(x) = (y \mapsto f(x,y))$. -For which $k,l \in \mathbb{N}$ does it hold that $\hat{f} \in C^k(M,C^l(N,P))$? - -REPLY [6 votes]: As far as I am aware, Kriegl and Michor's book deal only with the smooth case. To my knowledge the first full account on an exponential law for finite orders of differentiability was given in -Alzaareer, Schmeding: Differentiable mappings on products with different degrees of differentiability in the two factors, see https://arxiv.org/abs/1208.6510 for the preprint version. -there the linear case is dealt with. -Let me spell this out here: -If $M,N$ are open subsets of finite dimensional spaces(actually one can do much better then open in fin. dim. space) and $P = \mathbb{R}^n$, then the adjoint construction written out by you yields a (linear) topological isomorphism -$$C^k(M,C^l(N,P)) \rightarrow C^{k,l} (M\times N,P)$$ -where the space on the right hand side consists of all mappings which are $k$-times continuously differentiable with respect to the $M$-variable and every one of these derivatives is $l$-times continuously differentiable with respect to the $N$-variable (the cited paper contains a careful study of these mappings, chain-rules etc.). -So for example, if in your notation $r\geq l+k$ you will always obtain a map in $C^k(M,C^l(N,P))$, as one sees that $C^{l+k} (M\times N, P) \subseteq C^{k,l} (M\times N, P)$. -Now this is only linear theory. However, one can easily generalise this to the manifold setting you want (by the chain rules in the cited paper it makes sense to talk about these mappings on manifolds). The theory then generalises to manifold valued mappings as follows: First of all we need that $N$ is a compact manifold (otherwise $C^l(N,P)$ will not be a Banach manifold with the compact open $C^l$-topology, or any other of the usual function space topologies for that matter). Then the straight forward calculations in local charts should yield the same answer as above in the linear case. -Addendum 1: The topology on the space $C^{k,l} (M\times N,P)$ is a variant of the compact open $C^r$-topology. Here one controls of course the derivatives up to order $k$ in the $M$ direction and up to order $l$ in the $N$ direction on compact subsets. This topology was also discussed in detail in the cited paper. -Addendum 2: Mappings of type $C^{k,l}$ are not new per se, for example in the literature on ODE's one can find these maps, e.g. in Amann: Ordinary differential equations, mappings of type $C^{0,1}$ are considered and in Treves: Topological Vector Spaces, Distributions and Kernels, one finds in a side remark something about the compatibility of $C^{r,s}$ mappings with topological tensor products. However, in neither source is a treatment of the topological properties of these spaces (whence the remark in Treves' book is more in the direction of "it should work for a suitably defined space").<|endoftext|> -TITLE: the spectrum of the Laplacian and Dirac operator on $S^3$ -QUESTION [11 upvotes]: A paper on supersymmetry in 3-dimensions uses results on the spectra of elliptic operators on $S^3$: - -The eigenvalues of the vector Laplacian on divergenceless vector - fields is $(\ell + 1)^2$ with degeneracy $2\ell(\ell+2)$ with $\ell \in \mathbb{ Z}$. - -They came up with a similar statement for spinor fields on the 3-sphere - -The eigenvalues of the Dirac operator $D$ is $\pm (\ell + \frac{1}{2})$ with degeneracy $\ell(\ell+1)$ with $\ell \in \mathbb{ Z}$. - -On Math.StackExange it was suggested we can restrict the eigenspace decomposition the Laplacian on $SO(4)$ to $S^3$: -$$ L^2(SO(4)) = \bigoplus_{\pi \in \mathrm{Irr}[SO(4)]} \pi \otimes \overline{\pi}$$ -In our case, how Peter-Weyl Theorem play out when $SO(4)$ acts on bundles of divergenceless vector fields and spinor-fields $S^3$? - -REPLY [9 votes]: Let $M=G/K$ with $G$ compact and let $(\tau,W_\tau)$ be an irreducible representation of $K$, let $E_\tau$ be the associated $G$-homogeneous vector bundle of $M$. -Then, the space of $L^2$-sections of $E_\tau$ decomposes as -$$ -L^2(E_\tau) = \sum_{\pi\in\widehat G} V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau). -$$ -The proof follows by Peter-Weyl: -$\Gamma^\infty(E_\tau)\simeq C^\infty(G;\tau):=\{f:G\to W_\tau:f(gk)=\tau(k)^{-1}f(g)\}$. -For each $\pi\in\widehat G$, we have $V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau) \to C^\infty(G;\tau)$ given by $v\otimes L \mapsto (x\mapsto L(\pi(x^{-1}).v))$. This map is $G$-equivariant. -An other way, $C^\infty(G;\tau) = C^\infty(G,W_\tau)^K \simeq (C^\infty(G)\otimes W_\tau)^K$. Peter-Weyl theorem implies that -$$ -L^2(E_\tau) = \sum_{\pi\in\widehat G} (V_\pi\otimes V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes (V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes\operatorname{Hom}_K(V_\pi,W_\tau). -$$ -When $M$ is symmetric (e.g. $M=S^3$) and $G$ semisimple, the Laplace operator acts as the Casimir element of $\mathfrak g$ (the Lie algebra of $G$). Furthermore, the square of the Dirac operator $D^2$ coincides with the Casimir element plus $1/8$ the scalar curvature. See Section 3.5 of T. Friedrich, "Dirac operators in Riemannian geometry". -For example, let $S^{3}=\operatorname{Spin}(4)/\operatorname{Spin}(3)$, let $\tau$ be the spin representation, then $E_\tau$ is the spinor bundle. $\tau$ has highest weight $\frac12 \varepsilon_1$, thus $\operatorname{Hom}_K(V_\pi,W_\tau)\neq0$ if and only if $\pi$ has highest weight $\Lambda = \frac12((2k+1)\varepsilon_1\pm\varepsilon_2)$. The casimir element acts on $\pi_\Lambda$ as $\|\Lambda+\varepsilon_1\|-\|\varepsilon_1\|$ (see Wallach's book, Harmonic analysis on homogeneous vector bundles). Check that is equal to -$$ -\frac{(2k+1)(2k+5)+1}{4} -$$ -The sectional curvature is 6, then $D^2$ acts by $(k+3/2)^2$. Finally, the eigenvalues of $D$ are $\pm(k+3/2)$.<|endoftext|> -TITLE: Generalizing detropicalization -QUESTION [9 upvotes]: Given an identity in max,plus arithmetic, are there ways to turn it into an ordinary algebraic identity it other than by replacing addition by multiplication and replacing max by series-plus or by parallel-plus, where the series sum of $x$ and $y$ is $x+y$ and the parallel sum is $xy/(x+y)$? (See the related posts choosing between the two ways to tropicalize and Name and notation for a binary operation.) -As a related question, I ask: What are all the two-variable homogeneous rational functions over $\mathbb{C}$ such that $r(x,y)=r(y,x)$ and $r(r(x,y),z)=r(x,r(y,z))$? (Here I intend equality in the formal sense; e.g., I call $x/x$ the same rational function as 1, even though as functions they are not the same, since $x/x$ is not defined at 0.) -The only examples I know of are $r(x,y) = c$ (with $c$ an arbitrary constant), $r(x,y)=x+y$, $r(x,y)=xy/(x+y)$, and $r(x,y)=xy$, but I suspect that there are others I'm overlooking. Perhaps the theory of formal groups has an answer for me, but from what little I've seen, homogeneity does not play a role there. See the related post -Commutative associative rational binary operations. Now that I've played a bit with operations like $(x+y)/(1-xy)$ (thanks, Alexandre Eremenko!), I'm suspecting that I should be limiting myself to homogeneous operations. -(Note: The original version of the post used the word "detropicalize'' in a non-standard and unclear way, so I've changed the title of the post and the wording of the first paragraph accordingly. Thanks, John Mangual!) - -REPLY [6 votes]: Those are all. -Given a function $r$, by restricting to a particular value of $y$ (barring finitely many), we get a rational function, hence a map $\mathbb P^1 \to \mathbb P^1$. For all but finitely many values of $y$, this map will have the same degree, $d$. Assuming $r$ is nonconstant, let $y_1$ and $y_2$ be two such typical values such that $r(y_1,y_2)$ is also typical. Then the degree of $x => r(x,y_1) => r(r(x,y_1),y_2)$ is the degree of $x => r(x,r(y_1,y_2))$, so $d^2=d$,so $d=0$ or $1$. Clearly the $d=0$ case is the constant case. -For $d=1$, we get a rational inverse function to $r$, giving us an algebraic group structure on $\mathbb P^1$ minus finitely many points. (Or for each $y$-value but finitely many, we get an automorphism of $\mathbb P^1$, which is an element of $PGL_2$, so we have a curve in $PGL_2$ that is almost closed under composition. The closure of such a curve is always a subgroup.) There are only two of these: $\mathbb G_a$ or $\mathbb G_m$. But we still have to decide which isomorphism between the closure of these groups and $\mathbb P^1$ to take. The homogeneity means that the missing points can only be $0$ and $\infty$, which gives us three options: -$\mathbb G_a$, $\infty$ missing: $r(x,y)=x+y$ -$\mathbb G_a$, $0$ missing: $r(x,y) = 1/((1/x)+(1/y))=xy/(x+y)$ -$\mathbb G_m$, both $0$ and $\infty$ missing: $r(x,y)= xy$.<|endoftext|> -TITLE: Voronoi cells and the dual complexes in Riemannian manifolds -QUESTION [12 upvotes]: I would like to use some "intuitively clear" properties of Voronoi cells in general Riemannian manifolds, but I have trouble finding references. -Let $(X,d)$ be a connected Riemannian manifold and let $S$ be a discrete set of points. Define $Vor(s,S) = \{x\in X\colon \text{for any $s'\in S$ we have } d(x,s) \le d(x,s')\}$. -Assume $S$ is such that for any $s\in S$ there is a finite subset $N(s)\subset S$ such that $Vor(s,S) = Vor(s,N(s))$. - -Question 1. Is it true that $Vor(s,S)$ are submanifolds with a boundary, and the intersection of any number of them is a submanifold with a boundary? -Question 2. Under what general conditions $Vor(s,S)$ induce a structure of a CW-complex on $X$? - -In general, define the dual $2$-complex $\hat X_2$: the vertices are $S$, there is an edge between $s$ and $s'$ if $Vor(s,S)\cap Vor(s',S)$ is of codimension $1$ in X and there is a face spanned by a finite subset $F \subset S$ if $F$ is a maximal subset such that the cells $Vor(f,S), f\in F$ share a codimension-2 submanifold. -Perhaps this ad-hoc definition of the dual complex is obviously flawed, in which case I'd appreciate references to a better one. To me it is not even immediately clear $\hat X_2$ is a CW complex i.e. if the boundary of a face is in the $1$-skeleton. - -Question 3. Is it the case that if $X$ is contractible then $\pi_1(\hat X_2) =\{1\}$? (or if not, under what assumptions is it the case?) - -Remark: In the previous version of Question 1 I asked whether the intersections are a connected manifolds. Vidit Nanda below provided an example which shows that it's not always the case - -REPLY [3 votes]: Although this answers none of the OP's questions—and is in fact entirely tangential—some -might enjoy this image from an earlier MO question, Delaunay triangulations and convex hulls: - -           - - - -Image from a Mathematica demonstration project written by Maxim Rytin.<|endoftext|> -TITLE: Poisson Summation Formulas for Cut and Project Quasicrystals -QUESTION [5 upvotes]: In Lagarias' paper "Mathematical Quasicrystals and the Problem of Diffraction" http://www.math.lsa.umich.edu/~lagarias/doc/diffraction.pdf he discusses various ways one might get Poisson summation formulas for certain nonuniform point sets in the plane. More precisely, given a Delone set $\Lambda \subset \mathbb{R}^n$ obtained using a cut and project scheme, consider the tempered distribution -$$\mu_\Lambda = \sum_{x \in \Lambda} \delta_x.$$ -We would like to compute $\hat{\mu}_\Lambda$. In section 3 (specifically Thm 3.9 and the surrounding discussion) he seems to indicate that for cut and project sets, we know something about $\hat{\mu}_\Lambda$, but I can't tell how to interpret Thm 3.9 in that context. -Question 1: How close does Thm 3.9 get us to computing $\hat{\mu}_\Lambda$? -Question 2: Do we at least know that $\hat{\mu}_\Lambda$ is a tempered distribution of the form $$\sum_{x \in \Lambda'} c_x\delta_x$$ -for some Delone set $\Lambda'$? If so, how much do we know about the set $\Lambda'$? -Edit: In Meyer's paper "Quasicrystals, diophantine approximation and algebraic numbers" he computes a variety of weighted Poisson summation formulas, but falls short of a computation of $\hat{\mu}_\Lambda$. These computations appear to have been done in the 70's. Has anyone tried to improve his results since then? - -REPLY [2 votes]: Question 1 let $\gamma$ be the auto-correlation of $\Lambda$. -It can be proven that -$$ \lim_n \frac{\int_{x+A_n} \bar{\chi(t)} d \mu_\Lambda (t)}{vol(A_n)}(*)$$ -exists uniformly in $x$. In the case of Fourier Transformable measures, this limit is exactly $\widehat{\mu_\Lambda}(\{ \chi \})$. -It follows then from a result of Hof that -$$ \widehat{\gamma}(\{ \chi \}) = \left| \lim_n \frac{\int_{A_n} \bar{\chi(t)} d \mu_\Lambda (t)}{vol(A_n)} \right|^2 \,.$$ -I think the existence of the limit (*) is covered in a paper by Robert Moody (1). -Also the problem of when the two limits can be connected without the uniformity in $x$ (aka the Bombieri Taylor conjecture) is covered pretty well in a paper by Daniel Lenz (2) available here -Question 2 The answer is no in general. More precisely, the answer is yes if and only if $\Lambda=L+F$ where $L$ is lattice and $F$ is finite. -There is an old paper by Cordoba (3) which shows that this is not the case, and Lagarias mentions this paper. The proof is pretty technical, and I heard that there might be some issues with the proof though. -In this case, since $\Lambda$ is a Meyer set, there exists a much simpler proof of this claim, which can be found in my paper (Proposition 7.3 in (4))archiv link -A much stronger version of this claim was proven recently by Favorov (5) and independently by Kellendonk and Lenz (6) They basically proved that if $\Lambda$ is a Delone set with FLC, and $\delta_\Lambda$ is Fourier Transformable, with a discrete FT, then $\Lambda=L+F$ where $L$ is lattice and $F$ is finite. -I am actually pretty sure that in general, $\widehat{\mu_\Lambda}$ cannot even be a measure, you have to always treat it as a tempered distribution. -Bibliography - -R. V. Moody, Uniform distribution in model sets, Canad. Math. Bull. Vol. 45 (1), 123-130, 2002. -D. Lenz, Continuity of eigenfunctions of uniquely ergodic dynamical systems and intensity of Bragg peaks, Commun. Math. Phys. 287 , 225-258, 2009. -A. Cordoba, Dirac combs, Lett. Math. Phys., 17, 191-196, 1989. -N.Strungaru, On the Bragg Diffraction Spectra of a Meyer Set, Canadian Journal of Mathematics 65, no. 3, 675-701, 2013. -Favorov Bohr and Besicovitch almost periodic discrete sets and quasicrystals, Proc. Amer. Math. Soc. 140, 1761-1767, 2012. -J. Kellendonk, D. Lenz, Equicontinuous delone dynamical systems , Canadian Journal of Mathematics 65, 149--170, 2013.<|endoftext|> -TITLE: Symmetric black-hole curves -QUESTION [6 upvotes]: Is there a curve $C$ that connects $(0,1)$ to $(a,0)$ for some $a>0$, and, when reflected -to $C'$ in the $x$-axis, the shape $S=C \cup C'$ has the property that each horizontal -light ray entering $S$ from the left reflects off of $S$ (interpreted as a perfect mirror) -in such a way that it never emerges, i.e. it never again crosses $x=0$? -For example, a straight line $C$ fails to be such a curve: - -REPLY [11 votes]: I will do it since Fedya has no time. -An example of black hole for a strip of horizontal light rays can be constructed from two arcs of parabolas with common focus and vertical directrixes. - -You can add horizontal mirrors and take a pair of them with reflections in $x$-axis and get the curve you want, say as it shown on the following picture.<|endoftext|> -TITLE: C*-algebraic representation of observables vs self-adjoint operators one -QUESTION [7 upvotes]: I am trying to reconcile the "physicist" definition of an observable: self-adjoint operator on a Hilbert space, and the operational one as given by Strocchi in "An introduction to the mathematical structure of quantum mechanics" : an observable is an element of a C*-algebra. -Clearly one cannot represent the CCR as operators on finite dimensional Hilbert spaces, or as bounded operators on a Hilbert space (even infinite dimensional). -The only left option is to represent at least one of the two variables as an unbounded operator on a Hilbert space. This is incompatible with the operational definition of an observable as this unbounded element can't belong to a C*-algebra. -A possible solution to this is to consider bounded functions of $Q,P$ which is operationnally sensible, and people say that you get a C*-algebra, but I know no reference for this. -1) Do you know any? Do we still have self-adjoint operators? Does the commutation relation change? -Another solution is to take Weyl operators $e^{\imath t Q}$ and $e^{\imath s P}$ which are bounded and the Weyl form of CCR: these generate a C*-algebra. So one sees that by starting from an abstract Weyl C*-algebra as an observables algebra you can construct (through Stone theorem) self-adjoint densely defined operators on $\mathcal{H}$ under a small representation regularity assumption (so to verify Stone theorem strong continuity condition) [cf.p.63 Strocchi], with a 1-1 correspondance. -But now the self-adjoint operators that we can identify with an observable thanks to this 1-1 correspondance are defined only on a dense subset of $\mathcal{H}$, whereas no hypothesis was made on $Q,P$ domain in CCR. Hence Weyl operators are also defined on only part of the Hilbert space or you can't associate a self-adjoint generator to them on part of the Hilbert space, in either case you have a problem. -2) Was the starting Hilbert space too big? How to reduce its size so to have a generator for the Weyl operator defined on the whole Hilbert space? -So it seems to me that either there must be positive answers to my questions or one should abandon the identification between observables and self-adjoint operators: the operationally/mathematically correct association is between an observable and a Weyl operator, and you get under a representation regularity assumption a self-adjoint operator representation of the observable only on a dense part of the Hilbert space. -3) But then why bother with this kind of representation at all? Can't quantization be done directly in Weyl form? If not in what sense is Weyl-CCR harder to deduce during quantization than Heisenberg CCR? - -REPLY [3 votes]: I am basically repeating what Nick already said, but it was to long for a comment: -There is a one-to-one correspondence between self-adjoint operators and strongly continuous one-parameter groups of unitaries (Stone von Neumann Theorem). -The one-parameter group for an self-adjoint operator is given by $\{U_A(t):=e^{itA}\}_{t\in\mathbb R}$ and vice versa given $\{U(t)\}_t$ one obtains a self-adjoint (in generally only densely defined) operator $A$ called the generator of $U(t)$ by differentiating $U(t)$ wrt to $t$ and take the value at $t=0$, which can be made precise and can be found in any functional analysis book. -In the usual Fock representation of CCR, $t\to W(t f)$, with $W$ the Weyl operator is strongly continuous, and the generator of this one-parameter group is your self-adjoint operator. In your example $W(f)=e^{i (f_1 P+f_2 Q)}$ with $f=(f_1,f_2)\in\mathbb{R}^2$ is the Weyl operator and the self-adjoint operators $P$ and $Q$ are obtained as above. So in this (unique for finite degrees of freedom) representation of CCR, you see the two quantizations are exactly the same, after you checked that the Weyl relations "infinitesimal" give back the Heisenberg CCR relations. -In particular they act on the same Hilbert space.<|endoftext|> -TITLE: A version of Lusternik–Schnirelmann category for good open covers -QUESTION [12 upvotes]: Recall that the Lusternik–Schnirelmann category (or LS-category) of a space is the integer $n$ such that there is an open cover by $n+1$ open sets which have nullhomotopic inclusions, and no such cover by fewer open sets. This is quite hard to compute, but for instance, spheres clearly have LS-category equal to 1. -However, sometimes one is interested not just in open covers by (relatively) contractible open sets, but by covers where all finite intersections of such opens are also (relatively) contractible. This is especially true when dealing with higher stacks, if one is interested in efficient cofibrant replacements. -Thus, one might want to define something intermediate, like level-$n$ LS-category of a space, which is one more than the minimum number of open sets one needs to cover said space, such that intersections of at most $n$ opens is (relatively) contractible. Then a finite good open cover gives an upper bound on the level-$n$ LS-category for all $n$. There is also the analogue for good open covers, namely where all finite intersections are (relatively) contractible. -For instance, for spheres one can take a homeomorphism with the boundary of a topological simplex, then take the open cover by puffing up a face of the simplex by some small amount (this is the analogue of covering a circle by three arcs). Thus the level-$n$ LS-category of the $k$-sphere is bounded above by $k+1$. -Has something like this been considered in the literature before? -My interest stems from considering this problem for Lie groups, in particular matrix Lie groups $O(n,\mathbb{K})$, where for certain cases we have explicit covers of the minimum size. Even something like an upper bound would be good. - -REPLY [3 votes]: In some cases the nerve theorem might help (in the strong case without "relatively"; I also assume that some intersections in the question might be empty instead of contractible, hoping that this the intended definition of a good cover). Apropriate version can be found, for example in [A. Hatcher. Algebraic Topology. Cambridge University Press, Cambridge, -2001., Chapter 4.G]. In case of good covers where all intersections are contractible this yields that $X$ is homotopy equivalent to the nerve of the cover. Thus the number of sets in the cover is bounded from below by the number of vertices of the smallest simplicial complex homotopy equivalent to $X$. In case that the contractibility is assumed only for intersections up to $n$th level, then maybe some more refine versions of the Nerve theorem such as [G. Kalai and R. Meshulam. A topological colorful Helly theorem. -Adv. Math., 191(2):305–311, 2005., Theorem 3.2] can be used. (In particular, the nerves of good covers have been studied intensively in conections with Helly-type theorems.) -For example, the nerve theorem is already sufficient to get the exact bounds for spheres as described in the statement of the question via the following approach. One can get that $n+1$ sets are sufficient to cover the $(n-1)$-sphere as explained in the statement of the question. However, if we put the assumption on contractibility of intersections only up to level $n$, we get that $n+1$ sets are sufficient to cover arbitrary $k$-sphere with $k \geq n-1$ by considering $(k-n+1)$ suspensions of the example with $k=n-1$ (one has to be little careful to keep the sets open). Altogether we get that $Cat^n(S^k) \leq \min(k+1,n)$, where $Cat^n(X)$ is the minimum j such that there is a cover of $X$ by open sets $U_0,…,U_j$ which are contractible, borrowing the notation analogous to the notation of the answer of Mark Grant. -Using the Nerve theorem, one can deduce that -$Cat^n(S^k) \geq \min(k + 1,n)$: Indeed, for contradiction, let us assume that $Cat^n(S^k) = j$ where $j \leq k$ and $j < n$. Let $\mathcal{U} = \{U_0, \dots, U_j\}$ be the corresponding cover of $S^k$. Since $j < n$ we get that all intersections are contractible. Using the Nerve theorem, $S^k$ is homotopy equivalent to the nerve of $\mathcal{U}$. However, the simplicial complex with the smallest number of vertices homotopy equivalent to $S^k$ is the boundary of $k+1$-simplex, which has $k+2$ vertices, contradicting $j \leq k$.<|endoftext|> -TITLE: Kauffman's state model for the Alexander polynomial, via representation theory -QUESTION [5 upvotes]: I've been reading Oleg Viro's paper on "quantum relatives of the Alexander polynomial" (arXiv:math/0204290), which, among other more general things, derives state-sum formulas for the Alexander polynomial of a knot in terms of the representation theory of $\mathcal{U}_q(\mathfrak{gl}_{1|1})$ and (alternatively) $\mathcal{U}_q(\mathfrak{sl}_2)$ with $q$ a fourth root of unity. -But the state-sum model I know and love for the Alexander polynomial is Kauffman's original one (from his book "Formal Knot Theory"). Is this also derived from representation theory, like in Viro's paper (for one of these algebras or some other one)? -I'm interested in this question because the Kauffman state model is useful in knot Floer homology, which categorifies the Alexander polynomial. As far as I know (e.g. Sartori's recent preprint arXiv:1305.6162), a representation-theoretic categorification of the Alexander polynomial, thought of as a quantum $\mathfrak{gl}_{1|1}$-invariant, doesn't yet exist but may exist soon. I'd be very curious to know what sort of representation-theoretic structures knot Floer homology corresponds to, but as this seems to be a more difficult question, I'll stick with the Kauffman states for now! - -REPLY [2 votes]: Update: I've found at least a partial answer to this question, so I thought I'd write it up here and see if it's familiar, or if anyone can continue the story. -Kauffman's state sum model from "Formal Knot Theory" admits a generalization for tangles. I got this from Ozsvath and Szabo's new "bordered" construction of knot Floer homology, but I'm not sure whether this bit itself is new or whether someone's written it down before (I'd be curious to know). It goes as follows (restricted to braids for simplicity here, although it works in more generality): -To $n$ points on the $x$ axis, associate a vector space $\mathcal{I_n}$ of dimension $2^{n-1}$ over $\mathbb{C}(t)$. The space $\mathcal{I_n}$ can be naturally associated with $\wedge^*(R)$, where $R$ is the vector space of dimension $n-1$ generated by the line-intervals $r_1, \ldots, r_{n-1}$ between the $n$ points. A generator $r_{i_1} \wedge \ldots \wedge r_{i_k}$ is depicted by drawing $n$ small vertical lines over the $n$ points, creating $n-1$ small regions, and then putting dots in the regions corresponding to $r_{i_1}$ through $r_{i_k}$. -To a braid $B$ with $n$ strands, associate a map $f(B): \mathcal{I_n} \to \mathcal{I_n}$ defined by summing over "partial Kauffman states." These are assignments of a dot to one corner of each crossing of the braid such that no planar region has more than one dot. A partial Kauffman state $\sigma$ is compatible with a generator $r = r_{i_1} \wedge \ldots \wedge r_{i_k}$ of $\mathcal{I}_n$ on the input side (say the top of the braid) if all planar regions, adjacent to the top boundary, which are assigned dots in $\sigma$, are included among the regions $r_{i_1}, \ldots, r_{i_k}$. Then $f(B)$, applied to $r$, is a sum over all partial Kauffman states compatible with it. Each partial Kauffman state $\sigma$ determines a coefficient $c(\sigma)$ in $\mathbb{C}(t)$ (as a product over local contributions from corners) and another generator $r'$ of $\mathcal{I}_n$. The generator $r'$ has a region $r_i$ if either the corresponding planar region has no dot and does not intersect the top boundary, or the planar region intersects both boundaries, has no dot from $\sigma$, and $r_i$ is among the $r_{i_1}, \ldots, r_{i_k}$. Then we can write $f(B)(r) = \sum_{\sigma} c(\sigma) \cdot r'$, where the sum is over $\sigma$ compatible with $r$. -Surprisingly (to me), the data $\mathcal{I}_n$ and $f(B)$ can be related to quantum groups. Let $V$ denote the vector representation of $\mathcal{U}_q(\mathfrak{gl}_{1|1})$, with basis $\{v,w\}$ such that $|v| = 0$ and $|w| = 1$. Then, as a vector space over $\mathbb{C}(q)$, the tensor product $V^{\otimes n}$ can be identified with $\wedge^*(E)$, where $E$ is the $n$-dimensional vector space spanned by $\{w \otimes v \otimes \ldots \otimes v, v \otimes w \otimes \ldots \otimes v, v \otimes \ldots \otimes w\}$. Call these elements $e_1, \ldots, e_n$. Then consider the $n-1$-dimensional subspace $E_0$ of $E$, spanned by $\{e_1 - q^{-1}e_2, e_2 - q^{-1} e_3, \ldots, e_{n-1} - q^{-1} e_n\}$. The exterior product $\wedge^*(E_0)$ is a $2^{n-1}$-dimensional subspace of $V^{\otimes n}$ which is preserved by the action of the braid group via R-matrices. Furthermore, the braid group action on $V^{\otimes n}$ restricted to this subspace coincides with the map $f(B)$ under the identification $r_i \leftrightarrow (e_i - q^{-1} e_{i+1})$ and $t \leftrightarrow q^2$ (this is a computation which I didn't write here because this post is already long). -I'm very interested in understanding this connection more fully. As far as I can tell, the elements $e_i - q^{-1}e_{i+1}$ are part of the Lusztig canonical basis of $V^{\otimes n}$ at $q = \infty$ rather than $q = 0$. Is there a better way to describe these basis elements and the half-dimensional subspace they span? Does this picture have analogues for other groups (particularly $\mathfrak{sl}_2$)? And, of course, if these things are discussed in a paper somewhere, I'd be very glad to hear about it.<|endoftext|> -TITLE: Exercise in Kunen. Demonstrating a relative interpretation into a set -QUESTION [7 upvotes]: This is a question regarding exercise (II.2.15) in Kunen's Set Theory (2011): -The exercise reads: -"We shall see later that $ZFC\vdash $Con$(\Gamma)$ whenever $\Gamma$ is a finite subset of $ZFC$. Use this to define explicitly, in $ZFC$, a binary relation $E$ on $\omega$ such that $ZFC\vdash\varphi^{\omega,E}$ for each axiom $\varphi$ of $ZFC$". -Now the hint reads: -"List the axioms of $ZFC$ in some computable way, as {$\xi_i:i<\omega$}; let $ZFC_n=${$\xi_i:i -TITLE: Example of an unnatural isomorphism -QUESTION [59 upvotes]: Can anyone give an example of an unnatural isomorphism? Or, maybe, somebody can explain why unnatural isomorphisms do not exist. -Consider two functors $F,G: {\mathcal C} \rightarrow {\mathcal D}$. We say that they are unnaturally isomorphic if $F(x)\cong G(x)$ for every object $x$ of ${\mathcal C}$ but there exists no natural isomorphism between $F$ and $G$. Any examples? -Just to clarify the air, $V$ and $V^\ast$ for finite dimensional vector spaces ain't no gud: one functor is covariant, another contravariant, so they are not even functors between the same categories. A functor should mean a covariant functor here. - -REPLY [3 votes]: It seems that the functor on the category infinite sets that adds one disjoint point * to any set is not naturally isomorphic to the identity functor.<|endoftext|> -TITLE: Zoll sphere which is not a surface of revolution -QUESTION [7 upvotes]: Are there examples of Zoll spheres which are not the surfaces of revolution? - -REPLY [4 votes]: I'm repeating what was said above and adding some details for completeness. -A little chunk of historiography: Zoll found all the surfaces with periodic geodesic flow in the case of the isometry group isomorphic to $S^1$, i.e. the surfaces of revolution. It was done in his doctoral thesis in $1901$. First non-symmetrical results were found by Blaschke. The reference is W.Blaschke, Vorlesungen uber Differential geometrie, 1, Springer, Berlin, 1924. -By Korn-Licthenstein theorem (or simply by uniformization theorem) all the metrics on $S^2$ are conformal to a canonical one $g_0$. So any Riemanian metric $g$ on the sphere can be written in a form $g=\exp (\rho) g_0$. The space of Zoll metrics got people interested after the works of Zoll. -After Blaschke found the metrics which are not symmetrical by perturbation of Zoll's results, Hilbert asked the following question: what is a tangent space to the space of Zoll metrics at $g_0$? In other words, for what smooth functions on the sphere $\dot {\rho}$ do there exist Zoll deformations $g_t=\exp (\rho_t) g_0$ of the standart metric such that $\rho_0=-$ and $\frac{d \rho_t}{dt}=\dot{\rho}$ ar $t=0$? -His doctoral student Funk gave a necessary condition: $\dot{\rho}$ should be odd with respect to antipodal map. Victor Guillemin shows that this is actually sufficient using Radon transform. The methods are powerful, the work is hard to read but great. The reference is The Radon Transofrm on Zoll surfaces, Advances in Mathematics, 22, 1976.<|endoftext|> -TITLE: Decomposition theorem for semi-abelian varieties -QUESTION [5 upvotes]: Fact : - -Let $B$ an abelian subvariety of an abelian variety over a field $K$. We know that there exist an abelian subvariety $C$ of $A$ such that the restriction of addition gives an isogeny $B\times_K C\to A$. -The analogous result is true for an linear torus. - -Questions : -1)Is this result is true in the category of semiabelians varieties ? -2)Is there a good reference on semi-abelians varieties ? -Thanks ! - -REPLY [5 votes]: No. Take an extension of an elliptic curve by a torus which is of infinite order in the Ext group and take B the torus. (Reference: Serre, Groupes algébriques et corps de classes, Ch VII).<|endoftext|> -TITLE: Are there examples of Fano manifolds such that Tian's alpha invariant satisfies $\alpha_G(X)=\frac{n}{n+1}$ but without a Kähler-Einstein metric? -QUESTION [12 upvotes]: The alpha invariant $\alpha(X)$ of a Fano manifold $X$ of dimension $n$ is defined as the infimum of log canonical thresholds of (effective) $\mathbb{Q}$-divisors $D\sim_{\mathbb{Q}}-K_X$. Similarly, for $G\subset Aut(X)$ a compact subgroup of the automorphism group, one defines $\alpha_G(X)$ considering only $G$-invariant divisors. The alpha invariant has a corresponding analytic definition involving complex singularity exponents of singular hermitian metrics [2, Appendix]. -Tian introduced this invariant and proved that the lower bound $\alpha_G(X)>\frac{n}{n+1}$ implies the existence of a Kähler-Einstein metric [1] (in fact, even today it is one of the few sufficient conditions which is checkable in practice). I'd like to know if this theorem is sharp? That is: -Question: Are there examples Fano manifolds such that $\alpha_G(X)=\frac{n}{n+1}$ but without a Kähler-Einstein metric? -I'd also be interested in any partial results in the positive direction. -An example I know of with $\alpha(X)=\frac{n}{n+1}$ is a del Pezzo surface of degree $4$ (this is due to Cheltsov [3]), however by Tian's classification of Kähler-Einstein metrics on del Pezzo surfaces [4], such surfaces are known to admit Kähler-Einstein metrics. -References: -[1] G. Tian. On Kahler-Einstein metrics on certain K ̈ahler manifolds with $c_1(M)>0$. -[2] I. Cheltsov, C. Shramov, Appendix by J. P. Demailly. Log canonical thresholds of smooth Fano threefolds. -[3] I. Cheltsov. Log canonical thresholds of del Pezzo surfaces. -[4] Tian, G. On Calabi’s conjecture for complex surfaces with positive first Chern class. - -REPLY [9 votes]: This question was answered negatively by Kento Fujita today (at least when $G$ is trivial). -Theorem (Fujita): If $\alpha(X,-K_X)=\frac{n}{n+1}$, then $X$ is K-stable and hence admits a Kähler-Einstein metric.<|endoftext|> -TITLE: Maximize a weighed combinatorial sum -QUESTION [5 upvotes]: I am trying to maximize the function $$f_s(k)=\frac{1}{2k+s}\sum_{i=0}^k {2k+s\choose i}2^{-(2k+s)}$$ for both $\{s,k\}\in\mathbf N$, that is, for fixed $s$ what is the value of $k$ that maximizes $f_s(k)$. I have seen previous discussions in Sum of 'the first k' binomial coefficients for fixed n and Lower bound for sum of binomial coefficients? but I'm not seeing much light. I'd be happy enough with a partial result, for instance a lower bound on the maximum value and the $k$ achieves it, or the range in which the maximum $k$ is to be found. -Anyone any idea? - -REPLY [3 votes]: Using Stirling's formula one sees that $\binom{n}{n/2+\ell}$ is about -$$ -\sqrt{\frac{2}{\pi n}} 2^{n} e^{-2\ell^2/n}, -$$ -when $\ell$ is not too large. You can use this to evaluate your sum. -Let's look at the range when $k$ is of size about $s^2$ which is where the -maximum will be attained. Writing $n=2k+s = s^2/t$ your sum is approximately -$$ -\frac{1}{\sqrt{2\pi} s^2} \left( t \int_{\sqrt{t}}^{\infty} e^{-x^2/2} dx\right). -$$ -Choose $t$ such that the term in brackets is maximum -- this is not any nice -constant so far as I know, it is what it is. It then follows that your -maximum sum is roughly $C/s^2$ for some constant $C$. -A calculation shows that $t$ above is roughly $1.42$, and that $C$ is about $0.165$. -Note that then $k$ is about $s^2/2.84$. This is in rough agreement with the $s^2/3$ seen in Meyerowitz's answer; also when $s=100$ the asymptotic $0.165/s^2$ is very close to the $s^{-2.38}$ seen there.<|endoftext|> -TITLE: Has Reifegerste's Theorem on RSK and Knuth relations received a slick proof by now? -QUESTION [20 upvotes]: For the notations I am using, I refer to the Appendix at the end of this post. -Here is what, for the sake of this post, I consider to be Reifegerste's theorem: -Theorem 1. Let $n\in\mathbb N$ and $i\in\mathbb N$. Let $\sigma$ and $\tau$ be two permutations in $S_n$ such that $\sigma$ and $\tau$ differ by a Knuth transformation at positions $i-1,i,i+1$. Then, the recording tableaux $Q\left(\sigma\right)$ and $Q\left(\tau\right)$ differ only by the interchange of two entries. Specifically, one of the following two cases must hold: -Case 1: In one of the two tableaux $Q\left(\sigma\right)$ and $Q\left(\tau\right)$, the letter $i+1$ lies weakly northeast of $i-1$, which in turn lies weakly northeast of $i$. The other tableau is obtained from this one by switching $i$ with $i+1$. -Case 2: In one of the two tableaux $Q\left(\sigma\right)$ and $Q\left(\tau\right)$, the letter $i$ lies weakly northeast of $i+1$, which in turn lies weakly northeast of $i-1$. The other tableau is obtained from this one by switching $i-1$ with $i$. -Remark. Theorem 1 appeared first implicitly(!) in Astrid Reifegerste's paper "Permutation sign under the Robinson-Schensted correspondence" (also in an apparently older version as arXiv:0309266v1), where it serves as an auxiliary observation in the proof of Lemma 4.1 (formulated for dual Knuth transformations instead of Knuth transformations, but that is just a matter of inverting all permutations). Then, it was explicitly stated as Corollary 4.2.1 in Jacob Post's thesis "Combinatorics of arc diagrams, Ferrers fillings, Young tableaux and lattice paths. Both sources give rather low-level proofs which involve little theory but a lot of casework and handwaving (including references to pictures, sadly not in Zelevinsky's sense of this word). They seem to be very similar. I cannot say I fully understand either. Note that there are two cases depending on which of the Knuth relations is used, and one (the $bac$-$bca$ switch) is simpler than the other (the $acb$-$cab$ switch). -Question 1. Since Reifegerste and Post, has anyone found a slick proof of Theorem 1? I can't precisely define what I mean by "slick", but a good approximation would be "verifiable without a lot of pain" and possibly "memorable". I am not asking for it to be short or not use any theory. Indeed, I have a suspicion that an approach in the style of the proof of Corollary 1.11 in Sergey Fomin's appendix to EC1 could work: Since the recording tableau of a word $w$ is determined by the RSK shapes of the prefixes of $w$, it would be enough to show that for all but one $k$, the RSK shape of the $k$-prefix of $\sigma$ equals the RSK shape of the $k$-prefix of $\tau$. Due to Fomin's Theorem 11, this reduces to showing that for all but one $k$, for all $i$, the maximum size of a union of $i$ disjoint increasing subsequences of the $k$-prefix of $\sigma$ equals the maximum size of a union of $i$ disjoint increasing subsequences of the $k$-prefix of $\tau$. This is very easy to see when $\sigma$ and $\tau$ differ by a $bac$-$bca$ Knuth transformation. I'm unable to prove this for an $acb$-$cab$ Knuth transformation, though (I can only prove it with "all but two $k$" in that case). But the approach sounds very promising to me. Alternatively, the growth diagram view on RSK could help. -Question 2. What if we let $\sigma$ and $\tau$ be arbitrary words instead of permutations? Does the possibility of repeated letters break something? -Appendix: notations. -Here are definitions for some notations I am using above: - -A word over a set $A$ means an $n$-tuple of elements $A$ for some $n\in\mathbb N$ (where $0 \in \mathbb N$). The word $\left(a_1,a_2,...,a_n\right)$ is often written as $a_1a_2...a_n$. The entries of a word are called its letters. We identify every $a\in A$ with the one-letter word $\left(a\right)$. For any two words $u$ and $v$ over the same set $A$, we denote by $uv$ the concatenation of $u$ with $v$ (that is, the word $\left(a_1,a_2,...,a_n,b_1,b_2,...,b_m\right)$, where $\left(a_1,a_2,...,a_n\right)=u$ and $\left(b_1,b_2,...,b_m\right)=v$). -For any $n\in\mathbb N$, any permutation $\sigma \in S_n$ is identified with the $n$-letter word $\sigma\left(1\right) \sigma\left(2\right) ... \sigma\left(n\right)$ over $\mathbb Z$. -If $u$ and $v$ are two words over $\mathbb Z$, then we say that $u$ and $v$ differ by a Knuth transformation if either of the following four cases holds: - -Case 1: There exist words $p$ and $q$ and integers $a$, $b$, $c$ satisfying $a < b \leq c$ such that $u = p bac q$ and $v = p bca q$. -Case 2: There exist words $p$ and $q$ and integers $a$, $b$, $c$ satisfying $a < b \leq c$ such that $u = p bca q$ and $v = p bac q$. -Case 3: There exist words $p$ and $q$ and integers $a$, $b$, $c$ satisfying $a \leq b < c$ such that $u = p acb q$ and $v = p cab q$. -Case 4: There exist words $p$ and $q$ and integers $a$, $b$, $c$ satisfying $a \leq b < c$ such that $u = p cab q$ and $v = p acb q$. -In either case, we say that $u$ and $v$ differ by a Knuth transformation at positions $i-1,i,i+1$, where $i$ is the length of the word $p$ plus $2$. -(Note that instead of "Knuth transformation", some authors say "elementary Knuth transformation" or "elementary Knuth equivalence". Contrary to what the latter wording might suggest, the "differ by a Knuth transformation" relation itself is not an equivalence relation. The reflexive-and-transitive closure of this relation is the well-known Knuth equivalence. Note also that Reifegerste's theorem is only formulated for permutations (which, viewed as words, have no two equal letters), which allows one to ignore the distinction between $a < b \leq c$ and $a \leq b < c$; but I am keeping the $\leq$s apart from the $<$s because of Question 2 which sets them apart again.) - -If $w$ is a word, then $Q\left(w\right)$ denotes the recording tableau of $w$ under the Robinson-Schensted-Knuth correspondence. -Young diagrams and Young tableaux are written in English notation (so that the topmost row is the longest, and the leftmost column is the highest). Geographical terminology (like "northwest") refers to this orientation. - -REPLY [8 votes]: The equivalent result is well known (and made explicit) in the literature on dual Knuth transformations. From there, you can use the fact that $P(\sigma) = Q( \sigma^{-1})$ to complete the proof. Haiman's paper is the standard reference. -Haiman says two skew tableaux are dual equivalent if they are always of the same shape when acted on by a sequence of jeu de taquin moves. Theorem 2.6 shows this is equivalent to saying they differ by a sequence of elementary dual equivalences, which by Proposition 2.2 act as described in Case 1 or Case 2. Theorem 2.12 shows these elementary dual equivalences can be thought of as dual Knuth transformations. Since -$$P(\sigma) = Q(\sigma^{-1}) \ \ \text{and} \ \ (k_i\sigma)^{-1} = d_i\sigma^{-1} $$ -where $k_i$ is a Knuth transformation acting on positions $i-1, i, i+1$ and $d_i$ is a dual-Knuth transformation acting on values $i-1, i, i +1$, we can conclude that $k_i$'s action on $P(\sigma)$ is either Case 1 or Case 2. This can be seen directly by taking the row reading word of $P$ after applying a dual Knuth transformation. The arguments for the aforementioned results are relatively straightforward, and very similar to Marc's post. -I mention this to emphasize that there is no analysis of RSK in Haiman's approach. The only fact used is that the row reading word of a tableau inserts the same as the word itself. -Some additional comments: - -Haiman's work also characterizes the actions possible for the shifted Schensted correspondence. - -The proofs in Haiman's paper quite nice, especially when confined to the unshifted case. I've been told the goal of the paper was to get around the row-bumping lemmas you are trying to avoid. - -In the interest of self-promotion, my paper with Ben Young shows an analogous result for Edelman-Greene insertion (Commenters: has this been done earlier?). Work in progress shows the same for Kraskiewicz insertion, a shifted analogue of Edelman-Greene. Our proofs are based on the messy analysis of row-bumping.<|endoftext|> -TITLE: Extension of the Weyl dimension formula -QUESTION [9 upvotes]: Let $G$ be a compact semisimple group and let $\Gamma$ be a finite subgroup of $G$. I am interested, for $(\pi,V)\in \widehat G$ (irred rep of $G$), in a formula for $\mathrm{dim} V^\Gamma$, the dimension of the invariant space of $V$ by $\Gamma$. When $\Gamma$ is the trivial group, Weyl dimension formula says that -$$ -\mathrm{dim}(V) = \prod_{\alpha\in\Phi^+} \frac{\langle\lambda+\rho,\alpha\rangle}{\langle\rho,\alpha\rangle}, -$$ -where $\lambda$ is the highest weight of $\pi$ (I think the notation is usual). -One idea: let $P:V\to V$ given by -$$ -P(v)=\frac1{|\Gamma|} \sum_{\gamma\in\Gamma} \pi(\gamma)(v), -$$ -then $P$ is surjective onto $V^\Gamma$, thus $\mathrm{dim}(V^\Gamma)= \mathrm{Tr}(P) = |\Gamma|^{-1} \sum_{\gamma\in\Gamma} \chi_\pi(\gamma)$, where $\chi_\pi$ is the character of $\pi$. Hence, by applying the Weyl character formula, we obtain a formula for $\mathrm{dim}(V^\Gamma)$. However, I expect that there is a more explicit formula (like Weyl dimension formula, without a sum over the Weyl group) in terms of the root system of $\mathfrak g$. -Question: Is there a formula for $\mathrm{dim}(V^\Gamma)$ similar to Weyl Dimension Formula? - -REPLY [3 votes]: Your idea seems to have already been implemented, see Section 2 of Chenevier and Renard's "Level one algebraic cusp forms of classical groups of small ranks". But it appears to be not as straightforward as just applying the ordinary Weyl character formula - you need a certain "degenerate" version of it.<|endoftext|> -TITLE: How hard is it to compute the diameter and the growth function of a finite permutation group of small degree? -QUESTION [7 upvotes]: Let $G \leq {\rm S}_n$ be a finite permutation group, and let -$S = \{g_1, \dots, g_k\}$ be a generating set for $G$ which is closed -under inversion and which does not contain the identity. -The growth function of $G$ with respect to $S$ is the sequence -$(a_0, a_1, a_2, \dots)$, where $a_r$ is the number of elements of $G$ -which can be written as products of $r$, but no fewer, generators $g_i$. -The diameter of $G$ with respect to $S$ is the largest $r$ such that -$a_r > 0$. - -Question: How hard is it to compute the diameter and the - growth function of a given permutation group $G$ of degree $n$? - Or more specifically: using current computer technology, is it feasible - to do this for any given group $G$ of degree $n \leq 100$ and any given - sufficiently small generating set $S$? - -The motivation for this question is that while the Schreier-Sims algorithm -allows e.g. to compute the order of such groups and to perform element tests -instantaneously, even only computing the diameter of the Rubik's Cube Group -with respect to its natural generating set was a major effort -- -and its growth function is apparently not known in full so far. -My feeling goes in the direction that one can do essentially better, -i.e. that it should be possible to find an algorithm for computing -diameter and growth function which is by orders of magnitude more -efficient than enumerating group elements by brute force. -However maybe I am wrong, and somebody can point out reasons why -these problems cannot be solved efficiently? - -REPLY [5 votes]: I just came across this question and, even though I'm a bit late, I thought you might be interested in this reference: -Even, S.; Goldreich, O., The minimum-length generator sequence problem is NP-hard, J. Algorithms 2, 311-313 (1981). ZBL0467.68046. -This asserts that the calculation of the diameter of a Cayley graph of a permutation group is NP-Hard. I don't have access to the article myself but, from what I've read elsewhere, I believe this statement remains true even if you restrict to elementary-abelian 2-groups. (Which seems astonishing to me!)<|endoftext|> -TITLE: A constant associated to the character table of a finite group -QUESTION [6 upvotes]: Following on from some of myprevious MO questions on finite group theory... -$\newcommand{\Irr}{\operatorname{Irr}}\newcommand{\Conj}{\operatorname{Conj}}\newcommand{\AMZL}{{\rm AM}_{\rm Z}}$ -Let $G$ be a finite group, $\Irr$ the set of irreducible characters (working over $\mathbb C$) and $\Conj$ the set of conjugacy classes. Consider the following quantity, which seems to have first been introduced in work of Azimifard, Samei and Spronk -$$\alpha_G = \frac{1}{|G|^2} \sum_{\phi\in\Irr} \sum_{C\in\Conj} \phi(e)^2 |\phi(C)|^2 |C|^2 . $$ -It is not hard to show, just using basic character theory, that $\alpha_G \geq 1$ with equality if and only if $G$ is abelian. This constant/invariant arises as a minorant for a more interesting invariant of $G$ which was studied in the aforementioned paper. -Call this "more interesting invariant" $\AMZL(G)$. It turns out that $\AMZL(G)\geq \AMZL(G/N)$ for every normal subgroup $N$, a property which is vital to some work I am writing up that obtains lower bounds on $\AMZL(G)$. One branch of the case-by-case attack I'm using works by getting lower bounds on $\alpha_G$ when $G$ has trivial centre, and so if $\alpha_G$ never increases when we replace $G$ with a quotient then I could in fact get general lower bounds on $\alpha_G$ by inductively modding out by the centre. - -Question. Do there exists a finite group G and a normal subgroup N for which $\alpha_{(G/N)} > \alpha_G$ ? - -I did try for a while to show this can never happen but rapidly got stuck, so I'm hoping that MO readers who are more skilled/experienced with finite groups can either suggest counterexamples to try, or give some evidence that no such counterexamples exist. Please bear in mind that I have next to no experience with GAP or similar, so I don't mind concrete suggestions of code, but responses of the form "get GAP to look through all non-simple non-abelian examples of order < 100" may not be as helpful as their authors imagine. - -Just in case it helps to rule out certain avenues of attack: if $G$ is a group with two character degrees (i.e. there is an integer $m$ such that $\phi(e)\in \{1,m\}$ for all $\phi\in\Irr$) then one can evaluate $\alpha_G$ explicitly, using ideas similar to this paper: one obtains -$$ -\alpha_G - 1 = (m^2-1) \left( 1- \frac{|L|}{|G|^2} \sum_{C\in\Conj} |C|^2 \right) -$$ -where $|L|$ is the number of linear characters on $G$. -In particular, suppose if $G=H \rtimes C_2$ is a generalized dihedral group with $|H|=2n+1$ ($n$ a positive integer). Back of the envelope scribbling gives me $m=2$, $|L|=2$ and the conjugacy classes have sizes $1$, $2$ repeated $n$ times, and $2n+1$, so that -$$ \frac{\alpha_G -1}{3} = 1 - \frac{1}{2(2n+1)^2}(1+ 4n + (2n+1)^2) = \frac{1}{2}\left(1-\frac{1}{2n+1}\right)^2 $$ -It also seems to me that proper quotients of $G$ must either be abelian or generalized dihedral with smaller order, in which case we would have $\alpha_G \geq \alpha_{(G/N)}$ for every $N \lhd G$. - -REPLY [5 votes]: Let $G= \operatorname{SL}(2,5)$ and $N= \mathbf{Z}(G)$, so that $G/N = \operatorname{PSL}(2,5)\cong A_5$. If my code works correctly, then -$$ \alpha_{G/N} = 842/75 > 6661/600 = \alpha_G. $$ -My implementation of alpha (for character tables) in GAP: -alpha:= function( tbl ) - local clssizes, n, kg, phi, i; - clssizes:= SizesConjugacyClasses( tbl ); - n:= Size( tbl ); # group order - kg:= NrConjugacyClasses( tbl ); - return Sum( Irr( tbl ), - phi-> phi[1]^2 - * Sum( [1..kg], - i-> phi[i] * ComplexConjugate( phi[i] ) - * clssizes[i]^2 - ) - )/ n^2 ; -end;<|endoftext|> -TITLE: Is the definition of Gerstenhaber bracket related to operads? -QUESTION [7 upvotes]: I was reading these notes by Keller. On the page 19 he defines the Gerstenhaber bracket on the Hochschild cochain complex. But first of all he defines the operation $\bullet$ on the cochains by -$$ -\small -(f\bullet g)(a_1,\dots,a_{p+q-1})=\sum\limits_{i=0}^{p} (-1)^{i(q+1)} f(a_1,\dots,a_i,g(a_{i+1},\dots,a_{i+q}),a_{i+q+1},\dots,a_{p+q-1}) -$$ -where $f$ is a $p$-cochain and $g$ is a $q$-cochain. After that he defines the Gerstenhaber bracket by $[f,g]_G=f\bullet g-(-1)^{(p-1)(q-1)}g\bullet f$. -I have noticed that the summands in the operation $\bullet$ look a lot like the partial composition in operads, as if $f$ was an element of arity $p$ and $g$ was an element of arity $q$ in some operad. So my question is, is it just a coincidence, or there is indeed some way to see $f,q$ as operations in some operad? Or maybe is there a way to see the operation $\bullet$ coming from some operadic context? -I am a complete novice in these things, so I am sorry if my question is silly. Anyways I would be happy if someone would explain that to me. -Thank you for your help! - -REPLY [7 votes]: Yes, absolutely. I really think the best place for learning this is the book by Loday and Vallette. -This example fits more generally into the following context. Let $P$ be a Koszul operad and $A$ a $P$-algebra. There is attached to $P$ a cohomology theory for $P$-algebras. The two most well known examples is that when $P = \mathsf{Lie}$ we get ordinary Lie algebra cohomology, and that when $P = \mathsf{Ass}$ we get Hochschild homology. This is in general defined by the chain complex -$$ \mathrm{Hom}_{\mathbb S}(P^!,\mathrm{End}(A)) $$ -i.e. maps of $\mathbb S$-modules from the Koszul dual co-operad of $P$ to the endomorphism operad of $A$. This chain complex sits inside $$ \mathrm{Hom}_{\mathbb k}(P^!,\mathrm{End}(A)), $$ -i.e. maps which are not necessarily equivariant. The latter space is itself in a natural way the sum of all components of an operad, the convolution operad. Convolution gives an operad structure on maps from any co-operad to an operad. -Now on the sum of all components of an operad there is a pre-Lie structure given by operadic composition. This is the operation you called $\bullet$, and as you say, it is just given by composition. Antisymmetrizing this gives an honest Lie bracket. This bracket on $$ \mathrm{Hom}_{\mathbb k}(P^!,\mathrm{End}(A)), $$ leaves $$ \mathrm{Hom}_{\mathbb S}(P^!,\mathrm{End}(A)) $$ invariant, giving a Lie bracket on the chain complex for the operadic cohomology. When $P = \mathsf{Ass}$ and $A$ is an associative algebra we have defined the Gerstenhaber bracket on the Hochschild complex. This is what you are seeing. -There are some degree shifts involved here (as one sees from your formulas) that I'm not going to try to work out.<|endoftext|> -TITLE: Epimorphisms $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ are split -QUESTION [17 upvotes]: Consider the additive group of integer sequences $\mathbb{Z}^{\mathbb{N}}$. Why does every epimorphism of groups $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ split? $(\star)$ -Actually this claim is equivalent to the Whitehead problem for countable abelian groups: -"$\Rightarrow$": Recall Specker's result which states that $\mathbb{Z}^{(\mathbb{N})} \to \hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $e_i \mapsto \mathrm{pr}_i$ is an isomorphism, i.e. $\mathbb{Z}^{\oplus \mathbb{N}}$ is reflexive. Now assume that $A$ is countable and $\mathrm{Ext}^1(A,\mathbb{Z})=0$. Choose a presentation $0 \to P \to Q \to A \to 0$ with free abelian groups $P,Q$, w.l.o.g. of rank $\aleph_0$. By assumption $Q^* \to P^*$ is an epi, hence splits. Since $P,Q$ are reflexive, then also $P \to Q$ splits, and $A$ is free. -"$\Leftarrow$": If $f : \mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ is an epimorphism, then $f^* : \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ is a monomorphism, the cokernel $A$ of $f^*$ is countable and satisfies $\mathrm{Ext}^1(A,\mathbb{Z})=0$, since $f^{**} \cong f$ is epi. Hence $A$ is free, which implies that $f^*$ splits and therefore also $f^{**} \cong f$ splits. $~\square$ -The countable Whitehead problem was proved by Stein in 1950. He used injective resolutions, i.e. $\mathrm{Ext}^1(A,\mathbb{Z}) = \mathrm{coker}(\hom(A,\mathbb{Q}) \to \hom(A,\mathbb{Q}/\mathbb{Z}))$. In particular, $(\star)$ is true. On the other hand, the equivalence above suggests an alternative proof of the countable Whitehead problem. Therefore my question is: Is there a direct proof for $(\star)$? -By Specker's result an endomorphism of $\mathbb{Z}^{\mathbb{N}}$ corresponds to an endomorphism of $\mathbb{Z}^{\oplus \mathbb{N}}$, and therefore to a column-finite matrix. But I don't know how to characterize surjectivity. And this is where I get stuck. - -REPLY [4 votes]: I'll start by describing the notation that I'll use. -I'll think of elements of $\mathbb{Z}^\mathbb{N}$ as infinite column vectors -$${\bf x}=\begin{pmatrix} -x_0\\x_1\\x_2\\\vdots -\end{pmatrix}$$ -of integers, with rows indexed by $\mathbb{N}$. -The $i$th "unit vector" will be denoted by ${\bf e(i)}$. I.e., $e(i)_i=1$ and $e(i)_j=0$ for $j\neq i$. -As described in the question, Specker's result implies that endomorphisms of $\mathbb{Z}^\mathbb{N}$ are of the form ${\bf x}\mapsto A{\bf x}$ where -$$A=\begin{pmatrix} -a_{00}&a_{01}&a_{02}&\cdots\\ -a_{10}&a_{11}&a_{12}&\cdots\\ -a_{20}&a_{21}&a_{22}&\cdots\\ -\vdots&\vdots&\vdots&\ddots -\end{pmatrix}$$ -is an infinite matrix of integers, with rows and columns indexed by $\mathbb{N}$, that is row-finite (i.e., each row has only finitely many non-zero entries). -Now suppose that $\alpha:\mathbb{Z}^\mathbb{N}\to\mathbb{Z}^\mathbb{N}$ is a surjective group homomorphism, given by a row-finite matrix $A$. -To prove that $\alpha$ is split, we need to find a right inverse. In fact, it suffices to find another row-finite matrix $B$ such that $AB$ is lower unitriangular (i.e., diagonal entries are all $1$ and entries above the diagonal are all $0$), since such a matrix is invertible. -Fix $n\geq0$, and decompose $\mathbb{Z}^\mathbb{N}=G_n\oplus H_n$, where $G_n\cong\mathbb{Z}^n$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $i\geq n$, and $H_n\cong\mathbb{Z}^\mathbb{N}$ is the subgroup consisting of those ${\bf a}$ with $a_i=0$ for $ii$ with $\theta({\bf e(i')})=\theta({\bf e(i)})$, or equivalently ${\bf e(i)}-{\bf e(i')}\in\alpha(H_n)$. -So we can choose a sequence $0=t_0i$ and some ${\bf b(i)}\in H(n)$ with ${\bf e(i)}-{\bf e(i')}=\alpha\left({\bf b(i)}\right)$. For each $i$, do this with the largest $n$ such that $i\geq t_n$, and let $B$ be the matrix whose $i$th column is ${\bf b(i)}$. -Then $B$ is row-finite, and $AB$ is lower unitriangular.<|endoftext|> -TITLE: Sufficient condition for Riemann Hypothesis? -QUESTION [5 upvotes]: Is there an L-function ($L_s=\sum_{n =1}^{\infty} \frac{a_n}{n^s}$) having a functional equation coming from a relation of the type [1]: -$\sum_{n =1}^{\infty} a_n \; e^{-2\pi nx}= \frac{A}{x^k} \sum_{n =1}^{\infty} a_n'\; e^{-2\pi\frac{n}{Nx}}$ but not satisfying the Riemann Hypothesis? -Notice that in the above sum there is no term for $n=0$. -(I think Dirichlet L-function and some Modular forms L-function with Fricke involution satisfy such relation and RH, but is this possibly a sufficient condition? Any counter example?) -The L-fonction I know having a functional equation and satisfying RH have also an associated relation of the type [1] (coming from Poisson Summation formula or Fricke involution) sometimes they have a term in 0 (like for Zeta) sometimes not (like for Dirichlet L-functions). Now I know example of L-functions not satisfying RH with functional equation but it seems they appear only with combination of L-functions having a term n=0 in the Poisson summation like formula. -By the way do you know a L-function satisfying RH without a relation like [1] (with possibly a term n=0)? - -REPLY [12 votes]: There certainly are L-functions having a functional equation and violating the analogue of RH. (I am not completely sure I understand what type of functional equation you mean by your restriction, but the ones I mention satisfy a functional equation of so to say the usual form, so I hope this is fine.) -Eppstein zeta-functions are a classical example; Davenport and Heilbronn 'On the zeros of certain Dirichlet series' (JLMS) is a classical reference. -For recent investigations in such phenomena see for example Frank Thorne's Analytic properties of Shintani zeta functions. -You might also have a look at the closely related MO question Are there refuted analogues of the Riemann hypothesis? where some information related to this is to be found. -In particular, it seems that for RH one needs Euler product also (not just a functional equation) and then there might be an equivalent; see Frank Thorne's answer on the above mentioned question for more specific information.<|endoftext|> -TITLE: Coupling of non-probability/sub-probability measures -QUESTION [5 upvotes]: A coupling of two probability measures $P,\tilde P$ on a Borel space $X$ is any probability measure on $X^2$ whose one-dimensional marginals are $P$ and $\tilde P$. In particular, for any such coupling $\Bbb P$ we have -$$ - 2\cdot \Bbb P(X^2\setminus\Delta_X)\geq\|P - \tilde P\| \tag{1} -$$ -where $\Delta_X = \{(x,x):x\in X\}$ is the diagonal, and $\|\cdot\|$ denotes the total variation norm. At the same time, there always exists a maximal coupling $\Bbb P^*$ such that the equality holds in $(1)$. -Extension of these results to the case when $P$ and $\tilde P$ are non-probability measures, or at least sub-probabilities does not seem to be complicated, however I wonder whether it has been already addressed somewhere. So far a search on google did not help. - -REPLY [4 votes]: I thought, I could turn the comments into an answer… -The approach by couplings does not work without modifications and the reason is that couplings do not exist if the measures have different total mass: If $P$ and $Q$ are two measures on $X$ with different total masses which were coupled by $\mu$, then -$$ -\int_x\int_y d\mu(x,y) = \int_x dP(x) \neq \int_y d\tilde P(y) = \int_y\int_xd\mu(x,y). -$$ -However, for a given metric $d$ for probability measures one can build a metric for measures with different masses as follows: If $P$ has total mass $P(X) = p$ and $Q$ has total mass $Q(X) = q$, define -$$ -D(P,Q) = |p-q| + d(\tfrac{P}{p},\tfrac{Q}{q}) -$$ -(see Gromov's "Metric structures on Riemanian and Non-Riemannian Spaces", Chapter $3\tfrac12$.B). -There are also other approaches to metrics on measure spaces such as the Kantorovich-Rubinstein norm -$$ -W(P,Q) = \sup\bigg\{\int f\, dP - \int f\, dQ\ :\ f\ \text{Lipschitz with constant}\ \leq 1\bigg\} -$$ -and others (cf. Villani's "Optimal Transport - Old and New", Chapter 6).<|endoftext|> -TITLE: Discrete subgroups of products of SU(2) -QUESTION [6 upvotes]: It is known that there are -up to conjugation- 5 classes of discrete subgroups of SU(2). One way to show this is by means of the McKay correspondence. My question is more regarding products of $SU(2)$, say with $n$ factors. Well, since I am definitely not familiar with the correspondence above, I would like to ask whether I have any hope to classify all finite subgroups of the product of 2 or 3 copies of $SU(2)$ by following the same line of thought. Or in case someone knows a more economical alternative, I would be very happy to listen to it. - -REPLY [3 votes]: Incidentally, $SO(4)=SU(2)\times SU(2)/\{\pm (I\times I)\}$, so the classification of discrete subgroups of $SO(4)$ is (almost) equivalent to the classification of discrete subgroups of $SU(2)\times SU(2)$ (every discrete subgroup of $SU(2)\times SU(2)$ will correspond to a subgroup of $SO(4)$ containing $\pm I$). The discrete subgroups of $SO(4)$ correspond to orientable 3-dimensional spherical orbifolds, which have been explicitly worked out by Dunbar (see also here).<|endoftext|> -TITLE: why most of the angles are right -QUESTION [16 upvotes]: The Coxeter–Dynkin diagrams tell us that in a spherical Coxeter simplex most of the dihedral angles are right. Say among $\tfrac{n{\cdot}(n+1)}2$ dihedral angles we can have at most $n$ angles which are not right. - -Is it possible to see this statement without classification? - -REPLY [12 votes]: Let the root system be $v_1$, …, $v_n$ with all elements normalized to be length $1$. So $\langle v_i, v_i \rangle =1$, we have $\langle v_i, v_j \rangle \leq - \cos (\pi/3) = -1/2$ for at least $n$ pairs $(i,j)$, and we have $\langle v_i, v_j \rangle \leq 0$ for all $i \neq j$. -But then -$$\left\langle \sum_{i=1}^n v_i, \sum_{i=1}^n v_i \right\rangle \leq n + 2 n (-1/2) =0,$$ -a contradiction.<|endoftext|> -TITLE: Identity on Gauss sums -QUESTION [5 upvotes]: I'd like to ask how to prove or disprove the following identity (which I have to use on some computation regarding $\epsilon$-constant for constructing certain local Langlands parameter): -Let $\psi$ be any non-trivial additive character on $\mathbb{F}_q$, the field of $q$ elements. Then -$$\sum_{x\in\mathbb{F}_{q^d}}\left(\frac{x}{q^d}\right)\psi(\text{Tr}_{\mathbb{F}_{q^d}/\mathbb{F}_q}x)=\left(\sum_{x\in\mathbb{F}_{q}}\left(\frac{x}{q}\right)\psi(x)\right)^d,$$ -where $\left(\frac{\cdot}{q^d}\right)$ denote the quadratic symbol on $\mathbb{F}_{q^d}^*$. The identity is definitely correct up to sign, and I come up with it just via numerical checking. Thanks! - -REPLY [6 votes]: I think this is just the Davenport-Hasse Theorem, which Wikipedia calls the Hasse-Davenport lifting relation.<|endoftext|> -TITLE: If there is a non-constructible real, is there an $L$-generic real? -QUESTION [8 upvotes]: If we assume that $\Bbb R^V\neq\Bbb R^L$, can we deduce that there is some $x\in\Bbb R^V$ which is $L$-generic? -Of course if $V$ is a generic extension of $L$ this is true, but if $V=L[0^\#]$ this is also true (the existence of $0^\#$ implies the existence of $L$-Cohen generics), but $0^\#$ cannot be added by forcing. So it is possible to have reals which are not generic over $L$, but their existence does imply that a generic exist. -Is this always the case, or is there a counterexample? Namely, can we have $V=L[x]$ for some real number $x$ such that no real number in $L[x]$ is $L$-generic? -Edit: As Joel's answer shows, we can generate a counterexample using class forcing over $L$. To avoid these, we might as well require that the universe is not a class-generic extension of some $W$ for which $\Bbb R^W=\Bbb R^L$. - -REPLY [9 votes]: Great question! -In order to formalize the notion, let us understand the phrase "$x$ is $L$-generic" to mean: there is some partial order $\mathbb{P}\in L$ and some filter $G\subset\mathbb{P}$ that is $L$-generic, such that $x\in L[G]$. In particular, this refers to set-sized forcing only. -In this case, we can give a negative answer. Sy Friedman has a way to undertake the coding-the-universe forcing in such a way that the generic extension $V[G]=L[R]$ is minimal: Minimal Coding, Annals of Pure and Applied Logic, 1989, pp. 233-297. In particular, every real in the extension is either in $V$, or generates $R$, which is not set-generic (although it is generic for class forcing). Thus, if you start in $L$ and then undertake this forcing, you get a class forcing extension in which there are no $L$-generic reals for set forcing, and indeed, no $L$-generic sets of any kind. -In the linked paper, Friedman states the following: -Corollary. There is an $L$-definable forcing for producing a real $R$ which is minimal over $L$ but not set-generic over $L$. -It follows that this extension $L[R]$ has no set-generic objects at all not in $L$. That is, it is a strong counterexample, which applies not just to reals, but to sets of any size.<|endoftext|> -TITLE: C*-algebras and quantum fields -QUESTION [6 upvotes]: One can represent a quantum system by the Weyl algebra (which is a C*-algebra). For instance, a 1 degree of freedom system can be represented by the algebra generated by $e^{\imath t Q}, e^{\imath s P}$. My guess for modeling a quantum field would be to look at C*-algebras with an uncountably infinite number of generators, but in AQFT this is done thanks to a net of C*-algebras. Is there a link between the approach I think of as being intuitive and the AQFT one? - -REPLY [13 votes]: The Weyl algebra construction can be done abstractly for any real vector space (even infinite-dimensional) endowed with an antisymmetric bilinear form, thanks to B. Blackadar's universal C*-algebra construction using generators and relations ("Shape theory for C∗-algebras", Math. Scand. 56 (1985) 249–275). However, since you asked for a concrete correspondence, here is a more explicit description. In the case of (free) real scalar fields $\phi$ in a globally hyperbolic space-time $(M,g)$ subject to the Klein-Gordon equation $$ \Box_g\phi+m^2\phi=0\ ,$$ the Weyl unitaries are concretely given by the functionals $$ \mathscr{C}^\infty(M,\mathbb{R})\ni\phi\mapsto W_f(\phi)=e^{i\int_M f\phi\ d\mu_g}\ ,\quad f\in\mathscr{C}^\infty_c(M,\mathbb{R})\ ,$$ where $d\mu_g$ is the volume form associated to the Lorentzian metric $g$. One can understand the test functions $f$ as "component indices", just like we do in the case of the $2n$-dimensional symplectic (phase) space generated by $n$ positions $x_1,\ldots,x_n$ and $n$ momenta $p_1,\ldots,p_n$. Given any non-void open subset $O\subset M$ of the space-time manifold $M$, let $\tilde{\mathfrak{A}}(O)$ be the unital $*$-algebra generated by the Weyl unitaries $W_f$ as $f$ runs over the real-valued smooth functions compactly supported in $O$ (so we say that such $W_f$'s are localized in $O$), once we endow such functionals with the following operations: - -Vector space operations: $(\alpha W_f+\beta W_{f'})(\phi)=\alpha W_f(\phi)+\beta W_{f'}(\phi)$, $\alpha,\beta\in\mathbb{C}$; -Involution: $W^*_f=W_{-f}$; -Identity: $\mathbb{1}=W_0$; -Weyl product: $W_f W_{f'}=e^{\frac{i}{2}\Delta_{m,g}(f\otimes f')}W_{f+f'}$, where $\Delta_{m,g}$ is the distributional kernel of the difference between the retarded and the advanced fundamental solutions of the Klein-Gordon operator $\Box_g+m^2$. We recall that $\Delta_{m,g}$ provides our antisymmetric bilinear form, for $$\Delta_{m,g}(f\otimes f')=-\Delta_{m,g}(f'\otimes f)\ .$$ - -The above operations are then extended to general elements in the usual way, and they entail that the Weyl unitaries are worthy of their name, since we clearly have $W^*_f W_f=\mathbb{1}$. This $*$-algebra has non-trivial $*$-representations (namely, Fock representations associated to quasi-free states), hence it admits a minimal nontrivial C$*$-norm $\|\cdot\|$. The Weyl algebra $\mathfrak{A}(O)$ associated to $O$ is the C$*$-completion of $\tilde{\mathfrak{A}}(O)$ with respect to this C$*$-norm, and the correspondence $O\mapsto\mathfrak{A}(O)$ is obviously an isotonous net of C$*$-algebras. Moreover, due to the causal support of $\Delta_{m,g}$ entailed by the hyperbolicity of the Klein-Gordon operator, this net is also causal (i.e. elements localized in causally disjoint regions commute). Finally, uniqueness of retarded and advanced fundamental solutions for this operator guarantees that the isometry group of $(M,g)$ acts on this net as it should - namely, for any isometry $\psi$ of $(M,g)$, the action $$ \alpha_\psi(W_f)(\phi)=W_f(\psi^*\phi)=W_{\psi_*f}(\phi) $$ uniquely extends to unit-preserving *-isomorphisms satisfying $$\alpha_\psi\circ\alpha_{\psi'}=\alpha_{\psi\circ\psi'}\ ,\quad\alpha_\psi(\mathfrak{A}(O))=\mathfrak{A}(\psi(O))\ .$$ In the above formula, $\psi^*\phi=\phi\circ\psi$ is the pullback of the field configuration $\phi$, and $\psi_*f=f\circ\psi^{-1}$ is the pushforward of the test function $f$. To summarize, we have obtained a Haag-Kastler net of C$*$-algebras. -The above argument is a somewhat modernized version of a construction due to J. Dimock ("Algebras of local observables on a manifold", Comm. Math. Phys. 77 (1980), no. 3, 219–228). See also C. Bär, N. Ginoux and F. Pfäffle, "Wave equations on Lorentzian manifolds and quantization", ESI Lectures in Mathematics and Physics, European Mathematical Society (2007), arXiv:0806.1036 [math]. -One can perform a similar construction for Dirac fields (see for instance C. Dappiaggi, T.-P. Hack and N. Pinamonti, "The extended algebra of observables for Dirac fields and the trace anomaly of their stress-energy tensor", Rev. Math. Phys. 21 (2009) 1241-1312, arXiv:0904.0612 [math-ph]), with the simplification that for the CAR algebra it's not necessary to exponentiate the smeared fields to get C*-algebra elements. -For interacting fields, if one wants to keep close to what physicists do and steer clear of constructive methods a la Glimm-Jaffe (which are of course a better choice from the viewpoint of rigor but severely limit the models one may study in their present state of the art), one has to recourse to formal perturbation theory, which means one has to work with formal power series in the coupling constant and Planck's constant. This also means abandoning C$*$-algebras and working with more general *-algebras. Once one accepts this, perturbative renormalization can be dealt with in a rigorous way, using a language close to the one adopted above. See for instance R. Brunetti, M. Dütsch and K. Fredenhagen, "Perturbative algebraic quantum field theory and the renormalization groups", Adv. Theor. Math. Phys. 13 (2009) 1541–1599, arXiv:0901.2038 [math-ph].<|endoftext|> -TITLE: Stiefel manifolds and polar decompositions -QUESTION [6 upvotes]: The real Stiefel manifold $V_{n,k}$ of orthogonal $k$-frames in $\mathbb{R}^n$ can be viewed as the reductive homogeneous space $G/H=O(n)/O(n-k)$. If ${\frak{so}}(n)$ is the Lie algebra of $O(n)$, then we have the reductive decomposition -$$ -{\frak{so}}(n)={\frak{m}}+{\frak{h}} -$$ -where -$$ -{\frak{m}}=\left \{ -\begin{pmatrix} - A & B\\ - -B^T & O -\end{pmatrix}: A\>\> \text{is a}\>\> k\times k \>\>\text{skew-symmetric matrix} \right \} -$$ -and -$$ -{\frak{h}}=\left \{ -\begin{pmatrix} - O & O\\ - O & C -\end{pmatrix}: C\>\> \text{is a}\>\> (n-k)\times (n-k) \>\>\text{skew-symmetric matrix} \right \} -$$ -with ${\frak{so}}(n)$ and ${\frak{h}}$ a reductive pair. -According to Helgason, there is a local diffeomorphism -$$ -(\text{exp}X,h)\mapsto (\text{exp}X)h,\quad \text{where}\>\> X\in {\frak{m}}, h\in H. -$$ -However, there doesn't seem to be anything in the literature where this decomposition is calculated explicitly. I did find the following ``polar decomposition": -$$ -V_{n,k}\times P_k\rightarrow M_{n,k},\quad (v,r)\mapsto vr^{1/2} -$$ -where $P_k$ is the set of positive semi-definite symmetric matrices and $M_{n,k}$ is the set of all $n\times k$ real matrices. -Does this last decomposition have any connection with the polar decompositions of Lie groups? If not, is there an explicit description of the polar decomposition involving Stiefel manifolds somewhere? - -REPLY [8 votes]: It is easier for me to examine this geometrically, rather than from the point of view of Lie groups and algebras. -First, the identity matrix $I \in O(n)$ represents the standard orthonormal basis $e_1, \dots, e_n$ of $\mathbb{R}^n$, and its coset $I\cdot O(n-k)$ represents the $k$-plane spanned by the first $k$ basis vectors $e_1, \dots, e_k$. -The local splitting you're looking for is equivalent to extending the standard orthonormal basis to a local map from $V_{n,i}$ to $O(n)$, viewed as the space of orthonormal frames in $\mathbb{R}^n$, such that the span of the first $k$ vectors in each frame is equal to the corresponding $k$-plane in $V_{n,k}$. Helgason is basically pointing out that one way to do this is to use the connection naturally induced by the bi-invariant Riemannian metric on $O(n)$ to parallel transport the initial frame along geodesic rays starting at the identity matrix. -This local section can be described as a solution to a linear second first order system of ODE's along each geodesic (essentially the Jacobi equations along each geodesic), and in principle these ODE's could be integrated to give some kind of formula for this section. Unfortunately, I've never computed this and don't know what it looks like. And it's not clear to me how to transfer this description to the setting of Lie groups and algebras. My wild guess is that you can write the system of ODE's in terms of the Lie algebra, but you won't be able to integrate them explicitly. -One possibly easier way to do this is to using the dual description using invariant differential forms as presented in: -Griffiths, P. -On Cartan's method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry. -Duke Math. J. 41 (1974), 775–814.<|endoftext|> -TITLE: Naive question on adelic groups -QUESTION [6 upvotes]: The ever-reliable Wikipedia says: - -... an adelic algebraic group is a semitopological group defined by... - -No more details are given, and I was wondering if the multiplication only being separately continuous has any noticeable effect when working with adelic groups. I ask because there are some algebras which pop up in the study of von Neumann algebras where the multiplication is not jointly continuous, and this is mildly annoying. - -REPLY [8 votes]: A locally compact semitopological semigroup which is also a group is in fact a topological group, i.e. the existence of inverses plus separate continuity actually forces joint continuity. -This is a theorem of Ellis, see - -R. Ellis, Locally compact transformation groups. -Duke Math. J. 24 (1957) no. 2, 119–125 - -Quoting from the MathReview: - -Let G be a group of homeomorphisms on a locally compact space X. Suppose G has a Hausdorff topology such that multiplication is continuous in each variable separately. If the function π:G × X → X defined by π(g,x)=g(x) is continuous on the left, the author shows that π is jointly continuous. - -EDIT: it's been noted in the comments that the part I quoted does not explain in any way why we can deduce continuity of inversion. In an earlier paper Ellis had shown that in a topological semigroup whose underlying semigroup is a group, inversion is continuous: - -R. Ellis, A note on the continuity of the inverse. Proc. Amer. Math. Soc. 8 (1957), 372–373<|endoftext|> -TITLE: BSD conjecture for X_0(17) -QUESTION [7 upvotes]: I use Magma to calculate the L-value, yields -E:=EllipticCurve([1, -1, 1, -1, 0]); -E; -Evaluate(LSeries(E),1),RealPeriod(E),Evaluate(LSeries(E),1)/RealPeriod(E); -Elliptic Curve defined by y^2 + x*y + y = x^3 - x^2 - x over Rational Field -0.386769938387780043302394751243 3.09415950710224034641915800995 -0.125000000000000000000000000000 -$#torsionsubgroup = 4, c_{17}(E)=1.$ -But the strong BSD predicts that -$L(E,1)/\Omega_{\infty}$= $(#Sha(E)/#tor(E)^2)*c_{17}(E)$ -We will get $L(E,1)/\Omega_{\infty}=1/16$, not $1/8$. -Why does that happen? Thanks a lot. - -REPLY [16 votes]: Tim's answer is great, but I want to mention one other place where people have lost a power of 2. The canonical height is often defined relative to the divisor (O), as I do in my books. So it is given by -$$ \hat h(P) = \frac12 \lim_{n\to\infty} 4^{-n} h\bigl(x([2^n]P)\bigr). $$ -Here the $\frac12$ is inserted because $x$ has a double pole at $\infty$. However, in computing the height regulator for BSD, one should not use the $\frac12$, i.e., one should compute heights relative to the divisor $2(O)$. -Why, one might ask, use a weird divisor like $2(O)$. The answer is that when BSD is properly formulated for abelian varieties, it uses the height pairing $A(K)\times \hat A(K)$ that pairs points of $A$ with points on its dual, and the pairing is done relative to the Poincare divisor on $A\times \hat A$. If one traces through the definitions and identifies an elliptic curve with its dual, the Ponicare divisor on $E\times E$ is $(O)\times E + E\times (O)$, which eventually shows that one should use the height on $E$ relative to $2(O)$. -Note that this means that if you compute BSD using the wrong height on a curve of rank $r$, then your answer will be off by a factor of $2^r$. -(I learned about this potential error from Dick Gross many years ago.)<|endoftext|> -TITLE: Can one use the continuity method to show that the two dimensional hyperbolic space can be immersed in five dimensional Euclidean space? -QUESTION [5 upvotes]: First of all, I must clarify at the outset that I am simply asking if there is an alternative way to solve an already known problem. It is known that the answer to my question is yes. The problem is as follows: -Consider $(\mathbb{R}^2, g_{H})$ with the hyperbolic metric, i.e. -$$ ds^2 = e^{2y} dx^2 + dy^2 .$$ -Does there exist a smooth isometric immersion into $(\mathbb{R}^5, g_{flat})$? In other words we are looking for a smooth map $ u : \mathbb{R}^2 \rightarrow \mathbb{R}^5$ such that -$$ du_1(x,y)^2 + du_2(x,y)^2 + \ldots du_5(x,y)^2 = e^{2y} dx^2 + dy^2.$$ -My question is the following: Is it conceivable that one can solve this question using the continuity method? More precisely, consider the family of metrics $g_t$ on $\mathbb{R}^2$ given by -$$ ds^2 = e^{2ty} dx^2 + dy^2 .$$ -Clearly one can solve his pde when $t=0$. It would seem to me naively that the set of $t$ for which the pde is solvable is open (probably one can show this from the Implicit Function theorem). -Is it conceivable one can show this set is closed? Typically that is the difficult part of using the continuity method. -Remark: 1) There is an explicit solution to this immersion question. This can be found in the book "Isometric Embedding of Riemannian Manifolds in Euclidean Spaces" by Qing Han and Jia Xing Hong. The desired function $u$ is given explicitly. -2) Just in case the answer to my question is yes, i.e. one can actually use the continuity method to solve the immersion problem in $\mathbb{R}^5$, what are the obstructions to applying that same idea in $\mathbb{R}^4$? -I believe it is an open question whether the hyperbolic plane can be immersed in $\mathbb{R}^4$. - -REPLY [5 votes]: I think you are asking the wrong question. -By scaling you have that if you can solve for -$$ \mathrm{d}s^2 = e^{2ty} \mathrm{d}x^2 + \mathrm{d}y^2 $$ -for any non-zero $t$ you can also solve for -$$ \mathrm{d}s^2 = e^{2\tilde{t}y} \mathrm{d}x^2 + \mathrm{d}y^2 $$ -for any non-zero $\tilde{t}$. -Proof: Let $u'(x,y) = u(\lambda x,\lambda y)$, you have $|\mathrm{d}u'|^2 = e^{2\lambda y} \lambda^2(\mathrm{d} x^2 + \mathrm{d}y^2)$. So you just need to define $\tilde{u} = \lambda^{-1} u'$ to get the desired embedding. -This means that if the solution contains any point $t_0 \in \mathbb{R}\setminus\{0\}$ it will also contain the entirety of $\mathbb{R}\setminus\{0\}$. -So the continuity method doesn't do anything for you: regardless of whether the solution set is closed, what you really need to prove is that there exists some $t_0$ for which the differential equations are solvable. Or, in other words, the hard part lies in showing - -an open neighborhood of $\{0\}$ lies in the solution set, - -the part which you implicitly considered to be obvious. - -Another way to say the above is: the method of continuity is helpful if you want to get to "1" or "$\infty$" knowing that you have "0". If you can already reduce the problem to getting to "$\epsilon$", the method of continuity is not going to tell you much. - -Also, the question of immersion of hyperbolic plane in $E^4$ is partially known, at least if you allow piecewise smooth solutions. See - -http://www.ams.org/mathscinet-getitem?mr=1025303 -http://www.ams.org/mathscinet-getitem?mr=1334978 - -where immersions are constructed which are globally $C^{0,1}$ and real analytic away from some singular lines.<|endoftext|> -TITLE: Non-chaotic bouncing-ball curves -QUESTION [22 upvotes]: I was surprised to learn from two -Mathematica Demos by -Enrique Zeleny that an elastic ball bouncing in a V or in a sinusoidal channel -exhibits chaotic behavior: - -    -(The Poincaré map is shown above the red ball trajectories.) -I wonder: - -Q. Are there curves (other than a horizontal line!) - for which such a ball drop (from some range of horizontal - placements $x$ and vertical heights $y$) is not chaotic? In other words, are the - V and sine wave special? Or (more likely) is it that non-chaotic curves are special? - -(I cannot now access Zeleny's citation to Chaotic Dynamics: An Introduction Based on Classical Mechanics.) - -REPLY [7 votes]: This is really a collection of comments to Noam's wonderful parabola example, but a bit long... -The set of known integrable ordinary (non-gravitational) billiards consists of pieces of confocal conics, or their limits (straight lines). The internal angles are all 90 degrees apart from some exceptions I will ignore henceforth: The 60-60-60 and 30-60-90 triangles and circular wedges. The most common examples are the circle and ellipse, but also shapes such an elliptical and a hyperbolic arc, such that the two have the same foci. -In the case of a single conic, the equation to obtain the location of the next collision is linear: The intersection of a straight line and a conic is a quadratic equation, but we can divide by the known solution given by the initial point. - -Remark: This feature also holds for similarly oriented parabolas, and hence also the parabola gravitational billiard. -Question: Is a gravitational billiard consisting of two confocal parabolic arcs also integrable? - -Noam also conjectures that gravitational billiards with local minima cannot be chaotic: I have a recent preprint including some examples (circle, ellipse and oval gravitational billiards): http://www.arxiv.org/abs/1308.0362 Indeed, the low energy orbits appear regular. There is also interesting higher energy dynamics: We also ask (with some numerical justification) - -Conjecture: Are there any energies for which these gravitational billiards are ergodic? - - -Addendum: A generalisation of Noam's example is that of Newtonian gravity (ie inverse square force toward a centre), with the parabola replaced with a conic with a focus at -the centre. - -Here the gravitational attraction is to the black dot, billiard boundary is one branch of -a hyperbola with this as focus, and the caustics are arcs of an ellipse and a hyperbola confocal with the boundary hyperbola. As with the parabola, the caustics have a nice corner despite the smooth force field and billiard boundary. -Publishable? Alas it is a trivial consequence of the conclusion of Horsch and Lang. The second constant of motion for the two centre problem is preserved by billiard collisions with a confocal boundary. So, I suspect not at present, though would love to be contradicted...<|endoftext|> -TITLE: Does ZF imply a weak version of Hahn-Banach? -QUESTION [15 upvotes]: I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple consequence of Hahn-Banach theorem and enables one to convert the problem at hand into a problem in ordinary calculus. Now My question is: -Suppose we have a Banach space $X$ whose dual $X^*$ separates points, i.e. for every nonzero $x\in X$ there is $u\in X^*$ such that $u(x)\neq 0$. Can one prove in ZF that for all nonzero $x\in X$ there is $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$ ? -I know that Hahn-Banach theorem is strictly weaker than axiom of choice, but I'm looking for a proof without using any "choicy" argument. - -REPLY [12 votes]: Yemon Choi suggested a counterexample in the comments which can easily be made into an actual counterexample. -It is consistent with ZF (see Asaf Karagila's answer here for a few references) that the dual space of $\ell_\infty$ is $\ell_1$ with the usual norm. The duality pairing is -$$ -\langle x,y \rangle = \sum x_n y_n. -$$ -Since the coordinate functionals separate points in $\ell_\infty$, the hypothesis in the question is satisfied. -On the other hand, for the sequence $x_n = 1-1/n$ in $\ell_\infty$ there obviously is no sequence $y \in \ell_1$ of norm one such that $\sum x_n y_n = 1 = \lVert x_n \rVert_\infty$.<|endoftext|> -TITLE: A sufficient condition for a space to be an Eilenberg-Maclane space -QUESTION [8 upvotes]: I'm reading the paper "Configuration Spaces and Braid Groups on Graphs in Robotics" by Robert Ghrist, in which he states and proves the following theorem: -Theorem: Given a tree $T$, the configuration space $C^N(T)$ (of $N$ distint ordered points on $T$), and a connected subset $K\subset C^N(T)$, if the homomorphism $\pi_1(K)\rightarrow \pi_1(C^N(T))$ induced by inclusion is trivial, then $K$ is nullhomotopic in $C^N(T)$. -and then as a corollary states the following without proof or citation: -Corollary: The configuration space $C^N(T)$ is an Eilenberg-MacLane space of type $K(\pi_1,1)$: i.e., $\pi_k(C^N(T))=0$ for all $k>1$. -To get from the theorem to the corollary, I'm guessing he is using some sort of general sufficient condition for a space to be Eilenberg-Maclane, namely: If every inclusion of a subspace $K$ into a space $X$ is nullhomotopic whenever it induces a trivial map on $\pi_1$, then $X$ is an Eilenberg-Maclane space. I don't remember having seen anything like this, nor have I been able to find anything. I hope I'm not missing something obvious. Anyone have any ideas? - -REPLY [5 votes]: As I said in my comment, Ghrist's theorem is stated in too much generality to possibly be true. But if, say, you restrict to requiring $K$ to be a finite CW complex, here's a counterexample to your question. Let $X$ be $S^2$ with two points identified. Then it is not hard to see that any finite complex $K\subset X$ for which $\pi_1(K)\to \pi_1(X)$ is trivial must be nullhomotopic in $X$. But $X\simeq S^2\vee S^1$ is not a $K(\pi,1)$.<|endoftext|> -TITLE: A question on chiral rings and geometry of the vacuum bundle -QUESTION [7 upvotes]: I am reading "Mirror Symmetry" by Hori et al, and have a question on Chap.17 (Chiral rings and geometry of the vacuum bundle). On p.425 the authors say - -Consider the path-integral on the hemisphere. The boundary of the hemisphere is a circle on which our Hilbert space is based. The path-integral will give us a number, and so defines a functional from boundary filed configurations to numbers, equivalently, a state in the Hilbert space... - -Then they say - -To obtain a ground state at the boundary we consider the "neck" of the hemisphere to be infinitely stretched. In other words, we imagine connecting the hemisphere to a semi-infinite flat tube. Noe that on the flat tube the twisted and untwisted theories are equivalent. - -They continue - -... Similarly, if we consider the topological path-integral together with the insertion of the corresponding chiral fields, we obtain a correspondence between chiral fields and the ground state.... - - -I am lost because they suddenly introduce the hemisphere and identify the states with the Hilbert space on the boundary. -Question 1 -Should I think of this hemisphere as a Riemann surface? Or is this the operator formalism and manifolds with boundary? -Question 2 -Why are the twisted and untwisted theories equivalent on the flat tube? -Question 3 -What does it mean by inserting chiral field? I don't think this is explained anywhere in the book. Does the insertion mean that the operator acts on the field after some time corresponding to the position of the insertion? -I think I lack of firm understanding of the subject, so I would appreciate it if someone could kindly explain things from the very basic. - -REPLY [3 votes]: Warning: Your questions require answers using quantum field theory. Thus, in some cases in the following I refer to the functional integral and some of its properties. In the situation of interest for your question, these concepts may be made precise, though I do not do this here. -1) You should think of the hemisphere that they are considering a two-dimensional manifold with a metric. In general in a quantum field theory on a manifold $M,$ a metric is required to formulate the action functional. For instance the "kinetic terms" for a scalar field $\phi$ typically take the form $\int_{M} \phi \Delta \phi$ with $\Delta$ given by the usual Laplacian. -In a topological field theory, like the ones they are considering, it may turn out that certain quantities do not depend on the choice of the metric, or more generally depend only on an equivalence class of the metric under some notion of equivalence (like conformal equivalence). However, regardless one still picks a metric to formulate an action and a functional integral. -2) The twisted and untwisted theories are equivalent on any space, like the cylinder, which is metrically flat. -Starting from the ordinary field theory, we obtain the theory on a Riemannian manifold $M$ by replacing partial derivatives by covariant derivatives, promoting the Lebesgue measure to the volume form, and all other natural things associated to making the action well-defined on a general manifold. -The theory has a global $R$-symmetry, meaning that all fields also transform in representations of some group Lie Group $G_{R}$, and hence may be viewed as sections of vector bundles. $G_{R}$ also has the important property that enters the supersymmetry algebra non-trivially, with the supercharges also in representations of $G_{R}$. -To obtain the topologically twisted version, we now do an additional step utilizing $G_{R}$. We activate a connection (called an $R$ gauge field) for $G_{R}$ and covariantize derivatives with respect to this connection as well. -The value of this connection is fixed to be related to the value of the spin connection on $M$. More explicitly, the twisting construction requires a choice of Lie algebra homomorphism $\chi$ from the local holonomy algebra of $M$ (ie $so(d+1)$ if $M$ has dimension $d+1$) to the Lie algebra of the $R$-symmetry group. Then we set -$\chi$(spin connection)=$R$ gauge field. -If $M$ is flat, then the spin connection on $M$ is trivial so the second step does nothing, and the twisted theory on $M$ is the same as the ordinary untwisted theory on $M$. -3) The construction that they are doing is a special case of the state operator map in conformal field theory. In general given a conformal field theory in dimension $d+1$ it assigns a Hilbert space to any manifold of dimension $d$. -An important special case is the that of the sphere $S^{d}$. Given a formulation of the CFT in question in terms of fields, we consider all values of the fields on $S^{d}$ (satisfying some $L^{2}$ conditions which I suppress). We form a vector space $V$ whose basis elements are in one-to-one correspondence with these boundary conditions. An element of the Hilbert space associated to $S^{d}$ is a linear functional on these boundary conditions, ie an element of the dual space $Hom(V,\mathbb{C})$ (again satisfying certain normalizability conditions which I suppress). -There is a bijection between the set states on $S^{d}$ and the local operators $\mathcal{O}(p)$ at any fixed point $p.$ -To set up this isomorphism, one considers a functional integral on the $d+1$ dimensional ball with boundary $S^{d}$. To carry out the path integral one must specify boundary conditions for fields. The result of the functional integral produces a complex number, which depends upon these boundary conditions, and hence via the definition above results in a state. We refer to this state a $\Psi_{1}$, it is the state corresponding to the unit operator in the claimed bijection. -More, generally, for each local operator $\mathcal{O}(p)$ we obtain a state $\Psi_{\mathcal{O}}$ by considering the functional integral on the ball but now with the local operator $\mathcal{O}$ inserted at the center of the ball. This defines a map from local operators to states on $S^{d}$. If the theory is conformally invariant, then an inverse may be constructed by shrinking the $S^{d}$ to a point and constructing a local operator from any state. -(If you are new to CFT, I highly recommend that you work out the state operator map in detail for the case of a free scalar. The answer can be found in Polchinski's string theory book in chapter 2.) -Finally, let us return to the construction in question. We consider the path integral on a very long cigar. We view this as a flat cylinder of fixed circumference and length $L$, glued to a round cap where the cap has a fixed metric. We are interested in what happens as the length $L$ of the cylinder tends to infinity. -We perform the path integral in two steps. -First, by integrating over fields in the cap together with a local chiral operator $\mathcal{O}$ inserted at the tip of the cap, we obtain a state $\Psi_{\mathcal{O}}$ as above. -Next, we do the path integral over fields on the long cylinder. Although we could do this explicitly, it is simpler to instead note that the path integral on $S^{1}\times I$ computes the evolution operator $e^{-H L}$ where $H$ is the Hamiltonian of time evolution for states on $S^{1}$, and $L$ is the length of the interval $I$. Thus, the complete result of the path integral on the long cigar together with the insertion of the local operator is the state in the Hilbert space -$e^{-HL} \Psi_{\mathcal{O}}$ -Now we let $L$ tend to infinity. The operator $H$ is positive definite and bounded below by zero. Hence, as $L$ tends to infinity, $e^{-H L}$ is a projection operator onto the subspace of the Hilbert space with $H=0$. This is exactly the subspace of vacuum states.<|endoftext|> -TITLE: What are some deep theorems, and why are they considered deep? -QUESTION [73 upvotes]: All mathematicians are used to thinking that certain theorems are deep, and we would probably all point to examples such as Dirichlet's theorem on primes in arithmetic progressions, the prime number theorem, and the Poincaré conjecture. I am planning to give a talk on the history of "depth" in mathematics, and for -that reason I would like to have a longer list of examples and, if possible, some thoughts about what makes them deep. -Most examples, I expect, will be from after 1800, but I am also interested in examples before that date. -When it comes to the meaning of "depth," I am interested in both specific and general explanations. In specific cases, one might point to the introduction of unexpected methods, such as analysis in Dirichlet's theorem, or differential geometry in the Poincaré conjecture, which are not implicit in the statement of the theorem. -In most cases, it is probably not provable that these methods are necessary (e.g. there are "elementary" proofs of Dirichlet's theorem), but in some cases it is provable, by general theorems of logic. Both types of explanation are welcome. -Update. I am a little surprised that nobody mentioned reverse -mathematics, which seems to offer a precise sense in which certain -theorems are "equally deep." For example, on pages 36--37 of -Simpson's Subsystems of Second Order Arithmetic there is a list -of 14 theorems, including the Brouwer fixed point theorem and -Riemann integrability of continuous functions, which are equally -deep in a precise sense. Admittedly, these are not the deepest -theorems around, but they're not shallow either. Later in the book -one finds other results of equal, but greater, depth. How do MO members -view such results? - -REPLY [3 votes]: How about the compactness theorem which appears almost everywhere in Model Theory? It says a set of the first-order sentences has a model if and only if every finite subset of it has a model. There is a similarity between it and the finite intersection property for compact topological spaces too. This latter says a collection of closed sets in a compact space has a non-empty intersection if every finite subcollection has a non-empty intersection. The interesting thing that exists about that is one can find a topological proof for model theoretical compactness theorem. -Another interesting thing is that among the three most important topological features, I mean completeness, connectedness, and compactness, the compactness appears in many places. Something like a deep property.<|endoftext|> -TITLE: A characterization of the poset of filters on a set -QUESTION [7 upvotes]: For the lattices of all subsets of a given set it is known an axiomatic characterization: A poset is isomorphic to a set of all subsets of some set iff it is a complete atomic boolean algebra. -The question: How to characterize the sets of filters on a set? That is having a poset, how to check whether it is isomorphic to the set of filters on some set? -Note that we allow improper filters (An improper filter is also filter) to ensure that the set of filters is a complete lattice. - -REPLY [6 votes]: Under Stone-duality, the lattices which are isomorphic to the lattices of filters on a Boolean algebra are precisely the compact zero-dimensional frames. The lattices of filters on a complete Boolean algebra correspond to the compact zero-dimensional extremally disconnected frames. Therefore, by this answer, the lattices of filters on some set $X$ are precisely the compact zero-dimensional extremally disconnected frames where the isolated points in your frame form a dense subspace of the dual space in your frame. Translating this idea to a purely point-free context, the posets isomorphic to lattices of filters on a set are precisely the atomic compact zero-dimensional extremally disconnected frames.<|endoftext|> -TITLE: Compact objects in undercategories and filtered colimits -QUESTION [7 upvotes]: Let $\mathcal{C}$ be a compactly generated presentable $(\infty, 1)$-category. Consider the functor -$$ \Phi: \mathcal{C} \to \mathrm{Cat}_\infty, \quad x \mapsto (\mathcal{C}_{x/})^\omega,$$ -that sends an object $x \in \mathcal{C}$ to the $(\infty, 1)$-category of compact objects in the undercategory $(\mathcal{C}_{x/})$. -I am interested in when this functor commutes with filtered colimits. -I believe that if $\{x_i\}_{i \in I}$ is a filtered diagram of objects in $\mathcal{C}$, then the induced functor -$$\varinjlim -\Phi(x_i) \to \Phi( \varinjlim x_i),$$ -is (for formal reasons) always fully faithful, and that the image contains those objects of $\mathcal{C}_{\varinjlim x_i/}$ which are both compact and $n$-cotruncated. However, I do not know if this functor is essentially surjective in general: the problem is that the filtered colimit of idempotent complete categories need not be idempotent complete (I think?); the "image" of an idempotent is not a finite colimit. -For ordinary categories, the above displayed functor is an equivalence of categories (for instance, if $\mathcal{C}$ consists of commutative rings, and compact objects correspond to finitely presented algebras); I'm wondering if something goes wrong with $(\infty, 1)$-categories? - -REPLY [5 votes]: In fact, a filtered colimit of idempotent complete $\infty$-categories is idempotent complete. See Lemma 7.3.5.16 of Higher Algebra (in the 2014 edition). This gives an affirmative answer to the above question.<|endoftext|> -TITLE: Taylor expansion of a function of a matrix -QUESTION [6 upvotes]: Assume that ${\mathbf H}$ is a $N \times M$ matrix. The following parameter is called orthogonality deficiency and describes how much orthogonal the columns of ${\mathbf H}$ are. -$$ od({\mathbf H}) = 1 - \frac{\det({{\mathbf H}^H{\mathbf H})}}{\Pi_{n=1}^M\|{{\mathbf h}_n}\|^2}$$ -where ${\mathbf h}_n$ is the $n$th column of matrix ${\mathbf H}$. It can be seen that $0\leq od({\mathbf H})\leq 1$. If ${\mathbf H}$ is singular then $od({\mathbf H})=1$ and when $od({\mathbf H})=0$ the colums of mathrix ${\mathbf H}$ are orthogonal. -This is, indeed a very useful criterion for matrix orthogonality and has many applications in communications engineering and signal processing. -My question: -When the matrix ${\mathbf H}$ has the form ${\mathbf H}={\mathbf A}+e{\mathbf B}$, where $e$ is a scalar, how can we approximate $od({\mathbf H})$ when $e \ll 1$ and what is the approximation of $od({\mathbf H})$ when $e \ll 1$ (Probably using Taylor expansion)? - -REPLY [2 votes]: If $e$ is a small number, then $od(H)\approx od(A)+e(od)'_A(B)$. $od(A)=1-\dfrac{u(A)}{v(A)}$ -where $u(A)=\det(A^*A),v(A)=\Pi_i||Ae_i||^2$ and $(e_i)_i$ is the canonical basis. -$(od)'_A=-\dfrac{1}{v(A)}u'_A+\dfrac{u(A)}{v^2(A)}v'_A$. -$u'_A(K)=trace((A^*K+K^*A)adjoint(A^*A))$. -$v'_A(K)=\sum_i((e_i^*K^*Ae_i+e_i^*A^*Ke_i)\Pi_{j\not= i}||Ae_j||^2)$. -Edit: -1. The proof is based on this fact: -if $\phi:A\rightarrow \det(A)$, then $\phi'_A:K\rightarrow trace(K.adjoint(A))$. -2. Simplifications: $u(A)=|\det(A)|^2$ and $e_i^*K^*Ae_i+e_i^*A^*Ke_i=2Real((K^*A)_{i,i})$. -3. In the sake of simplicity, assume that $K$ is real. Let $(E_{i,j})$ be the canonical basis of $\mathcal{M}_n(\mathbb{R})$. Then the matrix of $(od)'_A$ is in the form $U=[u_1,\cdots,u_{n^2}]$. If you want a large variation of $od(A)$, then choose $B=U^T$ (in the orthogonal of $\ker(od'_A)$). Then $(od)'_A(B)=||U||^2$. -For instance, if $A=\begin{pmatrix}8&-5-6I&-3-I\\-5+4I&2+3I&7+7I\\6-I&-4+6I&8+2I\end{pmatrix}$, then $U\approx [-0.0029,-0.0154,-0.0120,-0.0010,-0.0064,-0.0119,-0.0110,-0.0111,0.0124]$. -$od(A)\approx 0.9489$ and $od(A-100.U^T)\approx 0.8147$.<|endoftext|> -TITLE: Casson invariant and signature -QUESTION [5 upvotes]: In W. Neumann, J. Wahl, "Casson invariant of links of singularities", -Comment. Math. Helv.,1990, Vol. 65, Issue 1, pp 58-78 some connection between the Casson invariant and the signature is formulated. Is there any heuristic explanation, why such a connection should be? The only reason I see is the Rokhlin invariant, which is defined by the signature and is reduction mod 2 of the Casson invariant. I would be happy to find some arguments in terms of Donaldson invariants or something like this. - -REPLY [4 votes]: Andras Nemethi and I have considerably generalized this result to rational homology $3$-spheres that are links of certain isolated surface singularities. -A rational homology $3$-sphere $M$ which is the link of an isolated surface singularity is equipped with a canonical spin-c structure. Associated to this spin-c structure is a Seiberg-Witten invariant which can be alternatively described in terms of the Casson-Walker invariant and the refined Reidemeister torsion as defined by Turaev; see this reference. If the $3$-manifold $M$ is an integral homology sphere, then it has a unique spin-c structure whose Seiberg-Witen invariant coincides with the Cason invariant. -In a sequence of papers (see here, here and here) A. Nemethi and I proved that for many classes of isolated singularities whose links are rational homology $3$-spheres, the Seiberg-Witten invariant of the above canonical spin-c structure of the link coincides with a certain algebraic geometric invariant of the singularity (essentially the geometric genus). -If the singularity happens to be a complete intersection, then this algebraic-geometric invariant can be expressed in terms of the signature of the Milnor fiber. Brieskorn integral homology spheres are of this type, and when specialized to this case, our result coincides Neumann and Wahl's older result. -I want to emphasize that our result applies even to situations when the singularity is not smoothable, i.e., Milnor fiber does not exist. (Most of the time the Milnor fiber does not exist.) -You can find a gentler introduction to this topic here.<|endoftext|> -TITLE: Applications of Lawvere's fixed point theorem -QUESTION [41 upvotes]: Lawvere's fixed point theorem states that in a cartesian closed category, if there is a morphism $A \to X^A$ which is point-surjective (meaning that $\hom(1,A) \to \hom(1,X^A)$ is surjective), then every endomorphism of $X$ has a fixed point (meaning a morphism $1 \to X$ which is fixed by the endomorphism). It unifies several (all?) diagonal arguments appearing in mathematical logic (Cantor's theorem, Russel's paradox, Tarski's non-definability of truth, the Recursion theorem, Gödel's first incompleteness theorem). See Yanovsky's paper for an expository account. Somehow it reminds me of the Yoneda Lemma because the proof is so short and simple, but the theorem unifies several theorems which are often regarded to be nontrivial. I wonder why it doesn't have a Wikipedia or at least an nlab entry yet (only Cantor's theorem alludes to it). Edit. Since March '14 there is an nlab entry. -The applications mentioned above all take place within the category of sets (or similar categories). But since the theorem applies to arbitrary cartesian closed categories, I wonder: -Question. Are there any interesting applications of Lawvere's fixed point theorem outside of mathematical logic by applying it to cartesian closed categories which are substantially different from the category of sets? -Qiaochu recently asked if the theorem implies Brouwer's fixed point theorem. But this seems to be a little bit too optimistic. Anyway, it would be nice to see at least some applications. Interesting examples of cartesian closed categories include the category $G\mathsf{-Set}$ for a group $G$ (here, a point of a $G$-set is already fixed by the $G$-action), the category $\mathrm{Sh}(X)$ of sheaves on a space $X$ (here a point is a global section), or the category of compactly generated (weak) Hausdorff spaces $\mathsf{CGHaus}$. What are interesting choices for the object $A$, and how can we use the resulting fixed point theorem? - -REPLY [8 votes]: A lot of programming languages (PHP,C++0x,Java...) were just added lambda expression facilities recently. -However their lambda expression definitions usually does not allow direct recursion. But we can always do this: - -Define a type WRAP that can -a) Be constructed from a function closure receives a WRAP and returns an generic type. this is equivalent to the surjective condition. -b) Be executed with another WRAP object and evaluate the function used to construct the object at this point. This is equivlent to a morphism $WRAP \to X^{WRAP}$ -Now apply the trick garanteed by Lawvere's fixed point theorem, which says every function X -> X have a fixed point (of type X). Since we don't even know whether X have an accessable constructor, the only way we can construct such an X is by recursion. Implement a function that receive such a function and return the fixed point, now we obtained the Y combinator for the programming language. - -See this code review post for an instance of how the above trick applies to C++1x.<|endoftext|> -TITLE: Group theory conjecture on hurwitz groups -QUESTION [11 upvotes]: Conjecture: Let $p$ be a prime. -Then the group -$G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^9, (([a,b]^4)b)^{2p} \rangle$ -has a composition series of the form -${\rm PSL}(2,8) - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p$. -Is there any literature on this subject, and if not, how can this conjecture be proved? - -REPLY [17 votes]: For some reason, people seem to be voting to close this, so I will give a quick reply. Your conjecture is true and it can be proved mainly by computer. -The group $G=\langle a,b \mid a^2, b^3, (ab)^7, [a,b]^9 \rangle$ has a homomorphism onto ${\rm PSL}(2,8)$. Let $K$ be the kernel. Then it can be shown that $K$ is nilpotent of class 2, with $|Z(K)|=2$ and $K/Z(K)$ free abelian of rank 7. The element $x:=([a,b]^4b)^2$ lies in $K$ and maps onto a generator of $K/Z(K)$. So factoring out the normal closure of $x^p$ in $G$ will result in an extension of an elementary abelian group of order $p^7$ by ${\rm PSL}(2,8)$.<|endoftext|> -TITLE: vector bundle trivial over every compact subset, then it is globally trivial -QUESTION [33 upvotes]: Let $X$ be a non-compact metric space (though if the answer to the question is positive, then it probably also holds for more general spaces like, e.g., paracompact Hausdorff) and $E \to X$ a vector bundle over it. - -Suppose that over every compact subset $K \subset X$ the restricted bundle $E|_K$ is trivial. Can we conclude that $E$ is globally trivial? - -REPLY [19 votes]: If you let $T=S^1 \times D^2$ be the solid torus and pick an embedding $i: T \to \mathrm{int}(T)$ which multiplies by 2 in $\pi_1$, the direct limit $X = \varinjlim(T \xrightarrow{i} T \xrightarrow{i} \dots)$ is a smooth 3-dimensional manifold (non-compact, but admitting a proper embedding into $\mathbb{R}^4$). Its homotopy type is $K(\mathbb{Z}[\frac12],1)$, so by the universal coefficient theorem $[X,\mathbb{C}P^\infty] = \mathrm{Ext}(\mathbb{Z}[\frac12],\mathbb{Z}) \neq 0$, so there exists a non-trivial complex line bundle $L \to X$. (We can even take $L \subset X \times \mathbb{C}^2$ since $[X,\mathbb{C}P^1] = [X,\mathbb{C}P^\infty]$.) Any compact $K \subset X$ is contained in a submanifold diffeomorphic to $T \simeq S^1$, so $L \vert_{K}$ is trivial. -EDIT: in fact, pick any homomorphism $\pi_1(X) = \mathbb{Z}[\frac12] \to \mathbb{C}^\times$ which doesn't factor through the exponential map $\mathbb{C} \to \mathbb{C}^\times$ and use that to define $L$.<|endoftext|> -TITLE: Does the boldface class $\Delta^1_2$ have the uniformization property? (assuming $V=L$) -QUESTION [5 upvotes]: DISCLAIMER: All pointclasses considered here are boldface. -Most of the time, when doing descriptive set theory, we want the projective sets to "behave well;" for example, maybe we don't want there to be nonmeasurable projective sets, or projective well orderings of $\mathbb{R}$, etc. Generally, this means making some (fairly conservative) large cardinal assumption, or equivalent. -At the far opposite end of things is the axiom that all sets are constructible, $V=L$. This axiom implies that there is a projective - in fact, $\Delta^1_2$ - well-ordering of the reals, and so projective sets become bad very early in the hierarchy. -My question is about the state of affairs when $V=L$ holds. My motivation is simply that I don't feel I have a good grasp on basic concepts in descriptive set theory, and the following seemed like a good test problem to assign myself; but I have thought about it for a while without making progress, so I'm asking here: -Let $\oplus$ be one of the usual pairing operators on $\omega^\omega$. For the purposes of this question, we say that a pointclass $\Gamma\subseteq \mathcal{P}(\omega^\omega)$ has the uniformization property if whenever $A\in \Gamma$, there is some $B\in \Gamma$ such that: - -$B\subseteq A$, and -Whenever $x\oplus y\in A$, there is a unique $z$ such that $x\oplus z\in B$. - -That is, we view $A$ as coding a relation on $\omega^\omega\times \omega^\omega$, and $B$ is the graph of a function contained in $A$. (This is not usually how uniformization is presented, but it's equivalent for all intents and purposes.) My question is then: - -Assume $V=L$. Let $D$ be the set of (boldface) $\Delta^1_2$ elements of $\omega^\omega$; does $D$ have the uniformization property? - -Now, it seems clear to me that $D$ should not have the uniformization property. [EDIT: As Joel's answer below shows, this is completely wrong.] The counterexample should be just the $\Delta^1_2$ well-ordering $\prec$ given by the assumption that $V=L$: uniformizing $\prec$ requires us to choose, for each real $r$, a real $s$ such that $r\prec s$; and although $\prec$ is $\Delta^1_2$, the usual way of doing this - choosing the immediate $\prec$-successor of $r$ - is no longer $\Delta^1_2$. -However, I don't know how to show that $\prec$ - or any other $\Delta^1_2$ set - cannot be uniformized in $\Delta^1_2$. I suspect I'm just missing something fairly simple. - -Note: it is known that the boldface pointclasses $\Pi^1_1$ and $\Sigma^1_2$ have the uniformization property, and assuming large cardinals, the uniformization property can be further propagated to every pointclass $\Pi^1_{2n+1}$, $\Sigma^1_{2n}$. On the other hand, the class $\Delta^1_1$ of Borel sets lacks the uniformization property, provably in $ZFC$. - -REPLY [3 votes]: Assuming $V=L$ then we have $AC$ and $CH$, so every set of reals is at most $\aleph_1$ Suslin. So we can find scales for them and uniformize them.In particular every $\Delta^1_2$ set of reals can be uniformized. As Joel said in the comment above this works for all $\Delta^1_n$ under $V=L$.<|endoftext|> -TITLE: Proof of the general expression for anomaly in a CFT and its partition function -QUESTION [5 upvotes]: I think the statement is that for any dimensional CFT the following is true, - -$$\langle T^{\mu}_\mu \rangle = \sum B_n I_n - 2(-1)^{d/2}AE_d,$$ -where $E_d$ is the `"Euler density" and $I_n$ are the independent "Weyl invariants of weight $-d$". -(...I am not sure of the definition of the geometric quantities coming on the R.H.S and I wonder if the notion of the "Euler density" and ``Weyl invariants" are related to the ideas of the Weyl tensor and the Euler tensor..) - - -For the special case of $3+1$ CFTs I am told that there is a relationship between the effective action and these anomaly coefficients as, - -$W = \frac {a_0}{\epsilon^4} + \frac {a_1}{\epsilon^2} + a_2 ln (\epsilon) + w(g)$ -where I guess $\epsilon$ is the UV regulator, $W$ is I guess the connected functional/free energy defined as $W = - ln (Z)$ ($Z$ being the partition function), $w(g)$ is the UV finite part dependent on the metric $g$ such that $w(\lambda^2 g) = w(g) - a_2ln(\lambda)$. -The most important and universal part of this answer is apparently that $a_2$ is universal and has the following properties that, -(1) $a_2(e^{-2\omega}g) = a_2(g)$ -(2) $a_2 = AE_4 + BI_4 -$ where $E_4 = \frac{1}{64}\int (R_{\alpha \beta \mu \nu}R^{\alpha \beta \mu \nu} - 4R_{\mu \nu}R^{\mu \nu} + R^2)$ and $I_4 = -\frac{1}{64}\int (R_{\alpha \beta \mu \nu}R^{\alpha \beta \mu \nu} - 2R_{\mu \nu}R^{\mu \nu} + \frac{R^2}{3})$ -Thus one says that the universal part of the free energy of a $3+1$ CFT is given by the integrated conformal anomaly. -I would like to know what is the proof of the above and its generalization to higher dimensions (if known!). - -It seems very mysterious to me that anomalies should determine the universal parts of a partition function! - -REPLY [3 votes]: The answer to your question is contained in the recent book by Spyros Alexakis ``The Decomposition of Global Conformal Invariants'' published last year by Princeton University Press. He answers the Deser-Schwimmer conjecture which (as far as I can understand your question) is the broader context of what you are asking.<|endoftext|> -TITLE: Effectiveness of the distinguished theta characteristic in characteristic 2 -QUESTION [9 upvotes]: Let $k$ be an algebraically closed field of characteristic 2. Let $C$ be a (smooth projective connected) curve over $k$. Can there exist a rational function on $C$ whose differential is holomorphic but nonzero? (Note that every perfect square has zero differential.) -Alternate formulation: at the end of his paper "Theta characteristics of an algebraic curve", Mumford points out that the canonical sheaf on $C$ has a distinguished square root $\mathcal{L}$: for any rational function $f$ on $C$ which is not a perfect square, the divisor $(df)$ is even and the class of $\frac{1}{2}(df)$ does not depend on $f$. The question is then whether $\mathcal{L}$ can admit a nonzero section. -For example, it is an entertaining exercise to check (from a Weierstrass model) that this can never occur for ordinary elliptic curves. - -REPLY [3 votes]: Let $F:X\to X$ be the relative Frobenius morphism. It is a finite morphism of degree $2$, and we have an exact sequence -$$0\to \mathcal{O}_X\to F_*\mathcal{O}_X \to L \to 0$$ -for some invertible sheaf $L$ on $X$. The standard formula for the canonical class of a finite morphism tells us that -$\omega_X = F^*(\omega_X\otimes L^{-1}) = \omega_X^{\otimes 2}\otimes -L^{\otimes -2}.$ -This implies that $L = \theta$ is a theta characteristic on $X$. I believe this is the theta characteristic discussed by Mumford. Now apply the cohomology to the first exact sequence to get that $H^1(X,\theta) \cong H^0(X,\theta)^*$ is the cokernel of the Frobenius map $H^1(X,\mathcal{O}_X) \to H^1(X,\mathcal{O}_X)$. -So, $\theta$ is effective if and only if the curve is not ordinary.<|endoftext|> -TITLE: Concerning Silver's result -QUESTION [10 upvotes]: Jack Silver proved that if $x$ is a real so that every $x$-admissible ordinal is a cardinal in $L$, then $0^{\sharp}$ exists. -I wonder whether various weaker or stronger versions of Silver's result have been considered in the literature. For example, -$\bf{Question \ 1.}$: -How strong is the statement that there is real $x$ so that every $x$-admissible ordinal is a recursively inaccessible? -$\bf{Question \ 2.}$: -How strong is the statement that there is real $x$ so that every $x$-admissible ordinal is inaccessible in L? - -REPLY [9 votes]: Re Q1: Sy Friedman has shown by class forcing over L that there can be consistently a real r -with the r-admissibles precisely the recursively inaccessibles. (S Friedman, "Strong Coding" APAL, vol 35,1987). -Re Q2: All $0^\sharp$-admissibles are limits (indeed fixed points in the enumeration) of Silver indiscernibles (as can be seen by iterating the $0^\sharp$-mouse inside the least admissible set containing it). So 2 is not a strengthening. -One obtains strengthenings really by changing the model $L$ to some other inner model, such as a core model, for example $L^\mu$ the least inner model with a measurable cardinal.<|endoftext|> -TITLE: Passing C through a slot -QUESTION [6 upvotes]: Question: Given a closed curve C, what will be the (bounds on) dimension of the interval it will pass through? -i.e. which are the necessary and sufficient conditions for a planar compact set C to pass through a closed interval in a plane? -The matter has been studied in the 1982 paper by Gilbert Strang, "The width of a chair," in Amer. Math. Monthly; but only ends in certain (interesting) conjectures in case of a non convex C (the problem in case of a convex C does get a definitive answer as its shortest orthogonal chord). -A development over this question would be the still open moving sofa problem (Leo Moser's 1966 problem, "Moving Furniture through a Hallway," Problem 66-11 in SIAM Review). - -Does the question have a definitive answer? Any references/latest work done with regards to the question are welcome. - -REPLY [7 votes]: Concerning nonconvex $C$, I posed it as an open-problem exercise in Computational Geometry in C to determine the worst polygon (or, equivalently, the worst polygonal -curve $C$), worst in the number of "moves," to get $C$ through the doorway interval -(p.321, Ex.4). I defined a "move" as monotonic $x$ and $y$ translation and $\theta$ rotation. The idea was to try to find a shape that requires many reversals. -At that time this was only known to lie between $\Omega(n)$ and $O(n^2)$ moves -for a polygon of $n$ vertices, and I am unaware of any subsequent improvements. -For the $O(n^2)$ upper bound, see: - -Yap, Chee-Keng. "How to move a chair through a door." IEEE Journal of Robotics and Automation (1987): 172-181. (IEEE link) - -This paper also computes the minimum door width for an $n$-gon with an $O(n^2)$-time algorithm.<|endoftext|> -TITLE: Is it decidable whether the support of a rational $\mathbb{Z}$-series is a regular language? -QUESTION [5 upvotes]: Let $S \in \mathbb{Z}\langle\langle A\rangle\rangle$ be a rational series in noncommutative variables. The support of $S$ is the set of all words $u \in A^*$ such that $(S, u) \not= 0$. It is undecidable to know whether the support of a given$^*$ rational series is cofinite (respectively equal to $A^*$). However, it is decidable whether the support is finite (respectively empty). See the exercises of Chapter III in [1]. -Question: is it decidable whether the support of a given rational series is a rational (= regular) language? -[1] J. Berstel and C. Reutenauer, Noncommutative rational series with applications. Encyclopedia of Mathematics and its Applications, 137. Cambridge University Press, Cambridge, 2011. xiv+248 pp. ISBN: 978-0-521-19022-0 -(*) The rational series can be given by a weighted automaton or by a finite linear representation. - -REPLY [6 votes]: Update. It is undecidable. Here is the proof. -If $f,g\colon A^*\to \{a,b\}^*$ are two morphisms, then one can construct a rational Z-series over A whose support is the complement of the equalizer of f,g. This is how Post correspondence is reduced to universality of $\mathbb{Z}$-series and is based on a faithful 2x2 rep of the free monoid over $\mathbb{N}$. See the proof of Thm 27 of http://www.infres.enst.fr/~jsaka/ENSG/MPRI/Files/References/JS-HWA.pdf -So it suffices to prove it is undecidable whether the equalizer of two free monoid morphisms is rational. -This is shown undecidable in Thm 5.2 here. It is also shown undecidable for context-free. -Update. Stefan Göllar has pointed out to me that this result is proved (in essentially the same way) in D. Kirsten and K. Quaas, Recognizablity of the support of recognizable series over the semiring of integers is undecidable, Inf. Process. Lett. 111(10):500-502 (2011). The main difference is that the authors of this paper Thm 5.2. -Update This is Exercise 1 of II.12 of the book of Salomaa et al and presumably was to use Hilbert's 10th problem.<|endoftext|> -TITLE: Effective Lang-Weil bounds for del Pezzo surfaces -QUESTION [5 upvotes]: Let $X$ be variety in $\mathbb{P}^N$ over $\mathbb{F}_q$ of dimension $n$ and degree $d$. -By the Lang-Weil bounds -$ |\# X(\mathbb{F}_q) - q^n| \le (d-1)(d-2)q^{n-1/2} + Cq^{n-1}$for a constant $C$ depending on $n$, $d$ and $N$. -Are there any bounds on $C$? Can you improve the estimate in special situations? I am interested in the case of del Pezzo surfaces where I want to use to test for local solubility. - -REPLY [7 votes]: I hope nobody minds me answering this (very old) question. -Me and my collaborators (Barinder Banwait and Francesc Fité) succeeded in completely answering this question in the paper: -Del Pezzo surfaces over finite fields and their Frobenius traces (https://arxiv.org/abs/1606.00300). -Specifically, let $S$ be a del Pezzo surface of degree $d$ over a finite field of size $q$. Then -$$\#S(\mathbb{F}_q) = q^2 + aq + 1$$ -for some $a \in \mathbb{Z}$. Then in our paper we completely classified which values of $a$ can occur for fixed $d$ and $q$. -For example for cubic surfaces we have $a \in \{-2,-1,0,1,2,3,4,5,7\}$, and each value occurs over every finite field, except for $a = 7$ which does not occur when $q = 2,3,5$. -Other degrees $d$ are similar, but the list of admissible values of $a$ and the number of exceptional cases in small finite fields is quite long.<|endoftext|> -TITLE: Numerically computing $\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$ -QUESTION [7 upvotes]: In the book, "Pi and the AGM" by Borwein and Borwein, it is mentioned that Gauss computed the following integral to the eleventh decimal palce. -$\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$ -How did he do it? Personnally, I looked at a Taylor expansion of -$\frac{1}{\sqrt{1-x}}$ -Where I substituted $t^4$ for $x$, and integrated term by term, but this gives a series that converges really slowly. Is there an obvious transform to make this computation faster? - -REPLY [8 votes]: A good place to look is pages 405 and 413 of the Nachlass section of Gauss's Werke III, which can be found online through Google Books. On page 405, he gives the following formula for "$\text{arc sin lemn }x$": -$$\text{arc sin lemn }x= x+{1\over2}\cdot{1\over5}x^5 + {1\cdot3\over2\cdot4}{1\over9}x^9+{1\cdot3\cdot5\over2\cdot4\cdot6}{1\over13}x^{13}+{1\cdot3\cdot5\cdot7\over2\cdot4\cdot6\cdot8}{1\over17}x^{17}+\cdots$$ -One page 413, he computes the value of $\int_0^1{dx\over\sqrt{1-x^4}}$, presumably using the expansion above, but explicitly citing the formula -$$\text{ arc sin lemn }{7\over23}+2\text{arc sin lemn }{1\over2}$$ -obtaining -$$1.3110287771\quad460599052\quad320.7$$ -He also notes there that Stirling had obtained the value $1.3110287771\ 4605987$. This is in reference to the calculations on pages 57-58 of Stirling's Methodus differentialis from 1730, which can also be found through Google Books. It might be worth noting that even Gauss was slightly off in the last couple of decimal places. A more accurate value, which I took from here, is -$$1.3110287771\quad460599052\quad324197949$$<|endoftext|> -TITLE: example of Local cohomology -QUESTION [8 upvotes]: Let $S=k[x_1,...,x_n]$ be a polynomial ring over field $k$ with maximal ideal $m=(x_1,...,x_n)$. I wanna make a $3$-dimensional $S$-module $M$ such that $H^0_m(M)=H^1_m(M)=0$ and $H^2_m(M)\neq 0$ be finitely generated (or in general case: $H^i_m(M)$ be finite for all $i=0,1,2$ ). Is there a simple way to create similar examples(for any dimension)? -Background: -$H^i_m(M)$ means $i$'th local cohomology module of $M$. - -REPLY [11 votes]: Take $M$ to be the second syzygy of $k$ over $S=k[x_1,x_2,x_3]$. Then a graded version of local duality tells us that $H^2_m(M)$ is dual to $Ext^1(M,R)= Ext^3(k,R)$, the last one is $k$ either by direct computation or duality again. -One can easily generalize this, the $j$ syzygy of $k$ in $n$ variables will have local cohomologies vanish up to degree $j-1$ and finitely generated up to $n-1$.<|endoftext|> -TITLE: cofinality of $(P(\kappa)/NS,\subseteq)$ -QUESTION [6 upvotes]: Let $\kappa$ be a cardinal (I'm most interested in $\kappa=\aleph_{\omega+1}$ but I suspect a general answer is known). What is the cofinality of $(P(\kappa)/NS,\subseteq)$? By this I mean the least cardinal $\lambda$ such that there exists a subcollection $X\subseteq P(\kappa)/NS$ of size $\lambda$ such that for any $A\in P(\kappa)$ there is $B\in X$ with $A\subseteq B$. -Here we consider $P(\kappa)/NS$ to be without the largest element. -This may be trivial but it's been bugging me all day! -Drake - -REPLY [10 votes]: I'm going to assume that when you write $\subseteq$, you really mean $\subseteq_{NS}$. We say $A \subseteq_{NS} B$ when $A$ is contained in $B$ except for a nonstationary set, i.e. $A \setminus B \in NS$. If we don't do this, then as Joel said, $\lambda = \kappa$, since the collection of "coatoms" is cofinal. -If we take the set of equivalence classes of stationary sets modulo $NS$, (i.e. $A \sim B$ when $A \triangle B \in NS$) we get an atomless boolean algebra under the set operations modulo $NS$, and this is what is commonly meant by $\mathcal{P}(\kappa)/NS$. Now in general an "upwardly dense" subset $A$ of a boolean algebra $B$ ("cofinal" as you call it) generates a (downwardly) dense set by just taking the complement of every element of $A$. -Now, it is known to be equiconsistent with infinitely many Woodin cardinals that $NS_{\omega_1}$ is $\omega_1$-dense. But by results of Gitik and Shelah, $\mathcal{P}(\kappa)/NS$ never has the $\kappa^+$-chain condition for regular $\kappa > \omega_1$. Hence the density of this algebra is always greater than $\kappa$. Under GCH, it must be equal to $\kappa^+$. -Woodin showed relative to an almost-huge cardinal that it is consistent for a successor cardinal $\kappa > \omega_1$ to have a stationary subset $S$ such that $\mathcal{P}(S)/NS$ has the $\kappa^+$-c.c. (See Foreman's article in Handbook of Set Theory.) I'm pretty sure that the stronger property of being $\kappa$-dense is not known to be consistent from any large cardinal assumption. Under a fragment of GCH, if $\kappa$ is the successor of singular cardinal (for example $\aleph_{\omega+1}$) $NS_\kappa$ is provably not $\kappa$-dense below any stationary set. (This is my result, to appear in my thesis soon.)<|endoftext|> -TITLE: Minkowski successive minima inequality for a lattice base? -QUESTION [6 upvotes]: Let $\Lambda$ be a lattice of $\mathbb{R}^n$, and $\lambda_i$ be the radius of the smallest ball containing $i$ linearly independent lattice vectors. -The Minkowski successive minima inequality says that: $ (\prod_{i=1}^n \lambda_i)^{1/n} \leq \sqrt{n} (\det \Lambda)^{1/n} $ -Can we also have an upper bound if in addition, we ask these independent vectors to be a basis of the lattice (and not only a basis of $\mathbb{R}^n$)? - -REPLY [4 votes]: Yes, a version of Minkowski's successive minima studied by Mahler and Weyl consists in letting $\lambda_i'$ to be the radius of the smallest ball containing $i$ linearly independent lattice vectors that can furthermore be completed to a basis of the lattice. The inequality of Minkowski holds with a somewhat worse constant. This can be found in Lekkerkerker's book or in Wey'ls paper.<|endoftext|> -TITLE: Universal $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set -QUESTION [9 upvotes]: Does anyone know of a reference for the fact that if $\lambda$ is a limit of Woodin cardinals, then the pointclass $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ is $\omega$-parameterized? By this I mean that there is a $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $U \subset \omega \times \omega^\omega$ that is universal in the sense that every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $A \subset \omega^\omega$ is equal to the section $U_n$ for some $n < \omega$. -As far as I am concerned it would be enough to have a complete $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $U \subset \omega^\omega$ such that every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set $A \subset \omega^\omega$ is equal to the pre-image $f^{-1}(C)$ by some recursive function $f$. -In either case, this seems not entirely trivial to me. Suppose that $A \in (\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$, say -$$x \in A \iff \exists B \in \text{Hom}_{\mathord{<}\lambda}\,(HC; \in, A) \models \varphi[x].$$ -It seems like we need to bound the complexity of the formulas $\varphi$ we are considering. One way to do this is to show that if there is a set $B \in \text{Hom}_{\mathord{<}\lambda}$ with $(HC; \in, B) \models \varphi[x]$ then there is also a set $B' \in \text{Hom}_{\mathord{<}\lambda}$ with $(HC; \in, B') \models \psi[x]$, where $\psi$ is the Skolem normal form of $\varphi$. We can get $B'$ using the fact that $\text{Hom}_{\mathord{<}\lambda}$ is projectively closed and every $\text{Hom}_{\mathord{<}\lambda}$ set admits a $\text{Hom}_{\mathord{<}\lambda}$ uniformization. -This might be more familiar in the equivalent context of $\mathsf{AD}^+$ where we would talk about $\Sigma^2_1$ rather than $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$, but I don't recall having seen any argument for it anywhere. Is there a published reference for this? Or perhaps an obvious argument that is simpler than the one I started to sketch above? -By the way, my motivation for asking this is the following. If $\lambda$ is a limit of Woodin cardinals and the derived model at $\lambda$ satisfies $\theta_0 < \Theta$ (or equivalently, every $\Pi^2_1$ set is Suslin) then every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set becomes $\lambda$-universally Baire in some $\mathord{<}\lambda$-generic extension $V[g]$. (More precisely, its re-interpretation in $V[g]$ becomes $\lambda$-universally Baire.) To get a single $\mathord{<}\lambda$-generic extension $V[g]$ in which every $(\Sigma^2_1)^{\text{Hom}_{\mathord{<}\lambda}}$ set is universally Baire, it seems like the simplest way would be to consider a universal set. - -REPLY [4 votes]: I don't know if this answer will address your specific problem directly, but under AD, Wadge's Lemma implies that every non-selfdual pointclass has a universal set. There is an argument in Jackson's article in the Handbook (p.1761): basically if you have a $\Gamma$-complete set $A$, then define for every product space $X=X_0 \times X_1 \times ...\times X_n$ some sets $U_X \subseteq \omega^{\omega} \times X$ by $U_x(y, x_0,...,x_n) \leftrightarrow f_y() \in A$ where $y \in \mathbb{R}$ code real-valued Lipschitz continuous functions on $\mathbb{R}$, i.e $y$ computes the output value of $f_y(a_0,...,a_n)$ given the input $(a_0,...,a_n)$ by $(y(),...,y())$. Then the sets $U_x$ are in $\Gamma$ because we assume $\Gamma$ is at least adequate (so we get recursive substitution) and whenever you take some set $B$ in $\Gamma$ then by Wadge's Lemma, it is Lipschitz reducible to $A$ so for some real $y$, $(U_X)_y=B$. -Now since $\Sigma^2_1$ is obviously adequate, and it is nonselfdual, then it has a universal set. I guess that makes it a Spector pointclass (it is normed, even has the scale property).<|endoftext|> -TITLE: Most significant results in motivic integration theory? -QUESTION [26 upvotes]: I am thinking about reading a course on motivic integration. I have already read certain introductions to the subject; yet I am not sure that they mention all the significant parts of the theory. So, my questions are: - -Which papers and results are the most important for the theory of motivic integrations? -What are the main applications of motivic integration? -Would it be wise to avoid all quantifier elimination and related things in the course? - -REPLY [3 votes]: Other very significant applications have to do with 'transfer theorems' of statements between local fields of positive and zero characteristic, in particular a different proof that the "fundamental lemma", as proved by Ngo in positive characteristic, is valid in characteristic zero: the first proof is by Waldspurger, but it can be proved using the (exponential) motivic integration theory of Cluckers and Loeser (see a survey of Cluckers, Hales and Loeser, e.g. on Cluckers's homepage; the foundational papers of Cluckers and Loeser are also there). -Even more recent applications of these ideas (due in particular to Cluckers, Gordon and Halupczok) involve transfering integrability statements and have applications to harmonic analysis over local fields. (Interestingly, in some cases, the transfer goes in the other direction here: from known results in characteristic zero to results in positive characteristic). -These results use a fair amount of model theory (in particular things known as the Denef-Pas language, and Presburger arithmetic).<|endoftext|> -TITLE: Making all cardinals countable and its HOD -QUESTION [8 upvotes]: Suppose that $G$ is $Col(\omega, -TITLE: What is the shape of mathematical universe? -QUESTION [8 upvotes]: Shape? At the usual mathematical literature when we can discuss about the shape of a "space" that we have a kind of "topography" on it. For example a topology, metric, geometry, etc. -Note that for example: we can define an "unformal" topology on proper class of $Ord$ which any continuous map $f:Ord\longrightarrow Ord$ in the sense of $\forall \alpha\in limord~~~f(\alpha)=\bigcup_{\beta \in \alpha} f(\beta)$ be continuous in the sense of this topology. Now the main question is: -Main Question: Is there any known "non trivial" metric, topology,... on the proper class of all sets or other proper classes like $HOD$, $L$? -Remark: Non trivial in the above question means "a topology, metric,... which the proof of being a topology or metric needs an essential use of the special properties of the ground proper class and this definition doesn't work for every collection of sets." - -REPLY [7 votes]: There is an answer to this for the universe of constructive mathematics. According to Martín Escardó, a universe of Martin-Löf type theory has indiscrete topology, constructively. He has a draft about it, and some slides at his web-page.<|endoftext|> -TITLE: Dimension of commutative subalgebras of a central simple algebra -QUESTION [8 upvotes]: let $k$ be a field, and let $A$ be a central simple $k$-algebra over $k$. -What is the maximal dimension of a commutative $k$-subalgebra of $A$? -If $A=M_r(D)$, where $D$ is a central division $k$-algebra of degree $d$, then we may find a commutative subalgebra of degree $([n^2/4]+1)d$, by choosing a commutative subalgebra $E$ of $M_r(k)$ of dimension $[n^2/4]+1$ and a maximal subfield $L$ of $D$, and forming the tensor product $E\otimes_k L$. -But is it the best possible dimension ? -Thanks in advance ! - -REPLY [2 votes]: This is a funny question, which has arisen as a critical component now and then. Yes, if we insisted that the subalgebra be semi-simple, then Schur's lemma gives the result, as in a comment. Perhaps some readers overlooked the possibility of getting $[n^2/4]+1$ (for the matrix case) by taking scalars together with upper-right-corner blocks, with everything else $0$. -The best reasonably-provable result I know was proven by D. Kazhdan as a part of J./I. Bernstein's proof of the admissibility of supercuspidal repns of p-adic reductive groups, and (for the matrix=split case) says that the dimension for a subalgebra with $\ell$ non-scalar generators is bounded by $(n^2)^{1-2^{-\ell}}$. -Since it's not obvious how to exceed $n^2/4$, this bound seems needlessly weak, but we can suspect that Kazhdan would have given a sharper bound if a simple one really held, at least as a meta-argument for no sharper bound holding. I've not tried to construct an algebra exceeding the $n^2/4$, although I have tried (and failed) to improve that bound. The argument is surprisingly non-trivial. A very sketchy version appears in Bernstein's original paper (in J. Fun. An. and Applications), a somewhat fuller version (revised after some comments from D. Renard) in my essay http://www.math.umn.edu/~garrett/m/v/proving_admissibility.pdf, and as an appendix in David Renard's relatively recent book on repn theory of p-adic groups.<|endoftext|> -TITLE: Distribution of roots of complex polynomials -QUESTION [29 upvotes]: I generated random quadratic and cubic polynomials with coefficients in $\mathbb{C}$ -uniformly distributed in the unit disk $|z| \le 1$. The distribution of the roots of 10000 -of these polynomials are shown below (left: quadratic; right: cubic). - -What explains these distributions? In particular: (1) Why the relative paucity of roots -near the origin. (2) Why is the density concentrated in $\frac{1}{2} \le |z| \le 1$? -(3) Why is the cubic distribution sharper than the quadratic? -(The roots of polynomials of higher degree distribute (visually) roughly like the cubic distribution.) - -In a comment to an earlier -posting of this question on MSE, Niels Diepeveen suggested I look at the $\log$ of -the roots instead, for they -"show a density that is independent of the imaginary part and symmetric w.r.t. the real part": -      -To (naive) me, this raises more questions than it answers: (4) Why the uniformity in -the imaginary direction? (5) Why does the $\log$ seemingly obliterate the distinction -between the quadratic and cubic distributions? - -REPLY [15 votes]: Letting $\mu_n$ be the distribution of a randomly chosen root of a random polynomial $f=c_0+c_1X+\cdots+c_nX^n$ in $\mathbb{C}[X]$ for IID random variables $c_i\in\mathbb{C}$, each chosen with some probability distribution $\lambda$ on $\mathbb{C}\setminus\{0\}$ we can show the following just about immediately. - -$\mu_n$ is invariant under the map $z\mapsto z^{-1}$.In particular, the distribution of the logarithm of a random root is invariant under reflection about the origin. -If $\lambda$ is rotationally invariant then so is $\mu_n$. In particular, the distribution of the logarithm of a random root is invariant under translation by an imaginary number. -If $\int\lvert z\rvert d\lambda(z)$ is finite, then as $n$ tends to infinity $\mu_n$ becomes concentrated on the unit circle. That is, for each $\epsilon\gt0$, $\mu_n(\{z\colon 1-\epsilon\lt\lvert z\rvert\lt1+\epsilon\})\to1$ as $n\to\infty$. - -In the case asked about here, $\lambda$ is uniform on the unit ball so that all of the conditions hold, and $\mu_n$ tends weakly to the uniform measure on the unit circle as $n\to\infty$. -For (1), note that the zeros of $g=c_n+c_{n-1}X+\cdots+c_0X^n$ are precisely $\alpha^{-1}$ as $\alpha$ runs through the zeros of $f$, but that $g$ has the same distribution as $f$. -For (2) note that for real $\theta$, the zeros of $g=c_0+c_1e^{-i\theta}X+\cdots+c_ne^{-in\theta}X^n$ are precisely $e^{i\theta}\alpha$ as $\alpha$ runs through the zeros of $f$, but $g$ has the same distribution as $f$. -For (3), note that if $c_0,c_1,c_2,\ldots$ is an infinite IID sequence, each with distribution $\lambda$, and if $0\lt r\lt 1$ then, -$$ -\sum_{k=0}^\infty\lvert c_k\rvert r^k -$$ -has finite mean $\int\lvert z\rvert d\lambda(z)/(1-r)$, so the sum is finite with probability 1. Hence the sequence of polynomials $f_n=c_0+c_1X+\cdots+c_nX^n$ converge uniformly on the ball of radius $r$ (with probability 1). So, the number of zeros of $f_n$ in the ball of radius $r$ is almost surely bounded as $n\to\infty$. This implies that $\mu_n(\{z\colon\lvert z\rvert\le r\})\to0$ as $n\to\infty$. Applying (1) to this also gives $\mu_n(\{z\colon\lvert z\rvert\ge1/r\})\to0$ as $n\to\infty$.<|endoftext|> -TITLE: Embedding a brouwerian lattice into a boolean lattice -QUESTION [6 upvotes]: I have already asked a similar question at -https://math.stackexchange.com/questions/470704/can-a-brouwerian-lattice-be-extended-into-a-boolean-algebra -but have received no answer. -Sorry, I ask a similar question second time, as it is very important for my research. -Is it true that every brouwerian lattice (=locale =frame) can be order-embedded into a boolean lattice? -If yes, can we warrant that this embedding preserves joins and finite meets? - -REPLY [12 votes]: $\def\pw{\mathcal P}$The answer to all the questions is yes. Here are some relevant embedding results: -Proposition 1: Every partial order $(P,\le)$ can be embedded in the powerset lattice $(\pw(P),\subseteq)$, i.e., an atomic complete Boolean algebra. The embedding can be made to preserve all existing meets, or all existing joins (but not both at the same time). -Proof: Map $a\in P$ to $\{b\in P:b\le a\}$ or to $\{b\in P:a\nleq b\}$. -Proposition 2 (Dedekind–MacNeille completion): Every partial order $(P,\le)$ can be embedded in a complete lattice $(L,\le)$ in such way that all existing joins and meets are preserved, and $P$ is dense in $L$ in the sense that every element of $L$ is the join of a subset of $P$, and the meet of a subset of $P$. If $P$ is a Heyting or Boolean algebra, then so is $L$, and the embedding preserves (relative) complements. -Proposition 3: Every distributive lattice $(L,\le)$ can be embedded in a powerset algebra so that all existing finite joins and meets are preserved. -Proof: Let $X$ be the set of prime filters on $L$, and define $f\colon L\to\pw(X)$ as $f(a)=\{f\in X:a\in F\}$. Here, a filter is an upward closed subset of $L$ which is closed under all existing finite meets, and it is prime if its complement is closed under all existing finite joins. Note that the join of the empty set is the smallest element of the lattice, if it exists, and dually for empty meets. -Proposition 4: For every Heyting algebra $(H,\le)$, there is an embedding of $H$ into a complete Boolean algebra (not necessarily atomic) that preserves finite meets and all existing joins (but not necessarily relative complements). -Proof: By Proposition 3, we may assume that $H$ is a bounded sublattice of some $\pw(X)$. Let $B$ be the Boolean subalgebra of $\pw(X)$ generated by the elements of $H$, and $\tilde B$ be the Dedekind–MacNeille completion of $B$. It suffices to show that the inclusion $H\subseteq B$ preserves all existing joins. -The key point is that for every $x\in B$, there exists $\hat x\in H$ such that -$$\tag{$*$} a\le x\iff a\le\hat x$$ -for every $a\in H$. This can be shown as follows: by expressing $x$ in the conjunctive normal form, and using the closure of $H$ under finite joins and meets, we can write $x=\bigwedge_{i -TITLE: Which Lie groups have adjoint representations that are bounded away from zero? -QUESTION [8 upvotes]: Studying stability of certain non-autonomous dynamical systems on Lie groups I have come across the following question: Exactly which finite-dimensional, real Lie groups have adjoint representations that are bounded away from zero? -Edit: by "bounded away from zero" I mean that the image of the adjoint representation avoids an open neighborhood of zero in End(g), where g is the Lie algebra. Equivalently, the closure of the image does not contain zero, or, the norm (pick your favorite one) of every element of the adjoint representation is bounded from below by one and the same positive number. By Hadamard's inequality, a determinant bound will do as well. -[end edit] -This should include compact Lie groups since for those there exists an inner product on the Lie algebra with respect to which all inner automorphisms are orthogonal, i.e. the elements of the adjoint representation have norm 1. Correct? -Also, for abelian Lie groups the adjoint representation is trivial, hence again bounded away from zero. -I believe that semisimple Lie groups should also be included but can not think of a valid argument. -Is there actually a counter example? I tried the general linear group GL(2) but the elements of the adjoint representation that I tried always have (some) unit eigenvalues. Is this an accident? I would have thought that GL(n) itself occurs as an adjoint representation somehow which would then not be bounded away from zero. But evidently I am not quite understanding the different dimensions here (the adjoint representation of GL(2) is a subgroup of GL(4)). -My apologies if this is trivial but I could not find anything that looked relevant in several books on Lie groups. - -REPLY [6 votes]: The adjoint rep is always bounded away from $0$. Let $\mathfrak{g}_0$ be a simple quotient of $\mathfrak{g}$. (I consider the $1$-dimensional Lie algebra to be simple, so there is always a simple quotient.) Let $\mathfrak{h}$ be the kernel of $\mathfrak{g} \to \mathfrak{g}_0$ and let $H = \exp(\mathfrak{h})$. -The adjoint action preserves $\mathfrak{h}$, so the adjoint representation is block upper triangular. The upper left block is the adjoint action of $G/H$ on its Lie algebra, which is $\mathfrak{g}_0$. Since $\mathfrak{g}_0$ is simple, it is unimodular, meaning that the adjoint rep has determinant $1$. This idea is taken from anton's answer. -In summary, the adjoint rep of $G$ can be put in block upper triangular form with the upper left block a matrix of determinant $1$ (and not a $0 \times 0$ matrix).<|endoftext|> -TITLE: Homotopy problem for infinite dimensional topological space II -QUESTION [5 upvotes]: This post here is a specification of this post. -Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of intrinsic metric spaces verifying : - -$X_{n}$ have topological dimension $n$. -$X_{n+1}$ is n-connected. -$X_{n} \subset X_{n+1}$ -The distance $d_{n}$ and $d_{n+1}$ generate the same topology on $X_{n}$. -$\forall x,y \in X_{n}$ : $d_{n+1}(x,y) \le d_{n}(x,y)$. -$(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$ - -Definition : Let $d$ be a distance on $ \bigcup{X_{n}}$, defined as follows : $d(x,y) = lim_{n \to \infty} d_{n}(x,y)$. -Remark : There is a small abuse in the previous definition because $d_{n}(x,y)$ is defined only for $x, y \in X_{n}$. But because we take $n \to \infty$, there is no problem. -Definition : Let $X:=\overline{\bigcup{X_{n}}}$, the complete metric space obtained as a completion of $\bigcup{X_{n}}$ for $d$. - -Question : Is $X$ weakly contractible ? - -Remark : If yes, perhaps some of these conditions are useless in the proof, and perhaps the useful conditions can be highly generalized. - -REPLY [2 votes]: The intristic metric assumption seems to be added to exclude $S^1$ from the previous post. But what about $S^d$? It can be approximated with bigger and bigger disks -$\{x\in S^d\ |\ x_1\leq 1-\frac{1}{n} \}$ (with the natural intristic metric) and in the limit you get the sphere. (Of course dimension of such a disk is $d$ so one needs some obvious bypasses for $n -TITLE: How close can closed geodesics be? -QUESTION [6 upvotes]: A consequence of the famous Jørgensen inequality is that there is a lower bound for the distance between closed geodesics in hyperbolic three-manifolds: for any $R>0$ there is a c>0 such that for any such manifold $N$ and any two distinct closed geodesics loops $c_1,c_2$ on $M$ both of length less than $R$ we have -$$ -d_H(c_1,c_2) \ge c \quad (\ast) -$$ -where $d_H$ is Hausdorff distance. -This is false for general locally symmetric spaces, since the higher-rank ones can contain flat tori. My question is the following: if we fix a real-rank one symmetric space $X$ (i.e. $X$ is a real, complex or quaternionic hyperbolic space, or octonionic hyperbolic plane) is there a c>0 (depending on the space) such that $(\ast)$ holds for all quotients of $X$ and all pairs of distinct closed geodesics of bounded length on such? (More generally one could ask the same question for Riemannian $n$-manifolds with sectional curvature between $-1$ and $\kappa<0$, with a $c$ depending on $\kappa$ and $n$ and the bound on the length). -Edit: As noted in Anton's comments the opening remark is false if one does not bound the length of the geodesics; I edited to add that. - -REPLY [7 votes]: I think such a bound exists (depending only on pinched curvature constant $\kappa$, dimension $n$, and $R$). Suppose one has an infinite sequence of pinched negatively curved manifolds where geodesics of length $\leq R$ have Hausdorff distance approaching $0$. By the generalized Margulis Lemma (see Ballmann-Schroeder), there is an $\epsilon$ so that in a $\kappa$-pinched $n$-dim. manifold, $\epsilon$-thin regions are tubular neighborhoods about short geodesics of radius $>R$. Thus, we may assume that the curves pass through the thick part of the manifold, otherwise they must be homotopic to short geodesics in the thin part, in which case it follows. One may take a pointed Cheeeger-Gromov limit (with basepoint on either curve in the thick part), to get a limit in which the Hausdorff distance between the two curves is zero. This gives a contradiction, since these curves are homotopic in the limit, but this implies they will be homotopic in the approximates.<|endoftext|> -TITLE: Bass's paper "Symplectic groups and modules", used in proof of the congruence subgroup property for Sp -QUESTION [13 upvotes]: Let $R$ be the ring of integers in a number field. While studying the congruence subgroup property for $\text{Sp}_{2g}(R)$ in -Bass, H.; Milnor, J.; Serre, J.-P. -Solution of the congruence subgroup problem for SLn(n≥3) and Sp2n(n≥2). -Inst. Hautes Études Sci. Publ. Math. No. 33 1967 59–137. -they quote a theorem of Bass that says the following. First, some notation. If $R$ is a commutative ring and $q$ is an ideal of $R$, then denote by $\text{Sp}_{2g}(R,q)$ the kernel of the map -$$\text{Sp}_{2g}(R) \longrightarrow \text{Sp}_{2g}(R/q).$$ -Also, denote by $\text{Ep}_{2g}(R,q)$ the normal subgroup of $\text{Sp}_{2g}(R)$ generated by the usual elementary symplectic matrices which happen to lie in $\text{Sp}_{2g}(R,q)$. -Theorem : For $R$ a Dedekind domain and $q$ an ideal of $R$ and $g \geq 2$, we have $\text{Sp}_{2g}(R,q) = \text{Sp}_{2}(R,q) \cdot \text{Ep}_{2g}(R,q)$. -This is part a of Proposition 13.2 in the above paper. The reference they give is -Bass, H, Symplectic modules and groups, in preparation. -However, this paper does not seem to have ever appeared. In a paper I'm writing right now, I need a fact which is a corollary of what I assume is the proof they have in mind (alas, it doesn't just follow from the statement). Does anyone know a published account of it? - -REPLY [6 votes]: Just after the limited work by Bass-Milnor-Serre was published, the work of Moore made possible a major improvement by H. Matsumoto in his Paris thesis here, treating uniformly all the split (Chevalley) group schemes including the somewhat exceptional case of symplectic groups relative to the congruence subgroup problem. I think Cor. 4.5 and the surrounding material cover the missing Bass paper, which he may have avoided completing due to Matsumoto's work. (But participants at the time could fill in that history much better than I can.) -The rank 1 group in your formulation comes from the unique long simple root for a symplectic group. -Probably the results for classical groups are also worked out in the later book by Hahn and O'Meara The Classical Groups and K-Theory (Springer, 1989), but I haven't yet tracked down an explicit reference there. Also, they don't include full details of the congruence subgroup problem, referring at times to the literature.<|endoftext|> -TITLE: How big can a commutative subalgebra of Weyl algebra be? -QUESTION [7 upvotes]: Consider the smallest Weyl algebra $A_1=\{q,p; qp-pq=1\}$. It is known that there exist pairs of commuting elements, say $L$ and $M$, that obey various polynomial relations, e.g. elliptic curves. I wonder how big the commutative subalgebra of the Weyl algebra can be? (of course, I mean how big the generating set of commuting elements can be since any product of commuting things trivially commutes with the generating set) - -REPLY [4 votes]: A maximal commutative subalgebra of the Weyl algebra can have arbitrarily many generators. (Though, as explained in my other answer, it must be finitely generated.) Let the Weyl algebra as $k\langle x,y \rangle / yx-xy-1$ (where $k$ is a field of characteristic zero) and define $z = xy$. Let $a$ be a rational non-integer. Set -$$\theta = \frac{\prod_{i=1}^{n} (z-a+i)(z-a-i)}{(z-a)^n} x,$$ -I will discuss below how to make sense of the division. Then I claim that $$k[\theta] \cap A_1 = k[\theta^{n+1}, \theta^{n+2}, \theta^{n+3}, \ldots],$$ -and that this is a maximal commutative subalgebra of $A_1$. This ring cannot be generated by fewer then $n+1$ generators. This example (slightly modified) for $n=1$ appears in Mikar-Limanov's paper, where he credits it to "someone in Moscow, Russia about 1968". -Thanks for making me finally figure out why this example works! - -We begin with the equations -$$zx = x(z+1) \quad zy=y(z-1)$$ -which may be checked directly. These maybe seen to imply -$$f(z) x = x f(z+1) \quad f(z) y = y f(z-1)$$ -for any polynomial $f$. Let $B$ be the ring generated by $k(z)$, $x$ and $y$, with relations -$$f(z) x = x f(z+1) \quad f(z) y = y f(z-1) \quad xy=z \quad yx= z+1$$ -for any rational function $f(z) \in k(z)$. So the Weyl algebra $A$ maps to $B$ and, using that $A$ is a domain, this can be seen to be an embedding. Note that, in $B$, we have $x^{-1} = (z+1)^{-1} y$ or, equivalently, $y = (z+1) x^{-1}$. So we can equally well describe $B$ as the ring generated by $k(z)$ and $x^{\pm 1}$, subject to the relation $zx=x(z+1)$. -Every element of the ring $B$ can be uniquely written as -$$\sum_{n=-\infty}^{\infty} f_n(z) x^n$$ -with the $f_n(z)$ in $k(z)$ and all but finitely many $f_n$ equal to zero. Multiplication is given by -$$\left( \sum_m f_m(z) x^m \right) \left( \sum_n g_n(z) x^n \right) = \sum_{m,n} f_m(z) g_n(z-m) x^{m+n}$$ -Define -$$ f^{(n)}(z) = -\begin{cases} f(z) f(z-1) f(z-2) \cdots f(z-n+1) & n > 0 \\ -1 & n=1 \\ -\frac{1}{f(z+1)f(z+2) \cdots f(z+(-n))} & n < 0 \\ -\end{cases}$$ -An easy induction establishes -$$\left( f(z) x \right)^n = f^{(n)}(z) x^n$$ -Lemma Let $\theta = f(z) x$ for some nonzero rational function $f(z)$. Then $\sum a_n(z) x^n$ commutes with $\theta$ if and only if $a_n$ is a scalar multiple of $f^{(n)}$ for all $n$. In other words, the elements of $B$ which commute with $\theta$ are all of the form $\sum c_n \theta^n$ for $c_n \in k$. -Proof Write $a_n(z)$. Comparing the coefficient of $x^{n+1}$ on both sides of $ \left( \sum a_n(z) x^n \right) \theta = \theta \left( \sum a_n(z) x^n \right)$ gives -$$a_n(z) f(z-n) = f(z) a_n(z-1).$$ -Writing $a_n(z) = f^{(n)}(z) b_n(z)$, we deduce that -$$b_n(z) = b_n(z-1).$$ -The only periodic rational functions are the constants. $\square$. -Set $f(z) = \prod_{i=1}^n (z-a+i)(z-a-i)/(z-a)^n$. I leave the following to you: -Lemma For $m>n$, the rational function $f^{(m)}(z)$ is in $k[z]$. For $m$ nonzero and $\leq n$, the rational function $f^{(m)}(z)$ has a pole at a noninteger. -Corollary The elements of $A_1$ which commute with $\theta$ are of the form $c_0 + \sum_{m >n } c_m \theta^m$. -The only detail to keep track of is the conversion from $x^{-m}$ to $\frac{1}{(z+1)(z+2) \cdots (z+m) } y^m$; this is why I made $a$ a noninteger. - -Having checked all this, I claim that $R=k[\theta^{n+1} , \theta^{n+2}, \cdots, ]$ is a maximal commutative subring of $A_1$. We showed above that it is contained in $A_1$. And, if $\alpha$ commutes with $\theta^{n+1}$ and $\theta^{n+2}$, then $\alpha$ commutes with $\theta$. So any element of $A_1$ which commutes with $R$ is itself in $R$. This shows that $R$ is maximal.<|endoftext|> -TITLE: Mellin transform between heat kernel and zeta-function -QUESTION [5 upvotes]: For some notion of a "positive operator" $D$ of "Laplacian type" one seems to be able to define a notion of a zeta-function as $\xi(s,f,D) = Tr_{L^2}(f D^{-s})$ where $f \in L^2$ (the space of square-integrable functions on the chosen manifold). Also one now defines the generalized heat-kernel corresponding to this as $K(t,f,D) = Tr_{L^2}(f\, exp(-tD))$. -Now I am faced with the following identity for which I would like to know the proof: -$$ \xi(s,f,D) = \frac{1}{\Gamma(s)}\int_0^{\infty} t^{s-1} K(t,f,D) \, dt $$ - -It would be helpful if someone could also elaborate on the notion of what is a "positive operator of Laplacian type" (..I have some rough idea based on specific examples..) and if someone could specify peculiarities in the above equations that are likely to come up if one goes to non-compact manifolds like hyperbolic spaces. -I am mostly interested in doing this on hyperbolic spaces. - -REPLY [9 votes]: Being of Laplacian type means that $D$ is a second order p.d.o. whose principal symbol coincides with that of a Laplacian of a metric $g$ on the manifold. Being positive signifies that that for any smooth functions with compact support $\newcommand{\bR}{\mathbb{R}}$ $f_0, f_1: M\to \bR$, $f_0\neq 0$, you have -$$ \int_M (Df_0)\cdot f_1 dV_g =\int_M f_0 \cdot (D f_1) dV_g,\;\; \int_M (Df_0) f_0 > 0. $$ -In case the manifold $M$ is compact, $\dim M=m$, then $D$ has a discrete spectrum -$$ 0<\lambda_0\leq \lambda_1\leq \cdots. $$ -Let $(\Psi_k)_{k\geq 0}$ denote an orthonormal basis of $L^2(M)$ consisting of eigenfunctions, $D\Psi_k=\lambda_k\Psi_k$. -For $s>-m$, then $fD^{-s}$ is an integral operator with integral kernel $\newcommand{\eK}{\mathscr{K}}$ -$$ \eK_{f D^{-s}}(x,y) =\sum_{k\geq 0} \lambda_k^{-s} f(x)\Psi_k(x) \Psi_k(y), $$ -while $fe^{-tD}$ is an integral operator with kernel -$$\eK_{fe^{-tD}}(x,y)= \sum_{k\geq 0} e^{-t\lambda_k} f(x)\Psi_k(x) \Psi_k(y). $$ -Then -$$ \xi(s,f,d)=\int_M \eK_{fD^{-s}} (x,x) dV_g(x)=\sum_{k\geq 0} \lambda_k^{-s} \int_M f(x) |\Psi_k(x)^2| dV_g (x), $$ -$$ K(t,f, D)=\int_M \eK_{fe^{-tD}} (x,x) dV_g(x)=\sum_{k\geq 0} e^{-\lambda_k t} \int_M f(x) |\Psi_k(x)^2| dV_g (x). $$ -Now use the elementary identity -$$ \lambda^{-s}=\frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} e^{-\lambda t} dt. $$<|endoftext|> -TITLE: Pullback-stability of internally projective objects -QUESTION [13 upvotes]: An object $X$ of a category $C$ is said to be projective if the hom-functor $C(X,-)$ preserves epimorphisms (or, in general, some restricted class of epimorphisms such as the regular or effective ones). The axiom of choice is equivalent to the assertion that all objects of Set are projective. In general, a "projective set" $X$ is one such that we can make $X$-indexed families of choices. -In section D4.5 of Sketches of an Elephant, an object $X$ of a topos $C$ is said to be internally projective if the right adjoint $\Pi_X : C/X \to C$ preserves epimorphisms, and $C$ is said to satisfy the internal axiom of choice if all objects are internally projective. The latter definition is found in many other places; I haven't seen the former elsewhere, but I don't know its actual origin. -My question is, if $X$ is internally projective in this sense in $C$, is $X\times Y$ internally projective in $C/Y$ for another object $Y$? This seems to be a necessary condition for the definition to be sufficiently "internal", but I haven't been able to prove it yet. -A similar question is, if $C$ satisfies the internal axiom of choice, does the functor $\Pi_f : C/X \to C/Y$ preserve epimorphisms for any morphism $f:X\to Y$? - -REPLY [7 votes]: For toposes, the stability property of your first question does hold. Suppose $X$ is internally projective in $\mathbb{C}$. And suppose $q: B \to A$ is an epimorphism from $v : B \to I$ to $u: A \to I$ in $\mathbb{C}/I$, hence an epimorphism in $\mathbb{C}$. Write $X^*$ for the object $X \times I$ of $\mathbb{C}/I$. The exponential $u^{X^*}$ in $\mathbb{C}/I$ has underlying object: -$$ \{(i,f) : I \times A^X \mid \forall x:X.\, u(f(x))=i\}$$ -And the map $q^{X^*}$ is a pullback in $\mathbb{C}$ of the map -$$I \times q^X: I \times B^X \to I \times A^X$$ along the embedding of $u^{X^*}$ in $I \times A^X$. Since the displayed map above is an epi by internal projectivity of $X$, so is its pullback $q^{X^*}$. -It is also true that the internal axiom of choice (IAC) is preserved by slicing. (I haven't managed to see that this follows as a direct consequence of the previous.) Suppose IAC holds in a topos $\mathbb{C}$. Consider an arbitrary object $w: Z \to I$ in $\mathbb{C}/I$, and an epimorphism $q$ as above. The exponential $u^w$ in $\mathbb{C}/I$ has underlying object -$$\{(i,f): I \times (1+A)^Z \mid \forall z:Z.\, (w(z)= i \to \exists a:A. f(z) = \text{inr}(a) \wedge u(a) = i) \wedge (w(z) \neq i \to f(z) = \text{inl}(*)) \}$$ -This uses a standard representation of slice-category exponentials in toposes in terms of partial map classifiers, with the simplification that, since $\mathbb{C}$ is boolean (using Diaconescu's theorem that IAC implies boolean), the partial map classifier for $A$ is $1 + A$. Similar to the previous argument, one now observes that $q^w$ is a pullback in $\mathbb{C}$ of the map -$$I \times (1+q)^Z : I \times (1+B)^Z \to I \times (1+A)^Z$$ -along the subobject inclusion of $u^w$. And once again the map displayed above is an epi, by internal projectivity of $Z$.<|endoftext|> -TITLE: The Ground Axiom for special statements of set theory -QUESTION [6 upvotes]: The Ground Axiom (GA), introduced by Hamkins and Reitz, asserts that the universe is not a nontrivial set forcing extension of any inner model, and it is known that GA is consistent relative to ZFC. My question is somehow related to GA. -First a definition: -Definition. Let $\Phi$ be a statement of set theory which is independent of ZFC (or ZFC+large cardinals). Call $\Phi$ forceable, if there is a model $W$ of ZFC and a set of forcing conditions $P$ in $W$ such that $\Phi$ is true in the generic extensions of $W$ by $P$ (we allow the existence of large cardinals in $W$). -Question. 1) Is there a characterization of all forceable sentences $\Phi$ such that the following holds: -If $\Phi$ holds in $V$, then there some inner model $W$ of $V$ in which $\Phi$ fails, and such that $V$ is a set generic extension of $W$. -2) Can you give at least one natural statement $\Phi$ as in (1). -Remark (suggested by Sy Friedman). In the paper "A Large $\Pi^1_2$ Set, Absolute for Set Forcings", Sy Friedman constructs a $\Pi^1_2$ formula $\phi$ which can have many solutions in a cardinal-preserving extension of $L$. $\phi$ also has the property that set-forcing will not add new solutions. -So no model in which $\phi$ has more than $\aleph_1$ solutions can be a set-generic extension of a model with only $\aleph_1$ solutions. -Also there are reals $r$ which are minimal over $L$ but which cannot be obtained by set forcing over $L$. Then in $V=L[r]$ the sentence $V=L$ fails, but there is no inner model $W$ of $V$ satisfying $V=L$ such that $V$ is a set forcing extension of $W$. - -REPLY [8 votes]: In your definition, probably you intend that the forcing is nontrivial, since otherwise any consistent $\Phi$ would qualify via trivial forcing. Note also that your forceable terminology conflicts with the terminology used elsewhere, where an assertion $\sigma$ is forceable in a model $V$, if it holds in some set-forcing extension $V[G]$. This is the possibility modal operator $\Diamond\sigma$ of The modal logic of forcing, whose dual $\Box\sigma$ asserts that $\sigma$ is necessary, that is, that $\sigma$ holds in all forcing extensions $V[G]$. Each of these operators is expressible in the first-order language of set theory. That is, $\Phi$ is forceable-in-your-sense if there is some model of set theory in which it is forceable. -Benedikt Löwe and I have also introduced the downward versions of these operators, where we say $\varphi$ is downward possible — let's write it as $\underline\Diamond\varphi$ here — if $\varphi$ holds in some ground model over which the universe was obtained by forcing. Although it is not obvious, the assertion, "there is an inner model $W$ for which $V=W[G]$ is a set-forcing extension and $W\models\varphi$" is first-order expressible in set theory. This kind of observation is the beginning of the subject known as Set-theoretic geology. The dual notion $\underline\Box\varphi$, which asserts that $\varphi$ holds in all ground models, is similarly expressible. The situation of these modal operators in set theory is investigated in our paper Moving up and down in the generic multiverse. In that paper, in order to determine the modal logic, we identify classes of statements, such as the downward switches, which can be turned on and off by going to deeper and deeper grounds, and the downward buttons, which are false in $V$ but true in some ground $W$ and all deeper grounds below $W$. The proofs required us to have large independent families of such buttons and switches, and other types of control statements. -Your statements $\Phi$ are similar to this kind of statement, but go just one step. Specifically, your statements $\Phi$ are precisely the statements for which $\Phi\to\underline\Diamond\neg\Phi$ holds. -I'm not sure what you want in terms of a "classification", but my point is that your property is first-order expressible in the language of set theory, even though it may appear to be second-order because of the quantification over inner models. Set-theoretic geology provides a first-order definable enumeration $W_r$ for $r\in V$ of the collection of all grounds $W_r$ of $V$, and so quantifying over grounds amounts to quantifying over the parameters $r$ used to define them and thus is a first-order quantifier. (You can see an account of the complexity of this definition in my recent paper Superstrong and other large cardinals are never Laver indestructible.)<|endoftext|> -TITLE: Exotic 4-manifolds with even positive partial betti number -QUESTION [8 upvotes]: It seems that usually smooth structures on compact 4-manifolds are distinguished by Seiberg-Witten/Donaldson invariants. And I don't know another direct way to do that. But in the case when $b_2^+$ is even the invariants vanish identically. -Does there exist an example of smooth compact exotic 4-manifold with $b_2^+$ even? - -REPLY [12 votes]: An often-overlooked invariant of a closed oriented 4-manifold $X$ is the closed oriented 4-manifold $\overline{X}$: the orientation-reversed manifold. Taking the Seiberg-Witten invariants of $\overline{X}$, one can distinguish smooth structures on manifolds $X$ with $b^->0$ odd, regardless of the parity of $b^+$. For instance, $\mathbb{C}P^2\# \overline{K3}$ has $b^+ = 20$ and infinitely many smooth structures (perform knot surgery on a K3 surface, blow up once, reverse orientation). -The real problem, then, is to understand what happens when both $b^+$ and $b^-$ are even. -I think one can distinguish infinitely many smooth structures on the connected sum of two K3 surfaces (so $b^+=6$, $b^-=38$) along the lines Tom Leness suggests in his answer. -The non-equivariant Bauer-Furuta invariant of a $\mathrm{Spin}^c$-structure $\mathfrak{s}$ on a closed oriented 4-manifold $X$ is a stable map $S^{(c_1(\mathfrak{s})^2−σ)/4}\to S^{b^+}$. When $X$ is a homotopy K3 surface, and $c_1(\mathfrak{s})^2=0$, we get the stable homotopy class of a map $S^4\to S^3$; this lives in $\mathbb{Z}/2$, and is actually in this case the mod 2 Seiberg-Witten invariant (I'm going on memory here, so this is a point to check). By Bauer's theorem, the invariant for a connected sum is the smash product of the invariants of the summands. Now, the smash-square of the Hopf map is the generator for $π^s_2(S^0)=\mathbb{Z}/2$. Hence, if $X_1$ and $X_2$ are homotopy K3's, and $s(X_i)$ is the number of $\mathrm{Spin}^c$-structures with $c_1(\mathfrak{s})^2=0$ and odd SW invariant, the product $s(X_1)s(X_2)$ is an invariant of $X_1\#X_2$. Via Fintushel-Stern knot surgery along on K3 along knots with suitable Alexander polynomials, one can make $s(X_i)$ arbitrarily large, and thereby distinguish infinitely many smooth structures on -$K3 \# K3$. -As so often in this subject, one has the feeling that information is being lost: maybe $X_1\#X_2$ actually knows $X_1$ and $X_2$.<|endoftext|> -TITLE: Can the hyperbolic plane be immersed in three dimensional Euclidean space, if we are only looking for a weak solution? -QUESTION [12 upvotes]: Consider the following question: -"Can the hyperbolic plane $(\mathbb{R}^2, g_H)$ be isometrically -immersed in three dimensional Eulidean space$(\mathbb{R}^3, g_{flat})$?" -I believe the answer to this question is no. Can someone give me a reference -for this theorem (in particular I want to look at the details of the -proof and understand why this is not possible). -My second question is as follows: I assume the answer is no if one is -asking for smooth (i.e. $\mathcal{C}^{\infty}$) immersion. Is anything -known if we relax this condition? More precisely, asking for an immersion -is asking whether a certain pde has a solution. What happens if I am just -looking for some "weak" solution to this pde? - -REPLY [9 votes]: Klotz-Milnor proved in 1972 that there is no $C^2$ isometric embedding of $H^2$ in $R^3$ so Nash-Kuiper can not be improved http://www.zentralblatt-math.org/zmath/scans.html?volume_=236&count_=348 -A practical realisation of an embedding is here: http://www.math.cornell.edu/~dwh/papers/crochet/crochet.html<|endoftext|> -TITLE: pullback of theta divisor -QUESTION [8 upvotes]: Let $C$ be a smooth curve and $J$ its Jacobian. Let $p$ be a point on $C$ and $j: C \to J$ be the map $x \mapsto x-p$. Let $\theta$ be the Theta divisor on $J$, i.e. the locus $\{ x_1 + \cdots + x_{g-1} - (g-1)p \mid x_i \in C\}$. It follows from Poincare formula that $j^* \mathcal{O}(\theta)$ is a degree $g$ line bundle on $C$. The question is: what is this line bundle? This seems to be something well-known, but I couldn't find any references. - -REPLY [7 votes]: Francesco's answer is completely general. As he emphasized, we should do everything in an intrinsic way so that there is a canonical choice of Theta. I hope I get this straight since it is very pretty. -Indeed this is Riemann's classical proof of Jacobi inversion. If $L$ is any degree $g$ line bundle, then $K-L$ has degree $g-2$, so the translate $C+(K-L)$ lies in $\mathrm{Pic}^{g-1}$. Then the pullback of $\mathscr O(\Theta) = \mathscr O(W(g-1))$ by this translation map $C\to\mathrm{Pic}^{g-1}(C)$ is $L$. (This also shows that this map induces an isomorphism from line bundles on $\mathrm{Pic}^{g-1}(C)$ of chern class $[\Theta]$, with line bundles on $C$ of degree $g$.) -I.e. consider this as the problem of intersecting the curve translate $C+(K-L)$ with the divisor $W(g-1)$. We know $L =\mathscr O(p_1+...+p_g)$ for some points $p_i$, and that the intersection has degree $g$. So if the intersection is proper, there exist $g$ points $p$ such that $(p+ K -p_1-...-p_g)$ lies on $W(g-1)$, i.e. such that -$p + K-p_1-...-p_g = q_1+...+q_{g-1}$, for some $q$'s. -But this is true for $p$ equal to any $p_i$, $i=1$,....,$g$, if for $p = p_{i_0}$ we take -the $q$'s to be such that $\mathscr O(q_1+...+q_{g-1}) = \mathscr O(K- \sum_{i\ne i_0}p_i)$. (Such $q$'s exist by Riemann-Roch.) -Thus intersecting $C+(K-L)$ with $W(g-1)$ was Riemann's way of finding a divisor $D$ such that $O(D) = L$. Your question is the same, but phrased in terms of line bundles. The common practice in many books of always working in $\mathrm{Pic}^0(C)$ via translating by a fixed point may obscure the intrinsic nature of the statements. In fact I seem to be guilty of this in my argument on pp.386-7 of Lectures on Riemann Surfaces, (World Scientific Publishers 1989, ed. by Cornalba, Gomez-Mont and Verjovsky), where I take the strange approach of deducing this intrinsic version from a clunky non intrinsic version, thus making it look less comprehensible. -If you look closely at Lemma 3.4 of Lange Birkenhake, p.335, on which the exercise Francesco cites is based, you may discern a translated version of the Riemann result proved here. -Let me enlarge a little on this. Classically the theta divisor is the zero locus of a certain holomorphic theta function. Computing its zeroes is done by the residue theorem. Recall there is a generalization of that theorem that allows one to compute not just the number of zeroes inside a given loop but also the sum of their values, (see Ahlfors,4.5.2 the argument principle, pp. 151-153, esp. eq.(47) p.153). -Your question is exactly this, i.e. not only how many intersection points exist, but what is their sum as a point of the Jacobian, i.e. as a line bundle. Traditionally this was a lemma used to prove Riemann's theorem that the zeroes of the theta divisor are, up to a precise translation, exactly the image of the Abel map on $C^{g-1}$. This is made very clear in Griffiths-Harris, where the number of zeroes is computed on pages 334-335, and their sum on pages 336-337, both by the residue theorem. These results are then applied to prove Riemann's theorem on pages 338-339. -This result appears in the classical form also in Mumford's Tata lectures on theta I p.149, or Siegel's Topics on complex function theory vol. 2, section 4.10. If one uses a modern approach to that theorem by equating the homology classes of those two divisors and proving this forces them to be translates, as in Lange and Birkenhake, one loses some explicitness of the classical approach. It does however then make available the relatively easy intrinsic proof given here for your question, and resembling that of their lemma 3.4 on p.335 of their book. I think in some sense this may be a more natural approach. -Note that this result is key to the construction of a "Poincare" bundle, since it gives a uniform way to construct a family of divisors on $\mathrm{Pic}^g(C)$, i.e. a line bundle on $C\times\mathrm{Pic}^g(C)$, whose restrictions to the curve $C$ induce all line bundles of degree $g$. Hence this construction too is essentially due to Riemann. This is clearly explained by George Kempf on pages 154-157, chapter 18, of his book Abelian Integrals, Monografias del instituto de matematicas, #13, Universidad Nacional Autonoma de Mexico, 1983.<|endoftext|> -TITLE: a question for the prime counting function -QUESTION [7 upvotes]: A famous inequality that has been proved by J.B. Rosser and L. Schoenfeld says that -$\frac{n}{\ln n-1/2}$ < $\pi(n)$<$\frac{n}{\ln n-3/2} , n\ge 67$. -Using this inequality we can prove that when $\pi(n)$ divides $n$ (and this happens infinitely often) then -$\pi(n)=\frac{n}{[\ln n-1/2]}$ (for $n\ge 67$) -(By $[\ln n-1/2]$ we denote the integer part of $\ln n-1/2$). -This is an exact formula for $\pi(n)$ that occurs infinitely often. -The question is: Do we have any knowledge about when $\pi(n)$ divides $n$ or anything in this direction? -Thank you for viewing. - -REPLY [10 votes]: Just to emphasize, $t/\lfloor \ln t-1/2 \rfloor$ is not, in any ordinary sense, as good an approximation to $\pi(t)$ as $t/(\ln t-1)$, let alone as good as $\int_e^t dx/\ln x$. It is simply, like a stopped clock, an approximation which is occasionally exactly right. -I'll define $R(u)$ to mean the closest integer to $u$, so your proposed approximation is $t/R(\ln t -1)$. Set $Li(t) = \int_e^t \frac{dx}{\ln x}$ (where the lower bound of the integral is set is irrelevant.) Then $\pi(t) - Li(t)$ is eventually smaller than $t/(\ln t)^N$ for any $N$. Integrating by parts shows that -$$Li(t) = \frac{t}{\ln t} + \frac{t}{(\ln t)^2} + \frac{2 t}{(\ln t)^3} + \frac{3! t}{(\ln t)^3} + \frac{4! t}{(\ln t)^4} + \cdots $$ -where $\cdots$ is meant in the sense of asymtotic series: If you stop the sum at the $N$-th term, the error will be bounded by a multiple of the $(N+1)$-st term. We have -$$\frac{t}{\ln t - 1} = \frac{t}{\ln t} \frac{1}{1-1/\ln t} = \frac{t}{\ln t} + \frac{t}{(\ln t)^2} + \frac{t}{(\ln t)^3} + \frac{t}{(\ln t)^4} + \cdots$$ -So $t/(\ln t -1)$ is pretty good, matching the first two terms of the asymptotic series. By comparison, $t/R(\ln t -1) = t/(\ln t - 1 + \theta)$, where $\theta$ oscillates between $\pm 1/2$. So $t/R(\ln t -1) = t/\ln t + (1-\theta) t/(\ln t)^2+\cdots$, only matching the first term of the series. -To emphasize the difference, the following figure plots -$$Li(t)- \pi(t) \ \mbox{(green)} \quad \frac{t}{\ln t -1} - \pi(t) \ \mbox{(blue)} \quad \frac{t}{R(\ln t -1)} - \pi(t) \ \mbox{(red)}$$ -for $t$ between $10^4$ and $10^6$. - -You can see that green is a little bit better than blue and both are in general far better than red, although red is occasionally exactly right.<|endoftext|> -TITLE: The independence of path induction -QUESTION [8 upvotes]: In §1.12 of the Homotopy type theory book, it is mentioned that indiscernibility of identicals is a consequence of path induction. More precisely, for each type $C$ dependent over a type $A$, there is a term -$$\mathsf{transport} : \prod_{a_0, a_1 : A} \prod_{p : a_0 =_A a_1} C (a_0) \to C (a_1)$$ -such that $\mathsf{transport} (a, a, \mathsf{refl}) \equiv \mathsf{id}_{C (a)}$, which is manifestly an instance of the general path induction principle, which constructs for each type $B$ dependent over $\sum_{a_0, a_1 : A} a_0 =_A a_1$ and each $s : \prod_{a : A} B (a, a, \mathsf{refl})$ a term -$$\mathsf{ind}(s) : \prod_{a_0, a_1 : A} \prod_{p : a_0 =_A a_1} B (a_0, a_1, p)$$ -such that $\mathsf{ind}(s, a, a, \mathsf{refl}) \equiv s (a)$. -Question. Is path induction strictly stronger than indiscernibility of identicals? (Or, does there exist a model of intensional type theory where the propositional equality satisfies indiscernibility of identicals but not path induction?) -I ask because the built-in eq type in Coq is defined as an inductive type whose induction principle is indiscernibility of identicals, rather than path induction. Coq is sufficiently rich to allow the path induction principle as well; however, it doesn't seem to be derived from the indiscernibility of identicals but rather by using pattern matching directly. - -REPLY [7 votes]: The problem with postulating only $\mathsf{transport}$ is that it is too weak to characterize the identity type up to equivalence. Let me define another type, called $\mathsf{doubleId}$, which has a $\mathsf{doubleRefl}$ and a $\mathsf{doubleTransport}$ satisfying the same rules as $\mathsf{refl}$ and $\mathsf{transport}$, but is not equivalent to the identity type. -We can take our type to be two copies of the identity type: -$$\mathsf{doubleId}(a,b) \mathrel{{:}{\equiv}} (a = b) + (a = b).$$ -There are two choices for $\mathsf{doubleRefl}$, so let us pick $\mathsf{inl}(\mathsf{refl})$. -This type has a $\mathsf{transport}$-like eliminator, defined by -$$\mathsf{doubleTransport}(a_0, a_1, \mathsf{inl}(p)) \mathrel{{:}{\equiv}} \mathsf{transport}(a_0, a_1, p)$$ -$$\mathsf{doubleTransport}(a_0, a_1, \mathsf{inr}(p)) \mathrel{{:}{\equiv}} \mathsf{transport}(a_0, a_1, p),$$ -but it does not have an $\mathsf{ind}$-like eliminator. For suppose we had a $\mathsf{doubleInd}$. Define $B$ over $\sum_{a_0, a_1 : A} \mathsf{doubleId}(a_0, a_1)$ by -$$B(a_0, a_1, \mathsf{inl}(p)) \mathrel{{:}{\equiv}} \mathsf{unit}$$ -$$B(a_0, a_1, \mathsf{inr}(p)) \mathrel{{:}{\equiv}} \mathsf{empty}$$ -Now we have $s : \prod_{a : A} B(a, a, \mathsf{doubleRefl})$, namely the map $\lambda a . \star$ (where $\star$ is the only element of $\mathsf{unit}$). If we had $\mathsf{doubleInd}$ then we could produce an element of $B(a, a, \mathsf{inr}(\mathsf{refl}))$, and thereby inhabit $\mathsf{empty}$. -Your question can be asked more generally: does the non-dependent eliminator (in your case $\mathsf{transport}$) suffice, or do we need the dependent one (in your case $\mathsf{ind}$)? The answer is: you need the dependent one, otherwise the type is not fixed up to equivalence. It is a good exercise to try this with the circle.<|endoftext|> -TITLE: Isometric embedding as a graph -QUESTION [7 upvotes]: Question -Let $M$ be a (finite dimensional) smooth manifold and $g,\bar{g}$ be Riemannian metrics on $M$. - -Under what conditions can we guarantee that there exists another finite dimensional Riemannian manifold $(N,h)$ and a smooth map $f:M\to N$ such that $(M,\bar{g})$ is realised as the graph of $f$ in the product manifold $(M\times N, g\oplus h)$? -To put it another way, when is it possible to write $\bar{g} = g + f^*h$? - -Is there a way to bound the dimension of $N$ required? - - -Comments - -Clearly by definition $\bar{g} - g$ must be positive semidefinite for this to work. But we can equally well ask the question in the context of pseudo-Riemannian manifolds where this requirement is unnecessary. - -There is a trivial lower bound on the dimension of $N$ from the fact that the maximal rank of $f^*h$ (equivalently of $\mathrm{d}f$) is bounded above by the dimension of $N$. So if in local coordinates $\bar{g} - g$ is a rank $k$ matrix somewhere, we know that $N$ has to be at least dimension $k$. - -The global question aside, what is the correct integrability condition for the local problem? This probably just requires a suitable rephrasing of the question, but I'm having a bit of problem seeing the right geometric picture. - -The rank 1 case is not too hard (I think). Without loss of too much generality we can let $N$ be $\mathbb{R}$ with the standard metric. Using that the gradient vector field is orthogonal to the level sets, we have additionally an integrability condition (roughly speaking, let $v$ be the smooth vector field of unit eigenvectors of $\bar{g} - g$ relative to $g$ with non-zero eigenvalue $\lambda^2$, then we need the vector field $\lambda v$ to be hypersurface orthogonal (in the metric $g$); this gives necessity. For sufficiency take a hypersurface orthogonal to $\lambda v$ and set $f = 0$ on there, and integrate along $\lambda v$ to get the desired function). - -REPLY [2 votes]: As Anton points out, in the case that $q = \bar g - g$ has constant rank $k>0$, it is necessary that $K = \ker(q)\subset TM$ be integrable in order that $q = f^*h$ for some smooth map $f:M\to N$, where $h$ is a Riemannian metric on $N$. -For local solvability, these conditions plus the (necessary) condition that $\mathcal{L}_X q = 0$ for all $X$ that are sections of $K$ is sufficient. (This latter integrability condition is the infinitesimal version of the invariance with respect local sections $S$ that Anton mentions.) This is because this latter condition ensures that locally one can write $q$ as a quadratic form in the variables that are constant on the leaves of $K$. -However, these necessary conditions are not sufficient for global solvability. For example, take $M$ to be the $2$-torus $\mathbb{R}^2/\mathbb{Z}^2$ with standard metric $g = dx^2 + dy^2$, let $q = (\cos\theta\ dx + \sin\theta\ dy)^2$ for some constant $\theta$ that is not a rational multiple of $\pi$, and let $\bar g = g + q$. This satisfies all of the local conditions but there is no rank $1$ mapping $f:M\to N$ for a Riemannian manifold $(N,h)$ such that $f^*h = q$ because each nonempty fiber of such an $f$ would have to be dense in $M$. -The case in which $q\ge0$ has variable rank is subtle because it is not at all obvious how to tell when such a $q$ can be written as a sum of squares of $1$-forms (or how many it would need).<|endoftext|> -TITLE: Line-preserving bijection of ${\mathbb{R}}^n$ onto itself -QUESTION [10 upvotes]: If $f:{\mathbb{R}}^n\to{\mathbb{R}}^n$ $(n\ge2)$ is a bijection such that the image of every line is a line (continuity of $f$ not assumed), must $f$ be an affinity? -Assuming continuity would certainly suffice, even assuming that $f$ is order-preserving on each line. Is there a counterexample if we drop the assumption that $f$ is a bijection? Any references? - -REPLY [7 votes]: As Francois Ziegler pointed out, the result that you need is the so-called fundamental theorem of geometry (FTG), which states the following: if a transformation of $\mathbb R^d$ with $d\ge2$ is bijective and maps lines onto lines, then it is affine. -The main result in AMS Proc. 1999 implies that, more generally, if a transformation of $\mathbb R^d$ with $d\ge2$ is surjective and maps lines into lines, then it is affine. As shown in Remark 11 in that paper, the surjectivity condition cannot be dropped. As the example given in Lev Borisov's answer shows, the surjectivity condition cannot be dropped even if the transformation is assumed to be continuous. -As far as the rendition of the FTG on page 52 in the book by Berger Geometry (cited in Francois Ziegler's answer) is concerned, it is explained on page 2737 in the mentioned paper AMS Proc. 1999 that the proof in Berger's book contains gaps/errors. Also, as noted above, one only needs the surjectivity, rather than the bijectivity condition assumed in Berger's book.<|endoftext|> -TITLE: A concrete realization of the nontrivial 2-sphere bundle over the 5-sphere? -QUESTION [16 upvotes]: Since $\pi_4 (PU(2)) = \pi_4 (SO(3)) = {\mathbb Z}_2$, the two-element group, -we know that half of the two-sphere bundles over the 5-sphere $S^5$ are trivial -and the other half are non-trivial and all isomorphic. Can you write an explicit concrete realization for this non-trivial bundle? I have in mind something along the lines of the Hirzebruch surface -(see eg. http://en.wikipedia.org/wiki/Hirzebruch_surface) -$\Sigma_1$ (or $\Sigma_k$, $k$ odd) which realizes the unique topologically non-trivial $S^2$ bundle over $S^2$. (Since $\pi_1 (SO(3) = {\mathbb Z}_2$ again ``half' the two-sphere bundles over the 2-sphere $S^2$ are trivial -and the other half are non-trivial and all isomorphic.) - -REPLY [28 votes]: Is this concrete enough? Recall that $\mathrm{SU}(3)$ fibers over $S^5$, with fibers equal to $\mathrm{SU}(2)$ and that this fibration is nontrivial. Let $S^1\subset \mathrm{SU}(2)$ be (any) subgroup and let $B = \mathrm{SU}(3)/S^1$. Then $B$ fibers over $S^5$ with fibers $S^2$. If $B$ were trivial, then $\mathrm{SU}(3)\to S^5$ would be trivial as well, but it is not, since $S^5$ is not parallelizable. -In more detail: Regard $\mathrm{SU}(3)$ as the set of triples $(e_1,e_2,e_3)$ of special unitary bases of $\mathbb{C}^3$. Define a map $\pi:\mathrm{SU}(3)\to S^5\subset\mathbb{C}^3$ by -$$ -\pi(e_1,e_2,e_3) = e_1\ . -$$ -This is a smooth submersion with fibers isomorphic to $\mathrm{SU}(2)$. Let $B$ be the set of pairs $(v,L)$, where $v\in S^5\subset\mathbb{C}^3$ and $L\in\mathbb{CP}^2$ is a line that is Hermitian orthogonal to the line spanned by $v$, i.e., $L$ is a line in $v^\perp\simeq\mathbb{C}^2$. Then $B\to S^5$ given by $(v,L)\mapsto v$ is a smooth $S^2$ bundle over $S^5$. If $B$ were trivial, there would be a section of $B$ over $S^5$ and hence a smooth mapping $\lambda:S^5\to\mathbb{CP}^2$ such that $\bigl(v,\lambda(v)\bigr)\in B$ for all $v\in S^5$. This would define a smooth complex line bundle $\Lambda$ over $S^5$, and, since every complex line bundle over $S^5$ is trivial, there would be a nonvanishing section of this line bundle, i.e., a mapping $\sigma:S^5\to S^5$ such that $\lambda(v) = \mathbb{C}\cdot\sigma(v)$ for all $v\in S^5$. However, then there would exist a unique mapping $\tau:S^5\to S^5$ such that $\zeta(v) = \bigl(v,\sigma(v),\tau(v)\bigr)$ is a special unitary frame for all $v\in S^5$, i.e., $\zeta:S^5\to \mathrm{SU}(3)$ would be a section of the nontrivial bundle $\mathrm{SU}(3)\to S^5$.<|endoftext|> -TITLE: Partial (or complete) flag varieties as GIT quotients of affine spaces -QUESTION [11 upvotes]: I am looking for presentations of partial or complete flag varieties as GIT quotients of affine varieties spaces. That is, for a choice of of dimensions $0=d_1 -TITLE: Uniquely geodesic and CAT(0) spaces? -QUESTION [6 upvotes]: Improvement after J-M Schlenker's comment below : -This post has been divided into two parts, the second part is here. - -Question : Is a finite dimensional metric space, uniquely geodesic if and only if it is CAT(0) ? - -In the case of a negative answer : -- Is CAT(0) assumption necessary ? Is it sufficient ? -- What are the classical counter-examples ? -- Is there a slight additive assumption for having a positive answer ? - -REPLY [5 votes]: A good counterexample is the Teichmuller space of a closed oriented surface $S$. It is uniquely geodesic by Teichmuller's theorem, but it is not $CAT(0)$.<|endoftext|> -TITLE: Axiomatizations of the real exponential field -QUESTION [17 upvotes]: According to Marker's "Model Theory: An Introduction", the real exponential field has a $\forall\exists$ axiomatization (because it is model complete) but no-one has any idea what such an axiomatization might look like. Has any progress been made on this since the early 2000s when Marker wrote this? - -REPLY [7 votes]: In her PhD thesis ``On the First Order Theory of Real Exponentiation'', Tamara Servi, has given a recursive subtheory -$T$ of $T_{exp}$ and has shown that if Schanuel's conjecture holds, then $T$ is complete and hence provides a recursive axiomatization of $T_{exp}$. There are some related results. -See section 4.7 of the above mentioned paper.<|endoftext|> -TITLE: Cohesive set with degree below non-high Martin-Löf random reals -QUESTION [8 upvotes]: A set A is cohesive if $A\subseteq ^* W_e$ or $A\subseteq^* \bar{W_e}$ for each $e\in \omega$ (standard enumeration of r.e. sets). By Jockusch and Stephan's 1993 paper 'A cohesive set which is not high.', this is equivalent to saying that A is r-cohesive, i.e. restricted to those recursive sets $W_e$. My question is if given a Martin-Löf random real R whose degree is not high, is it possible to have $A\leq_T R$ with A being (not high) cohesive? - -REPLY [2 votes]: I'm fairly certain I can construct such an $A$ and $R$ using a priority construction which builds them both simultaneously. It'd use $\Pi^0_2$-guessing about whether or not intersections are infinite to build $A$, and impose restraint to make $R$ non-high. I haven't read Jockusch and Stephan's construction, but I assume this is roughly how it went. -The added wrinkle is making $R$ random. To do that, we'd fix a small but sufficiently universal $\Sigma^0_1$-class $U$, and make $R$ run away from $U$. For example, make $U$ be the 5th element of a universal Martin-Löf test. Whenever $R$ looks to be in danger of being covered by $U$, we'd choose new measure and move $R$ there. -Unfortunately, it's a fairly involved construction, and there would need to be some calculations to show that there's always space available for $R$ to run to. So I can't give full details, and I can't promise that it's correct, but I think it is. Sorry I can't give anything more definite.<|endoftext|> -TITLE: Stabilization of $\infty$-categories versus SW stabilization -QUESTION [8 upvotes]: Spanier-Whitehead stabilization provides a way to extend a category $\bf E$ to a bigger one $\mathcal{SW}_\Omega(\bf E)$ where a given endofunctor $\Omega$ is invertible. The category $\mathcal{SW}_\Omega(\bf E)$ is constructed with - -Objects the pairs $(A,n)\in Ob(\mathbf C)\times\mathbb Z$; -The set of morphisms $(A,n)\to (B,m)$ corresponds to the colimit set -$$ - \varinjlim_{k\in\mathbb N} \hom_{\bf E}(\Omega^{n+k}A, \Omega^{m+k}B) -$$ - -It's a matter of bare computations to show that it defines a category, where $\bf E$ can be embedded via $A\mapsto (A,0)$, and where an entire family of functors $\bar\Omega^i\colon (A,n)\mapsto (A,n+1)$ can be defined; the functor $\bar\Omega^1$ plays the role of the initial endofunctor $\Omega$, and that's the end of the story. -But when you meet the formalism of stable $\infty$-categories you begin to wonder if there's a link between the two processes, the SW stabilization and what Lurie describes here (Def. 8.4). I'm aware that the SW construction is an "abstraction" (?) of the procedure exhibiting topological spectra, but I must confess I'm not able to go further (especially because I'm a "category theorist" slightly oriented to topology, not vice-versa). - -REPLY [12 votes]: I think it's more typical to use this construction to invert the suspension functor, rather than the loop functor. So let me write $\Sigma$ where you wrote $\Omega$. -1) The category $SW_{\Sigma}(\mathcal{C})$ can be identified with the direct limit of the sequence -$$ \cdots \rightarrow \mathcal{C} \stackrel{\Sigma}{\rightarrow} \mathcal{C} \stackrel{\Sigma}{\rightarrow} \mathcal{C} \stackrel{\Sigma}{\rightarrow} \cdots$$ -2) The construction makes sense for $\infty$-categories as well as ordinary categories. Moreover, it commutes with passage to homotopy categories. That is, if $\mathcal{C}$ is an $\infty$-category, then the homotopy category of $SW_{\Sigma}(\mathcal{C})$ is $SW_{\Sigma}( h \mathcal{C} )$. -3) Let $\mathcal{C}$ be a small pointed $\infty$-category which admits finite colimits, and let $\Sigma: \mathcal{C} \rightarrow \mathcal{C}$ be the suspension functor. Then $Ind( SW_{\Sigma}( \mathcal{C} ) )$ is the stabilization of $Ind( \mathcal{C} )$. -You'd typically put these together by taking $\mathcal{C}$ to be something like the $\infty$-category of pointed finite spaces. Then $Ind( \mathcal{C} )$ is the $\infty$-category of pointed spaces, and its stabilization is the $\infty$-category of spectra. Your conclusion is that $SW_{\Sigma}(\mathcal{C} )$ can be identified with the $\infty$-category of finite spectra, so that -$SW_{\Sigma}( h \mathcal{C} )$ is a model for the homotopy category of finite spectra (this is the classical construction).<|endoftext|> -TITLE: What are the local properties of schemes preserved under global sections? -QUESTION [15 upvotes]: $\newcommand{\Spec}{\mathrm{Spec}\ }$ -Let $(P)$ be a property of rings. I call $(P)$ local when $(P)$ satisfy these two -conditions: - -if $A$ is a ring satisfying $(P)$, then the distinguished rings $A_f$ also -satisfy $(P)$; -If $\Spec A$ is covered by distinguished open $\Spec A_i$ with the $A_i$ having -$(P)$, then $A$ satisfy $(P)$. - -Now if $(P)$ is local, then it is natural to extend the property $(P)$ to -schemes by saying that a scheme $X$ has property $(P)$ iff for all open affines -$\Spec A$ of $X$, $A$ has property $(P)$. -By definition of locality, then - -a ring $A$ satisfy $(P)$ iff $\Spec A$ satisfy $(P)$; -a scheme $X$ satisfy $(P)$ iff $X$ can be covered by affines open $\Spec A_i$ with -$A_i$ satisfying $(P)$. - -Likewise in the relative setting, local properties of morphisms of rings -allow to define a corresponding notion for morphisms of schemes. -[I have to point out that sometimes extending a local property $(P)$ of rings -to schemes this way is called $(\mathrm{locally}\ P)$, and a scheme $X$ is said to have -property $(P)$ when $X$ is locally P and satisfy some finiteness condition. -For instance $X$ is noetherian when it is locally noetherian and -quasi-compact; a morphism is of finite presentation when it is locally of -finite presentation and quasi-compact + quasi-separated.] -Now while this is a standard construction explained in all textbooks, it is -harder to find references for what happen to the global sections of non -affine open subschemes. -Indeed, if $X$ has a local property $(P)$, then an open scheme $U$ has also -property $(P)$, but $\Spec O_X(U)$ may not have $(P)$ when $U$ is not affine. -For instance: - -$X$ is noetherian does not mean that each ring of sections $O_X(U)$ is -noetherian -(An example is given by the union of two disjoints plane in projective -space $P^3_k$, -see http://sma.epfl.ch/~ojangure/nichtnoethersch.pdf); -$X$ is finitely generated (say over a field $k$) does not mean that each -ring of sections O_X(U) is finitely generated -(Same example as above, see also http://math.stanford.edu/~vakil/files/nonfg.pdf); -$X \to \Spec A$ is flat does not mean that $O_X(X)$ is flat over $A$. -(see Global sections of flat scheme also flat?). - -However there are properties that hold for sections over any open subschemes: - -If $X$ is reduced, then for every open subscheme $U$, $O_X(U)$ is reduced; -If $X$ is integral, then for every open subscheme $U$, $O_X(U)$ is integral. - -I am interested to what happens with other properties $(P)$. I am also -interested to what happens in the relative case: if a morphism $X \to Y$ has -property $(P)$, then does $\Spec_Y(X) \to Y$ also has $(P)$? - -REPLY [7 votes]: NormalityFor an integral scheme, being normal (integrally closed in ones own fraction field) satisfies this property. Indeed, suppose that $a, b \in A = \Gamma(X, O_X)$ are such that $a/b$ satisfy some polynomial $p(x) \in A[x]$. Then $a|_U, b|_U$ satisfy the same polynomial after restriction to each (affine) set $U \subseteq X$. -Of course this shows up in many applications of things like Stein factorization. - -Semi-normality -A reduced ring $R$ is seminormal if for any finite extension $R \subseteq S$ satisfying the following two properties is an isomorphism. - -The induced map on $\text{Spec}$'s is an isomorphism -The induced residue field extensions $k(r) \subseteq k(s)$ are isomorphisms for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$. - -The typical example of a seminormal ring is a node, the cusp $k[x^2,x^3]$ is not seminormal -Equivalently, $R$ is seminormal if and only if for any $a/b$ in the total ring of fractions of $R$, one has that if $(a/b)^2, (a/b)^3 \in R$ then $(a/b) \in R$ (see a paper by Swan, he might be assuming finitely many minimal primes, I forget the details). It follows similarly that seminormality satisfies this property. - -Weak normality -Weak normality is similar to semi-normality. A reduced ring is called weakly normal if for any finite birational extension $R \subseteq S$ satisfying the following properties is an isomorphism: - -The induced map on $\text{Spec}$'s is an isomorphism -The induced residue field extensions $k(r) \subseteq k(s)$ are purely inseparable for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$. - -This can also be phrased as requiring that every birational universal homeomorphism is an isomorphism. -I do NOT know if weakly normal rings satisfy the sort of property asked for. I do not think it is in the literature (but perhaps I am wrong). I remember I convinced myself that they did not several years ago, but never wrote down an example carefully.<|endoftext|> -TITLE: Iterated Milnor fibrations and Thom's a_f condition -QUESTION [7 upvotes]: Ok so there's a lot of litterature about nearby cycles functor since it was introduced by Grothendieck and Deligne but I couldn't find any clear answer to the following natural question: -Problem: Let $X$ be a reduced complex analytic space, $f = (f_1,f_2) : X \to \mathbb{C}^2$ a couple of functions and $K \in D^b_c(A_X)$ a constructible complex. When do we have a natural isomorphism between iterated nearby cycles: -$$ - \psi_{f_1}\psi_{f_2}(K) \simeq \psi_{f_2}\psi_{f_1}(K) -$$ -This is well known as part of the hypercube description of perverse sheaves when $f = id$ and $K$ is constructible with respect to the strict normal crossings divisor defined by the coordinates hyperplanes. -In general, I don't think a natural map between the two sides even exists but one might look for something like morphisms -$$ - \psi_{f_1}\psi_{f_2}(K) \leftarrow \psi_f(K) \to \psi_{f_2}\psi_{f_1}(K). -$$ -where $\psi_f(K)$ would be some sort of global or simultaneous nearby cycles that would induce the iterated nearby cycles under suitable hypothesis. -Actually I'm more interested in the algebraic case where $K$ is regular holonomic D-module but this kind of problems seems to have been studied a lot more by topologists in the spirit of Thom's isotopy lemmas to I'm trying to understand the Milnor fibration viewpoint. Please correct me if I'm wrong. -Dimenson 1 base: In the case $(X,x)$ is a germ of analytic space inside $U \subset \mathbb{C}^N$ and $g: X \to \mathbb{C}$ a single analytic function, the Milnor-Le fibration theorem states that for $0<\eta \ll \varepsilon \ll 1$ -$$ - g: \bar{B}(x,\varepsilon) \cap g^{-1}(D^*(f(x),\eta)) \to D^*(g(x),\eta) -$$ -is a locally trivial fibration over the punctured disc $D^*(g(x),\eta)$. The fiber $F_{g,x} = \bar{B}(x,\varepsilon) \cap g^{-1}(\eta)$ is the local Milnor fiber of $g$ at $x$. -Almost by definition, we have $\psi_g(K)_x = R\Gamma(F_{g,x},K)$. -First question: is the fibration independant of the local embedding $X \subset \mathbb C^N$. This seems to be well known but I've never seen an actual proof. It seems to me it could be proved quite easily if one can replace the ball $\bar{B}(x,\varepsilon)$ by a polydisk as in Le's "La monodromie n'a pas de point fixe" but I haven't written it down yet. -Dimenson > 1 base: -Consider $f = (f_1,f_2): X \to \mathbb{C}^2$. -In this case, Milnor's fibration theorem fails in general (classical exemples include simple blow-ups or Whitney's umbrella). -But, by iterating the usual one function construction for $f_i: X \to \mathbb{C}$, one can still define a Milnor fibration $X_{(f_1;f_2),x}(\varepsilon,\eta) \to S_{\eta_1}^1 \times S_{\eta_2}^1$ independant of $0 < \eta_1 \ll \eta_2 \ll \varepsilon$ with fiber $F_{(f_1,f_2),x}$. This is done for example in McCrory and Parusinky's "Complex monodromy and the topology of real algebraic sets". We have -$$ - \psi_{f_1}\psi_{f_2}(K)_x = R\Gamma(F_{(f_1,f_2),x};K) -$$ -But this fibration depends of the ordering we chose: $F_{(f_1;f_2),x} \neq F_{(f_2;f_1),x}$. -I expect the problem to disappear with Thom's $a_f$ condition. More precisely, in "Morphismes sans éclatement et cycles évanescent" Sabbah defines a morphism $f:X \to Y$ between stratified analytic spaces as being "sans éclatement" ("without blowup") if - -the stratification on $Y$ satisfies Whitney's conditions, -for each strata $Y_\beta$ subset $Y$, the stratification on $X$ induces a Whitney stratification on $f^{-1}(Y_\beta)$. -Thom's $a_f$ condition is satisfied. - -Let's stratify $\mathbb{C}^2$ by the coordinates hyperplanes and suppose there is a stratification $S$ of $X$ so that $K\in D^b(A_X)$ is $S$-constructible and $f:X\to \mathbb{C}^2$ is without blow-up. Then we have a locally trivial topological fibration $f: B(x,\varepsilon) \cap f^{-1}((\mathbb{C}^*)^2) \to (\mathbb{C}^*)^2$ with stratified fiber $F_{f,x}$. -Question: Am I right in thinking that the above fibration induces the iterated Milnor fibrations so that -$$ - \psi_{f_1}\psi_{f_2}(K)_x = R\Gamma(F_{f,x}, K) = \psi_{f_2}\psi_{f_1}(K)_x -$$ -Thanks - -REPLY [6 votes]: Why is the Milnor fibration (for dimension 1 base) independent of the embedding? Because the embedding just determines what you are using for balls, and Lê's old proof shows that the fiber-homotopy-type of the fibration doesn't depend on which balls you're using. If $B_1$ and $B_2$ are balls in one embedding and $P_1$ and $P_2$ are pull-backs of balls from the other, such that $B_2\subseteq P_2\subseteq B_1\subseteq P_1$, then the equivalence of the fibrations using $B_1$ and $B_2$, and using $P_1$ and $P_2$, shows that the fibrations using $P_2$ and $B_1$ are equivalent. -With a base of $\mathbb C^2$ or larger, $\mathbb C^d$, the problem is worse than just needing $a_f$ to control blowing up. In general, what one hopes for is that the discriminant of $f$ is a hypersurface in $\mathbb C^d$ and that, off of this hypersurface, the fibers are the same. This is where a_f gets used. But the discriminant (or Cerf diagram) is already an obstruction to a fibration over $(\mathbb C^*)^2$. -Under some manageable conditions, the stalks of $\psi_{f_1}\psi_{f_2}K$ and $\psi_{f_2}\psi_{f_1}K$ are isomorphic, but the complexes are not naturally isomorphic; the Milnor monodromy of $f_1$ would typically act very differently on these complexes. For instance, if our domain is affine space, $f_1$ has a critical point at the origin, and $f_2$ is a generic linear form, and the relative polar curve is not empty, then the $f_1$-monodromy is very different on the two complexes.<|endoftext|> -TITLE: hyperelliptic involution on a surface -QUESTION [5 upvotes]: What is the Dehn twist factorization of the hyperelliptic involution on an oriented surface of genus g (with one boundary component)? - -REPLY [8 votes]: This is almost covered by Proposition 4.12 of Farb and Margalit's book "The primer on mapping class groups". Another useful reference is the paper "Presentations for the punctured mapping class groups in terms of Artin groups" by Labruere and Paris. See the second displayed formula in Proposition 2.12 of that paper. -Here is an outline of the solution. Let $S$ be a surface of genus $g$ with one boundary component. Let $a_i$ be a "chain" of $2g$ curves in $S$. That is, $a_i$ and $a_{i+1}$ meet in exactly one point and otherwise the $a_i$ are disjoint from each other. This is not done cyclically - that is, $a_1$ and $a_{2g}$ are disjoint. Note that the boundary of a regular neighborhood of the union of the $a_i$ is isotopic to $\partial S$. Let's also use the symbol $a_i$ to denote a Dehn twist about the curve $a_i$. Consider the following product of Dehn twists: -$$H = ( a_1 a_2 \ldots a_{2g - 1} a_{2g} )^{2g + 1}$$ -Claim: $H$ is the desired mapping class. Proof: $H$ fixes the curve $a_i$ (up to isotopy) but reverses the orientation of $a_i$. Since the chain $\cup a_i$ fills the surface, we are done. QED -Proposition 4.12 of the primer and Proposition 2.12 of [LP] says that $H^2$ is isotopic to a full Dehn twist about $\partial S$. Since you allow isotopies that move boundary points, for you $H^2$ is isotopic to the identity. -All of these claims can be proved by drawing a few pictures, applying induction, and using the "Alexander method". Another approach (that is less general) is to consider $S$ as a double branched cover of the $2g + 1$ times marked disk, and then to think about the generator of the center of the braid group. -One final remark - there are many ways to choose the chain of curves $a_i$ and this might seem like a problem... however, any two ways of choosing the chain differ by a mapping class: thus the apparent choice is no choice at all. I believe that Farb and Margalit call this the "change of coordinates principle".<|endoftext|> -TITLE: Simplicity of infinite groups -QUESTION [7 upvotes]: Sorry about asking so many questions, but I am a bit further on in my classification, and I am up to the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^4b)^7 \rangle$. It has no small quotients (up to 500000), and I suspect that it is simple. Is there a way to check for simplicity in infinite groups? - -UPDATE (edit by YC) -The original question about the group $G$ has been answered below. Now I am up to the two groups $H := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^4b)^8 \rangle$ and -$I := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^4b)^9 \rangle$. I need to know if they are trivial, finite (the order would be good), or infinite. - -REPLY [18 votes]: In fact your group is trivial. Here are two different computations with Magma, the first using coset enumeration over the subgroup $\langle ab \rangle$, and the second using the Knuth-Bendix completion algorithm. - > G:=Group; - > H:=sub; - > time Index(G,H:CosetLimit:=100000000,Hard:=true); - 1 - Time: 27.730 - - > time R := RWSGroup(G:MaxRelations:=100000, TidyInt:=1000); - Time: 90.350 - > Order(R); - 1<|endoftext|> -TITLE: One-parameter subgroups of symplectic group associated to roots -QUESTION [6 upvotes]: I'm having trouble sorting out some basic definitions concerning Chevalley groups. The groups I'm interested in are the simply connected groups of type $C_n$, so the groups $\text{Sp}_{2n}$. The roots in $C_n$ are $\{\pm 2 \epsilon_i \text{ $|$ } 1 \leq i \leq n\} \cup \{\pm \epsilon_i \pm \epsilon_j \text{ $|$ } 1 \leq i < j \leq n\}$. Associated to every root in $C_n$ is a one-parameter subgroup of $\text{Sp}_{2n}$. Can someone tell me the actual matrices generating these one-parameter subgroups? Or give me a reference discussing this in a simple and concrete way? All the sources I've consulted only do things concretely for $\text{SL}_n$. - -REPLY [4 votes]: The actual matrices depend of course on the particular alternating bilinear form you use to define $\mathrm{Sp}_{2n}$. A standard choice is to use the form whose matrix relative to a basis $(e_1,\dots,e_n,e_{-n},\dots,e_{-1})$ is -$$ -J= -\begin{pmatrix} -& & & & & 1\\ -& & & & \cdots\\ -& & & 1\\ -& & -1\\ -& \cdots\\ --1 -\end{pmatrix}. -$$ -This has the advantage that 1º) the diagonal matrices in $\mathfrak g$ make a Cartan subalgebra, with basis the matrices $H_i=E_{i,i}-E_{-i,-i}$ where the $E_{i,j}$ are the standard matrix units, $(E_{i,j})_{kl}=\delta_{ik}\delta_{jl}$. And 2º) the infinitesimal generators of the root subgroups that you ask for are listed in Bourbaki's tables: they are -$$ -\begin{align} -X_{2\epsilon_i} &= E_{i,-i}\\ -X_{-2\epsilon_i} &= -E_{-i,i}\\ -X_{\epsilon_i-\epsilon_j} &= E_{i,j} - E_{-j,-i}\\ -X_{-\epsilon_i+\epsilon_j} &= -E_{j,i} + E_{-i,-j}\\ -X_{\epsilon_i+\epsilon_j} &= E_{i,-j} + E_{j,-i}\\ -X_{-\epsilon_i-\epsilon_j} &= -E_{-i,j} - E_{-j,i}. -\end{align} -$$<|endoftext|> -TITLE: Cover of a n-simplex with balls -QUESTION [5 upvotes]: Consider a n-simplex. For each edge (i,j), consider a n-ball, such that vertices i and j are antipodal on this ball. Is the simplex covered by the union of these balls? Thank you. - -REPLY [6 votes]: Yes. -If $P$ is our point, then it will be contained in the ball corresponding to edge $(i,j)$ if and only if the angle $\angle iPj$ is greater than or equal to $\frac{\pi}{2}$. If there is no such edge $(i,j)$, then every vertex $j$ is on a fixed side of the hyperplane through $P$ orthogonal to the line connecting vertex $1$ with $P$, so $P$ is outside the simplex.<|endoftext|> -TITLE: Configuration spaces of the torus -QUESTION [8 upvotes]: I would like a reference that calculates the rational homology of the unordered configuration spaces of the torus. - -REPLY [7 votes]: The calculation for even-dimensional manifolds, and in particular the torus, is carried out by Felix-Thomas in their paper "Rational Betti numbers of configuration spaces."<|endoftext|> -TITLE: Has Ribet's theorem been proved using only finite powers of primes? -QUESTION [7 upvotes]: Ribet proved the Serre epsilon conjecture using $p$-adic Galois representations (http://math.berkeley.edu/~ribet/Articles/invent_100.pdf). Can someone show how to replace all use of $p$-adics in this (or some other) proof of the theorem by arithmetic modulo some finite number of specified powers? -A word on the motivation: To actually find a proof of Fermat's Last Theorem in Peano Arithmetic would require eliminating $p$-adics from the proof, along with absolute Galois groups, and cohomological tools (all of which are un-interpretable in Peano Arithmetic), I expect that will be done someday. In fact I expect the absolute Galois groups can be easily removed in favor of Galois groups of specified number fields, given the particular way they serve in the existing proofs. This question looks toward removing $p$-adics from the proof of Ribet's theorem, which may not be easy at all. -Merely to show there is a PA proof of FLT, though, does not require eliminating any of these. The promising strategy for that today is to use $\mathsf{ACA}_0$ as in François's comment. In effect, you use numbers and sets of numbers but all sets of numbers must be defined by referring only to numbers and not also to sets of them. Large parts of Ribet's proof are already like that. It remains to check the whole in detail. While there is a good routine for turning $\mathsf{ACA}_)$ proofs of arithmetic statements into PA proofs, it seems unlikely to me that this route could lead to a humanly comprehensible PA proof. -Currently it seems that getting the Modularity Thesis into PA will be a substantially larger project. - -REPLY [6 votes]: This is a partial answer regarding the use of the use of the absolute Galois group of the rationals. In our paper Reverse Mathematics and Algebraic Field Extensions, Jeff Hirst, Paul Shafer and I analyze what is needed to have Galois theory of infinite extensions work as expected in the context of subsystems of second-order arithmetic. At no point do we need more than ACA0 to make things work as expected. (And when the ground field is $\mathbb{Q}$ we can often get by with a lot less.) Since ACA0 is conservative over PA, the use of the absolute Galois group of $\mathbb{Q}$ can always be eliminated in a proof of a purely first-order statement, so long as nothing is used that doesn't go through in ACA0. Since we haven't looked at the entire body of results in infinite Galois theory, we can't guarantee that the use of Galois theory that Ribet does can be eliminated. However, we would be very happy to look at any snag that you bump into while verifying this.<|endoftext|> -TITLE: Lower bound for Euler's totient for almost all integers -QUESTION [7 upvotes]: Let $\varphi(n)$ be the Euler's totient function. It is well know that $\liminf_{n \to \infty} \frac{\varphi(n)}{n / \log \log n} = e^{-\gamma}$, so that for $\varepsilon > 0$ it results $\frac{\varphi(n)}{n} \geq \frac{e^{-\gamma}-\varepsilon}{\log \log n}$ for large $n$. Actually, the "local minima" of $\frac{\varphi(n)}{n}$ are attached for $n = p_1 \cdots p_k$ (the product of the first $k$ primes) and the set of primorial is really sparse. I wonder if it is known a lower bound for $\varphi(n)$ like: "$\varphi(n) / n \geq f(n)$ for all $n$ but a set of null asymptotic density", where $f(n)$ is a function bigger then $\frac{e^{-\gamma}-\varepsilon}{\log \log n}$. -Thank you in advance for your help. - -REPLY [3 votes]: Your question has been answered by Lucia already, but you might also be interested in looking up the Erdős-Wintner theorem. A special case (proved already by Schoenberg) is that for each $u \geq 0$, the set of $n$ with $\phi(n)/n \leq u$ has an asymptotic density $D(u)$; moreover, $D(u)$ is continuous and increasing on $[0,1]$. -There are also estimates available for the size of $D(u)$ when $u$ is near zero, and of $1-D(u)$ when $u$ is near $1$. For this, see Erdős's paper "Some remarks about additive and multiplicative functions": -http://www.renyi.hu/~p_erdos/1946-11.pdf<|endoftext|> -TITLE: $\Sigma^0_1\wedge\Pi^0_1$-Determinacy holds in the second admissible above the game -QUESTION [9 upvotes]: Let $T$ be a game tree and $T\in N\in M$, where $N,M$ are the two least admissibles containing $T$. Let $A$ be a boolean combination of two lightface open sets in $[T]$, or alternatively, a boolean combination of two open sets that happens to be a $\Delta_1$-definable class of $N$ and $M$. (There may be further work needed to say "the right thing" here: maybe the same $\Delta_1$ relation specifies different classes over $N$ and $M$, I'm not sure if this will affect things.) -Then apparently $M\vDash G(A;T)$ is determined, and furthermore this is minimal, i.e. in general $N$ will not see that $A$ is determined. -Since I heard this as an aside from my supervisor (who's currently away) I was wondering if there is a reference out there for this. I am aware that $\Sigma^0_1\wedge\Pi^0_1$-Determinacy is proved all the way back by Gale and Stewart from ZFC, so I'm specifically interested in the version with the weaker hypothesis. - -REPLY [9 votes]: (As the absentee supervisor am I allowed to pitch in?) To my mind this is least confusing if we take $A$ to be $B\cap C$ the intersection of a lightface open with a lightface closed set in the full Baire space. Here then the `game tree' for our purposes can be defined to be the set of finite partial plays in $\omega^{<\omega}$ up to some finite round. (Martin for example defines game trees which may have finite maximal branches, but let us ignore that here, and we'll only allow trees where every node can be extended.) Then the tree $T = \omega^{<\omega }\in L_{\omega_1^{ck}}$. (Similar considerations hold for trees which are subspaces of other spaces $X^\omega$ for countable $X$, and 'double-admissible' sets containing $X$ as an element, which we shall ignore.) -(If the two sets are not lightface but are say open or closed in some real parameter $z$, then the tree may be in $L_{\omega_1^{ck}}$, but as Andreas points out the game (that is the payoff set $A$) may only be defined in $L_{\omega_1^{z ck}}[z]$. But let us continue to assume the param. $z=\emptyset$.) -Now a winning strategy for the (lightface) open game $B$ on this `game-tree' is definable over (but is not necessarily a member of) $L_{\omega_1^{ck}}$ (and similarly for the closed game with payoff $C$). -For $s \in \omega^{<\omega}$ let $N_s$ be the open neighbourhood of $s$: the set of all those $\omega$-sequences extending $s$. Suppose player I has no w.s for the game playing into $A$. -Let $z$ be : $$\{ s \mid s \mbox{ is a finite even length seq. with } N_s\subseteq B\wedge -\mbox{ I has a w.s in the game with starting position } s \mbox{ and playing into } C -\}$$ -This is a $\Sigma^1_1$ set and thus $z\in L_{\omega_2^{ck}}$ (but not necessarily $ L_{\omega_1^{ck}}$). The set, $D$, of $\omega$-sequences extending elements of $z$ is thus an open set in the param. $z$. By the usual argument for $\Sigma^0_1(z)$-Det. either I or II has a w.s to play into, or out of, $D$; moreover this strategy can be taken to be either in (in the case of I winning), or $\Pi_1$-definable over (in case of II), the next admissible set containing $z$: here $L_{\omega_2^{ck}}$. -Now argue that I cannot have the w.strat. in this latter game (since we assumed I had no w.s. for $G_A$). Hence II has the w.s. here. Call it $\tau$. -Let $$y = \{ s \mid s \mbox{ is a finite even length seq. with } N_s\subseteq B\wedge - s \notin z \} .$$ -For each $s\in y$ let $\tau(y)$ be the $L$-least w.s for II witnessing that $s\notin z$ (using $\Sigma^0_1$-Det.) We may assume the function $ s \rightarrow \tau(s)$ with domain $y$ is $\Sigma_1(L_{\omega_2^{ck}})$ (an essential use of $\Sigma_1$-Rep. here.) Now define a strategy $\tau^*$ by setting $\tau^*(s) =\tau(s)$ if $N_s\not\subseteq B$ and $\tau^*(s\smallfrown r) = \tau(s)(r)$ where $N_s \subseteq B$ (and $s$ is the shortest init. set. of $s \smallfrown r$ for which the latter is true). Now argue that $\tau^*$ is a w.s. for II in $G_A$. -This shows that Det$(\Sigma^0_1 \cap \Pi^0_1)$ holds definably over (rather than strictly within) the second admissible set over the tree.) -Lastly we point out that strategies for such games cannot all be elements of $L_{\omega_2^{ck}}$ - or in other words of $HYP(\mathcal{O})$: the complete $\Game(\Sigma^0_1 \cap \Pi^0_1)$ set of integers will be $\Pi_1(L_{\omega_2^{ck}})$.<|endoftext|> -TITLE: Feynman-Kac for jump-diffusion -QUESTION [9 upvotes]: I'm looking for a more general Feynman-Kac formula that works in the case of jump-diffusion processes. -I know that, given a pure diffusion process like -$$dS_t=\mu_tdt+\sigma_tdW_t,$$ if $u(t,s)$ satisfies the PDE -$$f_t(t,s)+\mu_tf_s(t,s)+\frac{\sigma_t^2}{2}f_{ss}(t,s)-V(s)f(t,s)=0$$ with terminal condition $f(T,s)=H(S_T)$ -then $u$ is of the form $$u(t,s)=\mathbb{E}\big[e^{-\int_t^T{V(S_x)dx}}H(S_T)|S_t=s\big].$$ -Is it possible to extend such a result when the process dynamics are given by -$$dS_t=\mu_tdt+\sigma_tdW_t+\gamma_tdN_t$$ with $N$ a Poisson process independent fom $W$? - -REPLY [7 votes]: Hi it is possible to get some Feynman-Kac formula in this case. The proof only use the martingale property and Itô's formula for jump-diffusion processes. -So let's have $X$ s.t. (I took the compensated version of your sde): -$dX_t=[\mu(t,X_t)+\lambda(t)\gamma(t,X_t)]dt + \sigma(t,X_t)dW_t+ \gamma(t,X_{t-})d\tilde{N}_t$ where $\tilde{N}_t$ is a compensated Poisson process of intensity $\lambda(t)$. -Please notice the explicit dependence in $t$ and $X_t$ of the above equation that is necessary to have Markov property for the solution which is necessary for the Feynman-Kac theorem to apply. -Now let's us be given $F(t,X_t)=e^{-\int_s^t V(X_r)dr}u(t,X_t)=e^{-IV(s,t)}.u(t,X_t)$ and apply Itô to this formula. You get : -$$dY_t=dF(t,X_t)=e^{-IV(s,t)}\Big(\big(\partial_t u(t,X_t)+\lambda(t)[u(t,X_t+\gamma(t,X_t))-u(t,X_t)]+\mu(t,X_t)\partial_x u(t,X_t)+\frac{\sigma^2(t,X_t)\partial_{xx}u(t,X_t)}{2}-V(t,X_t).u(t,X_t)\big)dt+ (u(t,X_t+\gamma(t,X_t))-u(t,X_t))d\tilde{N}_t+(\sigma(t,X_t)\partial_{x}u(t,X_t))dW_t\Big)$$ -Now if the $dt$ term is null then $Y_t$ is martingale and for $t=T$ : -$$Y_s=F(s,X_s=x)=E[Y_T|X_s=x]=E[e^{-\int_s^T V(X_r)dr}H(X_T)|X_s=x]$$ -So in this case the PIDE that solves the Feynman-Kac formula is : -$$\partial_t u(t,X_t)+\mu(t,x)\partial_x u(t,x)+\frac{\sigma^2(t,x)}{2}\partial_{xx}u(t,x)+\lambda(t)[u(t,x+\gamma(t,x))-u(t,x)]=V(t,x)u(t,x) $$ -With final condition $u(T,x)=H(x)$ -Best regards<|endoftext|> -TITLE: Shortest geodesic loop vs. shortest periodic geodesic -QUESTION [6 upvotes]: Are there simple conditions on a Riemannian metric on the two-sphere that imply that a geodesic loop of minimal length is actually a periodic geodesic? -For example, is this true for small perturbations of the round metric, for metrics with suitably pinched curvature, ... ? -Addendum (16/09/2013). When I asked this question I thought it might have an easy or well-known answer, but apparently that is not so. It may be helpful then if I provide some context. It will also be helpful for me since I'm now writing these things up in a paper with Balacheff. -An open question in Riemannian geometry that I learned from Florent Balacheff asks whether every Riemannian metric on the two-sphere that is sufficiently close to the round metric and whose area is $4\pi$ carries a closed geodesic of length at most $2 \pi$. What really caught my imagination is that this conjectured inequality is really local: there are metrics far from the round metric for which it is not true. However, it was a bit disappointing that the round metric or Zoll metrics did not come out to be the extremal metrics for this problem and I started to look for a similar inequality where it would be at least reasonable to conjecture that the extremal metrics are Zoll. I found the following variation, which I call the "figure 8 conjecture": -Figure 8 conjecture. A Riemannian metric on the two-sphere with area equal to $4\pi$ carries a closed geodesic that is regular homotopic to a figure 8 and whose length is at most $4\pi$. Moreover, if equality is attained at a (smooth) metric, the metric is Zoll. -Note that this result would imply that a Riemannian metric on the two-sphere with area equal to $4\pi$ carries a geodesic loop whose length is at most $2\pi$. The OP inquires when this implies that there is a periodic geodesic of length at most $2\pi$. -The reason the figure 8 conjecture may just be true is that it follows from a natural version of the Viterbo conjecture: -Let $\alpha$ be a contact form on the three-dimensional sphere $S^3$. If ${\rm Ker} \alpha$ is the standard (tight) contact structure on $S^3$, then the volume of $(S^3,\alpha)$, defined by -$$ -{\rm vol}(S^3,\alpha) = \int_{S^3} \alpha \wedge d\alpha , -$$ -is no less than the square of the smallest period of a closed Reeb orbit (i.e., a periodic integral curve of the vector field defined by $\alpha(X) = 1$, $d\alpha(X,\cdot) = 0$). -Michael Hutchings also arrived at this conjecture from his work on contact homology (see his blog) and actually at some point I thought I had found a counterexample and he kindly re-set me on the right track. I believe the only partial answers that are known are as follows (and this is joint work with Florent Balacheff): -1. If among all the smooth tight contact forms on the three sphere there is one for which the ratio (volume/square of the smallest period of a periodic Reeb orbit) is minimal, then the conjecture is true. In other words, the existence of a smooth minimizer for this ratio settles the conjecture. -2. If $\alpha_t$ is a smooth constant-volume deformation of the standard contact form on $S^3$ that is not formally trivial, then for each sufficiently small $t$ different from zero $(S^3, \alpha_t)$ carries a periodic Reeb orbit with period less than $\pi$. -A defomation $\alpha_t$ is formally trivial if for every positive integer $k$ there exists an isotopy $\phi_t$ (depending on $k$) such that $\alpha_t$ and $\phi_t^* \alpha$ agree to order $k$ at $t = 0$. -3. Let $K \subset \mathbb{R}^4$ be a convex body and let us identify $\mathbb{R}^4$ with the space of quaternions. If $K$ is invariant under left multiplication by the quaternions $i$, $j$, and $k$, then the boundary of $K$ carries a closed Reeb orbits whose period is less than the square root of its (contact) volume. -This last result follows from Sergei Ivanov's Finsler generalization of Pu's isosystolic inequality (a truly neat result, which has many beautiful corollaries -including Mahler's inequality for the volume product of symmetric convex bodies on the plane). - -REPLY [3 votes]: I am not sure this could be an answer or not, but it is too long to be a comment... -It has been known that if a loop $L$ realizes the injectivity radius at some point $p$, then it must be smooth at the antipodal point $q$ of $p$, i.e., the nearest cut point of $p$. Otherwise we can deform the corner to shorten the loop (cf. Cheeger-Ebin's book.) The only possible non-smooth corner of $L$ is $p$ itself. However, if $p$ is a point where the injectivity radius achieves it minimum, that is $inj(p)=\min_{M} inj(x)$, then we can take $q$ as a base point and consider all loops passing through $q$. Among these loop, $L$ must be the shortest one (because it realizes the minimal injectivity radius.) Hence it must be smooth at the antipodal point of $q$, that is, $p$. So the conclusion is: the loop which realizes the minimum of injectivity radius must be smooth. (We don't need to deal with conjugate points because the minimal loop collapses to a segment there.)<|endoftext|> -TITLE: Which limits commute with filtered colimits in the category of sets? -QUESTION [17 upvotes]: The question is in the title, but here is some background: -I previously asked for a general criterion to decide which colimits commute with which limits in the category of sets and received encouraging answers: (1) the question is already answered in some form by a paper of Foltz, (2) MO user Marie Bjerrum will soon provide what promise to be simpler to check criteria than Foltz's. -Now I'm curious about a special case of that question: for which categories $J$ do limits of shape $J$ commute with all filtered colimits? Certainly it is well known that finite $J$ satisfy this and it is also true that only filtered colimits commute with all finite limits [1], so one might initially guess that only finite limits commute with all filtered colimits. But this is wrong! For example, if $M$ is a finitely generated monoid (for example $\mathbb{N}$), limits of shape $BM$ commute with all filtered colimits (by $BM$ I just mean $M$ regarded as a one object category) [2]. A very similar argument actually shows that limits of shape $J$ commute with all filtered colimits as long as $J$ is "finitely generated", i.e., there is a finite set $F$ of morphisms such that every morphism in $J$ is a composition of some sequence of morphisms in $F$ (equivalently, there is a finite digraph and a functor from the free category on the digraph to $J$ which is surjective on morphisms). Are those all the $J$? EDIT: Answered: no! Since these work, any category with a finitely generated initial subcategory also works, and as Marc Hoyois said in his answer below, Paré proved that is all. -That is the end of the question but for the curious here are proofs of [1] and [2]: -[1] Assume colimits of shape $I$ commute with all finite limits in the category of sets. Let $F : J \to I$ be a finite diagram in $I$, we shall show that there is a cocone over it. Consider the functor $G : J^{\mathrm{op}} \times I \to \mathrm{Set}$ given by $G(j,i) = \hom_I(F(j),i)$. For any $i \in I$, $\lim_{j \in J^{\mathrm{op}}} G(j,i)$ is the set of cocones over $F$ with vertex $i$. If there were no cocones at all over $F$, $\mathrm{colim}_{i \in I} \lim_{j\in J^{\mathrm{op}}} G(j,i)$ would be empty. On the other hand $\mathrm{colim}_{i \in I} G(j,i) = \{\ast\}$ for any $j$ (because every element $f$ of $\hom_I(f(j),i)$ gets glued onto $id_{f(j)}$ by the map $G(j,f)$), so $\lim_{j\in J^{\mathrm{op}}}\mathrm{colim}_{i \in I} G(j,i) = \{\ast\}$. -Bonus mini-question: Is there a reference for the above result? I think it might be well known unwritten folklore. -[2] Let $F : I \to M\text{-Set}$ be a filtered diagram of sets with an action of the finitely generated monoid $M$. We want to show that $\mathrm{colim} _{i\in I}\;\mathrm{Fix}(F(i)) = \mathrm{Fix}(\mathrm{colim} _{i\in I}\; F(i))$ where $\mathrm{Fix}(X) = \{x \in X : x=mx$ for all $m\in M\}$. The canonical map from the LHS to the RHS is clearly injective (the notion of equality for fixed points in the colimit on the left is just that they are equal in the colimit of the underlying sets of the monoids, $F(i)$). If $M$ is generated by the finite set $m_1, \ldots, m_r$ we can see the canonical map is also surjective: given an element of the RHS represented by $x \in F(i)$, we know that for each $k=1,\ldots, r$, it is the case that $F(\alpha_k)(x) = F(\alpha_k)(m_k x)$ for some $j_k$ and some morphism $\alpha_k :i\to j_k$. Since $I$ is filtered we can equalize all the $\alpha_k$, and so for a single $j$ and a single $\alpha : i \to j$ we have that $F(\alpha)(x) = F(\alpha)(m_k x) = m_k F(\alpha)(x)$, so that $x$ comes from $F(\alpha)(x)$ in the LHS. - -REPLY [19 votes]: Categories $J$ such that limits of shape $J$ commute with filtered colimits in sets are called L-finite. There are several known characterization of them: see the nLab page about it. The page refers to Robert Paré, Simply connected limits (pdf). In particular, Proposition 7 says: -A category is L-finite iff it has an initial subcategory which is finitely generated. -There's always some ambiguity in the term "finite limit" because you can consider the source of a "diagram" to be either a graph or a category. This matters not for the notion of limit because of the free/forgetful adjunction between graphs and categories, but the free category on a finite graph may not be finite; it is however L-finite, as you'd expect. This is a situation where it seems more natural to view diagrams as graph morphisms rather than functors.<|endoftext|> -TITLE: Numbers that are generic w.r.t. exponentiation -QUESTION [42 upvotes]: This is a follow-up to my old question Number of distinct values taken by $x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x$ with parentheses inserted in all possible ways. -In the following let us assume $n$ to be a positive integer, and all other variables to be positive reals. Let $a\hat{\phantom{\hat{}}}b$ denote exponentiation $a^b$. -The number of distinct $\mathbb{R}^+\to\mathbb{R}^+$ functions obtained from the expression -$$\underbrace{x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x}_{n\text{ occurences of }x}\tag1$$ -by inserting parentheses in all possible ways depends on $n$ and is given by the OEIS sequence A000081. Note that different parenthesizations can result in the same function, e.g. -$$(x\hat{\phantom{\hat{}}}x)\hat{\phantom{\hat{}}}(x\hat{\phantom{\hat{}}}x)=(x\hat{\phantom{\hat{}}}(x\hat{\phantom{\hat{}}}x))\hat{\phantom{\hat{}}}x.\tag2$$ -If instead of considering functions, we fix some value of $x$, and ask about the number of distinct numeric outcomes of the expression $(1)$ for all possible parenthesizations, then, depending on the value of $x$ we fixed, the result can be either A000081 (in this case we call the value of $x$ generic), or a slower growing sequence. -For example, the number $2$ is not generic, because the corresponding sequence is A002845 due to some identities specific to the number $2$, e.g. -$$2\hat{\phantom{\hat{}}}(2\hat{\phantom{\hat{}}}2)=(2\hat{\phantom{\hat{}}}2)\hat{\phantom{\hat{}}}2.\tag3$$ -Actually, it is not difficult to see that no positive integer is generic. Likewise, $\sqrt2$ is not generic. Furthermore, it can be proved that no positive algebraic number is generic. -Questions: - - -Can we prove that $2^{\sqrt2}$ is generic? -Can we find an explicit$^*$ computable generic number? - - -One might think that a plausible candidate for a generic number could be $\pi$, but, unfortunately, we do not even know yet if $\pi^{\pi^{\pi^\pi}}$ is an integer. - -$^*$ By explicit I mean something that can be constructed from algebraic numbers and known $^{**}$ constants, elementary and known special functions, or an isolated root of an equation constructed from those constants and functions. -$^{**}$ known means they appeared in published books or reviewed papers. - -References: - -R. K. Guy and J. L. Selfridge, The nesting and roosting habits of the laddered parenthesis. Amer. Math. Monthly 80 (1973), 868-876. ᵖᵈᶠ -F. Göbel and R. P. Nederpelt, The number of numerical outcomes of iterated powers, Amer. Math. Monthly, 80 (1971), 1097-1103. ᵖᵈᶠ - -REPLY [13 votes]: Two observations: - -Not every transcendental is generic. Indeed the real solution $r$ of the equation $x^x=2$ is not generic because $r$ satisfies -$$x^{x^{x^x}}=\left( x^x\right)^x,$$ -and $r$ is transcendental by the Gelfond-Schneider theorem. -Here is an argument showing that no positive real algebraic is generic (by which I mean that every positive real algebraic satisfies an equation between two exponential towers that have the same number of $x$'s and that do not define the same functions on $\mathbb{R}^+$.) This is mentioned in the original problem but maybe it would be nice to put up a proof here. It seems quite significant to me that 'generic' is a strengthening of 'transcendental'. - -Rather than give a formal proof of the claim, I'll present an example that should make the general argument clear. Let $r$ be a solution of the polynomial equation $x^2-3x-2=0$. To prove that $r$ is not generic, first write the equation as -$$xx=x+x+x+1+1.$$ -(In general, construct an equation made from additions and multiplications, all of whose coefficients are 1.) -Equating $x$ raised to the left-hand-side and $x$ raised to the right-hand-side, we obtain the following non-identity (also satisfied by $r$) -$$(x^x)^x=(x^x)(x^x)(x^x)xx.$$ -Doing the same thing with the last equation, we get -$$x^{(x^x)^x}=\left(\left(\left(\left(x^{(x^x)}\right)^{(x^x)}\right)^{(x^x)}\right)^x\right)^x.$$ -Now call the left and right hand sides of the last equation $a$ and $b$. Then $r$ satisfies the equation $a^b=b^a$, and both $a^b$ and $b^a$ have the same number of $x$'s. Furthermore $a^b$ and $b^a$ cannot define the same function. This follows from the calculus problem to the effect that if $u$ and $v$ are greater than $e$ (the base of the natural logarithm) and if $u^v=v^u$, then $u=v$.<|endoftext|> -TITLE: Algorithm for finding inverse images of a local diffeomorphism -QUESTION [5 upvotes]: Let $F : R^k \to R^k$ be a smooth map whose Jacobian -$J(F): R^k \to R$ vanishes on a discrete set $S$, so that if -$O$ is the complement of $S$, then $f: O \to R^k$, the -restriction of $F$, is a local diffeomorphism, and in particular -there are only finitely many points $x$ inside the unit ball of -$R^k$ such that $f(x) = 0$. What I would like to know is if -there any a good algorithm for finding such points $x$. -Of course, if $k=1$ this is easy, using the Intermediate Value -Theorem; just partition the interval $[-1,1]$ into a reasonably large -number $N$ of subintervals of length $2/N$ and test on which -subintervals $f$ changes sign, and then use bisection to locate -a zero in those subintervals. If $k>1$ I suspect that some sort -of application of Marching Cubes may work, but the usual search -methods have not turned up anything. - -REPLY [3 votes]: With a Lipschitz bound on the derivative, Newton's method gives an algorithm to efficiently approximate $F^{-1}$. Hubbard's calculus textbook has a write-up using this perspective, viewing the result as a consequence of Kantorovich's Theorem. There's a fairly extensive analysis of the absolute error.<|endoftext|> -TITLE: On "super connected" graphs -QUESTION [8 upvotes]: A graph $G$ is called super connected if for every connected subgraph $H\subset G$ the graph $G-H$ obtained from $G$ after deletion of all vertices from $H$ is also connected. - -Conjecture: The only super connected graphs are $K_{n}$ and $C_{n}$. - -ADDED 1: Some related results can be found here http://www.sciencedirect.com/science/article/pii/S0012365X96003068 -ADDED 2: See also the question Graphs in which every spanning tree is an independency tree for another interesting theme. - -REPLY [5 votes]: There is another proof that gives an alternative description of super connected graphs in terms of spanning trees (hence perhaps of interest): - -The following statements are equivalent for any $n$-vertex graph $G$: -(i) $G$ is super connected. -(ii) $G$ is connected and in every spanning tree $T$ of $G$ every two leaves of $T$ are adjacent in $G$. -(iii) $G$ is the complete graph $K_n$ or the cycle $C_n$. - -(i) $\Rightarrow$ (ii): This is because $T-a-b$ is connected, whenever $a$ and $b$ are two leaves in $T$. -(ii) $\Rightarrow$ (iii): Let $T$ be a depth-first-search (spanning-)tree of $G$, rooted at any vertex $v_0$. As the leaves of any DFS tree are pairwise non-adjacent in $G$, $T$ must be a Hamiltonian path, and the endvertices of $T$ must be adjacent in $G$. Thus, $G$ has a Hamiltonian cycle $C$, say $C: v_0v_1\ldots v_{n-1}v_0$. Assuming $G\not=C$, $C$ has a chord, say $v_0v_i$ for some $i \ge 2$. Write $A=\{v_{1},\ldots, v_{i-1}\}$, $B=\{v_{i+1},\ldots, v_{n-1}\}$. Then, first, every vertex in $A$ is adjacent to every vertex in $B$ because any $a\in A$ and any $b\in B$ are leaves in the spanning tree $C+v_0v_i-aa'-bb'$ of $G$, where $a'$, resp., $b'$ is a neighbor of $a$, resp., $b$ on $C$. Next, every two vertices $v_s, v_t\in A\cup\{v_0,v_i\}$, $s\le t-2$, are adjacent because, by the fact above, we have a chord $v_pv_q$ for any $v_p\in A$, $s -TITLE: 2-layer tilings with a center-of-gravity constraint -QUESTION [10 upvotes]: I've encountered a tiling problem with a physical constraint that -might place it outside the literature on tiling. -"Tiling" is a bit of a misnomer; it is a special type of cover. -All the tiles are identical (congruent), convex shapes $S$. -Layer 1, $L_1$, is some infinite pattern of copies of $S$ -arranged on the plane with pairwise disjoint interiors. -Layer 2, $L_2$, is the exact same pattern as $L_1$, -but after some rigid motion of the whole arrangement -(translation, rotation, perhaps reflection). -Together $L_1$ and $L_2$ cover the plane: -$L_1 \cup L_2 = \mathbb{R}^2$. -But I want the center of gravity (c.g.) of each tile of $L_2$ to be on top -of a point of some tile in $L_1$. -Consider this last requirement an abstraction of a balance or -support constraint. -Finally, I would like to minimize double coverage, i.e., -$L_1 \cap L_2$. -Below are some examples. -Two penny packings of the plane (a), with $L_2$ shifted -by the radius of the disk, satisfy the constraints. -But if I've calculated correctly, 81% of the plane is -doubly covered—not very good. -The staggered squares in (b) improve the double coverage to $\frac{1}{3} =$ 33%. -Here the c.g. of each $L_2$ square sits on the meeting of two corners -of squares from $L_1$. -One can improve this tiling by clipping off a corner (1/8-th) -of each square, as in (c). Now the c.g. of each $L_2$ tile -sits over an interior point of an $L_1$ tile. -I calculate this reduces double coverage to $\frac{2}{9} =$ 22%. - -          - -Because the c.g. in (c) sits over an interior point, there is room -for improvement. Below in (d) I clip off a tiny portion of the opposite -corner (shown in green) to move the c.g. to the boundary, resulting in a tiny -improvement to 21.8%. - -      - -I have no reason to believe this is an optimal, or even a good tiling -under these constraints. -It seems there should be some fundamental positive lower bound -to the double coverage, but I am not seeing an argument to -establish such a bound. - -Update1. Improved by Yoav Kallus to 19.8% double-coverage with a simpler construction! -So the outstanding issue is: Lower bound?? -Update2. Surely Yoav's new 12.5% tiling made of overlapping equilateral triangles -is the optimal. - -REPLY [11 votes]: The pentagon $(0,0)(0,1)(1/2,1+x)(1,1),(1,0)$ with $x=(\sqrt{21}-3)/4$ gives a double coverage fraction of $x/2=0.1978\ldots$. - -UPDATE: The triangle can get you $1/8=0.125$<|endoftext|> -TITLE: A subgroup of the Weyl group -QUESTION [5 upvotes]: Let $D$ be a connected Dynkin diagram with an automorphism $\nu$ of order 2. -Let $Q=Q(D)$ denote the root lattice of $D$. -Let $W=W(D)$ denote the Weyl group, it acts effectively on $Q$ and it is generated by reflections $r_\alpha$ for $\alpha\in D$. -The automorphism $\nu$ acts on $Q$. -Let $W_0$ denote the centralizer of $\nu$ in $W\subset {\rm Aut}\, Q$. -I want to understand this group $W_0$. Let $D^\nu$ denote the subset of $\nu$-fixed vertices in $D$. -For $\beta\in D^\nu$ we have $r_\beta\in W_0$. -I assume that for all $\gamma\in D\smallsetminus D^\nu$, the vertices $\gamma$ and $\nu(\gamma)$ are not connected by an edge -(thus I exclude the case $D={\bf A}_{2n}$). -Then $r_\gamma$ and $r_{\nu(\gamma)}$ commute, and we have $r_\gamma r_{\nu(\gamma)}\in W_0$. - -Question. Is it true that $W_0$ is generated by $r_\beta$ for $\beta\in D^\nu$ - and by $r_\gamma r_{\nu(\gamma)}\in W_0$ for $\gamma\in D\smallsetminus D^\nu$? - -I am interested in the case $D={\bf D}_n$, but I would prefer to get a classification-free answer. - -REPLY [4 votes]: As indicated in my comments, the 1968 book Simple Groups of Lie Type by R.W. Carter has a good elementary treatment of your question (to which the answer is yes) in Chapter 13. All of this goes back pretty far in the history of Lie theory, with the Weyl group and root system arising from a simple Lie algebra (or Lie group) or from a simple algebraic group. Typically the discussion of symmetries of the Dynkin diagram occurs along with specific constructions in the Lie algebra or associated group. But Carter's exposition is clear and detailed, showing how to pass from the original Weyl group to its subgroup commuting with the given diagram symmetry. In particular, this subgroup has a natural set of generators as in your question. -While there are many treatments in textbooks or lecture notes, one online source may be useful: the 1967-68 Yale lecture notes on Chevalley groups by Steinberg -here. See his section 11, where he starts with standard examples and then treats the general theory. In all such sources, notation varies quite a bit but the ideas are pretty much the same.<|endoftext|> -TITLE: Is there a linearly Lindelöf non-Lindelöf $P$-space? -QUESTION [9 upvotes]: A completely regular topological space $(X,\tau)$ is a $P$-space, if every $G_\delta$-subset of $X$ is open (i.e $\tau$ is closed under countable intersections). -A topological space $X$ is linearly Lindelöf if every open cover of $X$ which is linearly ordered by the subset relation has a countable subcover. -Is there a linearly Lindelöf non-Lindelöf $P$-space? - -REPLY [6 votes]: It was shown by A.K. Misra in "A topological view of $P$-spaces" that any regular $\aleph_1$-Lindelöf $P$-space is normal (see Corollary 4.6). Also, it is not hard to see that any linearly Lindelöf space must be $\aleph_1$-Lindelöf. -Finding a normal linearly Lindelöf non-Lindelöf space is a very well known open problem in general topology. So now I´m pretty sure that your question is also open. -Edit: As pointed out by Robson (see comment below), the notion of $\aleph_1$-Lindelöf used by Misra in his paper doesn't correspond to modern terminology. So I don´t know if a regular linearly Lindelöf $P$-space must be normal. However the following is question 486 in "Open problems in topology II": - -(Arhangel´ski) Is the product of two linearly Lindelöf $P$-spaces Lindelöf? - -So I still think that the OP´s question is open, since product of two Lindelöf $P$-spaces is Lindelöf (so it is the same question after all).<|endoftext|> -TITLE: blow-up along singular variety -QUESTION [7 upvotes]: Can somebody give me a nice example of blow-up of a smooth algebraic variety along a singular subvariety? Something I can do some exercise on and check the differences with a smooth blow-up. Thanks! - -REPLY [15 votes]: Perhaps the easiest is to blow up a plane along a fat point: -Blow up $\mathbb A^2_k=\mathrm{Spec} k[x,y]$ at the ideal $(x^2,y^2)$. You should get a pinch point (Whitney's umbrella). This itself is an interesting singularity. It's simple normal crossing away from the pinch point, but not so simple there. If you want to practice blow-ups, then as a second step blow up the pinch point and marvel at the fact that you get a new pinch point. The only way to resolve or even make a pinch point any better is to normalize it. -Not all fat points lead to singular varieties. Blowing up $(x^2,xy,y^2)$ is the same as blowing up the simple point $(x,y)$.<|endoftext|> -TITLE: Can the Poisson summation formula break? -QUESTION [6 upvotes]: The Poisson summation formula states if $f: \mathbb{R} \to \mathbb{R}$ -then $\displaystyle \sum_{n \in \mathbb{Z}} f(n) = \sum_{n \in \mathbb{Z}} \hat{f}(n) $ where $$\hat{f}(\xi) = \int_{\mathbb{R}}dx \; e^{-2\pi i x \xi}\;f(x) $$ -The fine print is that $f$ needs to be Schwartz class or be a tempered distribution such as the Dirac-delta function -$$ \sum_{n \in \mathbb{Z}} \delta_n(x) = \sum_{n \in \mathbb{Z}} e^{2\pi i n x} $$ -It then says the Dirac comb is its own Fourier transform. -Are there counterexamples where the left and right sides converge yet these traces do not agree? - -Edit As discussed in the comments, it appears in Katznelson's textbook although I don't understand his example very well - convolving a function with the Fejer kernel many times at different scales. - -Edit Can anyone fill in details of Lucia's response? - -REPLY [88 votes]: Exercise 15 of Chapter VI (Section 1) of Katznelson's book "An introduction to Harmonic analysis" gives an example of a continuous $L^1$ function $f$ where both sides of the Poisson summation formula are absolutely convergent, but the formula does not hold. -To add a little to this, the usual proof of Poisson summation begins -by periodizing $f$ setting $F(x) = \sum_{n\in {\Bbb Z}} f(x+n)$ and then -computing the Fourier series of $F(x)$. The idea in Katznelson's example is -that even if $f$ is somewhat nice (e.g. continuous and $L^1$) it need not be the -case that $F$ is nice (e.g. it could be discontinuous). In particular Katznelson's -example constructs a nice $f$ for which $F$ turns out to be $1$ on all of ${\Bbb R}/{\Bbb Z}$ except for being discontinuous at $0$ where it takes the value $0$. So there -is a problem with the Fourier series for $F$ at $0$, and hence with the Poisson summation formula.<|endoftext|> -TITLE: Relation between Galois theory and Etale Cohomology -QUESTION [8 upvotes]: I am a graduate student working on category theory. I am familiar with categorical Galois theory, in the way developed by Janelidze - as described for example in "Galois Theories". I am now trying to comprehend the paper by Joyal and Tierney "An extension of the Galois theory of Grothendieck". It feels like there is a connection between Galois-theoretic results in category theory and Etale Cohomology, which is not clear to me. I do not have a very solid background on algebraic geometry. Can someone describe the connection between the two? I am quite sure that descent theory plays some role but as I said it is not clear to me. If someone could provide a way to look at this correctly, or some material that describes the connection, I would be very grateful - I have found some papers but the ones describing Etale Cohomology are using a very algebraic-geometric language that I can not really follow. There must be a way to look at etale cohomology from a pure category-theoretic point of view. Thanks for any help. - -REPLY [10 votes]: The relation between Galois theory and etale cohomology is simple : what Galois theory of a Field $K$ said is that the etale topos of $Spec K$ is the topos of (the category of) continuous $G$ set where $G$ is the Galois group of the separable closure of $K$ (over $K$) endowed with its profinite topology. Hence the etale cohomology of $Spec K$ is related to the group cohomology of the absolute galois goup of $K$. -Now the title of the paper you mentioned refer to a result of Grothendieck which give a characterization of categories which are the category of continuous $G$-set for some (topological) group $G$. (I don't remember the precise formulation of this theorem, but a more modern form of this result is the fact that a connected atomic topos with a point is the topos of continuous $G$-set for $G$ the localic group of automorphism of the point.) -Now the main result of the paper of Joyal and Tierney is that not only connected atomic topos, but any topos can be represented (relatively explicitly) as the topos of equivariant sheaf over a localic groupoid. Which in some sense generalize the result of grothendieck. -This can be used for example to represent the Etale topos of an arbitrary scheme by something geometrical, involving a space of geometric point of the scheme on which some galois groups acts locally.<|endoftext|> -TITLE: Does this infinite sum provide a new analytic continuation for $\zeta(s)$? -QUESTION [17 upvotes]: It is well known that the infinite sum: -$$\displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$ -only converges for $\Re(s)>1$. -The Dirichlet 'alternating' sum: -$$\displaystyle \zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$$ -allows for analytic continuation towards $\Re(s)>0$, but also does introduce additional zeros. -There are also nested sums see formulas 21 and 22 here that fully extend $\zeta(s)$ towards $s \in \mathbb{C}/1$. -However, I found this surprisingly simple infinite sum that Maple evaluates correctly for $\Re(s) >-1$: -$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(s+1+ \sum _{n=1}^{\infty } \left( {\frac {s-1-2\,n}{{n}^{s}}}+{\frac {s+1+2\,n}{\left( n+1 \right) ^{s}}}\right) \right)$$ -Following Peter's suggestion below, it can be easily shown that by splitting, re-indexing via $n \mapsto n-1$ and then recombining the sums, that for $\Re(s) > 1$ the formula correctly reduces to $\displaystyle \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$. -For the analytically continued domain $-1<\Re(s)<1$, the operations of re-indexing and recombining the split sums are no longer allowed, since the sums are now diverging. -I played a bit with the individual terms in the infinite sum and it is easy to see that for $s=0$ and $s=1$ all terms will be zero. Therefore $\zeta(0)$ will be fully determined by $\frac{s+1}{2\,(s-1)} = \frac{1}{-2}=-\frac12$. When $s\rightarrow 1$, there will be a pole induced by $\frac{s+1}{2(s-1)}$, however the pole $\frac{1}{2(s-1)}$ will be offset by the infinite sum approaching $0$ and converges to the nice result $\gamma -\frac12$ with $\gamma$ being the Euler-Mascheroni constant. -Did not have much time to go deeper, but did find a nice result for $s=2$ (n=1..7): -$s=2 \rightarrow \frac{1 }{4}+ \frac{1 }{36}+\frac{1}{144}+\frac{1}{400}+\frac{1}{900}+\frac{1}{1764}+\frac{1}{3136}+\dots$ = sum of 1/A035287 -For $s=3$ the following pattern emerges, however I did not find a good link to Sloane's. -$s=3 \rightarrow \frac{3 }{4}+\frac{5 }{108}+\frac{7}{864}+\frac{9}{4000}+\frac{11}{13500}+ -\frac{13}{37044}+\frac{15}{87808}+\dots$ -Questions: -(1) Has anyone seen this formula before? If so, I'd be grateful for a reference. -(2) To avoid this is just Maple playing a trick on me here, I am curious to learn whether the results can be reproduced in other CAS. (EDIT: now confirmed in Maple, Sage, Pari/GP and Mathematica that the formula works correctly, although numerical discrepancies are found for higher values $> 10^8$ of $\Im(s)$). -Final addition: -Paul Garrett´s reference below gives a good method to construe the formula above. However, I realized myself that I posted the formula on MO without any explanation on how I did find it in the first place. So, here it is to make the question complete before it vanishes into MO´s history: -Assume $s \in \mathbb{C}$, $\Re(s) \ge 0$ and take the well known expression: -$$\zeta(s) = \dfrac{s}{s-1} - \frac12+s \int_1^\infty \frac{1/2-\{x\}}{x^{s+1}}\,\mathrm{d}x$$ -and substitute the fractional part of $\{x\}$ by a closed form (see here): -$$\displaystyle \{x\} = x - \lfloor x \rfloor = -\frac{\arctan\left(\cot\left(\pi x\right)\right)}{\pi} + \frac{1}{2}$$ -which gives: -$$\displaystyle Z(s) = \dfrac{s}{s-1} - \frac12 +s \int_1^\infty \frac{\arctan\left(\cot\left(\pi x\right)\right)}{\pi x^{s+1}}\mathrm{d}x $$ -For all rational fractions $-1<\frac{k}{n}<1$, Maple consistently returns closed forms with an infinite sum e.g.: -$$Z \left(\frac23\right) =\displaystyle 1-\frac{11}{4}\,\sqrt [3]{2}+\frac{2}{3}\,\sum _{{\it n}=2}^{\infty }-\frac{3}{4}\,{\frac {- \left( {\it n}+1 \right) ^{2/3}-6\, \left( {\it n}+1 \right) ^ {2/3}{\it n}+5\,{{\it n}}^{2/3}+6\,{{\it n}}^{5/3}}{{{\it n}}^{2/3} \left( {\it n}+1 \right) ^{2/3}}}$$ -After experimenting with a few rationals, I saw a clear pattern emerge that I then could easily generalize into the formula above. To my surprise it also worked for irrational values and the domain $\Re(s) \ge 1$. Not sure whether Maple uses a similar technique as Paul Garrett´s suggestion below to transform the integral of the fractional parts into these infinite sums. - -REPLY [16 votes]: This can be construed as an application of Euler-Maclaurin summation, as carried out to an arbitrary number of stages in Appendix B of Montgomery-Vaughan's "Multiplicative Number Theory I: Classical Theory". -The general idea is to subtract the integral corresponding to a sum, breaking the integral into integrals over subintervals, as in the well-known idea -$$ -\zeta(s) - {1\over s-1} \;=\; \sum_n \Big({1\over n^s} - \int_n^{n+1}{dx\over x^s}\Big) \;=\; \sum_n \Big({1\over n^s} - {n\over (s-1)n^s} + {n+1\over (s-1)(n+1)^s}\Big) -$$ -Elementary estimates show that the new sum is absolutely and uniformly-on-compacts convergent on $\Re(s)>0$. One can repeat the approximation of sum by integral as many times as desired, although organization becomes an issue. The sum in the question is equivalent to application of this idea one more stage. Montgomery-Vaughan show how to systematize this and express an arbitrary (finite) number of iterates in terms of Bernoulli polynomials/numbers.<|endoftext|> -TITLE: How differentiable is the convolution of two continuous functions? -QUESTION [9 upvotes]: The question is really simple: -Given -$$ -f, g\in C^\alpha_c(\mathcal{R}^d) -$$ -is -$$ -f*g\in C^d_c? -$$ -I came up with a formal argument using the decay of the Fourier transform of continuous functions, but it is really formal and I would appreciate a reference. -What I have though so far, which has a few holes is the following: -1) Because the Fourier transform, by interpolation, goes from $L^p\rightarrow L^q$, for every $p<\infty$, and I am thinking about compactly supported continuous functions, their Fourier transform has to behave roughly like $L^1$ in the far field. Using that if a function is in $L^\infty$ its Fourier transform is uniformly continuous, I claim that $\hat{f}, \hat{g}\sim O(\frac{1}{|\chi|^d})$ when $\chi\to\infty$. -1*) This step is the one I don't trust, though I am looking for a continuous function, for which it's fourier transform does not have this decay. I boiled it down to find a continuous function that has no integrable derivatives at all, any ideas? Because if we take $f,g\in H^{\frac{d}{2}}$ borderline continuous, just by dividing the derivatives we get -$$ -\Delta^{\frac{d}{2}}(f*g)=\Delta^{\frac{d}{4}}f*\Delta^{\frac{d}{4}}g\sim L^2*L^2\in C^0 -$$ -Therefore, $f*g\in C^d$. -2) We use the Convolution Theorem, so $\widehat{f*g}\sim O(\frac{1}{|\chi|^{2d}})$. -3) Finally, if $h\in L^2$ and $\hat{h}\sim O(\frac{1}{|\chi|^{d+\alpha}})$. Then, $h\in C^{\alpha-\epsilon}$ $\forall \epsilon>0$. Therefore, $f*g\in C^{d-\epsilon}$ $\forall \epsilon>0$. -I know this argument is not completely rigorous, though it shows me that dimension should play a role. -Assuming nobody believes in the result (I don't think it's true, but it would be nice, what is in fact true), I have a similar question, which might be easier: -We know that if $\frac{1}{p}+\frac{1}{q}=1$ and $f\in L^p$ and $g\in L^q$ and assume they are compactly supported then, -$$ -f*g\in C^0_c -$$ -This can be easily seen by using than $L^q=(L^p)^*$, and the fact that convolution is translation invariant. -But, what happens if $p$ is better than the conjugate of $q$? -Lets say $f\in L^3$ and $g\in L^2$, is it true that -$$ -f*g\in C^\alpha_c, -$$ -for some $\alpha>0$? -Even simpler, take $f\in C^\alpha$ and $g\in L^2$, do you get -$$ -f*g\in C^{\alpha +\beta}_c, -$$ -for some $\beta>0$? -All of this assuming compact support of both functions. - -REPLY [7 votes]: $C^\alpha_c*C^\beta_c\subset C^{\alpha+\beta}_c$ but not much better than that. If you want it through Fourier analysis, the shortest route to go is to periodize and to use the Bernstein description of periodic $C^\alpha$ with non-integer $\alpha$ as the set of continuous functions $f$ such that for every $n$, there exists a trigonometric polynomial of degree $n$ such that $\|f-p\|_C\le Cn^{-\alpha}$. Then writing $f=p+r$ and $g=q+h$ and noticing that $p*g$ and $r*q$ are polynomials, you get away with the trivial bound $Cn^{-(\alpha+\beta)}$ for the term $r*h$, which you treat as an error term. The case of integer $\alpha$ (or $\beta$) is trivial: just differentiate $f$ all the way you can and only then switch to $g$.<|endoftext|> -TITLE: Cameron Martin space -QUESTION [5 upvotes]: I have seen two definitions of Cameron Martin space of a Gaussian measure $\nu$ on a Banach space (say $W$) and am unable to establish their equivalence. Any help would be appreciated. -1) It is the intersection of all linear subspaces of full measure. -2) Its elements are those which when seen as linear functionals on the dual of $W$ (say $W^\ast$) are continuous in the inner product induced on $W^\ast$ by $\nu$ (the Gaussian measure). - -REPLY [2 votes]: The proof of this equivalence can be found in Bogachev: Gaussian Measures, Theorem 2.4.7.<|endoftext|> -TITLE: Does the bundle of germs of functions $f:X\to \mathbb R$ have the same sheaf of sections as $X\times \mathbb R$? -QUESTION [5 upvotes]: I'm just starting to learn about sheaves, and I'm confused about a certain matter: -I've just learned, to my delight, that every sheaf $S$ on a space $X$ is the sheaf of sections of a particular bundle (by bundle I mean a surjective map $E\to X$), specifically the bundle of stalks of $S$. -So, for example, the sheaf of continuous functions on $X\to \mathbb R$ can be interpreted as the sheaf of sections of germs of such functions. -BUT: it seems to me that this particular sheaf is also the sheaf of sections of a simpler bundle, namely the projection $X\times \mathbb R\to X$, because sections of this bundle are precisely continuous functions $X\to \mathbb R$. -Now, I know these two bundles aren't the same space, because, as Mac Lane and Moerdijk point out in Sheaves in Geomtry and Logic, the former (at least for $X=\mathbb R$) isn't Hausdorff, while the latter plainly is. Also, I seem to understand that such bundles of stalks are always étale, while the latter bundle is clearly not étale. -So, am I missing something, or do the sections of these two very different bundles produce the same sheaf? - -REPLY [7 votes]: These bundles have the same sections. But which is simpler depends on your point of view. $X\times \mathbb R\to X$ is simpler set theoretically but is `more complicated' topologically in the the sense that it is not a local homeomorphism. The sheaf, taken as a bundle, is a local homeomorphism.<|endoftext|> -TITLE: Countable Maximal Ideals -QUESTION [10 upvotes]: This may be simple but I can not see a way. I am looking for an uncountable ring (with 1) containing a countable maximal left ideal which is not a direct summand (as a left ideal). - -REPLY [4 votes]: Eric's result for commutative rings, that an example must have the ring no bigger than the continuum and the maximal ideal infinitely generated, extends to left ideals for non-commutative rings, assuming that by "not a direct summand" you mean "not a direct summand as a left ideal". -Let $A$ be the ring, $I$ the countable maximal left ideal, and $S$ the simple left module $A/I$. Considering these all as left $A$-modules, there can be no non-zero maps $S\to I$ or $I\to S$. It follows that there can be no non-zero map $S\to A$, since if there were, then composing with the natural epimorphism $A\to S$ would give a non-zero map $S\to S$, which must be invertible since $S$ is simple, and so $A\cong I\oplus S$ as left $A$-modules. -Hence the map $A\to A$ given by right multiplication by any $0\neq a\in A$ must restrict to a non-zero map $I\to A$, and therefore to a non-zero map $I\to I$, since there are no non-zero maps $I\to S$. -So, as in Eric's argument, $A$ acts faithfully on $I$. -EDIT: By the way, this shows that $I$ is a two-sided ideal. The quotient ring $S=A/I$ must then be a skew field, since it is simple as a left module for itself. Also, $I/I^2$ is an $S$-module, but is at most countable, so must be zero. So $I$ is an idempotent ideal.<|endoftext|> -TITLE: Unicity of Yoneda isomorphism -QUESTION [5 upvotes]: I am wondering if there is only one unique Yoneda isomorphism, that is a natural isomorphism (natural in C and P, that is) between Hom(yC,P) and PC. -The Yoneda lemma says that there exists at least one, and its common proof is constructive, so we have an example. But is it the only one? -If it is unique, how do we prove that? If not, is there any counter-example? - -REPLY [8 votes]: If there is an isomorphism $\mathrm{Hom}(y C, P) \cong P (C)$ natural in $P$ and $C$, then by restricting to the representable functors one obtains an automorphism of the identity functor of the category $C$ comes from. There are categories for which the identity functor has a non-trivial automorphism group, e.g. $\mathbf{Ab}$, and also categories for which the identity functor has a trivial automorphism group, e.g. $\mathbf{Set}$. - -In more detail: let $P = y D$. Then $\mathrm{Hom}(y C, y D) \cong \mathcal{T}(C, D)$, and if we compose with the standard Yoneda isomorphism we get an isomorphism $y D \cong y D$, natural in $D$. The Yoneda embedding $y : \mathcal{T} \to [\mathcal{T}^\mathrm{op}, \mathbf{Set}]$ is fully faithful, so this determines an automorphism of the identity functor on $\mathcal{T}$. -Conversely, suppose $\theta : \mathrm{id}_{\mathcal{T}} \Rightarrow \mathrm{id}_{\mathcal{T}}$ is an automorphism. Then we can compose it with the Yoneda embedding and use functoriality of $\mathrm{Hom}$ to get a new natural bijection $\mathrm{Hom}(y C, P) \cong P (C)$: explicitly, it is the map defined by evaluating $P$ at $\theta_C$.<|endoftext|> -TITLE: Computing Thompson Series for the Monster Group -QUESTION [5 upvotes]: I am trying to do some experimentation with the values of Thompson series, but I have having a hard time finding a table that has these Thompson series with as many terms as I'd like. The tables I've seen only give me the required Head character values up to the $q^{10}$ term. -Can someone point me to a resource that either has more terms, or (even better), a resource that might help me compute them myself in sage? -Thanks a lot for any help. - -REPLY [5 votes]: Michael Somos (you can contact him at somos@cis.csuhio.edu) has a Pari-GP code which can help you calculate as many terms as you like, while D. Madore has also computed the first 3200 terms.<|endoftext|> -TITLE: Iterated Pieri's rule, Schur functors and intersection of subrepresentations -QUESTION [7 upvotes]: Let $\lambda$ and $\mu$ be two Young diagrams, such that $\lambda$ can be obtained from $\mu$ by extending one single column with additional $b$ boxes. Let $\Sigma^\lambda U$ and $\Sigma^\mu U$ denote the corresponding Schur functors (that we consider as representations of $GL(U)$) and let $a -\geq 0$ be a integer number. -By Pieri's rule $\Sigma^\lambda U$ can be considered as an irreducible subrepresentation of $\Sigma^\mu U\otimes\Lambda^b U$. Thus, one may think of $\Sigma^\lambda U\otimes\Lambda^a U$ as a subrepresentation of $\Sigma^\mu U\otimes\Lambda^b U\otimes\Lambda^a U$. -At the very same time, $\Lambda^{b+a}U$ is naturally an irreducible subrepresentation of $\Lambda^b U\otimes\Lambda^a U$. Thus, one may consider $\Sigma^\mu U\otimes\Lambda^{b+a} U$ as a subrepresentation of $\Sigma^\mu U\otimes\Lambda^b U\otimes\Lambda^a U$. -I'm interested in computing the intersection of these two subrepresentations. - -Let us decompose $\Sigma^\mu U\otimes\Lambda^{b+a} U = \oplus_{\nu\in P} \Sigma^\nu U$ into irreducibles. Here $\nu$ is running over the set of Young diagrams such that $\nu/\mu$ consists of $a+b$ boxes and is of width $1$ (every row of $\mu$ can be extended with at most one box). -Define the submodule $W=\oplus_{\nu\in Q}\Sigma^\nu U$ consisting of those $\nu\in P$, such that $\nu\supset\lambda$ (recall that $\lambda$ can be obtained from $\mu$ by extending $b$ consecutive rows with a single box). It's easy to see that every such $\nu\in Q$ appears in the decomposition of $\Sigma^\lambda U\otimes\Lambda^a U$ into irreducibles and these are the only ones that can. Thus, the intersection should be contained in $W$. -Conjecture: the intersection coincides with $W$. -The main problem is the following one: despite every irreducible factor $\Sigma^\nu U$ in $W$ being distinct and appearing in both $\Sigma^\lambda U\otimes\Lambda^a U$ and $\Sigma^\mu U\otimes\Lambda^{b+a} U$ with multiplicity one, most of the time its multiplicity in the ambient representation $\Sigma^\mu U\otimes\Lambda^b U\otimes\Lambda^a U$ is quite big. Thus, one should somehow use the actual form of Pieri's embedding as counting multiplicities is not enough. - -REPLY [3 votes]: It turned out to be a rather non-trivial statement and follows from an underknown work of Olver. An accessible reference would be Pieri resolutions for classical groups by Sam and Weyman (http://arxiv.org/abs/0907.4505).<|endoftext|> -TITLE: Do elements of the fundamental group give rise to isometries -QUESTION [8 upvotes]: Let $X$ be a complex algebraic variety, and let $\tilde X\to X$ be its universal cover. Suppose that there exists a Kahler-Einstein metric on $\tilde X$. Note that $\pi_1(X) \subset Aut(\tilde X)$. -Question. Assume $X$ is compact. Is $\pi_1(X)$ actually a subgroup of the group of isometries of $\tilde X$ with respect to the Kahler-Einstein metric? -I suspect it is, but I don't know enough about analytic geometry to be completely sure. - -REPLY [6 votes]: The answer to your question is no without further assumptions. -Here are three counterexamples (one for each curvature's sign): - -Positive curvature: take $X=\mathbb C/(\mathbb Z \oplus i \mathbb Z)$ the standard torus, and $\tilde X = \mathbb C$. If you endow $\mathbb C$ with the restriction of the Fubini-Study on $\mathbb P^ 1$ (under any embedding $\mathbb C \hookrightarrow \mathbb P^1$), then you get a metric with constant positive curvature on $\mathbb C$ which is clearly not invariant under the natural action of $\mathbb Z \oplus i \mathbb Z$ on $\mathbb C$. -Zero curvature: Take as before the torus, and choose on $\mathbb C$ a metric $e^u |dz|^2$ where $u$ is harmonic but not invariant under the action of $\mathbb Z \oplus i \mathbb Z$ (for example $u=\mathrm{Re}(z)$). -Negative curvature: Take $X$ to be any compact complex curve with genus $g\geqslant 2$. And put on the unit disk in $\mathbb C$ the restriction of Poincaré metric of the disk of radius $2$. Clearly it won't be invariant under the fundamental group of $X$ (at least if $X$ is well chosen). - -However, we can still say something in the last case. Indeed, if you assume that the Kähler-Einstein metric on $\tilde X$ is complete, then Yau's maximum principle shows that it is the unique such metric. Hence any automorphism of $\tilde X$ preserves this metric, so that $\pi_1(X)$ acts by isometries with respect to this metric.<|endoftext|> -TITLE: Arnold on Newton's anagram -QUESTION [26 upvotes]: Arnold, in his paper -The underestimated Poincaré, in Russian Math. Surveys 61 (2006), no. 1, 1–18 -wrote the following: - -``...Puiseux series, the theory which Newton, hundreds of years before Puiseaux, - considered as his main contribution to mathematics (and which he encoded as a second, - longer anagram, describing a method of asymptotic study and solution of all equations, - algebraic, functional, differential, integral etc.)...'' - -Arnold says this is several other places as well. -As I understand, the "first anagram" is this -6accdae13eff7i3l9n4o4qrr4s8t12ux -You can type this on Google to find out what this means. Or look in Arnold's other popular books and papers. - -Question: what is the "second anagram" Arnold refers to? - -P.S. This was my own translation from Arnold's original. The original is available free -on the Internet, but the translation is not accessible to me at this moment. I hope my translation is adequate. -P.P.S. I know the work of Newton where he described Puiseux series, probably it was unpublished. But there is no anagram there. - -REPLY [11 votes]: The two anagrams (first and second) are in the same letter of Oct 24, 1676 to Leibniz. -An insert of the second and more rare is here: - -and for the whole letter go here - lines -8, -7 and -6.<|endoftext|> -TITLE: Conditions for when an off-centre ellipsoid fits inside the unit ball -QUESTION [5 upvotes]: An ellipsoid $E$ has centre $\vec{c}=(c_1,c_2,c_3)$ and semiaxes $t_1$, $t_2$ and $t_3$ aligned with the $x$, $y$ and $z$ axes. What are the necessary and sufficient conditions on $\vec{c}$, $t_1$, $t_2$ and $t_3$ such that $E$ lies inside the unit ball? -For example, we can require that the set of points where the semiaxes meet the ellipsoid surface lie within the unit ball. This gives a set of inequalities such as $(c_1\pm t_1)^2+c_2^2+c_3^2\leq1$, which are necessary but not sufficient. How many points on the surface of the $E$ need to be checked before you know that the whole ellipsoid is within the unit ball? (Clearly, you only need to check the point on the surface of $E$ that is furthest from the origin, but then the problem becomes how to find that point.) -It is easy to formulate a set of inequalities for the special case that the ellipsoid is centred on an axis (i.e. only one $c_i$ is non-zero). However, the off-axis case seems impossible even in 2 dimensions: given an ellipse with centre $\vec c=(c_1,c_2)$ and semiaxes $t_1$ and $t_2$, aligned with the $x$ and $y$ axes, what are the conditions for this to lie within the closed unit disk? -For the 2D case, the solution might go along these lines (as suggested here amongst other places). Parameterise the perimeter of the ellipse as $x = t_1 \cos\phi + c_1$ and $y = t_2 \sin\phi + c_2$. Then any point on the ellipse perimeter has $x^2+y^2=(t_2^2-t_1^2)\sin^2\phi+2t_1c_1\cos\phi+2t_2 c_2 \sin\phi+\mathrm{const}$. Differentiate this expression to find the $\phi$ for which $x^2+y^2$ is maximal, and see if $x^2+y^2\leq1$ for that $\phi$. If any one of the four parameters $c_1$, $c_2$, $t_1$, $t_2$ is zero then this becomes a simple quadratic, but in general it seems to be not analytically possible. -So, my questions are as follows: - -What are the conditions for an off-axis ellipse to lie within the unit disk? -The on-axis ellipse can easily be extended to an on-axis ellipsoid (see Lemma 26 of http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=669165). Is this also true in the off-axis case? -(I'm not a mathematician so I don't know whether this is a very deep or a completely worthless question...) If the problem is not analytically tractable, why not? How can such an elementary question be so difficult? This question doesn't sound like Fermat's last theorem - it's just very basic Euclidean geometry. - -Thanks very much for any help! - -REPLY [3 votes]: This is just to elaborate on Robert's idea. Once you bring the question to this form, you can show that the matrix is positive definite iff $1\le \nu\le \min\rho_j$ and -$$ -\nu-1\ge\sum_j\frac{\lambda_j^2}{\rho_j-\nu} -$$ -(once you fix the first variable, the worst choice for every other one is obvious). -Assume $\rho_1\le\rho_2\le\rho_3$ -Note that when $\nu$ is near $1$ and near $\rho_1^-$, the RHS is larger, so the corresponding equation must have two roots (counting with multiplicity) on $(1,\rho_1)$. Since we are also guaranteed a root on each of the intervals $(\rho_j,\rho_{j+1})$ ($j=1,2$), we conclude that a certain quartic monic polynomial in $z=1-\nu$ must have four real negative roots, or, in other words, all roots must be real and all coefficients non-negative. The analytic condition for this event is known. There are several equivalent ways to state it. For programming purposes, the description in this paper seems to be the most convenient (you do not need anything beyond the front page; the definitions, etc. can be googled readily). So, you end up with a lot of elementary algebraic manipulations and 6 inequalities. Whether it beats the purely numeric approach is up to you to decide.<|endoftext|> -TITLE: "cov vs pp" problem -QUESTION [10 upvotes]: This is the problem $(\beta)$ of the section 14.7 in the "Analytical Guide" of the Shelah's book "Cardinal Arithmetic": - -$(\beta)$ Is $\operatorname{cov} ( \lambda , \lambda, \aleph_{1} , 2) =^{+} \operatorname{pp} ( \lambda )$ when $\operatorname{cf}( \lambda ) = \aleph_0$? - -Is this problem still open? -We know that $\operatorname{cov} ( \lambda , \lambda, {(\operatorname{cf}(\lambda))}^{+} , 2) = \operatorname{pp} (\lambda )$ for every singular $\lambda$ such that $\lambda < \aleph_\lambda$ (see for instance the Claim 3.7(1) in the Chapter IX of the Shelah's book). -Can the following statement be a theorem in ZFC? -$\operatorname{cov} ( \lambda , \lambda, {(\operatorname{cf}(\lambda))}^{+} , 2) = \operatorname{pp} (\lambda )$ for every infinite cardinal $\lambda$ with $\operatorname{cf}( \lambda ) < \lambda = \aleph_{\lambda}$. - -REPLY [11 votes]: Very much an open problem. I've been sporadically blogging about aspects of this question over the past few years, mainly in an effort to pin down exactly what's known and what's still unknown. There are some links to this work on my website. -With regard to your other question, the answer is yes under GCH, because both the covering number and the pp number must be equal to $\lambda^+$.<|endoftext|> -TITLE: A moment problem on $[0,1]$ in which infinitely many moments are equal -QUESTION [17 upvotes]: Suppose $\mu$ and $\nu$ are two probability measures on $[0,1]$. Let their $n$-th moments be denoted by $\mu_n$ and $\nu_n$, respectively, for $n \in \mathbb{N}$. -If we know that $\mu_n=\nu_n$ for infinitely many $n$, can we conclude that $\mu=\nu$? -One way to resolve this would be to see if the span of $\{ x^n \mid n \in S \}$ with $|S|=\infty$ is dense in the set of continuous functions $C[0,1]$. Is such a set always dense? - -REPLY [15 votes]: We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough. -Recall Müntz-Szász theorem, which states (in particular) the following: - -Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent: - -the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm. - -the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent. - - - -If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide. -If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.<|endoftext|> -TITLE: Isomorphic rings of functions -QUESTION [6 upvotes]: Let $X$ and $Y$ be two topological spaces with $C(X) \cong C(Y)$ (where $C(X)$ is the ring of all continuous real valued functions on $X$). I know that we can not conclude that $X$ and $Y$ are homeomorphic. But I wonder how independent $X$ and $Y$ could be ? For example is there any forced relation between their cardinality ? - -REPLY [13 votes]: Since we are talking about rings of continuous functions, I will only talk about completely regular spaces in this problem. It is well known that for completely regular spaces $X$, $C(X)\simeq C(Y)$ if and only if $\upsilon X\simeq\upsilon Y$ where $\upsilon X$ denotes the Hewitt-realcompactification of a space $X$. Of course, for information on rings of continuous functions, obviously one should consult the book Rings of Continuous Functions by Gillman and Jerison. -If $X$ is a completely regular space, then the Hewitt realcompactification $\upsilon X$ is defined to be the subset of the Stone-Cech compactification where $x\in\upsilon X$ if and only if whenever $f:X\rightarrow[0,1)$ is continuous and $\overline{f}:\beta X\rightarrow[0,1]$ is the unique continuous extension, then $f(x)<1$. -Of course, since $|\upsilon X|\leq 2^{2^{|X|}}$ for all spaces $X$, if $C(X)\simeq C(Y)$, then $|X|\leq 2^{2^{|Y|}}$ and $|Y|\leq 2^{2^{|X|}}$. -Assuming the existence of certain large cardinals, there are discrete spaces with $X$, $|\upsilon X|=2^{2^{|X|}}$. For example, if $\kappa$ is a strongly compact cardinal, and $X$ is a discrete space of cardinality $\kappa$, then $|\upsilon X|=2^{2^{|X|}}$, but -$C(X)\simeq C(\upsilon X)$. This is because the points in $\upsilon X$ are in a one-to-one correspondence with the $\sigma$-complete ultrafilters on $X$ for discrete spaces $X$, and if there is a compact cardinal $\kappa$, then there are many $\sigma$-complete ultrafilters on a set $X$ of cardinality $\kappa$. If $\kappa$ is a measurable cardinal, and $|X|=\kappa$ then $|\upsilon X|\geq 2^{|X|}$ and $C(\upsilon X)\simeq C(X)$. There are probably examples of spaces $X,Y$ of different cardinality that do not involve large cardinals where $C(X)\simeq C(Y)$, but $|X|\neq|Y|$, but in order to have a discrete space as such a counterexample, one needs the existence of measurable cardinals. -$\textbf{Added some time later}$ I will now give an example of a space $X$ with $|X|<|\upsilon X|$ without resorting to large cardinal hypotheses. In particular, we have $C(X)\simeq C(\upsilon X)$, but $|X|<|\upsilon X|$. -Let $\mathbb{N}$ denote the natural numbers. For each unbounded function $f:\mathbb{N}\rightarrow[0,\infty)$, let $\overline{f}:\beta\mathbb{N}\rightarrow[0,\infty]$ be the unique extension of the function $f$. Then there is some point $x_{f}\in\beta\mathbb{N}$ with $\overline{f}(x_{f})=\infty$. Let $X=\mathbb{N}\cup\{x_{f}|f:\mathbb{N}\rightarrow[0,\infty)\,\textrm{is unbounded}\}$. Then $X$ is a space of cardinality continuum. Furthermore, the space $X$ is pseudocompact. If $f:X\rightarrow[0,\infty)$ is continuous and unbounded, then $f|_{\mathbb{N}}$ is unbounded as well, so $f(x_{f|_{\mathbb{N}}})=\infty$, a contradiction. It is well known that a space $Y$ is pseudocompact if and only if $\upsilon Y=\beta Y$. Therefore, we have $\upsilon X=\beta X=\beta\mathbb{N}$. In particular, $|\upsilon X|=2^{2^{\aleph_{0}}}$ while $|X|\leq 2^{\aleph_{0}}$. - -REPLY [9 votes]: Let $X,Y$ be arbitrary sets (of arbitrary cardinals) armed with topologies $\tau_1 = \lbrace \emptyset, X\rbrace$ and $\tau_2 = \lbrace \emptyset, Y\rbrace$. Then it is clear that $C(X) \cong \Bbb{R} \cong C(Y)$. So there is no forced relation between the cardinal numbers.<|endoftext|> -TITLE: Cut elimination algorithms -QUESTION [9 upvotes]: Gentzen's Hauptsatz in first order logic includes an algorithm taking any proof in the sequent calculus with cut rule, and delivering a proof without cut rule (and with the subformula property). So far as I know cut elimination in the simple theory of types (STT) has no such algorithm. Rather, one proves non-constructively that the system without cut rule is complete for the same models as the system with. -But is there such an algorithm for STT, or is it known there cannot be? -And what about the proof that $\mathsf{ACA}_0$ is conservative over PA? (Compare Emil Jeřábek's answer to What metatheory proves $\mathsf{ACA}_0$ conservative over PA?) Does that come with an algorithm for producing PA proofs? - -REPLY [6 votes]: In some sense, the answer is trivially yes: if you know a theory $T$ proves $\varphi$ using only deduction rules from among $\mathcal{D}$, then (assuming $T$ and $\mathcal{D}$ are appropriately computable) you can just search through all appropriate proofs. This certainly holds in both cases you mention, and (I believe) in all cases of interest. -If you're asking about "feasible" algorithms, then of course things get more complicated. I don't know about STT and cut elimination, but in the case of $PA$ versus $ACA_0$, the conservativity proof is very straightforward - given $\mathcal{M}\models PA$, the structure $$(\mathcal{M}, \{X\subseteq\vert\mathcal{M}\vert: X\text{ is arithmetically definable in $\mathcal{M}$ with parameters}\})$$ is a model of $ACA_0$ - and that would seem to yield a very direct conversion of $ACA_0$-proofs to $PA$-proofs. (By "direct" here I just mean that there is a clear intuition behind the algorithm which is better than "unbounded search," not that it (a) converges quickly or (b) generates a proof of reasonable length.) -EDIT: I believe the paper "A Relative Consistency Proof" by Shoenfield addresses the issue of going from $ACA_0$-proofs to $PA$-proofs; also, see Reducing ACA₀ proof to First Order PA. - -REPLY [4 votes]: As Noah pointed out, if it's provable, there must be an algorithm for cut-elimination for STT because the underlying statement is $\Pi_2$ (for any deduction in STT, there is a corresponding deduction without cut). Moreover, if you can prove cut-elimination for STT in some theory T for which you do have an algorithm for cut-elimination, the algorithm for T "suffices" in some sense for STT. (For example, bounds on growth in T give you bounds in STT.) That said, I don't know the literature on cut-elimination for STT, so someone else will have to comment on what's actually in the literature about the algorithm. (For instance, if anyone has written it out explicitly.) -For your second question, the proof via cut-elimination Emil Jeřábek mentioned is constructive. One should be careful about terminology: the cut-elimination needed isn't the fancy stuff for $\mathrm{PA}$. The proof uses the much easier cut-elimination for pure first-order logic (which has a known algorithm!). In the presence of non-logical axioms, you can't eliminate all cuts, but you can eliminate all "free cuts" (cuts over formulas which are not parts of non-logical axioms). Then you add a new argument showing that cuts over the new axioms in $\mathrm{ACA}_0$ can be eliminated, basically in the obvious way: if you have a cut between -$$\exists X\forall x(\phi(x)\leftrightarrow x\in X),$$ -and -$$\forall X\neg\forall x(\phi(x)\leftrightarrow x\in X),\Gamma,$$ -you obtain a proof of $\Gamma$ by replacing $t\in X$ with $\phi(t)$ everywhere. (Details depend on the formulation and how you're handling induction axioms.)<|endoftext|> -TITLE: Birkhoff ergodic theorem and the measure of the bad points -QUESTION [12 upvotes]: In the Birkhoff ergodic theorem we have a PMPS $(X,B,\mu,T)$ and that for any $f\in L^1(X,\mu)$ $\frac{1}{N}\sum_{n=0}^{N-1}f(T^n x)\to \int f \, d\mu,$ in measure, in $L^1$-norm and $\mu$-a.e. -My question is: what is, given $\epsilon>0,$ the estimation of $\mu\left(x:\left|\frac{1}{N}\sum_{n=0}^{N-1}f(T^n x)-\int f \, d\mu\right|>\epsilon\right)$ when $N\in \mathbb{N}$ is big? -I am looking for a proof or reference that I have not yet found. - -REPLY [17 votes]: The key words here are "large deviations"; large deviations theory addresses exactly this question. The answer depends quite a bit on the specific measure and system in question, but roughly speaking one may say the following: if the system displays a sufficient amount of hyperbolic behaviour (for example, an Axiom A system, or a system with the specification property, which is a sort of uniform topological mixing) and if the measure $\mu$ has some sort of Gibbs property relative to a potential function $\phi$, then the measure of the set you describe decays exponentially in $N$, and the rate of exponential decay depends in a precise manner on the topological pressure function. -Let me state a concrete series of theorems to make the above more precise. Let $(X,T)$ be a transitive Axiom A system, and let $\phi\colon X\to \mathbb{R}$ be Hölder continuous. Then by a result of Bowen ("Some systems with unique equilibrium states", 1974/5, or if you prefer, "Equilibrium states and the ergodic theory of Axiom A diffeomorphisms"), there is a unique invariant measure $\mu$ that maximises the quantity $h_\mu(T) + \int\phi\,d\mu$. (This maximum value is the topological pressure $P(\phi)$.) Moreover, $\mu$ has the following Gibbs property: if $B(x,n,\delta)$ denotes the set of points $y$ such that $d(f^kx,f^ky)\leq \delta$ for all $0\leq k\leq n$, then there is a constant $K(\delta)$ such that -$$ -(*) \qquad -\frac 1{K(\delta)} \leq \frac{\mu(B(x,n,\delta))}{e^{-nP(\phi) + S_n\phi(x)}} \leq K(\delta) -$$ -for every $(x,n)\in X\times \mathbb{N}$. -Now a 1990 result of Lai-Sang Young on large deviations shows that for a system $(X,T)$ as above and a measure $\mu$ satisfying $(*)$, one has -$$ -\lim_{N\to\infty} \frac 1N \log \mu\{x\mid |\frac 1N S_N f(x) - \int f\,d\mu| \geq \epsilon\} = \sup \{ h_\nu(T) + \int \phi \,d\nu - P(\phi) \mid \nu\in \mathcal{M}_T(X), \left|\int f\,d\nu - \int f\,d\mu\right| \geq \epsilon\} < 0. -$$ -This behaviour is typical in the uniformly hyperbolic setting. In the non-uniformly hyperbolic setting, there are also examples where the rate of decay is slower than exponential. For example, this occurs in the case of the absolutely continuous invariant measure for the Manneville-Pomeau map. - -REPLY [15 votes]: There's no estimate that works in general. Krengel, "On the speed of convergence in the ergodic theorem", shows that for any ergodic transformation of $[0,1]$ and any sequence $(a_n)$ converging to $0$, no matter how slowly, there is a set $A$ such that -$$\limsup_{N\rightarrow\infty}\frac{1}{a_N}|A_N\chi_A(x)-\mu(A)|=\infty$$ -for almost every $x$, where $A_N=\frac{1}{N}\sum_{i -TITLE: $(\varphi, \Gamma)$-module of dimension 2 modulo $p$ -QUESTION [5 upvotes]: Let $p$ be a prime number $\geq 3$. Let $V$ be a representation of $Gal(\bar{\mathbb{Q}}_p/ \mathbb{Q}_p)$ with coefficients in $\mathbb{F}_p$. Assume $V$ is a non-split extension of two characters $\delta_1$ and $\delta_2$. -Let $D$ be the $(\varphi, \Gamma)$-module associated to $V$. It is a non-split extension of two $(\varphi, \Gamma)$ of rank $1$. Denote by $\Phi$ the matrix of $\varphi$ and by $G$ the matrix of a generator $\gamma$ of $\mathbb{Z}_p^*$. In a suitable basis, we have $\Phi = \begin{pmatrix} f_1 & f \\ 0 & f_2 \end{pmatrix}$ and $G = \begin{pmatrix} g_1 & g \\ 0 & g_2 \end{pmatrix}$ where $f_1, f_2, g_1$ and $g_2$ are elements of $\mathbb{F}_p^{\times}$ (see One dimensional (phi,Gamma)-modules in char p) and where $f,g \in \mathbb{F}_p ((X))$. -Can we find a basis where $f$ and $g$ are in $\mathbb{F}_p[[X]]$ ? If not, what is the "nicest" form possible for $f$ and $g$ ? (i.e. can we kill some of the denominators and which one ?) - -REPLY [4 votes]: The answer is "no" in general. If $\delta_2=1$ and $\delta_1$ is the mod $p$ cyclotomic character, then there are both peu and très ramifiées extensions. Let $res(g)$ denote the coefficient of $1/X$ in $g$. Theorem: in your notation, the extension is peu ramifiée iff $res(g)=0$. See for instance proposition 3.7.5 of Mathieu Vienney's PhD http://perso.ens-lyon.fr/laurent.berger/thesevienney.pdf -As to the "nicest" possible form: you can construct elements of $H^1(G_{Q_p},V)$ from elements of $D(V)^{\psi=1}$, see lemma I.5.5 of Cherbonnier and Colmez' JAMS paper. If you make this explicit, you'll have strong restrictions on $f$ and $g$.<|endoftext|> -TITLE: Book Recommendation - PDE's for geometricians / topologists -QUESTION [24 upvotes]: I am looking for recommendations for a book on partial differential equations, which is not written for applied mathematicians but rather focused on geometry and applications in topology, as well as more "qualitative" methods, rather then approximations or techniques for solutions. -I am looking for anything that might help me to study this, books, papers, surveys, etc. -What is your recommended road map for familiarizing myself with this subject? -Thank you. - -REPLY [8 votes]: I find your question too broad. I would recommend starting with a book that focuses on a particular question or area in differential geometry and presents the PDE theory needed. A very incomplete list of suggestions include the following: - -Aubin's book is a good way to learn the PDE theory required for the Yamabe problem. -A long list of books on Yang-Mills gauge theory can be found here: -http://www.amazon.com/Mathematical-Gauge-Theory-Books/lm/R2JAMJ9TVJ3DYU -Another list of books on Ricci flow here: -http://www.amazon.com/s/ref=nb_sb_noss_2?url=search-alias%3Daps&field-keywords=ricci%20flow - -There are other books on the Atiyah-Singer index theorem, harmonic maps, minimal surfaces, the complex Monge-Ampere equation, etc. -Books on elliptic PDE's used by many differential geometers include Gilbarg-Trudinger and Morrey.<|endoftext|> -TITLE: categorical characterization of large cardinals -QUESTION [25 upvotes]: Question 1. Is there a categorical representation of Kunen's inconsistency result? -Question 2. Is there a categorical characterization of very large cardinals (in particular for strong and supercompact cardinals)? -Question 3. Is there a categorical characterization of $0^{\sharp}$? -Remark 1. A categorical characterization of other large cardinals is also welcome. -Remark 2. Andreas Blass in the paper "Exact functors and measurable cardinals" has proved that the existence of a measurable cardinal is equivalent to the existence of a non-trivial exact functor from the category of sets to the category of sets. -Would you please give references for such matters. -Remark 3. The following papers may have some information about the relation between category theory and large cardinals: -1) Adequate subcategories-Isbell -2) Small adequate subcategories-Isbell, -3) Structure of categories-Isbell, -4) Exact functors and measurable cardinals-Blass, -5) Exact functors, local connectedness and measurable cardinals-Adelman & Blass. - -REPLY [4 votes]: In the paper Proof theory and set theory Takeuti has given such a characterization for measurable cardinals, strongly compact cardinals, supercompact cardinals and even large cardinals. -Let me first give the characterization for what Takeuti calls an $\omega$-huge cardinal. Call $\kappa$ is $\omega$-huge if there exists a non-trivial elementary embedding $j: V_\gamma \to V_\gamma$ with critical point $\kappa$ such that $\gamma=\sup_{n<\omega}\kappa_n$ with $\kappa_0=\kappa$ and -$\kappa_{n+1}=j(\kappa_n)$. -Let $\mathcal{C}$ be the category whose objects are the sets $V_\alpha,$ $\alpha$ an ordinal and whose arrows are elementary embeddings. Then the following are equivalent: - -There is an $\omega$-huge cardinal, - -there exists a functor $F: \mathcal{C} \to \mathcal{C}$ and a nontrivial natural transformation -$\eta: F \to F.$ - - -Now let give a characterization of measurable cardinals. $\kappa$ is measurable iff there exists a functor $F:\bf Set \to Set$ such that: - -$F$ commutes with direct limits whose cardinality is $<\kappa,$ - -$F$ commutes with pullbacks and pushouts, and - -$F$ does not commute with the following simplest direct limit with the cardinality -$\kappa$: -$F(\kappa) \neq \operatorname{dirlim} (F(\alpha), (F(i_{\alpha, \beta}))_{\alpha \leq \beta <\kappa}$, - - -where $i_{\alpha, \beta}$ is the identity map from $\alpha$ to $\beta.$<|endoftext|> -TITLE: Can an Einstein metric have the same Levi-Civita connection with a non-Einstein one? -QUESTION [22 upvotes]: We say that two metrics are affinely equivalent if their Levi-Civita connections coincide. Is it possible that an Einstein (=Ricci tensor is proporional to the metric) is affinely equivalent to a metric which is not Einstein? -Of course, since affinely equivalent metrics have the same Ricci tensor, -the question is equivalent to the question whether there exists a metric $g$ such that -(1) its Ricci tensor $Ric$ is nondegenerate (as a bilinear form), -(2) is parallel, $\nabla^g Ric= 0$, -(3) and is not proportional to the metric. -The answer is negative for metrics of Riemannian and Lorentzian signature. In both cases the result follows from the description of affinely equivalent metrics in these signatures and it is hard to generalize it for general signature. -The motivation comes from projective geometry. It is known (Mikes MR0603226 or Kiosak-M - arXiv:0806.3169) that if two metrics are projectively, but not affinely equivalent, and one of them is Einstein, then the second is Einstein as well. It is interesting and important to understand whether the assumption that the metrics are not affinely equivalent is necessary. -Edited after the comment of Robert Bryant: Let us assume in addition that the metric is not the direct product of Einstein metrics - -REPLY [22 votes]: There are trivial examples that arise just by taking products of irreducible Einstein metrics with different Einstein constants. Whether an irreducible example exists is a much harder question. I do not know the answer to that, but I can think about it. (However, see below, where I answer this question.) -Of course, no such irreducible example can be Riemannian, and I guess, from your statement about the Lorentzian case, it can't be Lorentzian either, though I don't see that immediately. -In light of this, I guess the first case to try would be to see whether or not there could be one in dimension $4$ that is of type $(2,2)$. (See Addition 1.) -Addition 1: Indeed, there is an irreducible example in dimension $4$ of type $(2,2)$. Consider the $6$-dimensional Lie group $G$ with a basis for left-invariant forms satisfying the structure equations -$$ -\begin{aligned} -d\omega^1 &= - \alpha\wedge\omega^1 - \beta \wedge \omega^2 \\ -d\omega^2 &= \phantom{-} \beta\wedge\omega^1 - \alpha \wedge \omega^2\\ -d\omega^3 &= \phantom{-} \alpha\wedge\omega^3 - \beta \wedge \omega^4 \\ -d\omega^4 &= \phantom{-} \beta\wedge\omega^3 + \alpha \wedge \omega^4\\ -d\alpha &= c\ \bigl(\omega^3\wedge\omega^1+\omega^4\wedge\omega^2\bigr)\\ -d\beta &= c\ \bigl(\omega^4\wedge\omega^1-\omega^3\wedge\omega^2\bigr)\\ -\end{aligned} -$$ -where $c\not=0$ is a constant. -Let $H$ be the subgroup defined by $\omega^1=\omega^2=\omega^3=\omega^4=0$ and let $M^4 = G/H$. Then the above structure equations define a torsion-free affine connection $\nabla$ on $M^4$ that satisfies -$$ -\mathrm{Ric}(\nabla) = -4c\ (\omega^1\circ\omega^3+\omega^2\circ\omega^4) -$$ -while both $h = \omega^1\circ\omega^3+\omega^2\circ\omega^4$ and $g = \omega^1\circ\omega^4-\omega^2\circ\omega^3$ are $\nabla$-parallel. Thus, $g + \lambda h$ is an example of the kind you want for any nonzero constant $\lambda$. -Addition 2: Upon further reflection, I realized that this example points the way to a large number of other examples, all of split type, and most having irreducibly acting holonomy as soon as the (real) dimension gets bigger than $4$. -The reason is that the above example is essentially a holomorphic Riemannian surface of nonzero (but real) constant curvature regarded as a real Riemannian manifold by taking the real part of the holomorphic quadratic form, i.e., $Q = (\omega^1+i\ \omega^2)\circ(\omega^3-i\ \omega^4) = h - i\ g$. (The group $G$ in the above example turns out to just be $\mathrm{SL}(2,\mathbb{C})$ and $H\simeq \mathbb{C}^\times$ is just a Cartan subgroup.) -Now, the same phenomenon occurs in all dimensions: Let $(M^{2n},Q)$ be a holomorphic Einstein manifold with a nonzero real Einstein constant and write $Q = h - i\ g$ where $h$ and $g$ are real quadratic forms. Then $(M,h)$ will be an Einstein manifold of type $(n,n)$ (with a nonzero Einstein constant) and $g$ will be parallel with respect to the Levi-Civita connection of $h$. Thus, all of the split metrics $g+\lambda\ h$ for $\lambda$ real will have the same Levi-Civita connection as $h$ and none of them will be Einstein. If the (holomorphic) holonomy of $Q$ is $\mathrm{SO}(n,\mathbb{C})$, then the holonomy of $h$ will be $\mathrm{SO}(n,\mathbb{C})\subset\mathrm{SO}(n,n)$, which acts $\mathbb{R}$-irreducibly on $\mathbb{R}^{2n}=\mathbb{C}^n$ when $n\ge3$. -(Constructing examples of non-split type might be interesting$\ldots$) -Addition 3: By examining the Berger classification (suitably corrected by later work), one can see that, if $M$ is simply connected and if $h$ is a non-symmetric pseudo-Riemannian metric on $M$ with irreducibly acting holonomy whose space of $\nabla$-parallel symmetric $2$-tensors has dimension greater than $1$, then the dimension of $M$ must be even, say, $2n$, and the holonomy of the metric $h$ must lie in $\mathrm{SO}(n,\mathbb{C})$. Of the possible irreducible holonomies in this case, only the subgroups $\mathrm{SO}(n,\mathbb{C})$ and $\mathrm{Sp}(m,\mathbb{C})\cdot\mathrm{SL}(2,\mathbb{C})$ (when $n=2m$) can occur if the metric is to be Einstein with a nonzero Einstein constant. Both of these cases do occur, and, in each case, the space of $\nabla$-parallel symmetric $2$-tensors has dimension exactly $2$. Thus, the construction outlined in Addition 2 gives all of the examples of desired pairs $(h,g)$ for which the holonomy is irreducible and that are not locally symmetric. -To make sure that we get the full list of examples with irreducible holonomy, we'd have to examine Berger's list of irreducible pseudo-Riemannian symmetric spaces for other possible candidates. (I suspect that, even there, the examples will turn out to be holomorphic metrics in disguise, but I have not yet checked Berger's list to be sure.) -The case in which the metric is irreducible but the holonomy is not remains, and it may not be easy to resolve with known technology.<|endoftext|> -TITLE: Topology on the set of analytic functions -QUESTION [33 upvotes]: Let $H(D)$ be the set of all analytic functions in a region $D$ in $C$ or in $C^n$. -Everyone who worked with this set knows that there is only one reasonable topology -on it: the uniform convergence on compact subsets of $D$. - -Is this statement correct? - -If yes, can one state and prove a theorem of this sort: - -If a topology on $H(D)$ has such - and such natural properties, then it must be the topology of uniform convergence on - compact subsets. - -One natural property which immediately comes in mind is that the point -evaluatons must be continuous. What else? -EDIT: 1. On my first question, I want to add that other topologies were also studied. -For example, pointwise convergence (Montel, Keldysh and others). Still we probably all feel -that the standard topology is the most natural one. - -If the topology is assumed to come from some metric, a natural assumption would be -that the space is complete. But I would not like to assume that this is a metric space -a priori. -As other desirable properties, the candidates are continuity of addition and multiplication. However it is better not to take these as axioms, because the topology -of uniform convergence on compacts is also the most natural one for the space of -meromorphic functions (of one variable), I mean uniform with respect to the spherical metric. - -REPLY [2 votes]: Given any topological space $(T,\tau)$ and a mapping -$\varphi :T\to T$, it is natural to ask - -What is the coarsest topology, finer than $\tau$, such - that $\varphi$ is continuous. - -This topology is the supremum of all $\tau_n$, which is constructed by adding the inverse images, through $\varphi$, of all open subsets of $\tau_{n-1}$. Note it $\hat{\tau}$. -If one takes the space $T_0$ of infinitely differentiable functions on $\mathbb{R}$, $\tau_0$, the topology of local uniform convergence (which is reasonable to preserve continuity) and $\varphi=\frac{d}{dx}$, then $\hat{\tau_0}$ is the topology of compact uniform convergence of functions and all their derivatives (one can check that the sequence $\tau_{n}$ is strictly increasing). The process is stationary iff $\varphi$ is already continuous (obvious). What is remarkable is that, due to Cauchy, $\varphi$ is continuous for $\tau_0$. The process can be applied to a set $(\varphi_i)_{i\in I}$ of functions, $\tau_n$ being constructed by adding the inverse images, through $\varphi_{i}$ of all open subsets of $\tau_{n-1}$ and taking their unions and finite intersections. If, on $C^{\infty}(D;\mathbb{C})$, one takes $\varphi_1=\frac{\partial}{\partial x}$ and $\varphi_2=\frac{\partial}{\partial y}$, one get the topology of compact uniform convergence of functions and all their (partial) derivatives. The process described above is, again, strictly increasing, but the restriction of it to $H(D)$ is stationary which is IMHO remarkable.<|endoftext|> -TITLE: Has Fermat's Last Theorem per se been used? -QUESTION [57 upvotes]: There is a long tradition of mathematicians remarking that FLT in itself is a rather isolated claim, attractive only because of its simplicity. And people often note a great thing about current proofs of FLT is their use of the modularity thesis which is just the opposite: arcane, and richly connected to a lot of results. -But have there been uses of FLT itself? Beyond implying simple variants of itself, are there any more serious uses yet? -I notice the discussion in Fermat's Last Theorem and Computability Theory concludes that one purported use is not seriously using FLT. - -REPLY [16 votes]: Recall that around 1977 Mazur has completely classified the possible torsion groups of elliptic curves over $\mathbb Q$. A few years prior, Kubert has worked on this problem and has established a number of partial results, including, in the paper "Universal bounds on the torsion of elliptic curves", the following statement (Main result 1, second part): - -If $\ell>3$ is a prime for which Fermat's last theorem is valid, then $\ell^2\nmid |E_\mathrm{tor}(\mathbb Q)|$. - -(let me remark that the proof splits into cases $\ell>5$, which substantially uses the assumption, and $\ell=5$ which doesn't and uses a complicated descent argument) -With this theorem we can, relatively easily, prove that if FLT holds, then $|E_\mathrm{tor}(\mathbb Q)|$ is a product of a squarefree number and a factor of $12$. -Of course, this result predates FLT by a long shot, and was quickly superseded by Mazur's theorem, but it is still noteworthy because it relies on full FLT and not just a single case.<|endoftext|> -TITLE: Is the product of ultrafilters cancellative? -QUESTION [9 upvotes]: Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for each first order structure $\mathcal{A}$, the ultrapower $\mathcal{A}^{\mathcal{U}}$ is elementarily embeddable in $\mathcal{A}^{\mathcal{V}}$, and $\mathcal{U}=_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler equivalent to $\mathcal{V}$) iff $\mathcal{A}^{\mathcal{U}}$ is isomorphic to $\mathcal{A}^{\mathcal{V}}$ for each first order structure $\mathcal{A}$. If $\mathcal{U},\mathcal{V}$ are ultrafilters, then there is a unique up-to Rudin-Keisler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ is isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ for all first order structures $\mathcal{A}$. I have a few questions about the product operation on ultrafilters. - - -If $\mathcal{U}\cdot\mathcal{V}=_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$? -If $\mathcal{U}\cdot\mathcal{V}\leq_{RK}\mathcal{U}\cdot\mathcal{W}$, - then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$? -If $\mathcal{V}\cdot\mathcal{U}=_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$? -If $\mathcal{V}\cdot\mathcal{U}\leq_{RK}\mathcal{W}\cdot\mathcal{U}$, - then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$? -If $\mathcal{U}\cdot\mathcal{U}=_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}=_{RK}\mathcal{V}$? -If $\mathcal{U}\cdot\mathcal{U}\leq_{RK}\mathcal{V}\cdot\mathcal{V}$, - then do we necessarily have $\mathcal{U}\leq_{RK}\mathcal{V}$? - - -I would also be interested in any proof or reference of similar results about the product of ultrafilters. - -REPLY [8 votes]: Convention: I'll use my favorite notation $\mathcal U\otimes\mathcal W$ for what is called $\mathcal U\cdot\mathcal W$ in Jonathan Verner's answer and $\mathcal W\cdot\mathcal U$ in the question. It follows from a theorem of Mary Ellen Rudin (in the 1960's if I remember correctly) that any RK-equivalence between two such products is either trivial or an instance of associativity. That is, if $\mathcal U\otimes\mathcal V=_{RK}\mathcal U'\otimes\mathcal V'$ then one of the following three alternatives must hold: -$\mathcal U=_{RK}\mathcal U'$ and $\mathcal V=_{RK}\mathcal V'$, or -for some $\mathcal W$, $\mathcal U=_{RK}\mathcal U'\otimes\mathcal W$ and $\mathcal V'=_{RK}\mathcal W\otimes\mathcal V$, or -for some $\mathcal W$, $\mathcal U'=_{RK}\mathcal U\otimes\mathcal W$ and $\mathcal V=_{RK}\mathcal W\otimes\mathcal V'$. -Combining this result with the fact that, for nonprincipal ultrafilters, $\mathcal U\otimes\mathcal V$ is never RK-equivalent to $\mathcal U$ or to $\mathcal V$, we find that the three parts of the question about $=_{RK}$ (parts 1, 3, and 5) have affirmative answers. I don't immediately see answers to the $\leq_{RK}$ parts 2, 4, and 6.<|endoftext|> -TITLE: A question about the Stone–Čech compactification of discrete spaces -QUESTION [7 upvotes]: Let $D(\kappa)$ be the discrete space of cardinality $\kappa$, and $\beta D(\kappa)$ its Stone–Čech compactification. -Is there, for every infinite cardinal $\kappa$, a subset $Y \in [\beta D(\kappa)]^{\kappa^+}$ such that $\psi (Y) = \kappa^+$? Can this be proved in ZFC? -For $\kappa = \aleph_0$, this is true, as a consequence of the Corollary 4.4.5 in the Chapter 11 of the "Handbook of Set-theoretic Topology": - -4.4.5. COROLLARY. There is a point $x \in \omega^*$ and a (relatively) discrete sequence $\{ x_{\alpha} : \alpha < \omega_1 \} \subseteq \omega^* \setminus \{ x \}$, such that each neighborhood of $x$ contains all but countably many of the $x_{\alpha}$'s. - -REPLY [4 votes]: Here´s a ZFC answer to your question. -All spaces are assumed to be Hausdorff. Recall that the tightness $t(x,X)$ of a point $x$ in the space $X$ is defined as the least cardinal $\kappa$ such that for every set $A \subset X$ such that $x \in \overline{A}$ there exists a $\kappa$-sized subset $B$ of $A$ such that $x \in \overline{B}$. The tightness $t(X)$ of the space $X$ is then defined by taking the supremum of the local tightness over all points $x \in X$. -Arhangel´skii proved that the tightness of a compact space can be characterized by means of a special kind of discrete set called "free sequence". Although we´re not gonna need to use its definition directly, let me recall you what it is. -A set $F \subset X$ is called a free sequence of length $\kappa$ if there exists a well-ordering $\{x_\alpha: \alpha <\kappa \}$ of $F$ such that $\overline{\{x_\alpha: \alpha < \beta \}} \cap \overline{\{x_\alpha: \alpha \geq \beta \}}=\emptyset$, for every $\beta < \kappa$. The freeness $F(X)$ of a space $X$ is the supremum of the cardinalities of its free sequences. - - -LEMMA 1 (Arhangel´skii) Let $X$ be a compact space. Then $F(x)=t(X)$. -LEMMA 2 (Juhasz and Szentmiklossy) Let $X$ be compact space. If $X$ contains a free sequence of size $\kappa$, then it contains also one that is convergent. - - -Now, the space $\beta D(\kappa)$ has tightness greater than $\kappa$ (see below). Hence it contains a free sequence of size $\kappa^+$. So it contains a free sequence $F$ of size $\kappa^+$ which converges to some point $x$. Convergence implies that $x$ has pseudocharacter at least $\kappa^+$ in $Y=F \cup \{x\}$. On the other hand, the pseudocharacter of $Y$ cannot exceed $\kappa^+$, because $|Y|=\kappa^+$. -It remains to prove that: - -THEOREM: $t(\beta D(\kappa)) \geq \kappa^+$. - -There might be a simpler way of proving this, but I could only think of this off the top of my head. -Proof: By a theorem of Nogura ("Tightness of compact Hausdorff spaces and normality of products", J. Math. Soc. Japan 28, No. 2, 1976), the tightness of a space $X$ does not exceed $\kappa$ if and only if $X \times \kappa^+$ is normal, where $\kappa^+$ is provided with its natural order topology. So we just need to prove that $\beta D(\kappa) \times \kappa^+$ is not normal. -To that aim, note first that $(\kappa^++1) \times \kappa^+$ is not normal (you can prove this directly or just use Nogura's theorem cited above). Now $\kappa^++1 \subset 2^{\kappa^+}$ and this last space has density $\kappa$, by the Hewitt-Marczewski-Pondiczery theorem. Let $f$ be any (necessarily continuous) bijection from $D(\kappa)$ to a dense subset $D$ of $2^{\kappa^+}$ such that $|D|=\kappa$. By Corollary 3.6.6 of Engelking's "General Topology" book, $f$ can be extended to a (necessarily perfect) map $F$ from $\beta D(\kappa)$ onto $2^{\kappa^+}$. Since product of perfect maps is perfect we obtain a perfect map from $\beta D(\kappa) \times \kappa^+$ onto $2^{\kappa^+} \times \kappa^+$, but this last space is not normal, because it contains the closed non-normal subspace $(\kappa^++1) \times \kappa^+$. Since normality is preserved by perfect maps, we get that $\beta D(\kappa) \times \kappa^+$ is not normal.<|endoftext|> -TITLE: Koszul (exterior/symmetric) duality for a 1-dim vector space -QUESTION [7 upvotes]: The simplest example of Koszul duality (see introduction of Beilinson, Ginzburg, and Soergel - Koszul Duality Patterns in Representation Theory) is as follows. -Let $V = \mathbb{C}x$ be a $1$ dimensional vector space. Then the exterior algebra is $A=\mathbb{C}[x]/(x^2)$, and the symmetric algebra is $A^! = \mathbb{C}[x]$. Koszul duality states that there exists an isomorphism (where "mod" means the category of finitely generated modules): -$F: D^b(A\mbox{-mod}) \rightarrow D^b(A^!\mbox{-mod})$ -The heart of the category on the left is $A\mbox{-mod}$; what is the image of this on the right? It should be possible to work this out by working through the above paper, but there are many things in that paper which I don't understand (and this toy example should be easy). - -REPLY [4 votes]: There's actually a general answer to this question, given by Mazorchuk, Ovsienko and Stroppel - Quadratic duals, Koszul dual functors, and applications (though I assume some version of it was known earlier). The category of graded representations of $A^!$ is the same as the (abelian!) category of linear projective complexes over $A$, and vice versa. So, in Dag Oskar Madsen's answer, I would write not $\mathbb{C}[n]\langle n\rangle$, but rather its projective resolution $A^![n+1]\langle n+1\rangle\overset{x}\to A^![n]\langle n\rangle$. Similarly, the $A^!$ modules $\mathbb{C}[x]/(x^{n+1})$ are sent by Koszul duality to the complexes -$$A[n]\langle n\rangle \overset{x} \to A[n-1]\langle n-1\rangle \overset{x} \to\dotsb \overset{x} \to A.$$<|endoftext|> -TITLE: Is there a Wall finiteness obstruction in other settings? -QUESTION [10 upvotes]: Let $\mathcal{S}$ be the $(\infty, 1)$-category of spaces. Then the compact objects of $\mathcal{S}$ are precisely the retracts of finite CW complexes. These are not the same as the finite CW complexes (i.e., the smallest subcategory of $\mathcal{S}$ containing $\ast$ and closed under finite homotopy colimits). Namely, given $X$ (let's say connected), the homology of the universal cover defines a perfect complex of $\mathbb{Z}[\pi_1 X]$-modules, which gives a class in $\widetilde{K}_0( \mathbb{Z}[\pi_1 X])$, which vanishes when $X$ is a finite CW complex. This class is the Wall finiteness obstruction; it is a classical result that this is the only obstruction for $X$ to be a finite CW complex. The reason for its existence is that taking the "image" of an idempotent is not a finite homotopy colimit (though it would be if one worked with $n$-categories for some $n$). -I'm curious about other examples of this type of finiteness obstruction. For instance, what happens if we do this in (for example) the $(\infty, 1)$-category of $E_\infty$-rings? The analogs of the finite CW complexes are the $E_\infty$-rings obtained from the free ones (free on a generator in some degree) via finite colimits, and the compact objects are these and the retracts. Is there an analog of the finiteness obstruction here? Other examples: algebras over some other operad, simplicial commutative rings, (pre)sheaves of spaces (resp. spectra), and equivariant spaces (resp. spectra). The only example that I am aware of considers the derived category of modules over a ring, where $\widetilde{K_0}$ measures the difference between "perfect complexes" and "those representable via finite complexes of frees." - -REPLY [6 votes]: My 1985 Math. Scand. paper The algebraic theory of the finiteness obstruction has a chain complex treatment of the Wall finiteness obstruction. -Chapter VII of Hans-Joachim Baues' 1999 Springer Monograph Combinatorial Foundation of Homology and Homotopy has an abstract homotopy theory treatment of the Wall finiteness obstruction.<|endoftext|> -TITLE: There are only finitely many varieties up to deformation -QUESTION [9 upvotes]: Let $h$ be a polynomial. Then results of several authors (including Chow, Grothendieck, Matsusaka, Mumford, Kollar and Viehweg) imply that the moduli space of polarized varieties with Hilbert polynomial $h$ is of finite type (even quasi-projective). -This means, in particular, that the moduli space has only finitely many connected components, and therefore, that there are only finitely many polarized varieties with Hilbert polynomial $h$ up to deformation. -Proving that the moduli space is of finite type is formally more difficult than the latter statement, I believe. -Question. Is there an "easy" proof for the finiteness of the number of deformation classes of polarized varieties with Hilbert polynomial $h$? - -REPLY [7 votes]: I believe the two things you are relating: -The finiteness of the - -number of deformation types -number of components of the moduli space - -of polarized varieties are essentially equivalent problems. -The way moduli spaces are usually constructed is that first one finds a projective space that contains all the objects in the class in question embedded by the given polarization (or possibly a fixed power). This requires a "Matsusaka's Big Theorem"-type result. -Next one considers the locus in the appropriate Hilbert scheme that parametrizes these objects. Then one needs various results, for instance that this locus is locally closed. This is usually difficult, but has little to do with finiteness. -Now, the deformation types correspond to the components of this locus. The reason this is not yet a moduli space is that the same object may be embedded several times depending on the choice of sections of the appropriate line bundle. So, the moduli space (if it exists) is constructed as the quotient of this by the action of the appropriate PGL (and of course one needs a theorem that says that it exists). Since this is an algebraic group, the quotient will be of finite type exactly when this locus in the Hilbert scheme is. -So, as Jason already said in a comment, to prove finiteness of deformation types, the difficult thing to prove is that there is some kind of boundedness. Being locally of finite type often follows directly from the same property of the Hilbert scheme (see comment of user36938), although I have to add that there are some situations when this last statement is not entirely true and in that case boundedness is really hard to prove. (Think of how $\mathbb Z$ is inside something of finite type over $\mathbb C$, say $\mathbb C$, but itself is not of finite type (over $\mathbb C$).) -So, I think the answer to your question seems to be "no". - -Added: I believe that those proofs you mention that these moduli spaces are of finite type actually prove the finiteness of deformation types. The main difference in difficulty is the existence. That's pretty tricky in general for the moduli space.<|endoftext|> -TITLE: Is this known alternating sum for Euler's constant? -QUESTION [5 upvotes]: This probably is known, but Wolfram Alpha doesn't recognize it -and couldn't find it in Mathworld (there is something close, -but using floor). -We have -$\lim_{s \to 1} (\zeta(s)-1/(s-1)) = \gamma$ -Also -$F(s) = \zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} $. -According to Maple 13: -$$\lim_{s \to 1} (F(s)-1/(s-1)) = \sum _{n=1}^{\infty }-{\frac { \left( -1 \right) ^{n-1}\ln \left( n \right) }{n}} \left( \ln \left( 2 \right) \right) ^{-1}+1/2\,\ln \left( 2 \right) = \gamma \qquad (1) $$ - -Is (1) known and/or trivial? - -I believe all terms and partial sums except the first of the sum are transcendental. - -Intuitive explanation how (1) could be hypothetically rational? - - - -Reference request? Was this known to Euler? - -Numerically (1) is correct to precision at least $500$ decimal digits. -Sage code: -nsu=1/2*mpmath.log(2)-mpmath.nsum(lambda n: (-1)**(n-1)*mpmath.log(n)/n ,[1, mpmath.inf])/ mpmath.log(2);nsu - -REPLY [11 votes]: Denote -$$ -f(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s). -$$ -Expanding into series and using $\zeta(s)=\frac1{s-1}+\gamma+O((s-1))$ leads to -$$ -f(s)=\log 2+(s-1) \left(\gamma \log 2-\frac{\log - ^22}{2}\right)+O\left((s-1)^2\right). -$$ -Differentiating both sides gives -$$ -f'(1)= --\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}\ln \left( n \right) }{n}}= -\gamma \log 2-\frac{\log - ^22}{2}. -$$<|endoftext|> -TITLE: Is there a monster behind the trees? -QUESTION [7 upvotes]: ‎First Fix the following notation:‎ -‎ -$‎‎\forall ‎\kappa‎\in Card~~~Tp(‎\kappa‎):="‎\kappa‎~has~tree~property"$‎ -‎‎‎‎ -The large cardinals as "monsters of heaven" live everywhere in the land of mathematics. Most of them appear in topology, measure theory, infinitary logic, category, model theory, etc. But some of them live in a strange misty place, "the forest of tall trees"! One of the most important problems of set theory is to discover that which one of these monsters have a forestial nature? The forestial nature of a monster gives us very important information about his possible behavior at many combinatorial situations in set theory and model theory. Now define the following concepts:‎ -‎ -‎ -Definition (1): ‎Let ‎‎$‎A‎$‎, ‎$‎‎B‎$ be ‎large ‎cardinal ‎axioms, $I‎$‎, ‎$‎J‎$‎ ‎be ‎sub ‎classes ‎of ‎‎$‎Ord‎$ , ‎‎$‎‎\lbrace ‎‎\alpha‎‎_{i}‎‎\rbrace‎_{i\in I}, ‎\lbrace ‎‎‎‎‎‎\beta‎‎_{j}‎‎\rbrace‎_{j\in J}\subseteq Ord$ , $\lbrace ‎‎‎‎‎‎\alpha‎‎_{i}‎‎\rbrace‎_{i\in I}\cap ‎‎\lbrace ‎‎‎‎‎‎\beta‎‎_{j}‎‎\rbrace‎_{j\in J}=‎\emptyset‎‎‎$ -‎then respectively every ‎statement ‎in ‎the ‎form ‎‎$(‎‎‎\star)‎$ and ‎$‎‎(‎\star‎‎\star‎)$‎ ‎called a‎ ‎"weak" and "strong" tree property "equation".‎ -‎ -‎‎ -$(‎\star‎)~~Con(ZFC+A+\bigwedge_{i\in I}Tp(\aleph_{‎\alpha‎_{i}}))\Longleftrightarrow‎ Con(ZFC+B+\bigwedge_{j\in J}Tp(\aleph_{‎\beta‎_{j}}))$ ‎‎ -‎ -$(‎\star\star‎)~~(A+\bigwedge_{i\in I}Tp(\aleph_{‎\alpha‎_{i}}))\Longleftrightarrow ‎(‎B+\bigwedge_{j\in J}Tp(\aleph_{‎\beta‎_{j}}))$ -‎ -‎ -Example (1): Note to the following ‎well ‎known ‎results:‎ -‎ -‎ -A‎ ‎wea‎k tree property equation: ‎ -‎‎ -‎ -$Con(ZFC+Tp(\aleph_{2}))\Longleftrightarrow ‎Con(ZFC+‎\exists~a~weakly~compact~cardinal‎‎‎)‎‎‎$‎‎ -‎ -‎‎ -A strong tree property equation: -‎ -‎ -$(‎\kappa‎‎~is~strongly~inaccessible~+~Tp(‎\kappa‎))‎\Longleftrightarrow (‎‎‎‎\kappa‎‎~is~weakly~compact)‎‎‎$‎‎ -‎‎‎ -‎ -Remark (1): ‎It ‎seems ‎there are ‎many weak ‎tree ‎property "‎unequalities" like the following results, ‎but strangely the weak and strong "equalities" are too rare. These equalities uncover some fundamental relations between large cardinal axioms and combinatorial properties of other cardinals and so could be very useful and valuable. -‎‎‎ -A‎ ‎weak ‎tree ‎property ‎unequality ‎discoverd by Menachem ‎Magidor:‎ -‎ -‎‎ -$‎Con(ZFC+ ‎‎‎\exists‎~\mathtt{0}^{\sharp}‎)‎\Longleftarrow‎ ‎‎‎Con(ZFC+Tp(\aleph_{2})+Tp(\aleph_{3}))‎‎$‎ -‎ -‎‎ -‎Another ‎weak ‎tree ‎property ‎unequality ‎discoverd ‎by ‎Uri ‎Abraham:‎ -‎ -‎‎‎ -‎$‎‎‎Con(ZFC+‎\exists‎~a~supercompact~cardinal~with~a~weakly~compact~cardinal~above~it‎) ‎‎\Longrightarrow‎ ‎Con(ZFC+Tp(\aleph_{2})+Tp(\aleph_{3}))‎$‎ -‎ -‎ ‎ -Question (1): ‎Is ‎there ‎any ‎known large ‎cardinal ‎axiom stronger than "$‎‎‎\mathtt{0}^{\sharp}‎$ exists" ‎and ‎weaker ‎than "there is a supercompact cardinal with a weakly compact cardinal above it " ‎‎like ‎$‎‎A‎$ ‎in ‎which ‎we have an ‎"weak tree property equality" ‎in ‎Magidor ‎and ‎Abraham's ‎results? ‎What ‎about ‎"strong tree property equality"? ‎‎In ‎the ‎other ‎words‎: is there any large cardinal axioms ‎$‎A‎$, ‎$B$, ‎‎$C$‎ ‎which ‎the ‎foll‎owing statements be true?‎ -‎ -‎ -$(1)~Con(ZFC+A)\Longleftrightarrow ‎Con(ZFC+Tp(\aleph_{2})+Tp(\aleph_{3}))‎‎‎‎$‎ -‎ -‎‎ -‎$‎‎(2)~B \Longleftrightarrow (C+Tp(\aleph_{2})+Tp(\aleph_{3}))$‎ -‎ -‎‎ -Question (2): Are there any other known weak or strong tree property equality different from statements in example ‎$‎‎(1)$?‎ - -REPLY [2 votes]: For a strong special tree property equation, here is Theorem 9.4 in Todorcevic´s "Coherent Sequences": - -$\kappa$ is strongly inaccessible + $sTp(\kappa) \Longleftrightarrow \kappa$ is Mahlo. - -Where $sTp(\kappa)$ would mean "there are no special $\kappa$-Aronszajn trees".<|endoftext|> -TITLE: Homotopes of simple Lie algebras -QUESTION [10 upvotes]: Let $\mathfrak{g}$ be a complex simple Lie algebra with bracket $[x,y]$. For which $z\in \mathfrak{g}$ does the formula -$$ -\mu(x,y)=ad (z)([x,y])=[z,[x,y]] -$$ -define another Lie bracket on the same vector space ? For $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$ this holds for every $z$. Explicit computation seems to show that for $\mathfrak{g}=\mathfrak{sl}(n,\mathbb{C})$ with $n\ge 3$ only $z=0$ is possible. Does this hold for all simple Lie algebras of rank at least $2$, and is there a proof somewhere ? -In the literature, often the Jacobi identity for $\mu(x,y)=[z,[x,y]]$ is studied for all $z$. Then it is known that only $\mathfrak{sl}(2,\mathbb{C})$ is possible. Another reference are simple Hom-Lie algebras, but this is different. -For nonassociative algebras a new multiplication $x\circ_{z} y$ on the same vector space induced by an element $z$ is called a homotope. -What is known for simple Lie algebras ? -EdiT: I saw that $[x,y]_R:=[R(x),y]+[x,R(y)]$ is a Lie bracket iff $R$ is a classical $R$-matrix. For $R=ad(z)$ we have $[x,y]_R=[z,[x,y]]$. So the question is, for which $z$ the map $R=ad(z)$ is a classical $R$-matrix of a simple Lie algebra. - -REPLY [7 votes]: In the paper Derivation Double Lie Algebras the following result is proved: -Theorem $3.2$: Let $\mathfrak{g}$ be a simple Lie algebra of rank $r\ge 2$ over an algebraically closed field $K$ of characteristic -zero, and $z\in \mathfrak{g}$. Suppose that $R={\rm ad}(z)$ is a classical $R$-matrix. Then $z=0$ and $R=0$. -The proof uses the Jordan-Chevalley decomposition for $z=n+s$, the Jacobson-Morozov theorem, arguments concerning roots and algebraic group actions etc. This gives $s=0$ and ${\rm ad}(z)^3=0$, and finally $z=0$. -For the case of rank $1$ we have the following result: -Proposition $3.1$: Let $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$. Then $R={\rm ad}(z)$ is a classical $R$-matrix for all $z\in \mathfrak{g}$. -The Lie algebra $\mathfrak{g}_R$ is isomorphic to $\mathfrak{r}_{3,1}(\mathbb{C})$ for all $z\neq 0$. -Here $\mathfrak{r}_{3,1}(\mathbb{C})$ is the $3$-dimensional solvable Lie algebra given by $[e_1,e_2]=e_2$ and $[e_1,e_3]=e_3$, and $\mathfrak{g}_R$ has bracket $[x,y]_R=[z,[x,y]]$.<|endoftext|> -TITLE: Quotients of l^infty -QUESTION [10 upvotes]: Let $M$ be a closed subspace of $l^\infty$. Suppose that the quotient $l^{\infty}/M$ is isomorphic to $l^\infty$. Is it true that $M$ is complemented in $l^\infty$? - -REPLY [10 votes]: Notice that the comment by Yemon gives a negative answer to the question. Bourgain constructs a short exact sequence -$0 \to \ell_1 \to \ell_1 \to X \to 0$ -that does not split. The space $X$ is not $\mathcal{L}_1$ for then (since the kernel of the quotient map is a dual space and the quotient mapping onto any $\mathcal{L}_1$ space locally lifts) the quotient map would lift by a classical result of J. Lindenstrauss. Dualizing the diagram, you get a quotient mapping from $\ell_\infty$ onto $\ell_\infty$ whose kernel is not $\mathcal{L}_\infty$ and thus is not complemented in $\ell_\infty$. -Incidentally, from such an $\ell_\infty$ example you can deduce the $\ell_1$ result by localization; that is, modulo standard localization arguments, your question is equivalent to the question Bourgain answered. The moral is "don't look for an easy proof".<|endoftext|> -TITLE: Chromatic number of induced subgraphs as upper bound to the chromatic number -QUESTION [7 upvotes]: Motivation: At the Erdős100 conference in Budapest András Gyárfás presented some interesting conjectures. One of them was the following: -Given that in a graph $G$, every subgraph $H$ formed by taking the induced subgraph on the vertices of a path, has chromatic number at most $r$. Is it true that then the chromatic number of $G$ is bounded by a function of $r$? -Question: Are there any results such that: If in a graph $G$, every induced subgraph with a special property has chromatic number at most $r$, then the chromatic number of $G$ is bounded by $f(r)$? -Of course i would like to avoid special properties that enable induced subgraphs with a lot of vertices, like the property that "This induced subgraph contains all but one vertex". -EDIT: Let $G$ be connected. - -REPLY [3 votes]: There is a very similar question to that of Gyarfas which is well-known, still open, and notoriously hard. -Question: Is it true that if every triangle-free induced subgraph of $G$ has chromatic number at most $r$, then $G$ has chromatic number at most $f(r,\omega)$, where $\omega$ is the clique number of $G$? -A positive answer to this question would directly imply the following conjecture of Gyarfas: -Conjecture: There is a function $g$ such that every graph with no odd induced cycle of length at least 5, and clique number at most $k$, has chromatic number at most $g(k)$.<|endoftext|> -TITLE: Double covers of the orthogonal groups -QUESTION [6 upvotes]: Let $S:=P\Omega_{2n}^+(q)$ with $n$ even and $q$ odd prime power be the simple orthogonal group. Then the Schur multiplier of $S$ is the Klein four-group $Z_2\times Z_2$. Therefore $S$ has three double covers. -Is there any relation between these three double covers? Are they isomorphic? When $n=4$ I know that the three involutions in $Z_2\times Z_2$ are permuted by an outer automorphism of $S$ of degree $3$ so that these three double covers are isomorphic. -Thank you in advance! - -REPLY [7 votes]: Sorry for editing this answer multiple times. However, as I managed to get the answer wrong I feel obliged to improve this answer and provide a few more details. I've broken this up into several parts, so you can read as much as you care about. - -My original (incorrect) answer: -The three double covers of the group you consider can be described as follows (I'll assume $n>4$ as you've already described the $n=4$ case). One is the special orthogonal group $\operatorname{SO}_{2n}^+(q)$ and the remaining two are the Half-Spin groups $\operatorname{HSpin}_{2n}(q)$. The half-spin and special orthogonal groups are not isomorphic but the two half-spin groups are isomorphic. Why this is the case is described in section 5 of this paper -http://arxiv.org/abs/1211.2551 -It essentially boils down to the fact that the Dynkin diagram of type $\mathrm{D}_n$ has an order 2 automorphism permuting the two nodes attached to the branch point. - -Why the first answer is wrong: -The mistake comes from thinking on the level of algebraic groups. Let $\mathbf{G}$ be a simple simply connected algebraic group of type $\operatorname{D}_n$ defined over an algebraic closure $\mathbb{K} = \overline{\mathbb{F}_p}$ of the finite field of odd prime order $p$, then $\mathbf{G}$ is the spin group $\operatorname{Spin}_{2n}(\mathbb{K})$. Under the restrictions applied in the original post (namely that $p$ is odd) one can obtain the remaining simple algebraic groups of type $\operatorname{D}_n$ as $\mathbf{G}/\mathbf{K}$ where $\mathbf{K} \leqslant Z(\mathbf{G})$ is a subgroup of the centre. -In particular, taking $\mathbf{K} = Z(\mathbf{G})$ we obtain the adjoint group $\mathbf{G}_{\mathrm{ad}}$ of type $\operatorname{D}_n$. Taking $\mathbf{K}$ to be one of subgroups of index 2 we obtain three groups, which are double covers of $\mathbf{G}_{\mathrm{ad}}$. One of these groups is the special orthogonal group $\operatorname{SO}_{2n}(\mathbb{K})$ and the remaining two are isomorphic and called the half-spin groups $\operatorname{HSpin}_{2n}(\mathbb{K})$. -We now let $F$ be a split Frobenius endomorphism of $\mathbf{G}$ defining an $\mathbb{F}_q$-rational structure $G = \mathbf{G}^F$, which is isomorphic to the finite spin group $\operatorname{Spin}_{2n}^+(q)$. The Frobenius endomorphism $F$ acts trivially on $Z(\mathbf{G})$ so stabilises every subgroup of $Z(\mathbf{G})$. In particular $F$ induces a Frobenius endomorphism on the quotients $\mathbf{G/K}$, where $\mathbf{K}$ is an index 2 subgroup of $Z(\mathbf{G})$, which we again denote by $F$. The important thing to realise here is that -$$(\mathbf{G/K})^F \not\cong \mathbf{G}^F/\mathbf{K}^F.$$ -This happens because $\mathbf{K}$ is not connected. In particular, the situation on the level of the algebraic groups does not directly descend to the corresponding finite groups. - -The correct answer: - -The finite simple group $S$ has three double covers. If $n=4$ then all - three double covers are isomorphic. If $n>4$ then two of these double - covers are isomorphic but they are not isomorphic to the third. - -Let $\mathbf{G}$ be the spin group (as above). Let us fix a maximal torus $\mathbf{T}_0$ of $\mathbf{G}$ and a Borel subgroup $\mathbf{B}_0$ containing $\mathbf{T}_0$. We denote by $(\Phi,X,\Phi^{\vee},X^{\vee})$ the root datum of $\mathbf{G}$ with respect to $\mathbf{T}_0$. In particular, $X = \operatorname{Hom}(\mathbf{T}_0,\mathbb{K}^{\times})$ and $X^{\vee} = \operatorname{Hom}(\mathbb{K}^{\times},\mathbf{T}_0)$ are the character and cocharacter groups respectively and $\Phi \subset X$ and $\Phi^{\vee} \subset X^{\vee}$ are the roots and coroots of $\mathbf{G}$. -The group $X^{\vee}$ can be viewed as a $\mathbb{Z}$-module, hence we can form the tensor product $V = \mathbb{R} \otimes_{\mathbb{Z}} X^{\vee}$ which is a real vector space. Recall that we have a perfect pairing $\langle -,-\rangle : X \times X^{\vee} \to \mathbb{Z}$, which can naturally be extended to the tensor products $\mathbb{R}\otimes_{\mathbb{Z}}X$ and $\mathbb{R}\otimes_{\mathbb{Z}}X^{\vee}$, then we define the coweight lattice to be -$$\Lambda^{\vee} = \{\gamma \in V \mid \langle \alpha,\gamma\rangle \in \mathbb{Z}\text{ for all }\alpha\in\Phi\}.$$ -Clearly $\mathbb{Z}\Phi^{\vee} \subset X^{\vee} \subset \Lambda^{\vee}$ (here $\mathbb{Z}\Phi^{\vee}$ denotes the $\mathbb{Z}$-span of $\Phi^{\vee}$ in $V$) but less clear is the fact that $\Lambda^{\vee}/\mathbb{Z}\Phi^{\vee}$ is isomorphic to the Klein four group. Note that, as $\mathbf{G}$ is simply connected we have $X^{\vee} = \mathbb{Z}\Phi^{\vee}$. The simple roots $\Delta$ also define a set of fundamental dominant coweights $\Omega^{\vee} = \{\omega^{\vee}_{\alpha} \mid \alpha\in \Delta\}$ by the condition that $\langle \beta,\omega^{\vee}_{\alpha} \rangle = \delta_{\alpha,\beta}$ (the Kronecker delta). -Recall that for each root $\alpha \in \Phi$ there is a minimal 1-dimensional unipotent subgroup $\mathbf{X}_{\alpha} \leqslant \mathbf{G}$ normalised by $\mathbf{T}_0$ called the root subgroup of $\alpha$. Let $\Phi^+$ be all $\alpha \in \Phi$ such that $\mathbf{X}_{\alpha} \leqslant \mathbf{B}_0$ then $\Phi^+$ forms a positive system of roots for $\Phi$, hence determines a unique system of simple roots $\Delta \subset \Phi^+$. -Each root subgroup $\mathbf{X}_{\alpha}$ is isomorphic to the additive group $\mathbb{K}^+$. We now choose an isomorphism $x_{\alpha} : \mathbb{K}^{\times} \to \mathbf{X}_{\alpha}$ for each $\alpha \in \Phi$ such that $tx_{\alpha}(c)t^{-1} = x_{\alpha}(\alpha(t)c)$ for all $t \in \mathbf{T}_0$ and $c \in \mathbb{K}^+$, then the set $\{x_{\alpha}(c) \mid \alpha \in \Delta, c \in \mathbb{K}^+\}$ forms a generating set for $\mathbf{G}$. We may now define the split Frobenius endomorphism $F$ on $\mathbf{G}$ by the condition that -$$F(x_{\alpha}(c)) = x_{\alpha}(c^q)$$ -for all $\alpha \in \Delta$ and $c \in \mathbb{K}^+$. Now let $b : \Delta \to \Delta$ be a bijection induced by a graph automorphism of the root system $\Phi$. We may now define an automorphism $\tau : \mathbf{G} \to \mathbf{G}$ by setting -$$\tau(x_{\alpha}(c)) = x_{b(\alpha)}(c)$$ -for all $\alpha \in \Delta$ and $c \in \mathbb{K}^{\times}$. It's important to note that $F$ and $\tau$ do not necessarily have this effect on $x_{\alpha}(c)$ for any $\alpha \in \Phi$. -Clearly $F$ and $\tau$ commute so $\tau$ restricts to an automorphism of the finite group $G = \mathbf{G}^F$. We wish to understand the action of $\tau$ on the centre of $\mathbf{G}$ and $G$. To do this we will need some more notation. Let us fix an isomorphism $(\mathbb{Q}/\mathbb{Z})_{p'} \to \mathbb{K}^{\times}$ then we obtain a surjective homomorphism $\iota : \mathbb{Q} \to \mathbb{K}^{\times}$ as the composition $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to (\mathbb{Q}/\mathbb{Z})_{p'} \to \mathbb{K}^{\times}$. We may now define a surjective homomorphism $\iota_{\mathbf{T}_0} : \mathbb{Q} \otimes_{\mathbb{Z}} X^{\vee} \to \mathbf{T}_0$ by setting -$$\iota_{\mathbf{T}_0}(r\otimes\gamma) = \gamma(\iota(r)).$$ -Noting that $\Lambda^{\vee} \subseteq \mathbb{Q}\otimes_{\mathbb{Z}}X^{\vee}$ we have the following lemma. -Lemma: The restriction of the homomorphism $\iota_{\mathbf{T}_0}$ to $\Lambda^{\vee}$ induces an isomorphism $(\Lambda^{\vee}/X^{\vee})_{p'} \to Z(\mathbf{G})$. -In fact, this gives us an isomorphism $\varphi : \Lambda^{\vee}/\mathbb{Z}\Phi^{\vee} \to Z(\mathbf{G})$ because $X^{\vee} = \mathbb{Z}\Phi^{\vee}$, the quotient $\Lambda^{\vee}/\mathbb{Z}\Phi^{\vee}$ is the Klein four group and $p$ is odd. Now, any element of the quotient $\Lambda^{\vee}/\mathbb{Z}\Phi^{\vee} $ can be represented by a fundamental dominant coweight $\omega_{\alpha}^{\vee}$ for some simple root $\alpha\in\Delta$. Furthermore, it can be shown that -$$\tau(\varphi(\omega^{\vee}_{\alpha} + \mathbb{Z}\Phi^{\vee})) = \varphi(\omega^{\vee}_{b(\alpha)} + \mathbb{Z}\Phi^{\vee}).$$ -Using this and the fact that $Z(G) = Z(\mathbf{G})^F$ one easily deduces the answer.<|endoftext|> -TITLE: Can a tangle of arcs interlock in plane? -QUESTION [7 upvotes]: This is a variation of the question Can a tangle of arcs interlock?, asked by Joseph O'Rourke, and solved. I reproduce the question here: - -Can a (finite) collection of disjoint circle arcs in $\mathbb{R}^3$ be - interlocked in the sense in that they cannot be separated, i.e. each - moved arbitrarily far from one another while remaining disjoint (or at - least never crossing) throughout? (Imagine the arcs are made of rigid - steel; but infinitely thin.) The arcs may have different radii; each - spans strictly less than $2 \pi$ in angle, so each has a positive - "gap" through which arcs may pass. - -My proposed variation is: - -Can they interlock in $\mathbb R^2$? - -I posted a comment at the original question, claiming that three circle arcs can be locked. - -And two cannot - -I soon realized that the examples with three arcs can in fact be unlocked, and I think Joseph O'Rourke did the same. -I reproduce here my solution to unlock them: - -So, the question is still open for two dimensions. - -REPLY [13 votes]: Surely if you take your first example, and put two more green arcs, one in the red circle but not in the blue, and one in the blue circle but not in the red, about as large as they possibly can be, that will hold the configuration fairly rigid and prevent that unlinking operation? -I don't see that anything significant can be done here except idly rotating the green circles in place.<|endoftext|> -TITLE: Do simple, multi-dimensional generalizations of this continued fraction formula exist? -QUESTION [8 upvotes]: Background -Let $x\in [0,1)\setminus \mathbb{Q}$ have regular continued fraction expansion given by -$$ -x = [a_1, a_2, a_3, \dots] = \cfrac{1}{a_1+\cfrac{1}{a_2+\dots}}, \qquad a_i \in \mathbb{N}. -$$ -Let -$$ -\frac{p_n}{q_n} = [a_1, a_2, \dots, a_n] -$$ -be the $n$th convergent, and let $T^n x = [a_{n+1}, a_{n+2}, a_{n+3},\dots]$. -A well-known formula gives -$$ - x- \frac{p_n}{q_n} = \frac{(-1)^n \cdot x \cdot Tx \cdot T^2 x \cdot \dots \cdot T^{n}x}{q_n} = \frac{(-1)^n }{q_n(q_{n+1}+q_nT^{n+1} x)}. -$$ -This formula holds (with very minor changes, if any) for most kinds of one-dimensional continued fractions, including the various real variants, such as even continued fractions, odd continued fractions, Nakada's $\alpha$-continued fractions, etc., and - I believe - for Hurwitz's complex continued fractions as well. -The question -On page 132 of his book on Multidimensional Continued Fractions, Schweiger gives a generalization of the formula as a corollary to Perron's Identity; however, this formula is comparatively much more complicated. -Are there specific multi-dimensional continued fraction algorithms for which this formula simplifies considerably to something closer to what is seen for one-dimensional continued fractions? Do there exist other analogs of this formula for multi-dimensional continued fractions? -Motivation -Anton Lukyanenko and I worked on a paper studying continued fractions on the Heisenberg group (a two-dimensional complex system), and were surprised to see a simple analog of the formula in our work (Theorem 3.23 on page 22). We believe this is unique among multidimensional continued fractions, but we were not sure if there existed other simple analogs that were just not as well known. - -REPLY [5 votes]: One of natural generalizations of continued fractions (CF) is 3D Voronoi-Minkowski CF. For Voronoi construction see Delone & Fadeev's The theory of irrationalities of the third degree for Minkowski see Hancock, H. Development of the Minkowski geometry of numbers. Vol. 1 (ch. X). This object is nice from different points of view. But it is essentially 2D object. -We can consider classical CF as a chain or as broken line, while 3D CF is a plane graph. We can walk along this graph in different directions. The answer depends on your problem. You can try to find all minima of a linear form $x+\alpha y+\beta z$ and you'll go in one direction, you can try to minimize simultaneously two forms $x+\alpha y$ and $x+\beta z$, and then you must go in another direction. -There is no universal formula because time is 2-dimesional and there is no universal direction. But one good news. We have universal invariant measure -$$dx_2\,dx_3\,dy_1\,dy_3\,dz_1\,dz_2\begin{vmatrix} - 1 & x_2 & x_3\\ -y_1 & 1 & y_3\\ - z_1&z_2&1\\ -\end{vmatrix}^{-3} -$$ -which is an analog of classical Gauss measure -$$dx_2\,dy_1\begin{vmatrix} - 1 & x_2 \\ -y_1 & 1 \\ - \end{vmatrix}^{-2}. -$$ -It arises in different contexts. -(See for example Kontsevich, M. L. & Suhov, Y. M. Statistics of Klein polyhedra and multidimensional continued fractions Pseudoperiodic topology, Amer. Math. Soc., 1999, 197, 9-27, -Karpenkov, O. On invariant Mobius measure and Gauss-Kuzmin face distribution ArXiv Mathematics e-prints, 2006 -A. A. Illarionov The average number of relative minima of three-dimensional integer lattices of a given determinant. Izvestiya: Mathematics, 2012, 76:3, 535–562.)<|endoftext|> -TITLE: What is the classification of characters in $p$-adic Hodge theory? -QUESTION [15 upvotes]: Let $K$ be a $p$-adic field and $\chi : Gal_K \rightarrow \mathbb{Q}_p^\times$ be a character. I know that $\chi$ is Hodge-Tate of weight $0$ iff $\chi(I_K)$ is finite (by Sen's theory), and that it is Hodge-Tate of weight $k$ iff $\chi.\chi_p^{-k}$ is HT of weight $0$. -Is there a similar description for De Rham, Semi-stable and Cristalline ? - -REPLY [18 votes]: The de Rham characters are the same as the Hodge-Tate ones. The semistable ones are the same as the crystalline ones, and in your notation they are the de Rham ones for which $(\chi \cdot \chi_p^{-k})(I_K)$ is trivial (and not merely finite). This can be found for example in Fontaine and Mazur's paper.<|endoftext|> -TITLE: Basis removal gives a basis -QUESTION [20 upvotes]: Let $V$ be a vector space. Let us say that a finite set $X$ of vectors in $V$ is harmonic if for $B \subseteq X$, -$$ -B \text{ is a basis of } V \implies X \setminus B \text{ is a basis of }V. -$$ -Let us say that the basis number of a harmonic set $X$ is the number of subsets of $X$ that are bases of $V$. - -Which integers arise as the basis number of some harmonic set? - -Clearly, the number of bases is even (unless $X$ is empty). Can it be an arbitrary even number? Can we get, e.g., 10 or 6? - -This set of integers is closed with respect to multiplication and contains 2. -But it consists of not only powers of 2. It is easy to obtain, e.g., 18 -(take 6 vectors in a 3-dimensional space: 3 vectors from a plane and 3 from another plane in general position). - -REPLY [5 votes]: We can generalize domotorp's argument. If a harmonic set has $2k$ vectors in some $k$-dimensional space, then each basis in the set must contain exactly $k$ vectors, so every other vector must live in some complement to this subspace, so the harmonic set is reducible. -If we restrict our attention to irreducible harmonic sets, we then have at most $2k-1$ in each $k$-dimensional subspace. So to choose a basis for an $n$-dimensional irreducible harmonic set, we can choose any vector, then any other vector, then any vector other than the at most $3$ in the $2$-dimensional space generated by those $2$, then any vector other than the at most $5$ in the $3$-dimensional space generated by those $3$, and so on. This gives a lower bound for the number of bases at: -$$ \frac{ 2n (2n-1) (2n-3) \dots 1}{n (n-1) (n-2) \dots 1}$$ -we also have the trivial upper bound of $\left( \begin{array}{c} 2n \\ n \end{array}\right)$. This gives us the possible intervals: -$n=1$: $[2,2]$ -$n=2$: $[4,6]$ -$n=3$: $[15,20]$ -$n=4$: $[35,70]$ -$n=5$: $[78.75,252]$ -$n=6$: $[173.25,924]$ -I think after $n=6$, these intervals continue to overlap, which means we cannot rule out any further numbers. But we at least rule out all numbers which do not appear in these intervals: $8,10,12,14,22,$ and so on, from being the number of bases of an irreducible harmonic set. Since every harmonic set whose number of bases is not a multiple of $4$ is irreducible, we can rule out $10$, $14$, $22$, $26$, $30$, $34$, $74$, and $78$ as being the number of bases of a harmonic set. We can also rule out $28$, because it cannot be independent and it can only be written as $2 \times 14$. -$\left( \begin{array}{c} 2n \\ n \end{array}\right)$ is achievable for each $n$ by $2n$ generic vectors in $n$-dimensional space. This means we can get $6$ and $12=2 \times 6$, so the first number I don't know if we can get is $18$. The next number is $36$. -EDIT: As domotorp points out, -$\left( \begin{array}{c} 2n \\ n \end{array}\right)-2$ is achievable, so $18$ and $36$ are. The next number I do not know is $38$.<|endoftext|> -TITLE: How to define compatible topology for first-order structures? -QUESTION [5 upvotes]: Background Because a bounded distributive lattice can be represented by the clopen sets of a Priestley space, I tried to learn some basics about Priestley spaces. After reading (on Wikipedia) - -A Priestley space is an ordered topological space with special properties. - -I googled for "ordered topological space", with confusing results. Even so I know that the Priestley separation axiom already ensures that the topology is compatible with the partial order, I still would like to know what it actually means for "the topology to be compatible with the partial order". I became confused when I read a reference ("On partially ordered sets possessing a unique order-compatible topology" by E. S. Wolk) which required that the topology must be coarser than the topology with the Dedekind-closed subsets as closed sets. It also required the topology to be finer than the interval topology, which is probably equivalent to the condition I would have guessed. - -Question What is wrong with saying that a partially ordered set is a universal Horn structure, and that a compatible topology just needs to satisfy the "natural" conditions for a topology compatible with a universal Horn structure? Here I would just require that all the functions $f^n$ (from the signature of the Horn structure) are continuous functions from the argument space with the product topology into the object space, and that for all predicates $P^n$ (from the signature of the Horn structure) the sets $\{(x_1,\ldots,x_n):P^n(x_1,\ldots,x_n)\}$ are closed sets for the product topology. (If equality is part of the language, also the binary predicate $=$ must be included among these predicates. This means that the topology must be Hausdorff, as I learned now.) -I'm restricting myself to universal Horn structures here, because this is sufficient for my problem, and I would be surprised if one could come up with a useful definition of compatible topology for general first-order structures (without any additional restrictions like being universal Horn structures). - -REPLY [4 votes]: Many notions of compatibility between a partially ordered set and a topology on its underlying set are analogous to separation axioms and other well known concepts from general topology. -One can generalize the separation axioms and other notions such as $T_{2}$,complete regularity, and zero-dimensionality to axioms on ordered spaces that basically say that the order is compatible with the topology in some sense. For instance, following this trend, the notion of Stone-duality becomes the notion of Priestley duality. In this answer, I will outline some of the separation axioms and related axioms for ordered topological spaces. One familiar with general topology should be able to see the analogy between these results regarding ordered topological spaces and general topology. -For example, a topological space $X$ with a partial order $\leq$ is said to be order-Hausdorff if whenever $x\not\leq y$, there is an upper set $U$ and an lower set $V$ with $U\cap V=\emptyset$ and where $x\in U^{\circ},y\in V^{\circ}$. One can show that a partially ordered topology is Hausdorff if and only if $\leq$ is closed in $X^{2}$. -In general topology, we usually want more than simply the Hausdorff separation axiom (remember that most interesting topological spaces are completely regular), and the case is not much different for ordered sets. For ordered spaces, we sometimes want our spaces to be more than just Hausdorff. -An ordered topological space $X$ is said to be completely order-regular if -i. whenever $x\not\leq y$, then there is a continuous order preserving map $f:X\rightarrow[0,1]$ with $f(y) -TITLE: Are maximal compact subgroups of connected groups connected? -QUESTION [13 upvotes]: Assume $G$ is a connected locally compact group and $M$ is a maximal compact subgroup of $G$. Is $M$ connected too? - -REPLY [18 votes]: Disclaimer: Locally compact groups are absolutely not my field of expertise. I hope an expert can check my statements below, and perhaps add some details and references. -The Malcev–Iwasawa theorem implies that any connected, locally compact group $G$ satisfies: - -$G$ has a maximal compact subgroup; -there exists $n\in\mathbb{N}$ such that for any maximal compact subgroup $K$ of $G$, the underlying space of $G$ is homeomorphic to $K\times\mathbb{R}^n$. - -In particular, every maximal compact subgroup of a connected, locally compact group is itself connected. -References: The following references state the necessary results without proof. - -Theorem 32.5 of Markus Stroppel's book "Locally compact groups". -The article "Compact subgroups of Lie groups and locally compact groups" (DOI: http://dx.doi.org/10.1090/S0002-9939-1994-1166357-9), published in the Proceedings of the American Mathematical Society, volume 120, number 2, in February 1994 (pages 623-634). See the statements of theorems A, B, and C in the introduction to this article. According to the discussion there, the theorems hold for connected, locally compact groups: they follow from the analogous results for Lie groups as soon as one knows that a connected, locally compact group is a projective limit of Lie groups.<|endoftext|> -TITLE: Uniformization of Kodaira fibered surfaces -QUESTION [10 upvotes]: Consider a Kodaira fibration. i.e. a smooth non-isotrivial fibration $X\rightarrow C$ with $X$ a smooth complex surface and $C$ a smooth complex curve, such that both the genus of $C$ and genus of the fibers (which are complex curves) are at least $2$. By abuse of notation I call $X$ a Kodaira fibered surface. What is known about the structure of the universal cover of $X$? In particular, when is it a bounded symmetric domain? I know that $X$ is a minimal algebraic surface of general type and that the universal cover of $X$ can never be a ball in $\mathbb{C}^{2}$. - -REPLY [6 votes]: If an algebraic surface has the bidisk as universal cover, then, by Hirzebruch's proportionality theorem, its topological index is 0 (the index is 1/3 (c_1^2 - 2 c_2)). -But Kodaira proved that for Kodaira fibrations the index is strictly positive. -To exclude the case that the universal covering is the ball, where -c_1^2 = 3 c_2, again by Hirzebruch's theorem, is harder and was done by Kefeng Liu. You may look in my article with Rollenske for more results on Kodaira fibrations. -Regards, Fabrizio Catanese.<|endoftext|> -TITLE: Is the category of group objects in an $(\infty,1)$-topos reflective as a subcategory of the groupoid objects? -QUESTION [5 upvotes]: Let $C$ be an $(\infty,1)$-topos. The $(\infty,1)$-category of group objects in $C$ is a full sub-$(\infty,1)$-category of groupoid objects in $C$: -$${\mathsf{Grp}}(C) \hookrightarrow {\mathsf{Grpd}}(C)$$ -Is this full subcategory reflective? Here is one way to go about constructing a $(\infty,1)$-functor in the other direction: -$${\mathsf{Grpd}}(C)\xrightarrow{{\mathbf{B}}} C \xrightarrow{R} {\mathsf{PointedConnected}}(C)\xrightarrow{\Omega} {\mathsf{Grp}}(C)$$ -The functor $R: C\to {\mathsf{PointedConnected}}(C)$ takes an ordinary object $X$ in $C$ and gives its pointed-connected-reflection, given, say, by taking the homotopy cofiber of the inclusion from the 0-truncation $X_0$ to $X$. Conjecturally, the above $(\infty,1)$-functor could be the group reflection of a groupoid object, but it's really a guess. -So, the question: is ${\mathsf{Grp}}(C)$ reflective in ${\mathsf{Grpd}}(C)$? How can we give the adjoint? - -REPLY [10 votes]: If $\mathcal{C}$ is an $\infty$-topos, the $\infty$-category of groupoid objects of $\mathcal{C}$ is equivalent to the full subcategory of $Fun( \Delta^1, \mathcal{C})$ spanned by the effective epimorphisms $X \rightarrow Y$. Under this equivalence, the group objects correspond to the full subcategory where $X$ is a final object. The inclusion has a left adjoint, which carries an effective epimorphism $f: X \rightarrow Y$ to the induced map $\mathbf{1} \rightarrow Y \amalg_{X} \mathbf{1}$.<|endoftext|> -TITLE: Minimal graphs of prescribed girth and chromatic number -QUESTION [6 upvotes]: The well known result of Erdős, states that - -Given integers $g > 2$ and $k > 1$ there exist a graph $G$ with $\chi(G) \geq k$ and girth at least $g.$ - -What I am wondering is - -When can we expect equality to hold? I.e for which parameters $(g,k)$ do we have graphs with girth $g$ and chromatic number $k$? - -Of course we cannot have $k=2$ and $g$ odd, but are there any other more interesting pairs $(g,k)$ for which there does not exist a graph of girth $g$ and chromatic number number $k$? In particular - -Has anyone thought of searching/classifying these graphs in an analogous way as cages for girth and degree? - -After what Jacob said I propose the following - -What is the smallest graph of girth $4$ and chromatic number $5.$ - -REPLY [3 votes]: The answer to your "new" question is known: the smallest order of a 5-chromatic graph of girth 4 has 22 vertices, and there are 80 such graphs. -Tommy Jensen and Gordon F. Royle, Small graphs with chromatic number 5: A computer search, Journal of Graph Theory 19(1):107-116, 1995. link<|endoftext|> -TITLE: Supersingular elliptic curves over $\mathbb{Q}$ -QUESTION [9 upvotes]: what are the examples of elliptic curves defined over $\mathbb{Q}$ with supersingular reduction at a prime $p$ and having a $p$-isogeny over $\mathbb{Q}$ ? - -REPLY [17 votes]: I felt that there couldn’t be such an isogeny, but when I saw @DavidLoeffler’s answer, I realized that I had an argument, too. The formal group of the elliptic curve would be of height two, defined over $\mathbb Z_p$, but such things don’t have $p$-isogenies: the quotient formal group is definable only over a suitably ramified extension.<|endoftext|> -TITLE: Main open computational problems in quantifier elimination? -QUESTION [21 upvotes]: A language is said to have quantifier elimination if every first-order-logic sentence in the language can be shown to be equivalent to a quantifier-free sentence, i.e., a sentence without any $\forall$s or $\exists$s. An example is the theory of real closed fields (such as $\mathbb{R}$), considered with the four basic operations, equality ($=$) and inequalities ($<$, $>$). -Question: how fast could an algorithm that does quantifier elimination be? By how much are current algorithms (such as the ones that proceed by cylindrical decomposition) worse than the best algorithms that are conceivably possible? What are, in brief, the main open computational problems in quantifier elimination? -(We can try to restrict the discussion to $\mathbb{R}$, though other "useful" theories also interest me.) -As far as I know, the situation is as follows: in general, there are first-order sentences on $k$ variables that are not equivalent to any quantifier-free sentences of length less than $\exp(\exp(C k))$; this means, in particular, that the worst-case performance of a quantifier elimination program has to be at least doubly exponential on $k$. This is matched by cylindrical-decomposition algorithms (correct me if I am wrong). At the same time, if the original formula contains only $\exists$s or only $\forall$s, then an algorithm that is singly exponential on $k$ is known. (I'm going by a very quick reading of Basu, Pollack, Roy, Algorithms in Real Algebraic Geometry; all errors are my own.) -The second case - on which exponential bounds are known - is important, since it covers all cases of the form "prove this formula holds for all $x_1, x_2,\cdots, x_k$". -Is this the end of the story, or is there a subarea where plenty of work could remain to be done? - -Well, there seems to be real interest in this question, but no answer as such yet. Let me suggest what would be very nice as an answer: a few open problems on the subject, hard but not impracticable, with statements that are neat enough for mathematicians yet also close enough to actual practice that their solution would likely be useful. -For example: would reducing the existential theory of $\mathbb{R}$ to $k$-SAT be such a problem? - -REPLY [14 votes]: The problem of the current approach in this area, exemplified by the book by Basu, Pollack, Roy, "Algorithms in Real Algebraic Geometry" is that one ends up with the simplest case: checking non-emptiness of a real algebraic set $S$, and the only well-analysed class of algorithms to solve it leads to chasing roots of a univariate polynomial of degree at least the number of connected components of $S$. Even worse, if $S=V_{\mathbb{R}}(f_1,\dots,f_k)$ then instead of looking at the ideal $(f_1,\dots,f_k)\in\mathbb{R}[x_1,\dots, x_n]$ one looks at $V_{\mathbb{R}}(\sum_{j=1}^k f_j^2)$, which leads to blowing up degrees by 2. -An interesting topic might be to try to bridge it with the sums of squares-based methods in real algebraic geometry. (Which are in practice computationally more feasible, by using semidefinite programming). The latter can be read about in e.g. "Semidefinite Optimization and Convex Algebraic Geometry". -There are many open problems in the latter itself, some of them of number-theoretic flavour. E.g., understand when a polynomial $f$ with rational coefficients $f=\sum_{j=1}^m h_j^2 \in\mathbb{Q}[x_1,\dots, x_n]$, with each $h_j\in\mathbb{R}[x_1,\dots, x_n]$ (i.e. $f$ is a sum of squares of polynomials with real coefficients) can be a sum of squares of polynomials with rational coefficients, $f=\sum_{j=1}^{m'} g_j^2$, with each $g_j\in\mathbb{Q}[x_1,\dots, x_n]$. The state of the art on this problem is here in the talk by C.Scheiderer. -A much more famous open problem there is the complexity of the feasibility problem for linear matrix inequalities: decide the emptiness of $$S(A_0,\dots,A_m):=\{x\in\mathbb{R}^m\mid A_0+\sum_{j=1}^m x_j A_j\succeq 0\},$$ where $A_j$ are symmetric matrices, and $B\succeq 0$ stands for "$B$ is positive semidefinite". See e.g. this. There has been no progress on this problem since Ramana's 1997 paper. The related problem is semidefinite programming, a.k.a. SDP: minimize a linear function on $S(A_0,\dots,A_m)$. It is a natural generalization of linear programming (LP), and has become very popular in various areas due to its expressive powers. Chapter 2 of this book is an accessible introduction to SDP. -The relation between the SDP and the feasibility problem for linear matrix inequalities is akin to the relation between LP and feasibility problem for "ordinary" linear inequalities. Namely, one can solve the minimization problem efficiently iff one can solve the feasibility problem efficiently, this is well-known as "equivalence of optimization and separation". -Khachiyan and Porkolab wrote a paper where one can find a number of constructions illustrating how much more delicate $S(A_0,\dots,A_m)$ are, compared to "ordinary" polyhedra.<|endoftext|> -TITLE: G-invariant differential forms on homogeneous space of Lie Groups -QUESTION [5 upvotes]: Let $G$ be a connected Lie Group and $K -TITLE: Physical meaning of the integral cohomology condition in Souriau-Kostant pre-quantization? -QUESTION [5 upvotes]: The question is in the title. The form of the condition looks like the Bohr-Sommerfeld quantization formula of angular momentum, is there a link between the two formulas? - -REPLY [4 votes]: For a closed 2-form $\omega$ on a manifold $M$, the integrality of the closed 2-form, that is, -$$ -\int_\sigma \omega \in a{\bf Z}, \quad \mbox{for all} \quad \sigma \in H_2(M,{\bf Z}), -$$ -for some real number $a$, ensures the existence of a principal circle-bundle $Y$ (and its associated line bundle $L$) over $M$ and a connexion $\lambda$ with curvature $\omega$. Then, it is possible to lift some groups of automorphisms of $\omega$ (subgroups of ${\rm Diff}(M,\omega)$) as groups of automorphisms of $(Y,\lambda)$. This procedure is called prequantization because it is the first step of an answer to the Dirac program of quantization consisting in representing symmetries in classical mechanics by unitary transformations in some Hilbert space (that is supposed to have a physical meaning). I would not want to develop why one needs this bundle, at the first place, to answer Dirac's program, and are not contented just with the automorphisms of $\omega$, because it will lead us too far. If you are happy with this answer I'm fine, else I'll try to say a more few words(*). -P.S. The fact that the number $a$ is required to be a multiple of $\hbar$ comes just from physics consideration. -(*) Edited: I added a few words here<|endoftext|> -TITLE: Fixed component of an $S^1$ action on $S^n$ -QUESTION [6 upvotes]: Suppose $S^1$ is acting smoothly on $S^n$ and $M$ is a connected component of the set of fixed points of the action. What can be said about $M$? -Is it true that $\pi_1(M)=0$? (sorry this first bit of the question is silly since any $S^k$ can appear) Is it true that $M$ has to be homeomorphic to a sphere? If not, what kind of manifolds can one get? - -REPLY [2 votes]: It is known that $M$ has the homology of a sphere, see [1] But you probably knew that. -What you also can describe pretty nicely is the cohomology of the orbit space $S^n/G$. To be more precise $S^n/G$ has the cohomology of the join of a sphere and a complex projective space and the dimension of this sphere is exactly the dimension of the fixed point set, see also [1] -[1] Cohomology ofS 1-orbit spaces of cohomology spheres and cohomology complex projective spaces, Mikiya Masudaco -Mathematische Zeitschrift -1981, Volume 176, Issue 3, pp 405-427<|endoftext|> -TITLE: Open Torelli problems -QUESTION [10 upvotes]: I just finished studying the proof of the Torelli Theorem for K3 surfaces made by Daniel Huybrechts (following the approach of Misha Verbitsky). -This theorem states that two K3 surfaces $X$ and $Y$ are isomorphic if and only if there is an isometry $\phi:H^2(X,\mathbb{Z})\rightarrow H^2(Y,\mathbb{Z})$ (with respect to the intersection form) which respects the Hodge decomposition. -Huybrechts (and Verbitsky before him) proved this theorem using the period map defined from the moduli space of K3 surfaces to the period domain. This approach is particularly interesting because it can be extended to the case of Hyperkahler manifolds. For this type of manifolds however there can be no strong result as for K3s, as Namikawa proved in this article. -The Torelli problem consists in knowing when the Hodge structure of a manifold determine the manifold. - -Question: Do you know other classes of manifolds for which the Torelli problem is an open problem? - -REPLY [10 votes]: There are versions of the Global Torelli Problem for Calabi-Yau threefolds which are open. Infinitesimal Torelli is true for CY3s, i.e. (families of) projective 3-folds with trivial canonical bundle and no holomorphic 1-forms, as proved already by Griffiths. The strongest possible form of global Torelli fails, as proved in the late 90s in my thesis and early papers, eg. "Calabi-Yau threefolds with a curve of singularities and counterexamples to the Torelli problem", Int. J. Math. 11 constructs explicit families of CY3s in which the polarized Hodge structure up to isomorphism does not determine the varieties up to isomorphism. But there are outstanding questions. Todorov, Yau and collaborators have studied a variant where the period map maps from the Teichmuller space to the period domain (without taking any discrete group quotient); this map could still be injective. There are several papers on the arXiv claiming results of this nature, such as http://xxx.lanl.gov/abs/1112.1163 and http://arxiv.org/abs/1205.4207. One can also ask whether in the simply-connected case, the variety up to birational equivalence could be recovered from the polarized Hodge structure; I don't know any counterexample to this. As a special case, one can ask Global Torelli for Calabi-Yau threefolds of Picard number one; I am not sure we know the result for any specific family, apart from the generic Torelli for the family of quintics in Voisin's "A generic Torelli theorem for the quintic threefold", in New trends in algebraic geometry (Warwick, 1996).<|endoftext|> -TITLE: Pseudofree $T^2$ actions on spheres -QUESTION [7 upvotes]: Is it possible to construct a smooth action of $S^1\times S^1$ on $S^{2n+1}$ ($n\ge 2$) such that no point on $S^{2n+1}$ has an infinite stabilizer? -Note that if such an action exists, it can not be linear. - -REPLY [4 votes]: This should only be considered as a comment to the answer above, unfortunately it does not fit in a comment. -One can show, that $\mathbb{Z}_p\times \mathbb{Z}_p$ cannot act freely on $S^n$ also via "elementary" topological methods as follows (I learned this in the book of Hatcher, but dont ask me where): -Assume $n\geq 2$ (all other cases are trivial) -Suppose there exists a free $\mathbb{Z}_p\times \mathbb{Z}_p$ action on $S^n$. Then the quotient $X:=S^n/(\mathbb{Z}_p\times \mathbb{Z}_p)$ can be made into a CW-complex. The long exact sequence for fibrations yiels -$$\pi_k(X)=\begin{cases}0 & 2\leq k\leq n \\ \mathbb{Z}_p\times \mathbb{Z}_p &k=1\end{cases}$$ -Now we construct a $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$ as follows. Via the quotient map $S^n\to X$ we attach one (n+1) cell, to form a space $X'$, and the attach cells of dimension greater than $n+1$ to kill all homotopy groups higher equals $n+1$. Call the resulting space $Y$. By definition $Y$ is a $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$ with one (n+1) cell. This means -$$H_{n+1}(Y,\mathbb{Z}_p)\in \{0,\mathbb{Z}_p\}$$ -On the other hand: It is well known that the infinite dimensional lens spaces $S^\infty/\mathbb{Z}_p$ is a $K(\mathbb{Z}_p,1)$, hence $S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p$ is a $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$. Now using Künneth we see that -$$H^{n+1}(S^{\infty}/\mathbb{Z}_p\times S^{\infty}/\mathbb{Z}_p,\mathbb{Z}_p)\notin \{\mathbb{Z}_p,1\}$$ -Since $K(\mathbb{Z}_p\times \mathbb{Z}_p,1)$ is unique up to homotopy this shows that such an action cannot exist.<|endoftext|> -TITLE: How complete is $\mathcal P(\kappa)/J_{bd}$? -QUESTION [5 upvotes]: While it is true that $\mathcal P(\kappa)$ is a complete Boolean algebra, it is not necessarily true that $\mathcal P(\kappa)/I$ is complete for an ideal $I$. In particular if we consider $I=J_{bd}$ the ideal of bounded subsets of $\kappa$. -For example, Hausdorff showed that there is an $(\omega_1,\omega_1)$ gap in $\mathcal P(\omega)/\textrm{Fin}$. That is, there is a sequence of order type $\omega_1+\omega_1^*$ which is strictly increasing, but there is no $A$ which realizes the cut between the $\omega_1$ and $\omega_1^*$ parts. Consider now the lower $\omega_1$ sets, if there was a supremum to this collection, it would have to be an element realizing the gap. - -Assuming $V=L$. How complete is $\mathcal P(\kappa)/J_{bd}$? If the answer is strictly less than $\operatorname{cf}(\kappa)$, what if we limit ourselves to strictly increasing sequences? Can we get $\operatorname{cf}(\kappa)$-completeness? - -That is, assuming that $\gamma<\kappa^+$ and $\langle A_\alpha\subseteq\kappa\mid\alpha<\gamma\rangle$ is strictly increasing modulo $J_{bd}$. Is there a least upper bound to the sequence? -I'd be interested to know (at least some partial) answer outside $V=L$, although that is less important. - -REPLY [6 votes]: If $\kappa$ is regular, then $\mathcal{P}(\kappa)/J_{bd}$ is a $\kappa$-complete boolean algebra. If $\langle A_\alpha : \alpha < \delta < \kappa \rangle$ is a sequence of subsets of $\kappa$, then the union of these is a least upper bound. This uses the $\kappa$-completeness of $J_{bd}$. It is not $\kappa^+$-complete for the same reason Paul mentioned for $\omega$. Take any partition of $\kappa$ into $\kappa$ many unbounded sets, $\langle A_\alpha : \alpha < \kappa \rangle$, let $B$ be an upper bound, and let $C \subseteq B$ have one point from each $A_\alpha \cap B$ removed. So $C$ is an upper bound and strictly below $B$. For singular $\kappa$, the algebra is $cf(\kappa)$-complete but not $cf(\kappa)^+$-complete: Take any unbounded $A$ set of size $cf(\kappa)$ and consider the algebra below $A$. It is isomorphic to $\mathcal{P}(cf(\kappa))/J_{bd}$.<|endoftext|> -TITLE: When is a groupoid the path groupoid of a graph? -QUESTION [5 upvotes]: I am actually interested in the $C^*$-algebras, so perhaps my question should be: How can you recognize whether a $C^*$-algebra $A$ is isomorphic to $C^*(\Lambda)$ for some (higher-rank) graph $\Lambda$, if $A$ is not presented to you via the Cuntz-Krieger relations? - -REPLY [5 votes]: This is an answer to the question: "Given a $C^\ast$-algebra $A$ that shares some properties of (higher-rank) graph algebras (such as separability, nuclearity, UCT, K-theoretic properties), can I conclude that $A$ is a (higher-rank) graph algebra?" -My answer is: this would require a classification of the class of $C^\ast$-algebras in question together with a description of the range of the classifying invariant on (higher-rank) graph algebras. This problem is open, even for purely infinite Cuntz--Krieger algebras; see Do phantom Cuntz-Krieger algebras exist? (Edit: this case was solved in http://arxiv.org/abs/1511.09463.) -In the simple purely infinite case, one can say the following: a unital UCT Kirchberg algebra $A$ is a graph algebra if and only if $K_1(A)$ is free. This theorem is due to Wojciech Szymanski.<|endoftext|> -TITLE: For $G=\mathbb{Z}^2\rtimes \mathbb{Z}$, $Spec(\mathbb{Z}G)$=? -QUESTION [8 upvotes]: Let $G$ be the group $\mathbb{Z}^2\rtimes_{\sigma} \mathbb{Z}=\langle y,z\rangle\rtimes_{\sigma}\langle x\rangle$, where $\sigma(x)=\begin{pmatrix}a, b\\c,d\end{pmatrix}\in SL_2(\mathbb{Z})$, which means that we have relations $xyx^{-1}=y^az^c, xzx^{-1}=y^bz^d$. Then we can form the group ring $R=\mathbb{Z}G$, note that since $G$ is a polycyclic-by-finite group, $R$ is a left Noetherian ring. -I am interested in what a general prime ideal $\wp$ in $R$ looks like, so I want to ask the following question: -What does Spec(R) generally look like? Especially when $\sigma(x)=\begin{pmatrix}1,0\\1,1\end{pmatrix}$ or $ ~ \sigma(x)=\begin{pmatrix}2,1\\1,1\end{pmatrix}$ - -REPLY [11 votes]: The natural map $\mathbb{Z} \to \mathbb{Z}G$ has central image and therefore induces a map between prime spectra $Spec(\mathbb{Z}G) \to Spec(\mathbb{Z})$. The preimage of the ideal generated by $(p)$ under this map is in a natural bijection with $Spec( kG )$ where $k = \mathbb{F}_p$ if $p$ is a prime number and $k = \mathbb{Q}$ if $p = 0$. So $Spec( \mathbb{Z}G )$ is the disjoint union of various prime spectra $Spec( kG )$, where $k$ is a field. -A prime $P$ of $kG$ is said to be faithful if $G$ embeds into the group of units of $kG/P$, or equivalently, if $G \cap (1 + P) = 1$. -If a prime of $kG$ is unfaithful, then $P^\dagger := G \cap (1 + P)$ is a non-trivial normal subgroup of $G$, and $P$ is completely determined by its image inside $k[G / P^\dagger]$, which is then a faithful prime of this smaller group ring. Thus it is enough to understand the faithful primes of $kG$. -James Roseblade published a paper in 1978 called "Prime ideals in group rings of polycyclic groups", which appeared in the Proceedings of the London Mathematical Society (I can send you a copy if you like). Although he did not completely settle the problem of classifying the faithful prime ideals in group rings $kG$ of polycyclic-by-finite groups $G$ over fields $k$, he made some serious breakthroughs in this paper. -I'd like to draw your attention to Theorem E, which says: -Let $A$ be the Zalesskii subgroup of the polycyclic group $G$. If $G$ is orbitally sound, then any faithful prime ideal of a group algebra $kG$ is controlled by $A$. -Being controlled by $A$ for a prime ideal $P$ means that $P$ is generated by its intersection with $kA$ and is therefore completely determined by an ideal in a smaller group ring. The Zalesskii subgroup $A$ is defined by the property that it contains the largest finite normal subgroup $F$ of $G$ and that $A/F$ is the centre of the Fitting (largest normal nilpotent) subgroup of $G/F$. Being orbitally sound is a technical condition that I won't go into here. -In particular, if $G$ happens to be torsion-free and nilpotent then $F$ is trivial and $A$ is just the centre of $G$, and then Roseblade's Theorem E essentially says that every faithful prime of $kG$ is controlled by the centre of $G$. This happens for example when $G = \mathbb{Z}^2 \rtimes \mathbb{Z}$ as in your question and when $\sigma$ is the matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$.<|endoftext|> -TITLE: Sets of evenly distributed points in the Euclidean plane -QUESTION [14 upvotes]: Is there a set $P \subset \mathbb{R}^2$ of points in the Euclidean plane whose intersection -with every convex subset of $\mathbb{R}^2$ of area $1$ is nonempty but finite? -If the answer is yes, can $P$ be chosen in such way that there is a constant $C_P$ with -the property that for every convex subset $S \subset \mathbb{R}^2$ of area $1$ we have -$1 \leq |S \cap P| \leq C_P$? -- And if yes, which $P$ admit the smallest $C_P$? -Remarks: - -Lattices are not examples as there is always an $\epsilon > 0$ such -that there are $\epsilon \times \frac{1}{\epsilon}$ rectangles which do not contain -a lattice point. -The question looks in some sense natural to me, and I wonder whether -it has already been considered before. Maybe someone knows a reference? - -REPLY [5 votes]: Your second problem is almost the same as the following old, open problem of Danzer and Rogers: -"What is the area of the largest convex region not containing in its interior any one of $n$ given points in a unit square?" Here the big question is whether the answer is $\Theta(\frac 1n)$ or not. -If the answer to the DR-problem is $\omega(\frac 1n)$, then there can be no bound in your problem for the number of points a unit convex set might contain. To see this, suppose by contradiction that you have a $P$ with some $C_P$ bound. Take a $\sqrt n \times \sqrt n$ size square, this contains $\Theta(n)$ points of $P$, to simplify calculations I suppose it contains exactly $n$. By scaling, the DR-problem gives us an empty convex set of size $\omega(1)$, which becomes bigger than $1$ if $n$ is big enough. -In the other direction, I am not sure if the implication holds, so I think your question is an excellent research topic. Can we prove that the two problems are equivalent?<|endoftext|> -TITLE: Other implications of Zhang's method -QUESTION [6 upvotes]: I have been reading a bit about Zhang's proof and the associated Polymath8 project. -Though Tao's high level summary -http://terrytao.wordpress.com/2013/06/30/bounded-gaps-between-primes-polymath8-a-progress-report/ -is interesting it still is rather technical. So without any knowledge of the techniques I am unable to get an intuition for the following related problems. -Let $T_k$ be the assertion that there are infinitely many prime pairs with distance $k$. -It is conjectured that this is true for any even $k$ and Zhang's theorem obviously implies that this is true for some $k$. That being said, there is no known $k$ for which this is true. - -Is there any hope that the method employed in the proof is able to show this for some explicit $k$, even if this $k$ is much larger than the bounds given so far. My inituition says no. -The goal of Polymath8 is to lower the bound given in Zhang's paper as far as possible. On the other hand it is believed that $T_k$ holds for any even $k$. Is there any hope that the methods can also be adapted to show the following: For any $n$ there is some $k \geq n$ such that $T_k$ is true. This seems much more in the style of the original theorem than question 1, so here I am more optimistic, but I guess some deeper knowledge is required here. - -REPLY [9 votes]: Zhang's strategy shows that any admissible tuple of size h contains at least 2 primes infinity often, as long as h is larger than some threshold (the primary theoretical thrust of the polymath project has been to reduce the required value of h). This approach seems unable to give (1), but immediately gives (2).<|endoftext|> -TITLE: Erdős, Harary, Tutte's "dimension of graph": Progress in last 48 yrs? -QUESTION [20 upvotes]: I just ran across this delightful paper by an amazing triumvirate: - -Paul Erdős, Frank Harary, and William Tutte. "On the dimension of a graph." Mathematika 12.118-122 (1965): 20. - (Cambridge link) - (PDF download link) -   - -They prove that $K_n$ has dimension $n-1$, $K_{n,n}$ has dimension $\le 4$, -and the dimension of the $n$-cube $Q_n$ is $2$. And the Petersen graph has dimension $2$: -      -Surely there must have been advances on characterizing graphs according to this concept in the last ~half-century(!). -Can anyone provide some updates the status on this notion? -Update. Here is a modern, metrically accurate drawing of the Petersen graph: -    -   (Image from Wikipedia: Unit Distance Graph) - -REPLY [22 votes]: There are several well-known and hard problems in discrete geometry that concern the dimension of a graph. For example, the unit distance problem asks for the maximum number of edges of a graph on $n$ vertices of dimension 2. Erdős conjectured in the 1940s that the answer is $n^{1+o(1)}$, but the best known bound is only $O(n^{4/3})$. -The chromatic number of the plane is another famous problem about graphs of dimension $2$. This question asks for the maximum chromatic number of any graph with dimension $2$. The answer is only known to be between $4$ and $7$, and it has been stuck that way for more than five decades. Shelah and Soifer speculate in a series of papers that the answer might depend on the axioms for set theory. This is indeed the case for some other distance graphs they construct. A related result from Paul O'Donnell's thesis is that there are graphs of dimension $2$, chromatic number $4$, and arbitrarily large girth. -It is unlikely that graphs of dimension $2$ will be characterized in the near future. - -REPLY [12 votes]: To supplement Jacob Fox's answer: A short survey on this topic is presented in The mathematical coloring book by Alexander Soifer, Springer, New York 2009. MR2458293 (2010a:05005). -Chapter 13 is Dimension of a graph, and begins with the results of the Erdős-Harary-Tutte paper. He then discusses a variant, that he calls "Euclidean dimension" of a graph, with numerous references. The results of Chapter 14, Embedding $4$-Chromatic Graphs in the Plane are also related and, in particular, O’Donnell's results are discussed. (And you want to keep reading past this chapter as well.) -If you are not familiar with this book, you will find it is written in a very unique, personal style, it surprises when one first sees it. The book describes not just mathematical results but also stories surrounding their discovery. You may enjoy it.<|endoftext|> -TITLE: smooth manifolds as real algebraic set (continued) -QUESTION [7 upvotes]: There are several ways of producing manifolds,say: -1.orbits space of group action -2.connected sum of manifolds -3.underlying topological space of nonsingular algebraic set -.... -here,i am interested in the 3rd one. -A well known theorem due to Nash and Tognoli says that Every compact smooth manifold is diffeomorphic to a nonsingular real algebraic set. -My question is: -Given a manifold $M$,How to find an algebraic set EXPLICITLY whose underlying topological space is diffeomorphic to $M$? -For example, can we find a specific -real algebraic set corresponding to an orientable riemann surface of genus $g$? - -Thanks for the very useful answers and comments. -I am a student interested in the topology of manifolds and just came to realize so many manifolds could come from algebraic varieties,so the topology of algebraic varieties would be an interesting topic. -A naive idea could be: -The topology of algebraic varieties should be determined by the algebra,i.e. the polynomials which generate the ideal corresponding to the variety. -In practice,i guess,it would be quite a hard problem to read topology from algebra,and in many cases,the other direction is also of interest,i.e.to get algebraic information of the variety by studying the topology of the underlying space. -In the very interesting survey on the topology of real and algebraic varieties given by Janos Kollar -https://web.math.princeton.edu/~kollar/FromMyHomePage/tanig.ps -He calls these two directions as Realization problem and Recognition Problem respectively. -To be more precise,realization problem studies which topological space could be realized as the underlying space of a projective algebraic variety,and recognition problem studies which algebraic properties are determined by its underlying topological space. -These two problems clearly summaries the interplay of algebra and topology in the algebraic variety. -Now my question is: -What's the status of the realization problem,i.e. -How much topology could be read from the algebra and how? - -REPLY [2 votes]: Speaking of compact orientable survaces, this is very simple, indeed: -hyperelliptic curves in $CP^2$ given by equations $w^2=P_n(z)$, where $P$ is a rational -function with $2n$ zeros and poles, all zeros and poles distinct. -of degree $n$, are smooth embedded surfaces with Euler characteristic $4-2n$. -EDIT. A gap in the previous argument was pointed in the comment of BS: only for genera -of the form $g=(d-1)(d-2)/2$ a smooth curve exist in $P^2$, and hyperelliptic curves usually have singular points. But they can be embedded to other spaces. -EDIT 2. Here is a way to construct a smooth compact orientable algebraic surface in $R^3$ -of any genus. $(x,y,t)$ are the coordinates in $R^3$. -$$t^2=(R^2-x^2-y^2)\prod_{j=1}^n((x-x_j)^2+(y-y_j)^2-r^2),$$ -where $R$ is large and the discs $(x-x_j)^2+(y-y_j)^2 -TITLE: Generalization of Levy-Solovay theorem to kappa-c.c. forcings -QUESTION [5 upvotes]: The Levy-Solovay theorem says that if $\kappa$ is measurable, then it remains measurable in the extension by a small forcing ($|\mathbb{P}|<\kappa$). Is still true if we replace $|\mathbb{P}|<\kappa$ with "$\mathbb{P}$ has the $\lambda-\textrm{c.c.}$ for some $\lambda<\kappa$"? Or even, can a c.c.c. forcing destroy measurability of $\kappa$? -This conjecture seems to check out with all the natural examples of forcings that I know. -Drake - -REPLY [10 votes]: The forcing to add $\kappa$ many Cohen reals is c.c.c and thereby preserves all cardinals but destroys the fact that $\kappa$ is even strongly inaccessible. (See for example Kunen "Set Theory: An Introduction to Independence Proofs" North-Holland, Ch. VII.)<|endoftext|> -TITLE: A frustrating cohomology class on the moduli of abelian surfaces -QUESTION [12 upvotes]: Here's a very frustrating question that I have been stuck on for some time. I believe that my question could fit in a general framework of what happens when you restrict $L^2$-cohomology classes on a Shimura variety to a sub-Shimura variety. However I formulate the question for the special case I am interested in. -Let $A_2$ be the moduli stack of principally polarized abelian surfaces. To the irreducible finite dimensional representation of $\mathrm{Sp}(4)$ of highest weight $a \geq b \geq 0$ we attach a a local system $V_{a,b}$ on $A_2$. -Suppose $(a,b) \neq (0,0)$. One can prove that $H^4_c(A_2,V_{a,b})$ vanishes unless $a=b$ is even, in which case $H^4_c(A_2,V_{2k,2k})$ is pure of Tate type and of the same dimension as the space of cusp forms of weight $4k+4$ for $\mathrm{SL}(2,\mathbf Z)$. The map $H^4_c \to H^4_{(2)}$ to the $L^2$-cohomology is an isomorphism. In terms of automorphic representations, these cohomology classes can be described as follows: for any level 1 cusp form $\pi$ on $\mathrm{GL}(2,\mathbf A)$ of weight $4k+4$ we consider the unique irreducible quotient of -$$ \mathrm{Ind}_{P(\mathbf A)}^{\mathrm{GSp}(4,\mathbf A)} \left( \vert \cdot \vert^{1/2} \pi \otimes \vert \cdot \vert^{-1/2} \right)$$ -where $P$ denotes the Siegel parabolic subgroup (whose Levi factor is $\mathrm{GL}(2) \times \mathrm{GL}(1)$); this is a discrete automorphic representation for $\mathrm{GSp}(4)$ which contributes a Tate type class to the $L^2$-cohomology in degrees $2$ and $4$. -There is a map $\mathrm{Sym}^2(A_1) \hookrightarrow A_2$ given by taking a pair of elliptic curves to their product. We can also restrict $V_{a,b}$ to $\mathrm{Sym}^2(A_1)$. By determining the branching formula for $\mathrm{SL}(2)^2 \rtimes S_2 \subset \mathrm{Sp}(4)$ we find that the trivial local system occurs as a summand in the restriction of $V_{a,b}$ to $\mathrm{Sym}^2(A_1)$ if and only if $a=b$ is even, in which case it appears with multiplicity $1$. So $H^4_c(\mathrm{Sym}^2(A_1),V_{2k,2k})$ is also pure of Tate type but $1$-dimensional. Again we could think about $L^2$-cohomology and it would not make a difference. -MAIN QUESTION: Is the restriction map $H^4_c(A_2,V_{2k,2k}) \to H^4_c(\mathrm{Sym}^2(A_1),V_{2k,2k})$ nonzero for $k \geq 2$? -Any ideas or pointers at all would be appreciated. I am very ignorant about automorphic representations, Shimura varieties etc. and I am naively hoping that there exists some general method for answering question of this form. -This question arose from the paper http://arxiv.org/abs/1210.5761 . A positive answer would imply that all even cohomology of $\mathcal{\overline{M}}_{2,n}$ is tautological for $n < 20$, and that the Gorenstein conjecture fails on $\mathcal{\overline{M}}_{2,20}$. - -REPLY [5 votes]: The answer to the main question above is indeed positive: the restriction map is nonzero for all $k \geq 2$. This was proved in Section 5 of my paper Tautological rings of spaces of pointed genus two curves of compact type. Compos. Math. 152 (2016), no. 7, 1398–1420. -The idea of the calculation is that these are Eisenstein cohomology classes for the Siegel parabolic subgroup, which corresponds to the 0-dimensional boundary strata in the Baily-Borel compactification. Therefore one can reduce to a calculation on a deleted neighborhood of the 0-dimensional part of the boundary. Such a deleted neighborhood can be described as a fiber bundle over a smaller locally symmetric space. In the end one ends up changing the problem from considering the submanifold $A_1 \times A_1 \subset A_2$ to considering the submanifold inside $A_1$ given by the image of the imaginary axis. This is equivalent to reformulating the problem as an assertion about classical modular symbols.<|endoftext|> -TITLE: Embedding of a "quotient graph" -QUESTION [8 upvotes]: Consider the simple undirected graph $G$ with natural equivalence relation $\sim$ on $V(G)$: -$u\sim v$ iff they are similar, i.e. iff there exists $\phi\in Aut(G)$ with $\phi(u)=v$. -Define a "quotient graph" $G_{Aut}$ in the following way: -$V(G_{Aut})=V(G)/\sim$ and there is an edge $A-B$ iff $\exists \ a\in A, b\in B$ with $ab\in E(G)$. -Conjecture: If in $G$ every pair of similar vertices are non-adjacent, then $G_{Aut}\subset G \ ?$ - -ADDED: As Anton showed, this conjecture is false. But what one can said if - $G$ is a tree? Does conjecture remains false? - -REPLY [10 votes]: No. Here is a simple counterexample: - -REPLY [3 votes]: The following seems to be a counterexample, verified by a Magma calcualtion. The vertex set is $\{1,2,\ldots,16\}$ and the edge set is $E$. The quotient graph is the complete graph on four vertices and has $S_4$ as automorphism group, but $|{\rm Aut}(G)|=4$. -> E:={ - {1,5}, {2,6}, {3,7},{4,8}, - - {1,9}, {1,10}, {1,11}, {1,12}, - {2,9}, {2,10}, {2,11}, {2,12}, - {3,9}, {3,10}, {3,11}, {3,12}, - {4,9}, {4,10}, {4,11}, {4,12}, - - {1,14}, {1,15}, {1,16}, - {2,13}, {2,15}, {2,16}, - {3,13}, {3,14}, {3,16}, - {4,13}, {4,14}, {4,15}, - - {5,9}, {5,10}, {6,10}, {6,11}, {7,11}, {7,12}, {8,12}, {8,9}, - - {5,13}, {5,14}, {6,14}, {6,15}, {7,15}, {7,16}, {8,16}, {8,13}, - - {9,13}, {10,14}, {11,15}, {12,16} }; - - > G:=Graph<16|E>; - > A:=AutomorphismGroup(G); - > Order(A); - 4 - > Orbits(A); - [ - GSet{@ 1, 2, 3, 4 @}, - GSet{@ 5, 6, 7, 8 @}, - GSet{@ 9, 10, 11, 12 @}, - GSet{@ 13, 14, 15, 16 @} - ]<|endoftext|> -TITLE: Does this quadratic form over a large field represent 1? -QUESTION [6 upvotes]: I have a field $K$ of transcendence degree two over $\mathbb{R}$, and elements $a_1,a_2,a_3\in K$. I would like to understand the set -$$ Q = \{ u\in K^3 : \sum_i a_iu_i^2 = 1\} $$ -In particular, I would like to know whether it is nonempty, and if so, I would like to find some examples of elements, and ideally some kind of formula for all elements. What general methods are known for this sort of question? My gut feeling is that $Q$ is likely to be empty for most choices of the parameters $a_i$; are there any theorems of that type? -Specifically, I am interested in the following case: -\begin{align*} - r &= y_2^3+(5y_1-9)y_2^2/5+(-20y_1+4y_1^2+24-16y_4)y_2/25-4(1-y_1)/25 \\ - A &= \mathbb{R}[y_1,y_4,y_2]/(r) \\ - K &= \text{field of fractions of } A \\ - a_1 &= y_2(1-y_1-y_2) \\ - a_2 &= y_4 \\ - a_3 &= y_2. -\end{align*} -Note that $A$ is a free module of rank three over the subring $A_1=\mathbb{R}[y_1,y_4]$, so $K$ is a free module of rank three over the purely transcendental subfield $K_1=\mathbb{R}(y_1,y_4)$. Using this, we could translate the question to one about a quadratic form in nine variables over $K_1$, which might or might not be more tractable. - -REPLY [3 votes]: Starting from Jason's observations, an obvious obstruction for the existence of a solution would be the existence of a point $M=(y_1,y_2)$ in $\mathbb{R}^2$ such that $b_1(M)>0$, $b_2(M)>0$, and $b_4(M)<0$. However, we have -$$16 b_2=(4y_1^2)\,b_4-(5y_2-2)^2\,b_1$$ -so whenever $b_1>0$ and $b_4<0$ we must have $b_2\leq0$ and this obstruction vanishes. -To put things differently, we have a smooth quadric fibration $f:Q\to U$ where $U$ is a suitable Zariski open subset of the real $(y_1,y_2)$-plane, and we are looking for a rational section of $f$. What I am saying is that, at least, for each $M\in U(\mathbb{R})$, the quadric $f^{-1}(M)$ has real points.<|endoftext|> -TITLE: Is there a nice way to "unravel" a great dodecahedron? -QUESTION [12 upvotes]: The great dodecahedron is a non-convex regular polyhedron bounded by 12 pentagonal faces, crossing each other, arranged in a star-shaped manner around each of its 12 vertices (see the Wikipedia page for a picture.) From a topological point of view, the abstract surface that you get if you glue the faces along the corresponding sides is a compact orientable surface; hence it is homeomorphic to a sphere with some number of handles. It is easy to check that it has genus 4. (You can also see it as a 3-sheeted Riemann surface with 12 branch-points.) -Thus it is possible to draw, on the surface of a 4-holed torus, a graph consisting of 12 vertices and 30 edges that subdivides the surface into 12 pentagonal regions, with 5 such regions meeting at each vertex. To put it more concisely: there exists a sort of 4-holed polyhedron which is regular in at least a topological (or, if you prefer, combinatorial) sense and whose Schläfli symbol would be {5 5}. -I was actually able to draw such a picture, but all I see is a tangle of lines twisting and wrapping around in all directions. My picture has no obvious symmetry. Is it possible to embed this object in ordinary 3-space without self-intersections, but with, let's say, at least an axis of 5-fold rotation? (Unfortunately, I suspect the answer is "no".) Or is it possible to at least make sure that there is a group of easily described transformations that coincides, at least partially, with the abstract symmetry group of the configuration (which has order 120)? - -REPLY [9 votes]: No, it is not possible to embed the genus 4 surface in $\mathbb R^3$ with 5-fold axial symmetry. The $2\pi/5$ rotation of the great dodecahedron has four fixed points. Two of these fixed points are at the center of a pentagonal face, and the other two are at a vertex where five pentagons meet. The combinatorial rotation angle at the center-of-face fixed points is, of course, $2\pi/5$. But the combinatorial rotation angle at the vertex fixed points is $-4\pi/5$. But there do not exist order-5 diffeomorphisms of $\mathbb R^3$ (or $S^3$) whose fixed points have two different rotation angles. (cf. http://en.wikipedia.org/wiki/Smith_conjecture ) -I think it is, however, possible to embed the genus 4 surface in the lens space $L(5,2)$ with 5-fold symmetry. - -[Added later] Here's how to embed the surface in $L(5,2)$ with 5-fold symmetry. (I apologize for the lack of pictures.) $L(5,2)$ is the union of two solid tori. Let $X$ and $X'$ be two meridian disks of the first solid torus, and let $Y$ and $Y'$ be two meridian disks of the second solid torus. The boundaries of $X,X',Y,Y'$ divide the torus $T^2$ into 20 parallelograms. Color these parallelograms black and white in an alternating, checkerboard fashion. Now consider $$X \cup X' \cup Y \cup Y' \cup \mbox{the 10 black parallelograms} .$$ This is a (piecewise linear) surface of genus 4 (which of course can be smoothed if one so desires). The surface has an obvious 5-fold symmetry, and this symmetry has four fixed points (the centers of $X,X',Y,Y'$) with rotation angles as described above. -All that remains is to divide this surface into 12 pentagons. The first two pentagons are easy: they are $X$ and $X'$. Now for the remaining ten pentagons. $Y$ and $X'$ intersect in five points on the boundary of $Y$. Draw lines from these five points to the center of $Y$. This divides $Y$ into five pieces $Y_0, Y_1, Y_2, Y_3, Y_4$. Similarly divide $Y'$ into five pieces $Y'_0,\ldots,Y'_4$ using intersections with $X$. Each of the ten (black) parallelograms intersects each of $X,X',Y,Y'$ on one side of the parallelogram. Cut each parallelogram into two triangles so that one triangle contains the $X$ and $Y$ sides of the parallelogram and the other triangle contains the $X'$ and $Y'$ sides of the parallelogram. Each $Y_i$ is adjacent to two of these triangles; add those two triangles to $Y_i$. Similarly add two triangles to each $Y'_i$. We have now divided the surface into 12 pieces, and one can check that combinatorially each piece is a pentagon, and furthermore that the combinatorics match the great dodecahedron.<|endoftext|> -TITLE: What is a good algorithm to measure similarity between two dynamic graphs? -QUESTION [13 upvotes]: I am using graphs to represent structure present in a scene. The vertices represent the objects in the scene and the edges represent the relationship between two nodes(touching, overlapping, none). Graphs are calculated for each frame. The structure of the graph changes when the objects are moved or modified in the video. -I have two graphs whose number of vertices and the edges between them keep changing with time. I want a similarity metric between two such graphs. -The method used currently is to encode the changes in graph structure in a string. So, we get two strings representing the change in graph structure with time. Now substring matching is done between the two strings and this is used to determine the similarity of the two videos. -I am not sure this is the right way to compare the similarity of two graphs. Something more mathematically concrete should exist. What are some techniques which might be helpful to me? I am not entirely sure whether this is the right place to post this but any pointers to what I should read to model this would be helpful. - -REPLY [3 votes]: There are two important survey papers that covers this topic that I highly recommend: -Conte, Donatello, et al. "Thirty years of graph matching in pattern recognition." International journal of pattern recognition and artificial intelligence 18.03 (2004): 265-298. -Foggia, Pasquale, Gennaro Percannella, and Mario Vento. "Graph matching and learning in pattern recognition in the last 10 years." International Journal of Pattern Recognition and Artificial Intelligence 28.01 (2014): 1450001. -Graph matching is also called "network alignment" in some papers.<|endoftext|> -TITLE: How does Thurston's Orbifold Geometrization imply that knots with meridional rank 2 are 2-bridge? -QUESTION [7 upvotes]: Problem 1.11 of Kirby's list asks whether every knot that has a knot group -which can be generated by n meridians, but not less than n, is an n-bridge -knot. There is a one-sentence update, saying that the Orbifold Geometrization Conjecture -implies that for n=2 the answer is yes. -How does this argument go? Is the 1985 result of Hodgson and Rubinstein, that the only knots with double branched cover a lens space are 2-bridge knots, relevant? - -REPLY [6 votes]: Let $\pi_1(K)$ be the fundamental group of the knot $K$. Let it be generated by two meridians $a$ and $b$ (what is always the case for a two-bridge knot). Consider the normaliser $N \langle a^2, b^2 \rangle$. The $\pi$-group $O(K)$ of $K$, according to Boileau and Zimmermann, in this case is $O(K) \cong \pi_1(K)/ N\langle a^2, b^2 \rangle$. -The proof by Boileau and Zimmermann then goes like that: in fact they show that for a two-bridge knot $O(K)$ is dihedral, once the knot is non-trivial, and if $O(K)$ is dihedral, then the knot is two-bridge and non-trivial. -In fact, if $\pi_1(K)$ is generated by $a$ and $b$, the group $O(K)$ could be either dihedral (finite or infinite) or cyclic of order $2$. The cyclic one corresponds to the trivial knot (B. and Z. refer to the proof of Smith's Conjecture). -The authors use the theorem by Hodgson and Rubinstein in the case of $O(K)$ is finite dihedral. Also they use Thurston's orbifold geometrisation to show that the double cover $V$ of the $\pi$-orbifold along $K$ will be a lens space. Then the theorem by H. and R. applies. -If $O(K)$ is infinite dihedral, B. and Z. consider the double cover $V$ of the $\pi$-orbifold along $K$ again, and deduce that it's $S^1\times S^2$ making use of Smith's conjecture and the $\mathbb{Z}_2$-invariant sphere theorem. A paper by Tollefson (Involutions on $S^1\times S^2$ and other 3-manifolds, Trans. AMS, (183), 139-152 (1973)) says that $K$ must be a trivial two-component link in this case. I don't know how Tollefson obtains this result. -Hope my reply helps. Cheers!<|endoftext|> -TITLE: Easier reference for material like Diaconis's "Group representations in probability and statistics" -QUESTION [20 upvotes]: I'm teaching a class on the representation theory of finite groups at the advanced undergrad level. One of the things I'd like to talk about, or possibly have a student do any independent project on is applications of finite groups in statistics and data analysis. Unfortunately, the only book I know on the subject is Persi Diaconis's Group representations in probability and statistics, which is lovely but nowhere close to the level that I would expect an undergrad to read (or for that matter, anyone from a field outside math). Are there any books or articles people know of which provide a gentler introduction? - -REPLY [7 votes]: This 74-page paper in Journal of Machine Learning Research (by Huang, Guestrin, and Guibas) -- -http://www.jmlr.org/papers/volume10/huang09a/huang09a.pdf --- is an amazingly useful and undergrad-friendly intro to the representation theory of the symmetric group (in a very surprising venue). Also, the fact that its applications are to machine learning and statistics looks very appropriate for OP's question. - -REPLY [3 votes]: The book Probability on Discrete Structures contains a chapter called "Random walks on finite groups" by Laurent Saloff-Coste which predates Benjamin Steinberg's book.<|endoftext|> -TITLE: Lingering foundational question about sheaves of abelian groups -QUESTION [5 upvotes]: Motivation for the question: -I have a standard working knowledge of sheaves. Given a scheme, a coherent module over its structure sheaf and a few hours I can compute things. Despite this I have always been unsure how to prove that the category of sheaves has all the properties which I know about. -Whenever I try and prove foundational results about sheaves, I end up producing circular arguments. The reason this happens is because for me, it is impossible to think about sheaves without my intuition kicking in. This is great for geometry, but not so great when I am trying to explain why you can check exactness at a local or stalk level to someone who is just starting Hartshorne. I always end up saying "it's just true!" or "its a standard exercise" which is never a good thing in mathematics. -I suspect the issue is that I don't really know how to prove that sheaves of abelian groups form an abelian category (at least i got stuck trying to do this). - - Question: I have always struggled proving that sheaves of abelian groups form an abelian category. Is there a slick way to do this? I am not afraid of category theory. - -REPLY [6 votes]: I take it you mean sheaves on topological spaces. I think it is valuable to grasp both of two approaches. One (which works essentially the same way for sheaves on sites) is to first see that presheaves of Abelian groups have biproducts, kernels, and cokernels trivially by naturality; and then see by adjointness that sheafification of presheaf coproducts and cokernels gives coproducts and cokernels of sheaves of groups. -The other way is to think of sheaves on a topological space as local homeomorphisms from espaces etales, verify that products and kernels are defined very simply and that products also have the coproduct property (inherited from Abelian groups pointwise) and that cokernels are defined by a relatively simple local criterion. -Maybe I am missing what your problem is, though. Can you point to a part of these steps that gives you trouble?<|endoftext|> -TITLE: Some variants of the Shelah's Weak Hypothesis -QUESTION [9 upvotes]: Are equivalent (in ZFC) the following two statements, for any infinite cardinal $\mu$? -(i) For every infinite cardinal $\kappa$, $|\{ \lambda \in \kappa : \lambda \textrm{ is a singular cardinal and} \operatorname{pp} (\lambda) \geq \kappa \}| \leq \mu$. -(ii) For every infinite cardinal $\kappa$, $|\{ \lambda \in \kappa : \lambda \textrm{ is a singular cardinal and} \operatorname{pp} (\lambda) > \kappa \}| \leq \mu$. -The statement (i), with $\mu = \aleph_0$, is the Shelah's Weak Hypothesis (SWH). See for instance page 360 on - -S. Shelah. Cardinal Arithmetic, - volume 29 of Oxford Logic Guides. Oxford University Press, New York, 1994. - -The statements (ii) appear in the final section of - -Moti Gitik. Short Extenders Forcings II, preprint, July 24, 2013. - -REPLY [7 votes]: They are equivalent, though it took me a while to see it. This is perhaps my fifth attempt at getting a proof, so caveat lector. -Clearly (i) implies (ii), so assume by way of contradiction that (i) fails while (ii) holds. -Choose $\kappa$ least such that -$$|\{\lambda<\kappa:\rm{pp}(\lambda)\geq\kappa\}|>\mu.$$ -We assume that (ii) holds, so we can show - -$\kappa$ is singular of cofinality $\mu^+$, and -$\{\lambda<\kappa:\rm{pp}(\lambda)=\kappa\}$ is unbounded in $\kappa$ of order-type $\mu^+$ - -Our next move is to recall some facts about $\rm{pp}_\theta(\lambda)$ where $\lambda$ is a singular cardinal. (Remember that $\rm{pp}_\theta(\lambda)$ is just like $\rm{pp}(\lambda)$ except that we are allowed to use sets of size $\theta$ instead of sets of size $\rm{cf}(\lambda)$; thus $\rm{pp}(\lambda)=\rm{pp}_{\rm{cf}(\lambda)}(\lambda)$.) -What we need is the following: -If $\rm{cf}(\lambda)\leq\theta<\lambda$, then - -$\rm{pp}(\lambda)\leq\rm{pp}_\theta(\lambda)$, -$\rm{cf}(\rm{pp}_\theta(\lambda))>\theta$, -if $\sigma<\lambda$ is singular of cofinality at most $\theta$ and $\rm{pp}_\theta(\sigma)\geq\lambda$, then $\rm{pp}_\theta(\lambda)\leq\rm{pp}_\theta(\sigma)$, and -$\rm{pp}_\theta(\lambda)=\rm{pp}(\lambda)$ if $\rm{pp}_\theta(\sigma)<\lambda$ for all sufficiently large singular $\sigma<\lambda$ of cofinality at most $\theta$ - -The first two of these are easy (famous last words...), the third is "Inverse Monotonicity" (see page 57 of Cardinal Arithmetic) while the last is one of the main conclusions of [Sh:371]: see $\otimes_2$ on page 313 of Cardinal Arithmetic. -Since $\kappa$ has cofinality $\mu^+$, we find -$$\rm{pp}(\lambda)=\kappa\Longrightarrow \rm{pp}(\lambda)<\rm{pp}_{\mu^+}(\lambda).$$ -and therefore $\{\lambda<\kappa:\kappa<\rm{pp}_{\mu^+}(\lambda)\}$ is unbounded in $\kappa$. -Suppose now that $\alpha<\kappa$, and let $\lambda_\alpha$ be the least $\lambda>\alpha$ such that $\rm{pp}_{\mu^+}(\lambda)>\kappa$. -If $\alpha<\sigma<\lambda_\alpha$ and $\sigma$ is singular, then -$$\rm{pp}_{\mu^+}(\sigma)<\kappa$$ -by our choice of $\lambda_\alpha$. -But then we know -$$\rm{pp}_{\mu^+}(\sigma)<\lambda_\alpha$$ -as otherwise inverse monotonicity would imply -$$\rm{pp}_{\mu^+}(\lambda_\alpha)\leq\rm{pp}_{\mu^+}(\sigma)<\kappa$$ -contradicting our choice of $\lambda_\alpha$. -But now [Sh:371] applies to $\lambda_\alpha$, and we conclude -$$\kappa<\rm{pp}(\lambda_\alpha)=\rm{pp}_{\mu^+}(\lambda_\alpha)$$. -Thus, $\{\lambda<\kappa:\kappa<\rm{pp}(\lambda)\}$ is unbounded in $\kappa$ hence of cardinality at least $\mu^+$. This violation of (ii) gives us our contradiction.<|endoftext|> -TITLE: Solution of linear ODE -QUESTION [10 upvotes]: Let $A=A(t)$ be a smooth one parameter family of $n\times n$-matrices, $n\ge 2$. -It seems that the solution of linear ODE -$$\dot x= Ax$$ -can not be written in a closed form using $\int$, $A$, $x(0)$ and the standard functions. - -Question 1. Is it a theorem? - -If "no". - -Question 2. Any idea how to prove such statement? - -REPLY [19 votes]: This is known as Differential Galois Theory, first developed by Picard and Vessiot. In your case you should look for authors such as Kolchin or Singer and Van Der Put. Some systems definitely admit solutions in "closed form" (you can build them!), but most won't. - The ingredient is the "monodromy group", measuring the multivaluedness of the analytic continuation (in the complex line) of local solutions, whose Zariski closure is an algebraic subgroup of $GL(n,\mathbb C)$ (assuming your equation starts with coefficient in $\mathbb R$ or $\mathbb C$). The solvability of the (connected component of the identity) of this algebraic group is the criterion for solvability in "closed form". -A comprehensive reference is: -van der Put, Marius; Singer, Michael F -"Galois theory of linear differential equations" -Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 328. Springer-Verlag, Berlin, 2003. -MathSciNet : MR1960772<|endoftext|> -TITLE: Reconstructing the number of Hamiltonian cycles -QUESTION [5 upvotes]: As is common terminology in graph reconstruction, given a graph $G$, we call a vertex deleted subgraph of $G$, a card, and call the multiset of all cards, the deck of $G$. The graph reconstruction problem is to determine the isomorphism type of $G$ by only looking at its deck. -Question: Is there a slick argument for reconstructing the number of Hamiltonian cycles of $G$ from its deck? -I have seen a paper of Tutte where he solves this using some machinery he uses to reconstruct several polynomial invariants of $G$. I don't have access to that paper right now, and I was wondering if a simpler solution exists when we restrict our attention just to counting Hamiltonian cycles. - -REPLY [7 votes]: Bill Kocay found a more direct combinatorial method to reconstruct the number of hamiltonian cycles and some other spanning subgraphs. It is in his paper "Some new methods in reconstruction theory", Lecture Notes in Mathematics Volume 952, 1982, pp 89-114. You can get it here if you have sufficient Springer access.<|endoftext|> -TITLE: Kolmogorov superposition for smooth functions -QUESTION [16 upvotes]: Kolmogorov superposition theorem states that a continuous function $f(x_1,\ldots,x_n)$ can be written as -$$f(x_1,\ldots,x_n)=\sum_{q=0}^{2n}\Phi_q\left(\sum_{p=1}^{n}\phi_{q,p}(x_p)\right)$$ -for certain continuous functions $\Phi_q$ and $\phi_{q,p}$. -For smooth, $f\in C^{\infty}$, can we obtain such a representation with the $\Phi_q$ and $\phi_{q,p}$ smooth too? Moreover, I am particularly interested in the case in which more than smooth, $f$ belongs to some Denjoy-Carleman quasianalytic class, but the question about smoothness would be a point to start. -I don't assume independence of $\Phi_q$ and $\phi_{q,p}$ of their parameters or of $f$. - -REPLY [6 votes]: The inner functions $\phi_{qp}$ are independent in Kolmogorov's superposition theorem and they are very "bad" functions like cantor function. The smoother $f$ is, the more singular $\Phi_q$ has to be to cancel out the singularities of $\phi_{pq}$ except in some special cases, such as $f$ is a constant. There is no linear relation between the smoothness of $f$ and it representing function $\Phi_q$. The smoothness of $f$ mainly depends on the behaviour of $\Phi_q$ on the image of singular points of $\sum_p \phi_{qp}$ .<|endoftext|> -TITLE: Hopf algebra structure on the ring of quasisymmetric functions -QUESTION [5 upvotes]: I'm looking for a particular description of the Hopf algebra structure on the ring of quasisymmetric functions. -Let me illustrate by giving this kind of description for the Hopf algebra of symmetric functions. -Fix a ground field $k$. -Edit: I'm happy to assume $k=\mathbb{Q}$. As darijgrinberg pointed out in the comments, $\mathrm{QSym}$ only has the Lyndon monomial basis when $\mathrm{char}(k)=0$. -Let $\Lambda$ be the ring of symmetric functions over $k$. Putting a Hopf algebra structure on a $k$-algebra $R$ is the same as putting a group structure on $\mathrm{Hom}_{k-alg}(R,S)$ for each $k$-algebras $S$ such that postcomposition is a group homomorphism. Since $\Lambda$ is a polynomial algebra in the elementary symmetric functions $e_1,e_2,\ldots$, we can identify $\mathrm{Hom}_{k-alg}(\Lambda,S)$ with the set of sequences $(s_1,s_2,\ldots)$ of elements of $S$ (where $f:\Lambda\to S$ is identified with $(f(e_2),f(e_2),\ldots)$). If we further identify the sequence $(s_1,s_2,\ldots)$ with the power series $1+s_1T+s_2T^2+\ldots\in 1+TS[[T]]$, then the group structure (coming from the Hopf algebra structure on $\Lambda$) is multiplication of power series. -I am hoping for something similar for the ring of quasisymmetric functions. Specifically, I would like, for each $k$-algebra $S$, a group $G(S)$ (which won't generally be abelian) and a bijection $\mathrm{Hom}_{k-alg}(\mathrm{QSym}_k,S)\leftrightarrow G(S)$, such that the group multiplication $G(S)\times G(S) \to G(S)$ is induced by the comultiplication on $\mathrm{QSym}$. I'm hoping the description of $G(S)$ is as explicit as possible (like the group $1+TS[[T]]$ in the case of symmetric functions). -It is known that $\mathrm{QSym}$ is a polynomial algebra. One possible free generating set is the monomial quasi-symmetric functions -$$ -M_{(s_1,\dots,s_k)}:=\sum_{i_1<\ldots -TITLE: How universal is operadic approach to studying algebras? -QUESTION [10 upvotes]: I have just started to read about operads, so this question might be silly. -So it seems to me that any "reasonable" class of algebras can actually be defined as a class of all algebras over a certain operad. For example, associative algebras are algebras over the associative operad $\mathcal{As}$, commutative algebras - over the commutative operad $\mathcal{Com}$. The same is true for Lie, Poisson, Gerstenhaber, $A_{\infty}$-algebras. -So my question is: is there a (interesting) class of algebras that can't be viewed as a class of algebras over a certain operad? I mean, how universal is the operadic approach to studying algebras? -If there are algebras that are not algebras over an operad, then what are the obstructions? -Thank you very much! - -REPLY [2 votes]: I put in a remark about this question back in 2013. I'm going to expand on that for emphasis. Let me shamelessly quote myself, from the expository paper http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf -where I describe the coining of the word operad. -``To these ends, I consciously sacrificed all-embracing generality: many types of algebras defined by identities are deliberately excluded". In particular, algebras defined by identities with repeated variables are deliberately excluded despite their evident interest. I knew about alternating algebras and Jordan algebras from a course at Yale with Nathan Jacobson and about graded Lie algebras (char 2 or 3) from my thesis, and they are deliberately excluded.<|endoftext|> -TITLE: Where should I learn about Kolmogorov complexity of overlapping substrings? -QUESTION [9 upvotes]: I would like to know more about the relationship between the Kolmogorov complexity of a string and that of its substrings. The relation that up to an additive constant, $K(x,y) = K(x) + K(y\ |\ x, K(x))$ begins to address this issue, but I am hoping there are results on substring complexity as opposed to the complexity of pairs of strings. For example, suppose that it is possible to "cover" an infinite binary sequence $X$ with (possibly overlapping, possibly length-bounded) substrings $\sigma$ such that each $\sigma$ has $K(\sigma) < s|\sigma|$ (with $s$ being some fixed constant). What can be said, if anything, about $K(\sigma)$ for arbitrary substrings $\sigma$ of $X$? While an answer to that specific question is interesting if you have it, I am most hoping for pointers on where to learn about substring complexity more generally. - -REPLY [3 votes]: Although I don't have the references you seek, I did notice that the following observation answers your specific question, in the case you don't impose any bound on the covering substrings. -Claim. There is an infinite binary string $X$, which is covered by non-overlapping substrings $\sigma$, with $K(\sigma)\lt \frac12|\sigma|$, or indeed less than $s|\sigma|$ for any fixed desired $s$, such that every finite string arises infinitely often as a substring of $X$. -The idea is simply to build $X$ as the concatenation of strings of the form $\sigma=\tau^\frown0000\cdots000$, where an arbitrary string $\tau$ is simply padded with an enormous number of $0$s so as to ensure $K(\sigma)$ is small in comparison with $|\sigma|$. Each such string has low complexity compared with its length. Consequently, if we construct $X$ as the concatenation of all such strings arising with all possible finite binary strings $\tau$, then $X$ will satisfy your covering property, but it will exhibit every possible finite string as a substring. -In particular, we cannot deduce anything special about $K(\sigma)$ for an arbitrary substring of $X$ under the circumstances you describe, since every finite string is a substring of $X$. -In this example, the covering strings get longer and longer, but a similar argument works -even when you impose a bound on the lengths. -Claim. For any fixed even length $k$, there is an infinite binary string $X$ that is covered by non-overlapping finite strings $\sigma$ of length at most $k$, such that $K(\sigma)\leq\frac12|\sigma|+1$, such that every string of length $k$ arises as a substring of $X$. -Build $X$ as the concatenation of all possible strings $\sigma$ of the form $$\tau^\frown000\cdots000\qquad\text{ or }\qquad000\cdots000^\frown\tau,$$ -where the string $\tau$ is half of the string and the rest is made of $0$s. Any such string has $K(\sigma)\leq \frac12|\sigma|+1$, since you can just specify $\tau$ and then indicate whether the $0$s come before or after. But now the point is that if you build $X$ by sometimes putting the zeros first and sometimes by putting them last, then you can arrange that every possible string $\eta$ of length $k$ arises as a substring of $X$, since if $\eta=\eta_0^\frown\eta_1$ is cut in half, then $000\cdots000^\frown\eta_0^\frown\eta_1^\frown000\cdots000$ arises as a substring of $X$, where $\eta_0$ appears as the end of the first half and $\eta_1$ arises as the beginning of the next part.<|endoftext|> -TITLE: cohen-macaulayness of reduced and non-reduced schemes -QUESTION [9 upvotes]: Let $X$ be a Cohen-Macaulay scheme (let's say of finite type over a field). -Let $X_{red}$ be the corresponding reduced scheme. Is it true that $X_{red}$ is also -Cohen-Macaulay? - -REPLY [8 votes]: The simplest counter-example I know is the following: Hartshorne showed that if $k$ has positive characteristic, $k[s^4, s^3t, st^3,t^4]$ (which will be $X_{red}$) is a set-theoretic complete intersection (said complete intersection will be $X$). The former is well-known to be not CM (cheapest proof: $s^4,t^4$ form a s.o.p but not a regular sequence). -There are more examples of projective curves which are set-theoretic c.i. (you can find quite a few papers). Among them the ones which are not arithmetically CM give counter examples via taking the affine cone. -EDIT (to address the OP's new question below): this new situation is discussed in my answer quoted above by Vesselin, so you may want to take a look. To use that answer's notation, you need at least two height one primes, say $P,Q$, which are Cohen-Macaulay, and $a[P]+b[Q]=0$ in the class group of $Y$ (this takes care of the assumption that $X$ is set theoretically principal), but $[P]+[Q]$ is not CM. Such examples probably still exist (for example there are torsion classes which are not CM), but we may need a lot of luck (or hard work) to write one down.<|endoftext|> -TITLE: Random walk with positive uniformly distributed steps -QUESTION [6 upvotes]: Let $U_1,U_2,\ldots$ be iid random variables distributed uniformly on $[0,1]$. I am interested in the random walk $X_i = \sum_{j \leq i} U_j$. In particular, - -What is the expected number of points appearing in an interval $[x,x+1]$? - -Experimentally, this seems to converge to $2$. -Here is an intuitive explanation. Suppose that $x+1-\delta$ is the first point within the interval $[x,x+1]$ which the random walk hits. One can calculate that the expected number of points within the interval is $e^\delta$. To estimate the distribution of $\delta$, consider the point $x-\epsilon$ just preceding $x+1-\delta$. Assuming that $\epsilon$ is uniformly distributed (which is close to the truth when $x$ is large), the density of $\delta$ at a point $\delta'$ is proportional to $\Pr[\epsilon \leq \delta'] = \delta'$, and so it is $2\delta'$. Therefore the expected number of points in $[x,x+1]$ is -$$ -\int_0^1 2\delta e^\delta \, \mathrm{d}\delta = 2. -$$ - -REPLY [9 votes]: Let $X_t$ be the number of points in $[0,t]$. Then, $X_t$ is a renewal process. Let $m(t) = E[X_t]$. Then renewal theorem says -$$ - m(t+h) -m(t) \stackrel{t \to \infty}{\longrightarrow} \frac{h}{\mu} -$$ -where $\mu = E[U_1]$ is the expected increment. -With $\mu = 1/2$ and $h = 1$ in this case, one gets $2$ for the limit.<|endoftext|> -TITLE: Automorphisms of Hyperbolic groups and Graphs of Groups -QUESTION [5 upvotes]: I have been reading Levitt's paper Automorphisms of Hyperbolic groups and Graphs of Groups. I am having some trouble trying to fit all the bits together, and would appreciate some help with this last step. -In the paper, Levitt considers minimal graphs of groups, and gives results regarding (a specific subgroup of) the outer automorphism group of the fundamental group of an arbitrary graph of groups. A graph of groups is minimal if edge groups are proper subgroups of adjacent vertex groups. -Levitt applies this work to the (canonical) JSJ-decomposition of a (one-ended) hyperbolic group. The "canonical" means that we can forget that we are looking at a specific subgroup of the outer automorphism group: his analysis can be applied to the whole outer automorphism group. Which is good, and is the point. -Now, I (believe that I) understand all of the fiddly bits of this paper. However, I have come across a stumbling block when trying to pin it all together. You see, the canonical JSJ-decomposition Levitt is using is due to Bowditch (from Bowditch's paper Cut points and canonical splittings of hyperbolic groups). This decomposition is entirely canonical, unlike the one Sela obtains for the torsion-free case (where it is canonical up to certain "moves"). Bowditch's decomposition is canonical because of the addition of "elementary" vertices, that is, vertices whose stabiliser is virtually cyclic. My problem is as follows: I believe that these vertices imply that the graph of groups is not necessarily minimal. -For example, pin together two hyperbolic triangle groups $H$ and $K$ across an infinite cyclic subgroup, so $G=H_{C_1=C_2}K$, and ensure that these subgroups are malnormal infinite cyclic. Then the JSJ-decomposition of this (one-ended hyperbolic group) looks like $H-C-K$, where the edge stabilisers are $C_1$ and $C_2$, and where $C_1=C=C_2$. A contradiction. No? -I am sure I am just missing something silly, something obvious, something I should have seen a long time ago. However, I have no idea what that could possibly be! - -REPLY [6 votes]: This has been dealt with in comments, but since MO works better if answers are given, I'll elaborate a little bit here. -A graph of groups satisfying your definition of 'minimal' is usually called 'reduced'. You're absolutely correct that Bowditch's JSJ is often not reduced (though note that, in your example, the graph of groups that you write down is only the JSJ if neither $H$ nor $K$ splits over a 0- or 2-ended subgroup relative to $C$). -The relevant definition in Levitt's paper is on page 53: - -We assume that $\Gamma$ is minimal: $G_e$ is a proper subgroup of $G_v$ for every terminal vertex $v$ (equivalently, $\pi_1(\Gamma')$ is a proper subgroup of $\pi_1(\Gamma)$ for every proper connected subgraph $\Gamma'$). - -The non-parenthetical part of the sentence is indeed very easy to misunderstand in the way you have; I suppose by 'terminal vertex' he must (unusually) mean 'leaf'. -The parenthetical part, though, makes it absolutely clear: his definition of `minimal' coincides with the usual one. That is, a splitting is minimal if the the Bass--Serre tree has no proper invariant subtrees. Bowditch's JSJ is always minimal, as can be seen, for instance, from the fact that orbits in the boundary are dense.<|endoftext|> -TITLE: Papers better than books? -QUESTION [23 upvotes]: Not so long ago I took a class called "Discrete analysis". I remember that I couldn't find any "novice" level material on Mobius functions in combinatorics. So then I went to the roots and read Rota's original paper "On the foundations of combinatorial theory I" and it really impressed me. So I wonder is there other mathematical subjects that it would be better for novice to get started with by reading rather original papers than actual books? -ADDED: Thanks for your answers. That's really interesting! - -REPLY [45 votes]: Very recently I and Misha Sodin had a strong incentive to learn the Ito-Nisio lemma (which, roughly speaking, says that weak convergence in probability of a series of symmetric independent random variables with values in a separable Banach space implies almost sure norm convergence to the same limit). The textbooks we could find fell into 2 categories: those that didn't present the proof at all and those presenting it on page 2xx as a combination of theorems 3.x.x, 4.x.x, 5.x.x, etc. The original paper is less than 10 pages long, essentially self-contained, and very easy to read and understand. -The moral is the same as Boris put forth: the books are there to optimize the time you need to spend to learn the whole theory. However, for every particular implication A->B the approach they usually take is something like E->F->G, G->F, (F and Q)->B; since A->E, then A->G; once we know G, we have F, so it suffices to prove that A->Q to show that A->B; we show that Q,R,S,T,U are equivalent, with the trivial implication S->Q left to the reader as an exercise; finally, we prove that A->S. So if all you need is A->B, you may be much better off reading the paper whose only purpose is to prove exactly that.<|endoftext|> -TITLE: Methods to approximate the betweenness centrality on large networks -QUESTION [5 upvotes]: To calculate the between centrality wiki def: -$g(v) = \sum_{s\neq v \neq t} \frac{\sigma_{st}(v)}{\sigma_{st}}$ - of a node in a graph/network;$\sigma_{st}$ is the total number of shortest paths from node to node and the $\sigma_{st}(v)$ are the paths including the node of concern. -that is very computationally intensive due to the large number of shortest paths that must be calculated. Is there a stochastic method to approximate it? Can non-reversing truncated random walkers traverse the graph (before being ergodic) to sample the hit count for when the node $v$ is encountered? In a way it is a monte carlo approach where the paths are sampled from random walk paths taken. -are there any references for this as well? - -REPLY [5 votes]: Stochastic approximation methods for betweenness centrality have been studied by many people. A good reference is "Centrality estimation in large networks" by Brandes and Pich (2007) -For large sparse networks, exact and approximation algorithms can benefit significantly from the exploitation of structural features of the network. Decomposition into bi-connected components, for example, or the collapsing of groups of structurally equivalent vertices, parallel edges, etc. These techniques don't change the overall complexity of the problem, can speed the calculations dramatically. Details can be found in this paper<|endoftext|> -TITLE: Does the existence of the von Neumann hierarchy in models of Zermelo set theory with foundation imply that every set has ordinal rank? -QUESTION [18 upvotes]: Let $T$ be the theory consisting of Zermelo's original set theoretic axioms (extensionality, empty set, pairing, union, powerset, infinity, separation, choice) together with foundation. Put more succinctly, $T$ consists of ${\rm ZFC}$ axioms without the replacement axiom scheme. The theory $T$ is too weak for most set theoretic purposes because, for instance, it cannot even prove the existence of transitive closures. We consider strengthening $T$ in the following two ways. Let $A$ be the axiom asserting that $V_\alpha$ exists for every $\alpha$ and let $A^*$ be the axiom asserting that every $x$ is an element of a $V_\alpha$. For example, $V_{\omega+\omega}$ is a model of $T+A^*$. Does $T$ together with $A$ imply $A^*$? Or is it possible that $V_\alpha$ exists for every ordinal $\alpha$, but there is a set without an ordinal rank? - -REPLY [12 votes]: Take the Zermelo ordinals to be defined by $Z(0) = 0$, $Z(\alpha+1) = \{Z(\alpha)\}$, and $Z(\lambda) = \{Z(\alpha): \alpha<\lambda\}$ (where $\alpha, \lambda$ are von Neumann ordinals). Then if we add $Z(\omega+ \omega)$ to $V_{\omega +\omega}$ and close under pairing, union, subsets, and powersets, we get a model of $T$ - Choice + $A$ which is not a model of $A^*$. -More precisely, let $D_0 = V_{\omega+\omega} \cup\{Z(\omega+\omega)\}$ and $D_{n+1}$ be the result of adding pairs, unions, subsets, and powersets of element of $D_n$ to $D_n$. Clearly, $M = \bigcup_{n <\omega}D_n$ is transitive and models $T$ - Choice. A simple induction shows that: -For every $x\in D_n$ there is an $\alpha<\omega+\omega$ such that the (von Neumann) ordinals in $tc(x)$ are less than $\alpha$. -Since $V_{\omega+\omega}\subseteq M$, it follows that the von Neumann ordinals in $M$ are just those in $\omega+\omega$. So it models $A$. Since the rank of $Z(\omega+\omega)$ is $\omega+\omega$, it doesn't model $A^*$. -(To get Choice in the form ``every set is well-orderable" we just throw in $x \times x$ at $D_{n+1}$ for $x\in D_n$).<|endoftext|> -TITLE: Formal series convergence in deformation quantization and $C^*$-condition -QUESTION [7 upvotes]: A link between formal series convergence in deformation quantization (strict deformation quantization) and producing $C^*$-algebras instead of mere $*$-algebras (which $(\mathcal{C}^{\infty}(M)[[t]],\star)$ is) after deforming the initial commutative $C^*$-algebra of observables is evoked here: http://ncatlab.org/nlab/show/C-star+algebraic+deformation+quantization -However this is done very briefly and in the language of higher symplectic geometry which I'm not familiar with. -Can anyone give simpler explanations and arguments/theorems justifying the link between convergence and the $C^*$-condition? Or some reference for an introduction? - -REPLY [6 votes]: OK, let me give a try on this question. There are several problems hidden underneath which one has to address. -First, for physical reasons a formal deformation is not sufficient. $\hbar$ is a constant of nature but not a formal parameter... More severely, the formal star product algebras do not allow for a reasonable notion of spectra and thus can not produce physically reliable prediction concerning possible values of measurements. For this (and many related reasons) formal deformation quantization is not the final answer. -Second, the $C^*$-world provides all we need for a good quantum mechanical interpretation: good notion of spectra compatible with a spectral calculus etc. -So in some sense, this is the situation we all want to reach: finding a $C^*$-algebra containing particular elements which have a physical interpretation. Note that just saying "this $C^*$-algebra is the algebra of observables of my system" is physically still meaningless. You have to specify an interpretation of which algebra element corresponds to which (physical) observable (ie. which, at least in principle, realisable measurement apparatus). -Third, and this is the bad side of the story: except for very simple situations (CCRs) it is very very hard to write down a $C^*$-algebra corresponding to a certain quantum system of which one only knows its classical counter part. So this is the problem of quantization: given a classical physical system, one wants to guess its quantum description. Of course, there are physically relevant situations where this is known and well-understood, but I'm taking here of more general classical systems: eg. for systems with gauge degrees of freedom, the physically relevant degrees of freedom form the "reduced phase space" which can carry an complicated geometry such that there are simply -no physical observables which allow for a simply CCR quantization. -Together, this indicates that the formal DQ might be not a solution but a first step: since for formal DQ one has very well-understood and powerful existence and classification results, one tries to use them and, as a second step, guess/construct the desired $C^*$-algebraic framework. But this is not at all easy. -So one way to go is to take the formal power series in the star product and ask for their convergence (in a mathematically meaningful way). This is tricky and has been achieved ony in (few) examples. In fact, I have some current projects in this directions. Ideally, one obtains a topological non-commutative algebra afterwards, which might not yet be $C^*$. But, and this is the first non-trivial step, it will be an algbebra over $\mathbb{C}$ and not just over $\mathbb{C}[[\hbar]]$. Of this algebra, one can then study HIlbert space representations by, in general, still unbounded operators. Arriving at this stage there are several standard techniques to build nice $C^*$-algebras out of it. -Why can one hope that this works? In those examples where one has a $C^*$-algebraic deformation in one of the many senses of "strictness", -it always involves some reasonable behaviour for $\hbar \to 0$. If one requires not just continuity, but some weak sort of smoothness, then one can "differentiate" the continuous field of $C^*$-algebras (from the right) at $hbar = 0$. This will give then a formal star product. As usual, one has only smoothness but not analyticity and hence the formal star product will not directly sum up to the continuous field of $C^*$-algebras. So in my opinion, this is the relation one can hopefully expect in quite some generality.<|endoftext|> -TITLE: Is Taylor expansion related to Helmholtz decomposition? -QUESTION [5 upvotes]: The Taylor expansion of a vector field $f(x)$ to the order of one is -$$f(x)=f(x_0)+Jf(x_0)\cdot\Delta x+o(\Delta x)$$ -where $Jf$ is Jacobian of the vector field and $\Delta x=x-x_0$. -Suppose we decompose it into symmetric and skew-symmetric part such that -$$S=\frac{Jf+(Jf)^T}{2}$$ -$$A=\frac{Jf-(Jf)^T}{2}$$ -Then -$$f(x)\approx f(x_0)+S\Delta x+A\Delta x$$ -Could we regard $f(x_0)$ as a translation, $S\Delta x$ as an expansion or a contradiction along axes of eigen directions for it resembles strain tensor and $A\Delta x$ a rotation since skew-symmetric matrix is infinitesimal of rotation matrix? -If it is correct, I'm a bit confused with, say operator $A$, what object it acts on? -Otherwise, does this decomposition have any practical meaning in terms of vector field or velocity field? - -REPLY [4 votes]: you ask whether the decomposition of the Jacobi tensor $J{f}$ into symmetric and antisymmetric parts $S$ and $A$ produces a Helmholtz decomposition of $\Delta f=J{f}\cdot\Delta{x}$ into a curl-free part $\chi=S\cdot\Delta x$ and a divergence-free part $\omega=A\cdot\Delta x$. -this is indeed correct, I will try to write it out explicitly (trying to avoid the errors I made in my first answer) -notation: the subscript $0$ indicates the dependence on $x_0$ and derivatives with respect to $x_0$; I abbreviate $\xi=\Delta x=x-x_0$ and denote by $\nabla_\xi$ the derivative with respect to $\xi$ (or $x$). -$${\omega}(\xi)={A}(x_0)\cdot\xi=\tfrac{1}{2}\xi\cdot(\nabla_0 f_0)-\tfrac{1}{2}(\nabla_0 f_0)\cdot\xi=\tfrac{1}{2}(\nabla_0\times{f}_0)\times\xi$$ -$$\Rightarrow\nabla_\xi\cdot{\omega}(\xi)=-\tfrac{1}{2}(\nabla_0\times f_0)\cdot(\nabla_\xi\times\xi)=0$$ -$$\chi(\xi)=S(x_0)\cdot\xi=\tfrac{1}{2}(\nabla_0 f_0)\cdot\xi+\tfrac{1}{2}\xi\cdot(\nabla_0 f_0)$$ -$$\Rightarrow\nabla_\xi\times\chi(\xi)=-\tfrac{1}{2}\nabla_0\times f_0+\tfrac{1}{2}\nabla_0\times f_0=0$$ - -vector identities used: -$$a\times(b\times c)=b(a\cdot c)-c(a\cdot b),\;\; -a\cdot(b\times c)=c\cdot(a\times b)$$<|endoftext|> -TITLE: Distribution mod 1 of exponential growth sequences -QUESTION [10 upvotes]: Let $t_n$ be a sequence of real numbers and $C,r>1.$ Suppose that for every $n\geq 1$ we have $\frac{1}{C}r^n\leq t_n \leq Cr^n.$ Does there exist a real number $\xi$ and an $\varepsilon>0$ such that $|| \xi t_n ||\geq \varepsilon$ for every $n\geq1$? -Here $|| x ||$ denotes the distance between $x$ and the nearest integer. -If $r/C^2>1$, then the sequence is lacunary and the answer is yes (by a result discovered independently by Khintchine, Pollington and De Mathan). This is not my area so I'm neither familiar with the literature nor adept at such arguments. Basic Mathscinetting turned up intersting realted results but nothing I could use to answer the above question. - -REPLY [10 votes]: I think the answer is yes. For any natural number $n$, let $P(n)$ denote the assertion that there exists an interval $I_n$ of length $\sqrt{\varepsilon}/t_n$ such that $\| \xi t_m \| \geq \varepsilon$ for all $\xi \in I_n$ and $m \leq n$. If $\varepsilon$ is small enough, it appears that $P(n)$ implies $P(n+A)$ for some suitably large constant $A$ (depending on $C,r$, and with $\varepsilon$ sufficiently small depending on $A,C,r$), basically by taking $I_{n+A}$ to be a random subinterval of $I_n$ and using the union bound to upper bound the probability that this fails to work. By completeness we can find a $\xi$ inside an infinite nested sequence of the $I_n$ which should then do the job.<|endoftext|> -TITLE: Extend a representation of a stabilizer group on a smooth DM stack to a locally free sheaf? -QUESTION [5 upvotes]: Consider a smooth tame Deligne-Mumford stack $[Y/G]$, a point $[p]$ on it with stabilizer group $H$. Is it true that every representation of $H$ can be extended to a locally free sheaf on $[Y/G]$? -Alternatively, consider a smooth scheme (or algebraic space) $Y$ with a $G$ action. Let $p$ be a point on $Y$ with a finite stabilizer group $H$. Is the restriction map from the Grothendieck group of $G$-equivariant locally free sheaves on $Y$ to the representation ring of $H$ -$$K^0_G(Y)\to Rep(H)$$ -$$E\mapsto E|_p$$ -surjective? (An affirmative answer with possibly some additional conditions, or a negative answer with an explicit counterexample are both very welcome.) - -REPLY [9 votes]: Unfortunately that is not always possible. For instance, let $Y$ be $\mathbb{A}^n$, and let $\rho:G\times \mathbb{A}^n\to \mathbb{A}^n$ be a faithful, linear representation. Then the locally free sheaves on $[Y/G]$ are the same as $G$-representations. If there exists a point $p$ of $Y$ whose stabilizer is $H$, then you are asking whether every linear representation of $H$ is the restriction of a linear representation of $G$. -So let $\rho$ be the standard linear representation of $\mathfrak{S}_n$ on $\mathbb{A}^n$, let $G$ be the alternating subgroup $\mathfrak{A}_n$, and let $p$ be the vector $(0,0,1,1,x_5,\dots,x_n)$, where $0,1,x_5,\dots,x_n$ are all distinct. The stabilizer subgroup of $p$ is the cyclic subgroup generated by the involution $(12)(34)$. There is a nontrivial character of this group. Yet there is only the trivial character of $\mathfrak{A}_n$.<|endoftext|> -TITLE: How to understand a solenoid? -QUESTION [7 upvotes]: Consider the invertible extension of the circle-doubling map $T(x)=2x \pmod 1$, the new system can be represented as $X=\{(x_k)\ | \ x_{k+1}=T(x_k)\}$ (see GTM 259 M. Eindiedler & T. Ward Exercise 2.1.9) dynamically. Another representation can be described algebraically (see also GTM259 2.1.9) as the quotient of $\mathbb{R}×\mathbb{Q}_2$. A third representation may be the dual of 2-adic rational field. My question is how to image it intuitively? - -REPLY [3 votes]: In my opinion, the best way to understand the solenoid intuitively is using nested intersections. -Let $\mathbb{T}_3 \overset{def}{=} D^2 \times S^1$ where $S^1 = [0,1] \mod 1$ and $D^2 = \{z \in \mathbb{C} \mid \left|z\right| < 1\}$. Consider $f:\mathbb{T}_3 \to \mathbb{T}_3$ given by $$f(z,\theta) = (\dfrac{1}{10}z + \dfrac{1}{2}e^{i \theta},2\theta \mod 1).$$ -Note that $f$ is well defined because $$\left| \dfrac{z}{10} + \dfrac{1}{2}e^{i \theta}\right| \leq \left|\dfrac{z}{10}\right| + \dfrac{1}{2}\left|e^{i\theta}\right| \leq \dfrac{1}{2} + \dfrac{1}{10} \leq 1.$$ Actually this implies that $f(\overline{\mathbb{T}_3}) \subset \mathrm{int} \mbox{ } \mathbb{T}_3$. -This map gives you an extension of $2x \mod 1$. Then the solenoid is the nested intersection $\Lambda = \mathop{\bigcap}\limits_{n=0}^\infty f^n(\mathbb{T}_3)$, which is the attractor of the dynamical system $(\mathbb{T}_3, f)$. Moreover $\Lambda$ is the set where $f$ is bijective. Observe that you have a contraction in one direction and expansion in the other. Actually the map acts in the following way: Note $S^1$ spins twice inside $\mathbb{T}_3$, and the discs $D^2 \times \{\theta\}$ go to $D^2 \times \{2\theta\}$, i.e.: $$f(D^2 \times \{\theta\}) \subset D^2 \times \{2\theta\}.$$ -Here you can see some pictures: http://www.matcuer.unam.mx/~aubin/vista/index4.html -This book can be illustrative as well: C. Christenson and W. Voxman, Aspects of topology. 2nd Edition, BSC Associates, Moscow, Idaho, USA, 1988, pp. 167-170. Also, it contains the inverse limit construction that @Asaf mentioned.<|endoftext|> -TITLE: Duality between singularities and non-compactness in the yoga of weights -QUESTION [8 upvotes]: According to Deligne's "yoga of weights", the cohomology of an algebraic variety should have a weight filtration. For concreteness we can consider the rational cohomology of complex varieties, with their mixed Hodge structure. -It seems to me that in the yoga of weights there is a kind of duality between singularities and non-compactness. The simplest example should be: - -if $X$ is smooth, then $H^n(X)$ has weights at least $n$ -if $X$ is compact, then $H^n(X)$ has weights at most $n$ - -We also have the following: - -let $X$ be a smooth variety and $X \to Y$ a smooth compactification. Then $W_n H^n(X) = \mathrm{Im}(H^n(Y) \to H^n(X))$. -let $X$ be a compact variety and $Y \to X$ a resolution of singularities. Then -$H^n(X)/W_{n-1}H^n(X) = \mathrm{Im}(H^n(X) \to H^n(Y))$. - -Are there more examples of this "duality"? Is there a unifying principle here? - -REPLY [6 votes]: This may be already clear to you, but from my perspective, the clearest manifestation of this duality is in the setting of mixed Hodge modules (or some other version of ``mixed sheaves''). -Let $f: X \to Y$ be a morphism of complex algebraic varieties, and let $D_m(X)$ and $D_m(Y)$ refer to the derived categories of mixed Hodge modules. Then there are functors -$$ -f^!, f^\ast : D_m(Y) \to D_m(X) -$$ -$$ -f_\ast, f_! : D_m(X) \to D_m(Y) -$$ -Let me also define the shifted functor $f^\dagger = f^! [\dim Y - \dim X]$. -We also have the Verider duality functors $\mathbb D_X$ and $\mathbb D_Y$. The functors are related as follows: -$ -f^! \mathbb D_Y = \mathbb D_X f^\ast$, and $f_! \mathbb D_X = \mathbb D_Y f_\ast$. -We have that: $f_\ast$ and $f^!$ increase weights, whereas $f_!$ and $f^\ast$ decrease weights. -In this language: -If $f$ is smooth (i.e. submersive) then $f^\dagger$ commutes with $\mathbb D$, i.e. $\mathbb D_X f^\dagger \simeq f^\dagger \mathbb D_Y$ -If f is proper then $f_\ast$ commutes with $\mathbb D$. -Thus smoothness gives a relationship between relative dualizing sheaf and constant sheaf, and properness relates relative cohomology with relative compactly supported cohomology. -For example, in the case $f: X \to pt$, we have -$$ f_\ast f^\ast \mathbb Q \simeq H^\ast (X) $$ -$$ f_! f^! \mathbb Q \simeq H_\ast(X)$$ -$$ f_! f^\ast \mathbb Q \simeq H^\ast _c(X)$$ -$$ f_\ast f^! \mathbb Q \simeq H_\ast ^{BM}(X)$$. -Smoothness of $X$ means the dualizing sheaf ($f^! \mathbb Q$) is isomorphic to the constant sheaf ($f^\ast \mathbb Q$) up to a shift. Properness means that compactly supported cohomology (with coefficients in some sheaf) is isomorphic to ordinary cohomology. This recovers your first observation.<|endoftext|> -TITLE: An interaction between prime numbers -QUESTION [13 upvotes]: Let   $p_1\ p_2\ \ldots$ be the sequence of all natural prime numbers. There is a slight (just slight) but clear tendency for imitating the number of primes in an interval $(p_k;\ p_n)$   by the number of primes in the double interval   $(p_k\!+p_{k+1};\ p_{n-1}\!+p_n)$; possibly by   $(2\cdot p_k; 2\cdot p_n)$   too. Let me ask two open questions along this line. The first one will be most likely hopeless while the second one may lead to a discussion and at least to numerical computations. -P1.   Does there exist a natural number   $d$   such that for every natural number   $n$   the real interval -$$ (2\cdot p_n;\ 2\cdot p_{n+d})$$ -contains at least one prime? -P2.   (when P1 fails): Given a natural number   $d$,   let   $w(d)$   be the least natural number such that the interval of P1 (see above) does not contain any prime number. What is the growth of the sequence -$$w(1)\ \ w(2)\ \ w(3)\ \ldots$$ - -The above notions got shifted from my original definition by a half of a prime. The question Q1 below is still equivalent to question P1 above: -Q1.   Does there exist a natural number   $d>1$   such that for every natural number   $n$   the real interval -$$ (p_n\!+p_{n+1};\ p_{n+d-1}\!+p_{n+d})$$ -contains at least one prime? -Q2.   (when Q1 fails): Given a natural number   $d>1$,   let   $v(d)$   be the least natural number such that the interval of Q1 (see above) does not contain any prime number. What is the growth of the sequence -$$v(1)\ \ v(2)\ \ v(3)\ \ldots$$ -EXAMPLE   Consider the consecutive primes -$$p_{360} = 1901 \qquad p_{361}=1907 \qquad p_{362}=1913$$ -Then the real interval -$$(p_{360}\!+p_{361};\ p_{361}\!+p_{362})\ \ =\ \ (3802; 3820)$$ -contains no primes, i.e.   $v(2)\le 360$. - -In general, I'd be interested in similar relative properties of primes, where primes are studied in relations to other primes, and the relation is not trivial, meaning not reduced to general properties between integers. - -REPLY [22 votes]: The answer to P1 is negative, thanks to the recent work of Maynard on bounded gaps between primes. -What Maynard shows is that given any $d$, there exists a k-tuple $h_1,\dots,h_k$ such that for infinitely many $n$, at least $d+1$ of $n+h_1,\dots,n+h_k$ are prime. In fact, the argument shows that for sufficiently large $x$, the number of $n \in [x,2x]$ such that at least $m$ of $n+h_1,\dots,n+h_k$ are prime, and the rest are almost prime in the sense that they have no prime factor less than $x^\varepsilon$ for some small fixed $\varepsilon>0$, is $\gg \frac{x}{\log^k x}$. (Such strengthenings of Zhang/Maynard type theorems are discussed in this paper of Pintz, and also in this Polymath8b preprint.) A standard upper bound sieve (e.g. Selberg sieve or beta sieve) then shows that after removing about $O( x/\log^{k+1} x)$ of these $n$, one can also ensure that none of the numbers between $2(n+h_1)$ and $2(n+h_k)$ are prime. If we let $p_i$ be the first prime greater than or equal to $n+h_1$, then we have $2(n+h_1) \leq 2p_i \leq 2p_{i+d} \leq 2(n+h_k)$, and so we obtain a counterexample to P1 for any $d$. -Unfortunately we don't get an effective rate on P2 this way due to the reliance on the Bombieri-Vinogradov theorem in the work of Maynard etc. However this can likely be removed by following the ideas mentioned near the end of this paper of Pintz, though I did not attempt this. [EDIT: it looks likely that the quantitative version of Maynard's results in this recent paper of Banks, Freiberg, and Maynard will do the trick, after some small modification.]<|endoftext|> -TITLE: What Turing-Complete models of computation carry a notion of time complexity that "agrees" with that of Turing Machines? -QUESTION [10 upvotes]: Certain models of computation are technically Turing-Complete, but cannot feasibly simulate a Turing Machine within the usual time constraints we hope for. One example of this is Godel's recursive functions: the computable function $f(x) = 2x$ is implemented by calling the successor function $2x$ times, which intuitively takes $O(2^x)$ time. -So, my question: - -What Turing-complete models of computation, or simplistic programming languages (Turing tar pits), can compute every computable function with only a worst-case polynomial-time blowup in time complexity over the fastest Turing machine that computes that same function? - -REPLY [6 votes]: One of the simplest model that has recently been proved to be an efficient simulator (polynomial time slowdown) of Turing machines are 2-tag systems: -Damien Woods, Turlough Neary, "On the time complexity of 2-tag systems and small universal Turing machines" (2006) -Abstract: We show that 2-tag systems efficiently simulate Turing machines. As a corollary we find that the small universal Turing machines of Rogozhin, Minsky and others simulate Turing machines in polynomial time. This is an exponential improvement on the previously known simulation time overhead and improves a forty year old result in the area of small universal Turing machines.<|endoftext|> -TITLE: Bound on the number of minimal vertex covers of a graph -QUESTION [5 upvotes]: Can the number of minimal vertex covers of a graph be super-polynomial (like exponential)? I suspect it can, but can't think of any examples. -Vertex cover $C$ of a graph $G$ is a subset of its vertices that any edge has an incident vertex in that set. That is: -$$ -C\subseteq V(G) -\hspace{1cm} -\text{s.t.} -\hspace{1cm} -\forall_{xy\in E(G)} x\in C \vee y\in C -$$ -Minimal vertex cover $C$ of a graph $G$ is a vertex cover of $G$ such that the set $C'$obtained by removal of any vertex from $C$ is not a vertex cover of $G$. That is: -$$ -C \text{ is a vertex cover} -\hspace{1cm} -\wedge -\hspace{1cm} -\forall_{v\in C}C':= C-v \text{ is not a vertex cover} -$$ - -REPLY [11 votes]: The union of $k$ triangles has $3^k$ minimum vertex covers. You can easily find connected examples.<|endoftext|> -TITLE: How strong are large cardinal properties of Ord? -QUESTION [12 upvotes]: Ordinal numbers are generalizations of natural numbers. In this sense the "proper class" of all ordinals ($Ord$) is very similar to "infinite" set of all natural numbers ($\omega$). In the other direction we know that many large cardinal axioms are generalizations of the properties of $\omega$ and without assumption of uncountablity one can prove (in $ZFC$) that $\omega$ is a large cardinal in many types. For example $\omega$ is a strongly compact cardinal because $\mathcal{L}_{\omega , \omega}$ is a compact logic. This fact shows that the nature of $\omega$ is very adequate to be a large cardinal. Now a natural question arises for the $Ord$ like this: -Question (1): Is the nature of $Ord$ adequate to be a large cardinal? More precisely if $A$ be a large cardinal type, then how strong is the statement: "$Ord$ is a large cardinal of type $A$"? -It seems that because of the similarity between $\omega$ and $Ord$ these statements must be very close to a provable statement in $ZFC$ or be a "very weak large cardinal axiom". For example it is provable in $ZFC$ that "$Ord$ is strongly inaccessible" and the statement "$Ord$ is Mahlo" is weaker than existence of an "uncountable" Mahlo cardinal (see the Cantor's upper attic for more details). Even there are some problems during investigating large cardinal properties of $Ord$, for example in defining some types of large cardinality for $Ord$ particularly in large cardinals based on -elementary embedding definitions. So: -Question (2): Is there an equivalent definition for any large cardinal axiom $A$ which the statement "$Ord$ is a large cardinal of type $A$" be meaningful? -Question (3): Is there a large cardinal axiom like $A$ such that: -$ZFC\vdash \neg~(Ord~is~a~large~cardinal~of~type~A)$ - -REPLY [18 votes]: $\newcommand{\Ord}{\text{Ord}} -\newcommand{\ZFC}{\text{ZFC}}$ -Here is one way to formalize your concept a little more tightly, -which provides the answers to your questions. For any large -cardinal property $P$, let's take the phrase "$\Ord$ is $P$" to -be the theory asserting $\sigma$, for any sentence $\sigma$ that -ZFC proves is true in $V_\delta$ under the assumption that -$\delta$ has property $P$ in $V$. -With this formalization, "$\Ord$ is $P$" asserts that the universe is just like we would expect, if we were living inside $V_\delta$ for an actual $P$ cardinal $\delta$. For example, "$\Ord$ is measurable" implies that there are a proper -class of weakly compact cardinals, since this is true in -$V_\delta$, whenever $\delta$ is measurable, and "$\Ord$ is -supercompact" implies that there are many partially supercompact -cardinals, with nice limit properties. For example, they would -form a stationary class in the sense that every definable class -club would contain one of them. This notion seems to capture what -one would want to mean by saying $\Ord$ has property $P$ as a -purely first-order theory about sets. -With this idea, the point I would like to make is that assuming -$\Ord$ is $P$ is essentially equivalent to assuming that what you -have is $V_\delta$, where $\delta$ has property $P$ in a larger -universe. -Theorem. For any large cardinal property $P$, a model of set -theory $M$ satisfies "$\Ord$ is $P$" if and only if $M\prec -V_\delta^N$ for some taller model of set theory $N$ with a -cardinal $\delta$ having property $P$ in $N$. -Proof. The backward direction is immediate, since $\delta$ having -property $P$ implies that $V_\delta$ satisfies every assertion of -$\Ord$ is $P$. For the forward direction, suppose $M$ satisfies -$\Ord$ is $P$. Let $T$ be the theory consisting of $\ZFC$, plus the -assertion "$\delta$ is $P$", using a new constant symbol $\delta$, -plus the assertions $\varphi^{V_\delta}$, for any $\varphi$ in the -elementary diagram of $M$, using constants for elements of $M$. -This theory is finitely consistent, since otherwise there would be -finitely many assertions in it that are contradictory, and so -there would be a statement $\varphi$ true in $M$ that provably -could not hold in $V_\delta$ for any cardinal $\delta$ with -property $P$. But that would contradict our assumption that -$M\models\Ord$ is $P$. -If $N$ is any model of the theory, then $\delta$ has property $P$ -in $N$, and we get $M\prec V_\delta^N$, because $V_\delta^N$ -satisfies the elementary diagram of $M$. Another way to say this -is that there is an elementary embedding $j:M\to V_\delta^N$, -mapping every element of $M$ to the interpretation of its constant -in $N$. QED -Thus, if one is inclined to assume $\Ord$ is $P$, then why not go ahead and make the full move to a model with an actual $P$ cardinal $\delta$, such that our old -world looks exactly like $V_\delta$ in this new world. In particular, under this terminology, the theory $\ZFC+\Ord$ is $P$ is equiconsistent with $\ZFC+\exists \delta$ with property $P$. -Corollary. The following theories are equiconsistent: - -$\ZFC+\Ord$ is $P$. -$\ZFC+\exists \kappa$ with property $P$. - -Lastly, I would like to point out that there is some variance in -the literature about what "$\Ord$ is $P$" should mean. For -example, one often finds the phrase "$\Ord$ is Mahlo" to mean only -the weaker assertion, that every definable closed unbounded class -of cardinals contains a regular cardinal. This is what one finds, for example, -at Cantor's Attic. But this is strictly weaker -in consistency strength than ZFC+$\exists\kappa$ Mahlo, since this latter theory implies -the consistency of the former, as it is true in $V_\kappa$ -whenever $\kappa$ is Mahlo.<|endoftext|> -TITLE: Representing SU(3) with 3 ropes in 3 dimensions -QUESTION [12 upvotes]: The short question is: how exactly is SU(3) realized with ropes? -The long question: There is this idea that deformations of a configuration of three infinitely long, flexible ropes that cross each other can be mapped -to the eight Gell-Matrices, the generators of SU(3). -SU(3) appears in the quantum harmonic oscillator and in qutrits. So a visualization of SU(3) with ropes is useful to quantum information theory. -The SU(3) idea is mentioned in http://arxiv.org/abs/0905.3905 on page 35. It seems that the idea started in this way: deformations of configurations of TWO ropes reproduce Dirac's string trick and behave like the Pauli matrices of SU(2). Deformations of configurations of ONE rope reproduce U(1). -THREE ropes apparently yield a relation between eight different versions of the third Reidemeister move and the eight Gell-Mann matrices. But the paper is too terse for me to see the relation in detail. A literature search does not bring up anything related to this idea. And I got no answer to my email. -Can anybody help to understand the details? -Added points: -Peter's answer below mentions a relation between the braid group $B_3$ and SU(3). A Google search does not yield anything about this topic. Can anybody provide a reference? -A graph similar to that of Joseph's answer below is also part of the paper. But I am not interested in QCD or unification: I'd like to understand how the deformations in that graph yield or correspond to the Gell-Mann matrices $\lambda_1$ to $\lambda_8$. I can see that the deformations of the graph correspond to $F_1$, $F_2$ and $F_3$, where $F_i=e^{i \pi \lambda_i / 2}$. This gives $F_1^4=F_2^4=F_3^4=1$, the unit matrix, as it should. Also the SU(2) subgroup is generated as it should. Next, the $\lambda_8$ deformation behaves as expected. But I cannot see (so far) that $\lambda_1 \lambda_4= \lambda_6/2 + i \lambda_7/2$. Can anybody provide a hint? - -REPLY [5 votes]: This doesn't exactly answer the question, but it seems like it should be closely related. In general, if $G$ is a complex semi-simple Lie group, there is an associated braid group $B_{\mathfrak g}$ (edit: (the pure braid group) is the fundamental group of the complement in $\mathbb{C}[\mathfrak h]$ of the reflection hyperplanes of the Weyl group $W$ of $G$). (edit: The braid group is the fundamental group of the quotient of this space by $W$.) I have been told that there is a group homomorphism $B_{\mathfrak g} \to G$, although I unfortunately don't know a reference. -For $G = SL_n(\mathbb C)$, things can be made more explicit. The braid group $B_{sl_n}$ is the standard braid group (see https://en.wikipedia.org/wiki/Braid_group), and I believe the map $B_n \to SL_n(\mathbb C)$ is given by assigning the generator $\sigma_i \in B_n$ the identity matrix, with the $2\times 2$ submatrix in row $i,i+1$ and columns $i,i+1$ is replaced by -$\left[\begin{array}{cc} 0&-1\\1&0\end{array}\right]$. -(Sorry if the formatting is off - for some reason my computer isn't displaying things correctly at the moment.)<|endoftext|> -TITLE: Inference using Topological Data Analysis: Is it worth it for a regular statistician to learn TDA? -QUESTION [36 upvotes]: After having read Gunnar Carlsson's Topology and Data I feel enthusiastic to use some topological data analysis (TDA) methods in my current research, mostly in social sciences. We often handle huge databases and I think it can be an interesting exercise to do TDA. I wonder two things about the current state and purpose of TDA (I do understand TDA's main advantages are not for the social sciences though): - - -Are there TDA methods than can be used to establish a relationship among a variable of interest and a set of (possible) explanatory variables? That is, to do some sort of statistical inference? -Can we make predictions based on TDA tools? - - -Any bibliographical reference or explanation about why this is not possible is appreciated. - -REPLY [6 votes]: There was a workshop on topological data analysis (TDA) at the Institute for Mathematics and its Applications in October 2013. -Link: http://www.ima.umn.edu/2013-2014/W10.7-11.13/ -Note in particular the tutorial on statistical inference for TDA by Alessandro Rinaldo: http://ima.umn.edu/videos/?id=2443, and see the CMU TopStat page at http://www.stat.cmu.edu/topstat/. -Update: there is an R package available for TDA - see http://www.stat.cmu.edu/~flecci/software/index.html. See also the tutorial at http://www.stat.cmu.edu/topstat/Talks/files/Jisu_150623_TDA_tutorial.pdf.<|endoftext|> -TITLE: Is there an agreed-upon name for this type of subgroup? -QUESTION [6 upvotes]: In Embedding theorems for groups, (J. London Math. Soc. 34 1959 465–479.) Neumann and Neumann (NB: this is not the Higman-Neumann-Neumann paper of the same name) make the following definition. -Definition: A subgroup $H$ of a group $G$ is an E-subgroup of $G$ if for every normal subgroup $N\triangleleft H$, there is a normal subgroup $S\triangleleft G$ such that $S\cap H=N$. Equivalently, for every normal subgroup $R\triangleleft H$, the normal closure $R^G$ of $R$ in $G$ is such that $R^G\cap H=R$. -This comes up again in The SQ-universality of hyperbolic groups, by Olshanskii, wherein he simply refers to this situation as the normal structure of $H$ being a restriction of the normal structure of $G$. -Presumably this notion has come up elsewhere, although I don't know of other places where it is explicitly named. It's a natural notion to deal with when working to show a group is SQ-universal, for example. Lately I've been using it in my research, and I would like to know if there is an agreed-upon name for this type of subgroup. As far as I can tell, "E-subgroup" is not widely known. Is there something that is? - -REPLY [6 votes]: I think the magic acronym is CEP. See http://en.wikipedia.org/wiki/CEP_subgroup -Update I've just noticed that the question is also tagged reference-request, so here's one (Google reveals many, I'm not sure if there is a canonical one): Non-amenable finitely presented torsion-by-cyclic groups by Ol'shanskii and Sapir. See the introductory Section 1.2 on Page 3 of the pdf.<|endoftext|> -TITLE: The state of the art in the rectification of homotopy-coherent structures -QUESTION [15 upvotes]: My question concerns rectification theorems for homotopy-coherent structures. As the meaning of this may be unclear, let me list a few examples of what I am thinking of: - -Cordier and Porter proved a rectification theorem for diagrams in Kan-enriched categories $\mathcal{M}$ satisfying certain additional conditions: for every small category $\mathbb{D}$, if $\mathbf{Coh}(\mathbb{D}, \mathcal{M})$ is the homotopy category of homotopy-coherent diagrams $\mathbb{D} \to \mathcal{M}$ and $\operatorname{Ho} {[\mathbb{D}, \mathcal{M}]}$ is the homotopy category of strict diagrams $\mathbb{D} \to \mathcal{M}$, then the obvious functor $\operatorname{Ho} {[\mathbb{D}, \mathcal{M}]} \to \mathbf{Coh}(\mathbb{D}, \mathcal{M})$ is an equivalence of categories. This holds in particular for $\mathcal{M} = \mathbf{Top}$ (originally proved by Vogt) and $\mathcal{M} = \mathbf{Kan}$. -It is well-known that every monoidal category is monoidally equivalent to a strict monoidal category: this is one of the consequences of Mac Lane's coherence theorem. -A. J. Power proved a rectification theorem for algebras for 2-monads: given a 2-monad $\mathsf{T}$ on the 2-category $\mathfrak{Cat}$ that preserves bijective-on-objects functors, every pseudo-$\mathsf{T}$-algebra is equivalent to a strict $\mathsf{T}$-algebra in the 2-category of pseudo-$\mathsf{T}$-algebras and strong homomorphisms. - -What is the state of the art in the rectification of homotopy-coherent structures, in the general sense suggested above? -I am especially interested to hear about results concerning the rectification of homotopy-coherent diagrams in (not necessarily simplicial) model categories. This appears to be a crucial step toward showing that homotopy limits and colimits in the sense of derived functors agrees with homotopy limits and colimits in the Joyal–Lurie sense. (See e.g. §4.2.4 in [Higher topos theory].) - -REPLY [5 votes]: One general rectification result is that for algebras over an operads, as in - -Clemens Berger, Ieke Moerdijk, Resolution of coloured operads and rectification of homotopy algebras (arXiv:math/0512576) - -which is briefly summarized on the nLab at model structure on algebras over an operad -- Properties -- Rectification of algebras.<|endoftext|> -TITLE: Looking for concrete description of a category derived from abelian groups -QUESTION [11 upvotes]: The category of abelian groups $\mathsf{Ab}$ is the $\mathcal{Ind}$-completion of the full subcategory of finitely presentable abelian groups $\mathsf{Ab}_{fp}$. This is not so special, since the analogous statement holds for any finitary variety e.g. groups, rings, boolean algebras etc. -However one nice property of finitely presentable abelian groups is that they are closed under finite limits, in fact under finite products and subalgebras. Therefore $(\mathsf{Ab}_{fp})^{op}$ has finite colimits, which implies that it forms (essentially) the finitely presentable objects of the locally finitely presentable category $\mathcal{Ind}((\mathsf{Ab}_{fp})^{op})$. -To put it another way, $\mathcal{Ind}((\mathsf{Ab}_{fp})^{op})$ is essentially algebraic and in a formal sense has a presentation by certain restricted "partial" operations and equations. - - -My question is whether a concrete description of $\mathcal{Ind}((\mathsf{Ab}_{fp})^{op})$ is known. - - -Note that its dual is not the category of profinite abelian groups, since I am considering the finitely presentable rather than the finite abelian groups. -There are many examples where the analogous construction leads to a well-known category e.g. starting with $\mathsf{Set}$ we obtain $\mathcal{Ind}(\mathsf{Set}_{fp}^{op}) \cong \mathsf{BA}$, starting with $\mathsf{DL}$ we obtain $\mathsf{Poset}$, starting with $\mathsf{Vect}(\mathbb{F})$ we obtain the same again, and likewise for join-semilattices with bottom. -Many thanks. - -REPLY [7 votes]: Very interesting problem! I'd not seen it before, but from what I can make out, it looks as though this category can be described concretely as having for its objects triples $(A, T, i: A \otimes \mathbb{Q}/\mathbb{Z} \to T)$ where $A$ is an abelian group, $T$ is a torsion abelian group, and $i$ is an injective homomorphism. Morphisms are given by pairs of homomorphisms $A \to A'$, $T \to T'$ that are compatible with the injective homomorphisms. (Hence, something like a gluing construction.) -The $Ind$-completion of the opposite of finitely presented abelian groups is equivalent to the category of left exact functors $\Phi: \text{Ab}_{fp} \to \text{Set}$, so I'll start there. -Finitely presentable abelian groups are the same as finitely generated $\mathbb{Z}$-modules. Every such module $M$ is a product of a finite power $\mathbb{Z}^n$ and a finite abelian group $F$, so that by left exactness $\Phi(M)$ can be written in the form $\Phi(\mathbb{Z})^n \times \Phi(F)$. It makes sense to consider the restriction of $\Phi$ to the full subcategories $\text{Ab}_{\text{fin}}$ (of finite abelian groups) and $\text{Pow}(\mathbb{Z})$ (of finite powers of $\mathbb{Z}$). Each of these subcategories is finitely complete, and the full inclusions are left exact, so that $\Phi$ restricts to left exact functors on each of these two subcategories. -Interestingly, both of these subcategories happen to be self-dual. Thus in the first case we may as well consider the category of left exact functors $\text{Ab}_{\text{fin}}^{op} \to Set$, that is to say the $Ind$-completion of $\text{Ab}_{fin}$, which is the category of torsion abelian groups. Hence the first restricted functor is given by $\hom(-, T)$, where $T$ is a torsion abelian group. -In the second case, we may as well consider the category of left exact functors $\text{Pow}(\mathbb{Z})^{op} \to Set$. The domain is the Lawvere theory of abelian groups; left exact functors are product-preserving functors, so such a functor must be of the form $\hom(-, A)$ for some abelian group $A$. In fact, in this case left exact functors coincide with product-preserving functors, essentially because exact sequences in $\text{Pow}(\mathbb{Z})$ split. -The data $(T, A)$ we have thus extracted do not completely characterize $\Phi$ because we have not taken into account how $\text{Pow}(\mathbb{Z})$ and $\text{Ab}_{\text{fin}}$ interact in $\text{Ab}_{fp}$. We need to consider that $\Phi$ preserves the kernel pair of a morphism $f: \mathbb{Z}^n \to F$ mapping to a finite abelian group. It is easy to convince oneself that $\Phi(f): A^n \to \hom(F, T)$ is a group homomorphism. Since $F$ can be decomposed further as a product of cyclic groups, ultimately we find that the left exactness requirement boils down to the fact that $\Phi$ should preserve exact sequences of the form -$$0 \to \mathbb{Z} \stackrel{- \cdot n}{\to} \mathbb{Z} \to \mathbb{Z}_{n}$$ -so that we require exactness of -$$0 \to A \stackrel{- \cdot n}{\to} A \to T_{n}$$ -where $T_n$ is the subgroup of elements of $T$ annihilated by $n$. This can be equivalently rephrased as saying that the induced map $A \otimes \mathbb{Z}/(n) \to T \otimes \mathbb{Z}/(n)$ is injective (naturally over all $\mathbb{Z}_n$), and the neatest way of encapsulating all is by passing to the (filtered) colimit over all $\mathbb{Z}/(n)$, which leads to the statement that we have an injective map -$$A \otimes \mathbb{Q}/\mathbb{Z} \to T \otimes \mathbb{Q}/\mathbb{Z} \cong T.$$ -There are quite a few details to be checked in this analysis, and I don't claim I've checked every one, but I should imagine this is in the right direction.<|endoftext|> -TITLE: The prerequisites for Deligne's Théorie de Hodge I, II, III -QUESTION [23 upvotes]: I am an undergraduate student. I am not sure if it's OK to ask this question here. -I want to learn Hodge theory. But I do not know how to start it, and how much mathematics I should need before I read Deligne's paper. -Is there an elementary book or note on Hodge theory for undergraduate students? Is it worthy to read Hodge's book The theory and applications of harmonic integrals? - -REPLY [14 votes]: I might be repeating what others have said, but I think the question is whether you want to learn classical Hodge theory or mixed Hodge theory. Deligne's papers are about the latter. In principle it's very possible to learn one of them and not much of the other - a complex analyst can make use of classical Hodge theory without ever caring about weight filtrations and so on, and an algebraic geometer can use mixed Hodge theory as a tool without learning the transcendental parts that go into the construction. (If you are interested in computing the cohomology of algebraic varieties then mixed Hodge theory is an amazingly powerful tool.) -In any case, if it really is mixed Hodge theory you want to learn, then Deligne's papers are not a bad start. There's also the book of Peters and Steenbrink, Deligne's ICM address "Poids dans la cohomologie des variétés algébriques" and Brylinski and Zucker's "An overview of recent advances in Hodge theory". But you should be aware that none of these papers are easy to read! They will assume knowledge of algebraic geometry and a large amount of homological algebra (derived categories, simplicial techniques, et cetera). To make things worse, a large part of the motivation for the theory comes from étale cohomology and the Weil conjectures. -A better way into mixed Hodge theory might be to learn some applications first and figure out why it's useful. For instance, you could learn about the Hodge-Deligne polynomial of an algebraic variety. In simple cases you can think very concretely about the Hodge-Deligne polynomial as a gadget that counts the numbers of points of your variety over various finite fields. Once you have understood how one can use it as a tool and why it's absolutely amazing that a Hodge-Deligne polynomial even exists, you can read about mixed Hodge theory to understand why it's well defined.<|endoftext|> -TITLE: Lower Algebra: Modules over the monoidal category of abelian groups -QUESTION [15 upvotes]: Proposition 6.3.2.18 of Higher Algebra identifies $Mod_{Sp}(Pr^L)$, the symmetric monoidal category of right modules over the monoidal category $Sp$ of spectra in $Pr^L$ the category of presentable categories, with the full subcategory $Pr^L_{St}$ of stable presentable infinity categories. In particular Lurie proves that the functor $-\otimes Sp: Pr^L \to Pr^L_{St}$ is a localization functor. -Question 1: Is there an analogue for ordinary (non $\infty$) categories? In other words, is $-\otimes Ab: Pr^L \to Mod_{Ab}(Pr^L)\cong Pr^L_{Ab}$ a localization functor from the (ordinary) category of presentable categories to the (ordinary) category of presentable abelian categories? -Proof strategy: I believe that Lurie's argument can be modified to prove that $-\otimes Ab$ is a localization functor if I could prove that every presentable category that is a right $Ab$-module is also abelian. -I can verify that $C$ is pre-abelian: - -Since $C$ is presentable we know that that for any $a \in Ab$ the functor $- \otimes a$ has a right adjoint $[a, -]$. Thanks to the appendix of this paper (shared with me by Zhen Lin) we have that $C$ is enriched and tensored in $Ab$. Moreover, the $Ab$ action corresponds to the tensor. -Since $C$ is presentable, it is complete and cocomplete. In particular it has a $0$ object, biproducts, kernels, and cokernels. - -I cannot verify that $C$ is actually abelian which requires that: - -Every monomorphism is a kernel and every epimorphism is a cokernel. - -Alternate strategy: -In Lurie's proof (summarized well in 5.3.1 of Groth's notes) he uses the description of $C \otimes Sp = Fun^R(C^{op}, Sp)$ and the fact that $Sp = lim(S \xrightarrow{\Omega} S \xrightarrow{\Omega} \ldots)$ to show that $C \otimes Sp$ is the stabilization $Sp(C)$. Then he shows stabilization of a stable category is itself. -Mimicking this, I can show that $C \otimes \mathbb{Z} \cong Fun^R(C^{op}, \mathbb{Z})$ which is obviously pre-abelian since it is presentable and $Ab$-enriched. - -Is $C \otimes \mathbb{Z}$ abelian? I think it might be because it is a subcategory of a presheaf category. -If $C$ is already abelian does the tensor map $C \otimes \mathbb{Z} \to C$ give an equivalence? - -REPLY [19 votes]: A locally presentable category $\mathcal{C}$ has a (unique) structure of an $Ab$-module if and only if it is additive. Such a category need not be abelian. -This is one reason to prefer the setting of stable $\infty$-categories to the theory of abelian categories. The identification of presentable stable infty-categories with $Sp$-modules implies (among other things) that there is a robust theory of tensor products of presentable stable infty-categories. In the abelian setting, developing an analogous theory requires some fairly restrictive assumptions. -Edit: Actually, I think it's not as bad as I suggested; there's a well-behaved tensor product on Grothendieck abelian categories. But that's a smaller collection of categories than the $Ab$-modules.<|endoftext|> -TITLE: How many Lagrangian submanifolds? -QUESTION [8 upvotes]: An $n$-dimensional submanifold $L$ of a symplectic manifold $(M^{2n}, \omega)$ is called Lagrangian if $\omega|_L = 0$. I want to get some feeling about how many Lagrangian submanifolds are. -For each $\alpha \in H_n(M)$, is there a Lagrangian submanifold representing $\alpha$? Maybe it's not a good idea to distinguish Lagrangians by its homology classes. Floer homology is the same if we move Lagrangians by Hamiltonian isotopy. Is there a notion of space of Lagrangian submanifolds modulo Hamiltonian isotopy? - -REPLY [4 votes]: There's an idea of Hitchin's that (all Lagrangian submanifolds)/(Hamiltonian isotopy) should be, at least approximately, the same as (all "special" Lagrangian submanifolds). "Special" is defined with respect to some background structure on the symplectic manifold, like a metric. Special Lagrangians make a finite-dimensional space at least locally. -A good analogy is the Hodge theorem (but Hitchin's idea doesn't work as well): the space (closed differential forms)/(exact differential forms) is the same as (harmonic differential forms), again the meaning of "harmonic" depends on some background structure, and also harmonic forms make a finite-dimensional space. -The Hitchin paper is here: -http://arxiv.org/abs/dg-ga/9711002<|endoftext|> -TITLE: Is an ideal generated by multilinear, irreducible, homogeneous polynomials of different degrees always radical? -QUESTION [7 upvotes]: I asked this question on math.se and someone even put a bounty on it, yet there was no answer. Hence, I am asking here. Assume $\Bbbk$ to be a field of characteristic zero. - -Definition. A polynomial $f\in\Bbbk[x_0,\ldots,x_n]$ is called multilinear if $\deg_{x_i}(f)=1$ for each $0\le i \le n$. In other words, $f$ is linear in each variable. If $f$ is homogeneous of degree $d$, then $f$ is a linear combination of monomials of the form $x_{i_1}\cdots x_{i_d}$ with $0\le i_1 -TITLE: Construction of Thom-Spectrum for G_2-Structures -QUESTION [9 upvotes]: The motivation to this question is the paper of Crowley and Nordstrøm "A New Invariant of $G_2$-Structures". I am trying to find a homotopy theoretic interpretation of the following geometric situation: The exceptional Lie-group $G_2$ can be regarded as the stabilizer of a vector in $S^7$ under the action of $Spin(7)$ on $\mathbb{R}^8$ (this action can be depicted by octonionic multiplication). On a 7-dimensional $Spin$-manifold a $G_2$-structure on the tangent bundle $TM$ is equivalent to a homotopy class of non-vanishing unit sections $s\colon M \rightarrow \Sigma M$ into the associated real spinor bundle $\Sigma M$, giving an isomorphism of oriented Riemannian vector bundles $\underline{\mathbb{R}} \oplus TM \cong \Sigma M$. -Crowley and Nordstrøm show, that every $G_2$-structure is bounded by a $Spin(7)$-structure, i. e. an 8-dimensional spin manifold $W$ with a unit section $\bar{s} \colon W \rightarrow \Sigma ^+ W$ into the positive half-spinor bundle, extending $s$. It follows that the Euler class of the positive half-spinor bundle vanishes. -Some thoughts so far have been: We are looking at sections into bundles associated to tangential structures, while bordism works in the stable normal direction. Thus, to fix a trivialization between normal and tangential structure, it could be worthwhile taking the homotopy fibre $B$ of the map $BG_2 \times BSpin(7) \rightarrow BSpin$ as an underlying space and construct the Thom-spectrum MB over B by the map $\xi \colon B \rightarrow BG_2 \times BSpin(7) \rightarrow BO(7) \rightarrow BO$. But if every $G_2$-structure is bounded by a $Spin(7)$-structure then the group $\pi _7 (MB)$ should vanish, correct? -Another idea would be the classifying space $BG_2$ and its Thomspectrum. At least it is known that the bordism group is $\Omega ^{G_2}_7 \cong \mathbb{Z}/3\mathbb{Z}$. But in that situation I do not see how to transfer this normal $G_2$-structure to a tangential structure. -So my question is: Which choice of underlying space and Thomspectrum seems reasonable to deal with the explained geometric situation? - -REPLY [6 votes]: The second version of the arXiv post of the Crowley-Nordström paper goes into some detail in giving an answer to this question: see Section 7 of version 2. The out-line is as follows: -1) First define stable $G_2$-structures and stable tangential $G_2$-bordism. -2) Then make the standard identification of stable tangential bordism and normal bordism. -3) Finally, use the Thom spectrum for normal $G_2$ bordism - this is the suspension spectrum of the Thom space of the inverse to the canonical vector bundle $V$ over $BG$, where $V$ is obtained from the standard representation of $G_2$ on $\mathbb{R}^7$. i.e. our candidate for $MG_2$ is the suspension spectrum of $T(-V)$, where $-V$ is the K-theory inverse of $V$ and $T(-V)$ is its Thom space: of course, you have to take some care to make the rank of the representative of $-V$ and the indexing of the suspension spectrum match up. -Whether this is what you want, I am not sure, but did allows us to compare $G_2$ bordism and $SU(2)$ bordism and to give a geometric interpretation of the $\nu$-invariant modulo 3.<|endoftext|> -TITLE: Occurrences of D. H. Lehmer's 10-th degree polynomial -QUESTION [13 upvotes]: Salem numbers and Lehmer's minimum height problem are venerated not only in number theory and diophantine analysis, where they are considered naturally interesting for their own sake, but also in fields such as hyperbolic geometry and holomorphic dynamics. As is so well known, the least known Salem number is a root $1.176280\ldots$ of the following monic reciprocal $10$-th degree $\{-1,0,1\}-$polynomial discovered way back in 1933 by D. H. Lehmer in his work on primality testing: -$$ -x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1. -$$ -I am interested in seeing any mathematical contexts or computations through which this particular polynomial shows up, apparently accidental occurrences included (not to say preferred). -Here is an example from topology: this is the Alexander polynomial of infinitely many knots, including the $(-2,3,7)$-pretzel knot. - -REPLY [7 votes]: Here's a paper of McMullen where the Lehmer polynomial shows up (see Theorem 1.2 there): -http://www.math.harvard.edu/~ctm/papers/home/text/papers/blowup/blowup.pdf - -REPLY [5 votes]: well, this link will direct you to just under 300 scholarly articles on Lehmer's polynomial: -http://scholar.google.com/scholar?q=polynomial+lehmer&btnG=&hl=en&as_sdt=2005&sciodt=0%2C5&cites=18047953845360790641&scipsc=1 -(oh, and I added the big-list tag to this question, I guess that is appropriate :) - -a small selection: - -The -Lehmer polynomial and pretzel links -What is Lehmer's Number? -Heights of polynomials and entropy in algebraic dynamics -On the -distribution of the roots of a polynomial with integral -coefficients -Primes in sequences associated to polynomials (after Lehmer) -Lehmer's -problem for compact Abelian groups -Higher -Mahler measure for cyclotomic polynomials and Lehmer's question -Lehmer's Conjecture for -Hermitian Matrices over the Eisenstein and Gaussian Integers -Solved -and unsolved problems on polynomials -Polynomials with restricted coefficients and prescribed noncyclotomic factors -A seventeenth-order polylogarithm ladder -Lehmer's question, knots and surface dynamics<|endoftext|> -TITLE: If an oracle Turing machine halts with every infinite arithmetic oracle, can it fail to halt with some non-arithmetic oracle? -QUESTION [10 upvotes]: Let $e$ be an index of an oracle Turing machine program and $k$ be some natural number. Let us say that a subset of $\mathbb N$ is arithmetic if it is definable in the model $\langle \mathbb N,+,\cdot,<,0,1\rangle$. Now suppose that there is a non-arithmetic oracle $A$ such that $\Psi_e^A(k)\uparrow$. Is it possible that $\Psi_e^B(k)\downarrow$ with every infinite arithmetic oracle $B$? -Note that it is not possible if we remove the key assumption that $B$ must be infinite. Suppose there is some non-arithmetic oracle $A$ so that $\Psi_e^A(k)\uparrow$. We can then consider the infinite binary tree $T$ consisting of all finite sequences $s$ such that $\Psi_e^B(k)$ does not halt with any oracle $B$ extending $s$ in less than length of $s$ many steps. The tree $T$ is arithmetic and therefore must have an arithmetic path $P$ (this follows from the proof of König's Lemma). Clearly $\Psi_e^P(k)\uparrow$. But of course $P$ can be finite! In this argument, there is no obvious way to guarantee that $P$ is infinite because it is possible to have an infinite arithmetic binary tree without an arithmetic path with infinitely many 1s. - -REPLY [7 votes]: This is a really great question! -Here is a different way to think about Emil's example. -Consider the notion of a annotated truth predicate. This is a -labeling of every arithmetic sentence as true or false, in -accordance with Tarski's recursive truth requirements, but -annotated in the sense that whenever an existential sentence -$\exists x\, \varphi(x)$ is labeled as true, then a satisfying -witness $n$ is annotated right there, for which $\varphi(n)$ is -also labeled as true. Thus, an annotated truth predicate is a truth predicate, where the Skolem witnesses are provided, and you don't have to go search to see if the witnesses are really there. -Now, consider the Turing machine, which on any input begins to -inspect the oracle, to see if it is an annotated truth predicate. -Thus, on any input the program begins to check that all the -Tarskian truth conditions are met, that the atomic formulas are -labeled true or false correctly, that the Boolean combinations are -labeled correctly, and then, for the existential assertions, it -checks whether the annotations follow the rules. As long as these -conditions are being met, the program continues operating, but as -soon as a violation is found, then the program halts. -The point, now, is that because there is no arithmetic truth -predicate, there also is no arithmetic annotated truth predicate, -and so on any arithmetic oracle the program will eventually find a -flaw and therefore halt. -But meanwhile, there are annotated truth predicates (for example, -of hyperarithmetic complexity), and on these, the program will -never halt. -So this is a program with your desired features.<|endoftext|> -TITLE: Why does the degree of the variety of rank at most $r$ $n\times n$ matrices equal dim$S_{(n-r)^{n-r}}C^n$? -QUESTION [10 upvotes]: Let $X_r\subset Mat_{n\times n}$ denote the matrices of rank at most -$r$, and let $S_{\pi}C^n$ denote the irreducible $GL_n$-module corresponding -to the partition $\pi$. -One can check that -degree($X_r$)=dim($S_{(n-r)^{n-r}}C^n)$=$\Pi_{i=0}^{n-r-1}\frac{(n+i)!i!}{(r+i)!(n-r+i)!}$ - Does anyone have a geometric (or any) explanation for this equality? Had it been observed previously? -Here $(n-r)^{n-r}=(n-r,...,n-r)$ has Young diagram the square box. - -REPLY [7 votes]: We can do this more generally for any determinantal variety, so just consider $n \times m$ matrices where we don't need to assume $n=m$. Naively, we could calculate the degree from a free resolution of the determinantal variety, but unfortunately that's a complicated procedure. For just the purposes of getting the degree, it suffices to use any rank $1$ module supported in the determinantal variety. It turns out there are such modules which are both Cohen-Macaulay and have a linear free resolution $F_\bullet$ (without the rank assumption, these are known as Ulrich, or maximally generated maximal Cohen-Macaulay (MGMCM), modules in the literature). This is good because then ${\rm rank}(F_i) = {\rm rank}(F_0) \cdot \binom{c}{i}$ where $c = (n-r)(m-r)$ is the codimension of the determinantal variety. But then an easy computation with Hilbert series shows that the degree of the determinantal variety must be ${\rm rank}(F_0)$. -The complex is described in Exercise 6.34 of Weyman, Cohomology of Vector Bundles and Syzygies. Here's a brief construction: let $k^n \to k^m$ be the generic matrix thought of as a 2-term chain complex with $k^m$ in homological degree $0$. Apply the Schur functor $S_{(m-r) \times (n-r)}$ (here $(m-r) \times (n-r)$ means the rectangular partition $(n-r, n-r, \dots, n-r)$ repeated $m-r$ times). Then we get a complex (Schur functors can be used in any symmetric monoidal Abelian category, here our category is chain complexes) which has homology only in degree $0$ (and this is the rank $1$ module we were looking for) and the $0$th term is the representation $S_{(m-r) \times (n-r)}(k^m)$, which specializes to your example. The homology vanishing requires some work, but it can be proven using standard commutative algebra techniques (the Buchsbaum-Eisenbud acyclicity criterion) and some properties of Schur complexes. A bonus is that all of what I said is independent of characteristic (unlike the resolution of the coordinate ring of the determinantal variety!). Now, to advertise my own work, this has been generalized from determinantal varieties to matrix Schubert varieties: http://arxiv.org/abs/1006.5514 using the Schubert functors of Kraskiewicz and Pragacz: http://dx.doi.org/10.1016/j.ejc.2003.09.016 -Finally, let me address the comment asking about (skew-)symmetric matrices. A similar technique works here, and is partially discussed in Exercises 6.35 annd 6.36 of Weyman's book. -Edit: I wrote something about Schur complexes but it doesn't work as stated in the (skew-)symmetric cases. I think a certain modification will work, but I haven't looked into it. -The formulas for (skew-)symmetric matrices can also be found in a paper of Harris and Tu: http://dx.doi.org/10.1016/0040-9383(84)90026-0 though it's stated in a different, but closely related, form. There are also some formulas in Enright and Hunziker (section 6) which don't just give the degree, but the numerator of the Hilbert series in terms of dimensions of some representations: http://dx.doi.org/10.1016/S0021-8693(03)00159-5 but I don't see how to specialize those formulas to get the ones discussed here.<|endoftext|> -TITLE: Looking for "large knot" examples -QUESTION [16 upvotes]: This question is about knots and links in the 3-sphere. I want to find an example of a "large" knot or link with some special properties. I'm looking for some fairly specific examples, but I'm also interested in related examples that don't precisely address my concern, so "near answers" are welcome. -Here is an example of something I would call a "large" knot. - -It is hyperbolic, but it also has an incompressible genus 2 surface in the knot complement. The genus 2 incompressible surface has a rather nice property, that it bounds a handlebody on one side (in $S^3$ rather than in the knot complement). -I'd like to find a similar example, but what I specifically want is a knot (or link) in $S^3$ which contains two genus 2 closed incompressible surface in the knot/link exterior, such that both surfaces bound handlebodies in $S^3$. The key thing I would like is for the two surfaces to intersect -- they can not be isotoped to be disjoint. -Is this possible? I presume it is, but I haven't seen examples of this type, and examples aren't rapidly coming to mind. -Other similar examples I'd like to see (although not "the" question here) would be knot or link exteriors that have infinitely many genus 2 closed incompressible surfaces -- preferrably bounding handlebodies on one side in $S^3$. -I'd also love to see a knot in $S^3$ with a genus 2 incompressible surface that does not bound a handlebody in $S^3$. Maybe this isn't possible? - -REPLY [15 votes]: As Ian Agol mentioned, manifolds with Conway spheres will have genus 2 incompressible surfaces. I'd just like to point out that an easy source of knots whose complements contain Conway spheres are pretzel knots, or more generally, Montesinos knots. -Here is a specific example of the pretzel knot $P(5, -7, 5, -4)$ with two incompressible Conway spheres in blue and red. -Form a pair of closed genus two surfaces by tubing along the arcs to join the boundary components pairwise on each Conway sphere. The resulting genus two surfaces each bound the handlebodies on the side indicated by the colors. (In the image, only one handle has been added to each Conway sphere.)<|endoftext|> -TITLE: $(BU,f)$ structures on manifolds via stable normal bundles and stable tangent bundles -QUESTION [5 upvotes]: Background -Consider $BU=colim \, BU_k$ where we take $BU_k$ to be the specific model of classifying space for the group $U(k)\subseteq O(2k)$ given by the quotient space of the infinite real Stiefel manifold $V_{2k}$ by the action of $U(k)$. The spaces $BU_k$ as described come with maps $f_k : BU_k \rightarrow BO_{2k}$ that are fibrations. -With the above setup, we can define a $(BU,f)$-structure on a stable vector bundle $\xi : X \rightarrow BO_{2k}$ as a particular lift $\tilde{\xi} : X \rightarrow BU_k$. We consider lifts $\tilde{\xi}_0, \tilde{\xi}_1$ equivalent if there is $k>>0$ and a fiberwise homotopy $H:X\times [0,1] \rightarrow BU_k$ between the two lifts. (Fiberwise homotopy means $f_k \circ H = \xi$). -There is a map $I: BO \rightarrow BO$ given by sending a subspace $A\subseteq \mathbb{R}^n$ to its orthogonal complement $A^{\perp} \subseteq \mathbb{R}^n$. -The question -In Stong's notes on cobordism theory, he shows that a $(BU,f)$-structure on the stable normal bundle is equivalent to an $(I^*BU,f^*)$-structure on the stable tangent bundle; this is OK. Is it possible, though, to construct a bijection between $(BU,f)$-structures on the stable normal bundle and $(BU,f)$-structures on the stable tangent bundle? -Some thoughts -For other kinds of $(B,f)$-structures it is doable, I believe. Certainly for $(BO,1)$-structures you can do it. Also, for $(BSO,f)$-structures you can do it. If $TX$ is the tangent bundle of $X$ and $N$ is the normal bundle to $X$ for some embedding in $\mathbb{R}^{n+k}$, $k>>0$, one has a canonical trivialization $TX \oplus N \cong \epsilon^{n+k}$. As the trivial bundle has a canonical choice of orientation, given an orientation of $TX$, we can get an orientation on $N$ by requiring the induced orientation on $\epsilon^{n+k}$ agrees with the canonical one. One can do the same in the other direction. -A note in Davis & Kirk claims you can do it for complex structures (Exercise 137), but I don't think the discussion is correct. It works for complex vector bundles, but that is weaker than having complex structures. E.g. the case of $X=pt$, with a trivial 2-dimensional bundle $\epsilon^2$. There are two possible (inequivalent) lifts of the bundle to $BU_1$ as defined above, but only one lift to $G_1(\mathbb{C})$. - -REPLY [4 votes]: This is a question about general vector bundles, not tangent/normal bundles. So let $X$ be a finite complex, and $V,W \to X$ be two real vector bundles such that $V \oplus W$ is trivialized and $n$-dimensional. -If a stable complex structure on $V$ is given, pick an embedding $\iota:V \oplus \mathbb{R}^r \to X \times \mathbb{C}^N$ of complex vector bundles for some large $N$. The orthogonal complement $V^{\bot}$ of the image of $\iota$ is a complex vector bundle, and we have a specific isomorphism of real bundles -$$W\oplus \mathbb{C}^N \cong W \oplus V \oplus \mathbb{R}^r \oplus V^{\bot} \cong \mathbb{R}^{n+r} \oplus V^{\bot},$$ -giving the bundle $W$ a stable complex structure. If $N$ is large enough, then two such embeddings are isotopic, and so the orthogonal complements are concordant, hence isomorphic (as complex vector bundles). If you compose $\iota$ with the embedding $\mathbb{C}^N \to \mathbb{C}^{N+1}$, you add a trivial line bundle to the orthogonal complement. This shows that the stable complex structure on $W$ is uniquely determined. -The whole construction is symmetric in $V$ and $W$, and therefore induces the desired bijection.<|endoftext|> -TITLE: Oloid and sphericon: rolling develops entire surface -QUESTION [9 upvotes]: Wikipedia says that, - -"The oloid is one of the only known objects, along with some members of the sphericon family, that while rolling, develops its entire surface." - -Below are illustrations of these two objects: - -    -   Link to video of oloid rolling. -    -   Link to video of sphericon rolling. - -Two questions: -Q1. The phrasing of the Wikipedia sentence quoted above seems to -allow the interpretation that -it is unknown if there are other objects with the same rolling property. -Is this the case? Or is there a proof that these are the only such objects? -Q2. Does anyone know where I can find a proof of this rolling property for -either of these objects? - -REPLY [8 votes]: Here is a more general construction, which contains both the sphericon and the oloid as particular cases. - -Take two intersecting planes, $\alpha$ and $\beta$. -Take two compact convex regions $A\subset\alpha$ and $B\subset\beta$. -Make sure that $A$ is split in two non-empty regions by the plane $\beta$, and $B$ by the plane $\alpha$. -Let $a=A\cap\beta$, and $b=B\cap\alpha$. Both sets $a-b$ and $b-a$ are nonempty. -The desired object is the convex hull of the union of the two planar convex regions, $A\cup B$. Let $S$ be its boundary. - -Of course, because we the convex hull, in condition 2 we can drop the condition that the two planar regions are convex, but this doesn't increase generality. -Proof that this works. -Let $\partial A$ be the boundary of $A$ in the plane $\alpha$, and $\partial B$ the boundary of $B$ in the plane $\beta$. -A. Condition 3 implies that for any point $P$ on $\partial A$, there are two points $Q_1$ and $Q_2$ on $\partial B$, which are separated by the plane $\alpha$, hence the segments $PQ_1$ and $PQ_2$ are also separated by $\alpha$. This also holds for $B$. -B. From A, follows that if $P\in\partial A\cap S$, there are two points $Q_1,Q_2\in\partial B\cap S$, so that the segments $PQ_1,PQ_2\subset S$. -C. From condition 3, and remarks A and B, and the definition of convex hull, it follows that $S$ is the union of segments $PQ$, where $P\in\partial A\cap S$, and $Q\in\partial B\cap S$. If both $P$ and $Q$ would belong to $\partial A\cap S$, then by condition 3, the segment $PQ$ would be in the interior of the convex hull, so we can rule this possibility out. -D. From condition 4, $\partial A\cap S$ and $\partial B\cap S$ are open curves. -E. From D, and since the curves $\partial A\cap S$ and $\partial B\cap S$ don't intersect, the surface $S-(\partial A\cup\partial B)$ is connected. -F. By using my answer to another question, -When is the hull of a space curve composed of developable patches?, the surface $S$ is made of developable patches. To apply that result, join the two open curves $\partial A\cap S$ and $\partial B\cap S$ by two curves which lie on $S$, so that the union is not self-intersected. -G. From C, E and F, the surface $S-PQ$ is developable, where $PQ$ is as in C. -H. During complete rolling cycle, the curve $\partial A\cap S$ is covered twice, once from one end to the other on one side, and once back, but on the other side. Similarly for $\partial B\cap S$. -Therefore, $S$ has the property that, when rolling, it develops its entire surface. -Examples. -For both the sphericon and the oloid, we take $\alpha\perp\beta$. -For oloid, take $A$ and $B$ two disks, so that each disk has its center on the boundary of the other disk. -For sphericon, take $A$ and $B$ two identical hemidisks, each being symmetric w.r.t. $\alpha\cap\beta$, and so that their centers coincide with their intersection. -The generalization made by Manfred Weis in his comment is also included here. -It is easy to morph the oloid and the sphericon. Take $A',B'$ two identical disks, each being symmetric w.r.t. $\alpha\cap\beta$. Take a plane $\gamma$ through their mass center, perpendicular on $\alpha$ and $\beta$. Cut the circles by $\gamma$, and take $A$ as the part of $A'$ which contains the center of $A'$, and similarly for $B$. It is easy to see that the oloid and sphericon are particular cases of this construction, and by varying the distance between the centers of the disks, we can morph one into the other. -Answers. -Q1. No, the sphericon and the oloid are not unique with these properties, there is an infinity of such objects. Perhaps not many were known. -Q2. In this answer, hopefully. - -Update 1. -If the curves $\partial A\cap S$ and $\partial B\cap S$ are line segments, we obtain a tetrahedron. If they are each made of two segments, they can be chosen so that we get an octahedron. -In general, a convex polyhedron whose faces admit a Hamiltonian path, can be rolled so that it develops its entire surface, but is not necessarily obtained by the method I proposed. -One may want to rule out such cases, in which, after the rolling starts, the rolling is not unique, hence for which the contact between the object and the plane surface on which it is rolled is a (sur)face. One may want that at any point, the contact is along a segment. In this case, we have to add the condition that the curves $\partial A\cap S$ and $\partial B\cap S$ don't contain line segments.<|endoftext|> -TITLE: Knots and Dynamics. Recent breakthroughs? -QUESTION [7 upvotes]: I recently started reading Étienne Ghys slides on knots and dynamics which seem very interesting. I know this approach to knots and dynamics is not entirely new, for instance, this is from the 1980's . This post gave me the impression that it is a promising topic http://terrytao.wordpress.com/2007/08/03/2006-icm-etienne-ghys-knots-and-dynamics/. Two questions: (the second one is pretty general and I don't expect to have a response if someone finds it not too appropriate for this site, although an answer is always tremendously appreciated) - -Are there any recent breakthroughs after the release of these Ghys's slides that I should pay attention to? -Can anyone think of a research opportunity related to this topic for a not too long master's thesis project? - -In general I am more interested in the theoretical results rather than its physical applications. In particular those involving geometry and topology. Thanks in advance. - -REPLY [4 votes]: It's not clear when Ghys made the slides to which you have linked. The only date I could find in those was 1963 (referring to the Lorenz equations), which would make the bound on "recent" rather generous. Here's a quick summary of relatively recent concrete activity in this area that might be interesting to you. -In 1983, various reasonable conjectures were made by Birman and Williams (see BW1 and BW2) after considerable experimentation with the Lorenz equation at various parameter values. The basic line of investigation aimed to find answers to the following question: - -What types of knots and links can occur as periodic orbits of stable flows on $S^3$? - -For instance, it was conjectured in BW1 that no flow supported all knots as periodic orbits. The conjecture stood uncontested for just over a decade, until Ghrist constructed a flow supporting all links as periodic orbits! His paper G is merely six pages long, picture-filled, and eminently readable. -The entire story is summarized in Bob Williams' article W which contains a wealth of other information including references. - -This is really not my area of expertise, but here are two potential research problems. I'm not sure how tractable or elementary this would be, but that's the nature of the beast... - -Given a triangulated knot $K$, construct small triangulations $T$ of $S^3$ and piecewise-linear flows $\phi:T \to T$ which contain $K$ as a periodic orbit. - -And somewhat harder-seeming, - -Given a family $F$ of knots, construct a flow on $S^3$ which supports all knots as periodic orbits except the ones in $F$. - - -References -BW1: Birman and Williams, Knotted periodic orbits in dynamical systems-I: Lorenz's equations, Topology 22 , 1 (1983), 47 - 82. -BW2: Birman and Williams, Knotted periodic orbits II: Knot holders for fibered knots, Low Dimensional Topology, Contemporary Mathematics 20, A.M.S. (1983), 1-60. -G: Ghrist, Flows on $S^3$ supporting all links as orbits, Electronic Research Announcements of the AMS, 1(2), 91-97 (1995). -W: Williams, The universal templates of Ghrist, BAMS 35, No. 2, 145-156 (1998).<|endoftext|> -TITLE: a measure of difference for arrangements of sphere points -QUESTION [5 upvotes]: Suppose one has a distribution of $N$ points on the sphere. Is there an agreed upon metric for the difference of this distribution and $N$ equidistant points on the sphere? To me entropy seems like the right concept, but I'm not sure how to go about defining it. - -REPLY [3 votes]: there exists a great variety of metrics for the uniformity of points on a (hyper)sphere; the reference list of this 2010 paper will point you to them. For a comparison of the different metrics: - -A. Figueiredo, Comparison of tests of uniformity defined on the hypersphere, Statistics and Probability Letters, 77, 329 (2007). -P.J. Diggle, N.I. Fisher, and A.J. Lee, A comparison of tests of uniformity for spherical data, Australian Journal of Statistics, 27, 53 (1985).<|endoftext|> -TITLE: Is projective morphism with projective fiber flat? -QUESTION [5 upvotes]: Let $X, Y$ be quasi-projective Noetherian schemes and $f:X \to Y$ be a projective surjective morphism. Assume that every fiber of $f$ is isomorphic to a projective space $\mathbb{P}^n$ for a fixed $n$. Is it then true that $f$ is flat? - -REPLY [9 votes]: If $Y$ is non-reduced, then $X = Y_{red} \times \mathbf{P}^{n}$ is a counterexample.<|endoftext|> -TITLE: Is $\mathcal{D} \bigl( \mathrm{QCoh}(\mathfrak{X}) \bigr)$ compactly generated? -QUESTION [14 upvotes]: An object $E$ in a triangulated category $\mathcal{T}$ with (small) coproducts is called compact if the functor $\mathrm{Hom}_{\mathcal{T}}(E,-)$ commutes with arbitrary coproducts or, equivalently, if any morphism from $E$ to some coproduct factors through a finite subcoproduct. -Neeman (The Grothendieck Duality Theorem..., Prop 2.5) showed that $\mathcal{D} \bigl( \mathrm{QCoh}(X) \bigr)$ is compactly generated if $X$ is a quasi-compact and separated scheme and the proof uses only the semi-separatedness of $X$. Note that compact = perfect. -Bondal-Van den Bergh (Generators and representability of functors, Cor 3.1.2) prove this statement for $X$ quasi-compact and quasi-separated. -Whenever it comes to semi-separated and quasi-compact schemes I look for connections to algebraic stacks $\mathfrak{X}$ in the sense of Goerss, Naumann, ..., i.e., a stack in the flat topology with affine diagonal and some faithfully flat atlas from some affine scheme. Those schemes which are algebraic stacks are precisely the semi-separated and quasi-compact ones. Has anyone ever tried to extend the proof of Neeman to show that $\mathcal{D} \bigl(\mathrm{QCoh}(\mathfrak{X})\bigr)$ is compactly generated? -The 2-category of algebraic stacks with some fixed atlas is equivalent to the 2-category of flat Hopf algebroids (Naumann, The stack of formal groups..., §3.3). Under this equivalence quasi-coherent sheaves correspond to comodules over the associated Hopf algebroid. So a reformulation is: Is $\mathcal{D} \bigl( \mathrm{Comod}(\Gamma) \bigr)$ compactly generated for every flat Hopf algebroid $(A,\Gamma)$? - -REPLY [5 votes]: Neeman found a counterexample in a 2014 paper, $B\mathbb{G}_a,$ the classifying stack of the additive group scheme in positive characteristic. The original reference is rather inexplicit, but this followup paper by Hall, Neeman, and Rydh proves an even stronger fact, that $D_{qc}(B\mathbb{G}_a)$ in positive characteristic has no nonzero compact objects whatsoever.<|endoftext|> -TITLE: Loops and suspensions of higher categories -QUESTION [9 upvotes]: Given a pointed $(\infty,n)$-category $\mathcal{C}$, one can define the suspension of $\mathcal{C}$, $\Sigma\mathcal{C}$, via the homotopy pushout of $$\ast\leftarrow \mathcal{C}\rightarrow \ast.$$ Dually one can define $\Omega\mathcal{C}$. Can one explicitly identify these $(\infty,n)$-categories in terms of $\mathcal{C}$? -My vague intuition, based on the case $n=0$, says that the $\Omega\mathcal{C}$ should be the endomorphisms of the distinguished object and $\Sigma\mathcal{C}$ should be what you get when you take the free monoidal $(\infty,n)$-category on $\mathcal{C}$, regard it as an $(\infty,n+1)$-category with one object and then invert the $n+1$-morphisms. However, my understanding of (homotopy) limits and colimits in this setting is pretty poor. -Feel free to use any model you wish. - -REPLY [4 votes]: First let me thank Urs, Karol, and Rune Haugseng for helpful comments. -Now note that the inclusion, $i$, of $\infty$-groupoids into $(\infty,n)$-categories has an $\infty$-categorical left adjoint, $L$ (for lack of a better name), and a right adjoint $(-)^\prime$. -Given a pointed $(\infty,n)$-category $\mathcal{C}$, the loop category $\Omega \mathcal{C}$ is defined by the following (homotopy) pullback diagram: -\begin{array}{ccc} -\Omega\mathcal{C} & \rightarrow & \ast\\ -\downarrow & & \downarrow \\ -\ast & \rightarrow & \mathcal{C} -\end{array} -Now $\mathcal{C}^{\prime}$ is the maximal sub-$\infty$-groupoid of $\mathcal{C}$ (the core). Since $\ast$ is an $\infty$-groupoid, the inclusion $\ast\rightarrow \mathcal{C}$ factors canonically through $\mathcal{C}^\prime$. By a standard finality argument we see that $\Omega \mathcal{C}$ is equivalent to the homotopy pullback: -\begin{array}{ccc} -\Omega\mathcal{C} \simeq \Omega^{Top}\mathcal{C}^\prime& \rightarrow & \ast\\ -\downarrow & & \downarrow \\ -\ast & \rightarrow & \mathcal{C}^\prime -\end{array} -Regarding $\mathcal{C}^\prime$ as a space (since it is an $\infty$-groupoid), we see that $\Omega\mathcal{C}$ is equivalent to the space of topological (based) loops on $\mathcal{C}^\prime$ since $i$ preserves (homotopy) limits. -Unraveling this a bit, we see that $\Omega(-)$ is naturally equivalent to $i\Omega^{Top}(-)^\prime$ which is a composite of right adjoints. It follows that the left adjoint, $\Sigma(-)$, is naturally equivalent to $i\Sigma^{Top}L(-)$. -As a consequence, an $(\infty,n)$-category which is a loop category is necessarily a loop space. This shows that the inclusion of spectra objects in $(\infty,0)$-categories (i.e., spectra) into spectra objects in $(\infty,n)$-categories is an equivalence. So the two categories have the same stailizations.<|endoftext|> -TITLE: Do partitions of unity exist if we impose additional conditions on the derivatives? -QUESTION [9 upvotes]: Let $ ~~\cup_{k=-1}^{\infty} U_k = \mathbb{R} $ be an open covering of -$\mathbb{R}$. It is a well known fact that partitions of unity subbordinate to -the cover exists, i.e. there exists smooth -functions $ \varphi_{k} : U_k \rightarrow \mathbb{R} $ with compact -support such that -$$ \sum_{k=-1}^{\infty} \varphi_k(x)^2 \equiv 1.$$ -My first question is vague: Do there exist partitions of unity -subbordinate to the cover -if we impose some additional conditions on the derivatives (and the -nature of the conditions are in terms of equalities, not inequalities.) ? -A more precise question is as follows: -Given a smooth function $f: \mathbb{R} \rightarrow \mathbb{R}$, -do there exist functions -$ \varphi_{k} : U_k \rightarrow \mathbb{R} $ with compact support -such that in addition to being a partition of unity subbordinate -to the cover, it also satisfies -$$ \sum_{k=-1}^{\infty} \varphi_{k}^{\prime}(x)^2 \equiv f(x)$$ -? Here prime denotes derivative with respect to $x$. -I assume the answer should depend on what $f(x)$ is. Ideally -$\varphi_k$ should be smooth functions, but at the very least they ought -to be $C^1$. -I would prefer if we do not assume $f$ is nowhere vanishing, -but if there is an answer assuming that, I would still like to see it. -Remark: The answer probably also depends on the open covering. -$\textbf{Very specific question:}$ Choose some number -$\tau \in (\sqrt{2}, 2) $. Say $\tau = 1.5$. Now define -$$ U_k = \{ x \in \mathbb{R}: \frac{2^k}{\tau} < |x| < 2^k \tau \} \qquad k=0,1,2, \ldots $$ -$$ U_{-1} = (-1,1).$$ -The collection $\{U_k\}_{k=-1}^{\infty} $ is an open covering of $\mathbb{R}$. -I want smooth functions $\varphi_k: U_k \rightarrow \mathbb{R}$ and -numbers $n_k$ such that -$$ \sum_{k= -1}^{\infty} \varphi_k(x)^2 \equiv 1 $$ -and -$$ \sum_{k=-1}^{\infty} n_k^2 (\varphi_k(x)^2 + \varphi_k^{\prime}(x)^2) -\equiv e^x$$ -It is easy to see that if I set $n_k =1$ and $f(x) = e^x-1$, then it is -the previous question I had asked. This question is "easier", because one -is allowed to choose the numbers $n_k$ (in other words there is a better -chance that the answer here might be yes, because of the freedom in -choosing $n_k$). - -REPLY [6 votes]: I hope the following construction will give you what you really need. If not, you'll have to explain why. -Take any nice locally finite covering $\mathbb R\subset\cup_j U_j$ and take any smooth partition of unity $1=\sum_j\psi_j^2$ subordinated to this covering. Take any smooth positive function $F$ on $\mathbb R$. Choose the numbers $n_j>0$ so that $\sum_j n_j^2(\psi_j^2+(\psi_j')^2)\le F/2$. Define the smooth function $\theta$ by -$$ -(\theta^{\,\prime})^2\sum_j n_j^2\psi_j^2=F-\sum_j n_j^2(\psi_j^2+(\psi_j')^2) -$$ -Finally, associate with each $U_j$ two functions $\varphi_{j,0}=\psi_j\cos\theta$ and $\varphi_{j,1}=\psi_j\sin\theta$ and enjoy the identities -$$ -\sum_{j,k}\varphi_{j,k}^2=1, \quad \sum_{j,k}n_j^2(\varphi_{j,k}^2+(\varphi_{j,k}')^2)=F\,. -$$<|endoftext|> -TITLE: Function with zeros plus/minus the primes -QUESTION [5 upvotes]: While playing with Cohen's pari script prodeulerrat found a function. -For $s \in \mathbb{C}$ define -$$ f(s) = \prod_{p \text{ prime}} (1-\frac{s^2}{p^2})$$ -The product converges everywhere, no poles and the zeros -are $\pm p$. -At integers one can tell if $f(n)=0$ via primality testing. -Cohen's script computes $f(s)$ in $O(|s|)$ and it -iterates over primes. - -Q1 Is there an alternative way to compute $f(s)$? -Q2 An explicit series converging to $f(s)$? - -$f(1)=1/\zeta(2)$. - -Q3 Is there closed form for $f$ at integers? - -Xray: - -Complex plot: - -REPLY [2 votes]: A standard method to improve the speed of convergence is to look for approximations, which can be explicitly evaluated. In this case one would take $f(s)\approx\zeta(2)^{-s^2}$. We have -$$ -f(s)\zeta(2)^{s^2} = \exp\left(\sum_p s^2\log (1-\frac{1}{p^2}) + \log(1-\frac{s^2}{p^2})\right). -$$ -Each summand is now of magnitude $\frac{s^4}{p^4}$, thus the speed of convergence has improved, since you have to add up primes substantially larger then $s$ anyway, unless you only look for very rough bounds. -If this is not enough, you can develop $\log(1-\frac{1}{p^2})$ and $\log(1-\frac{s^2}{p^2})$ into a Taylor series, and pull out another power of $\zeta$. The next term will probably give $\zeta(4)^{(s^4-s^2)/2}$, and the series should then converge like $\frac{s^6}{p^6}$. -At some point you will have to ask yourself whether the algebraic manipulations necessary to improve the convergence are worth the saving in computation time, the answer to this question of course depends on your application or interest.<|endoftext|> -TITLE: property of trace modulo $n$ -QUESTION [8 upvotes]: I recently noticed an interesting (at least to me) property of the trace but have been unable to prove it. -Let ${\mathbb K}$ is an algebraic number field with ${\mathcal O}$ as its ring of integers, $n$ is any positive integer relatively prime to the discriminant of ${\mathcal O}$ and -$$ -T_{n} = \left\{ x \in {\mathcal O}: {\rm Tr}(x) \equiv 0 \bmod n \right\}. -$$ -Then the elements $y \in {\mathcal O}$ such that $xy \in T_{n}$ for all $x \in T_{n}$ seem to always be of the form a rational integer plus $n$ times an element of ${\mathcal O}$. -Does anyone know how to prove such a result (or a counterexample too, of course)? -References to such results in the literature would also be welcome. -Thank you. - -REPLY [3 votes]: For a finite degree field extension $E/K$ the trace is a $K$-linear map $E\to K$ and it determines the symmetric bilinear form $Tr(xy)$ on the $K$ vector space $E$. If the extension is separable (= etale) then the trace is not identically zero, so its kernel has codimension one, so the orthogonal complement of its kernel is one-dimensional and contains $K$ and so must be exactly $K$. -What you're looking at is exactly this, except that instead of a field extension $E$ over $K$ it's a ring extension $\mathcal O/n$ over $\mathbb Z/n$. (It's etale because the discriminant is prime to $n$.) I believe the same principles basically apply.<|endoftext|> -TITLE: Minor theorems of Pappus and Desargues in "old school" geometry? -QUESTION [7 upvotes]: My question concerns the dependence relations between the minor theorem of Pappus which, following Heyting, I will denote by $P_9$, and (one of the) minor theorems of Desargues, $D_9$. -$P_9$ states that: "If in a hexagon two diagonal points are on the corresponding diagonals and the diagonals are concurrent, then the third diagonal point is also on its corresponding diagonal too." -$D_9$ on the other hand (is equivalent to): "Given two triangles $A_1A_2A_3$ and $B_1B_2B_3$, such that $A_i \neq B_i$, $A_iA_j \neq B_iB_j$; $A_iB_i$ $i=1,2,3$ are concurrent, $A_1 \in B_2B_3$ and $B_1 \in A_2A_3$, then the points $C_i = A_jA_k \cap B_jB_k$ ($i \neq j \neq k \neq i$) are colinear." -It is easy to show that $D_9 \Rightarrow P_9$. My question is whether the converse holds, Heyting claims this is unknown but his book "Axiomatic Projective Geometry" dates from 1980 (second ed.) so this might no longer be true, though a quick internet search failed to point me to anything useful. Does anybody know what is the status of $P_9 \Rightarrow D_9$? It would be nice considering Hessenberg's theorem showing that full Pappus implies full Desargues. -Hopefully someone can help me out, thanks in advance! - -REPLY [3 votes]: A. Seidenberg in Pappus implies Desargues (1976) claims to correct Hessenberg's incomplete proof. From a recent review by Marchisotto (2002), I gather that Seidenberg's proof is "for real". - -REPLY [3 votes]: The converse of the Hessenberg’s theorem is not true. In the quaternionic projective plane the Desargues' theorem is true but the Pappus's is false. See historical notes in http://www.sciencedirect.com/science/article/pii/S0024379501002877# (On Pappus' configuration in non-commutative projective geometry, by Giorgio Donati).<|endoftext|> -TITLE: Graphs in which every spanning tree is an independency tree -QUESTION [12 upvotes]: It follows from this question -and the corresponding answers, that the complete graphs and the cycles are precisely the graphs -$G$ having the property that, for every spanning tree $T$ of $G$, the set of leaves of $T$ is a -clique in $G$. -Motivated by this fact, I am looking for a characterization of all (connected) graphs $G$ -having the property that, for every spanning tree $T$ of $G$, the set of leaves of $T$ is an -independent set in $G$. -(Such spanning trees are called independency trees -in the literature.) - -REPLY [3 votes]: Here's something more about these graphs. - -Theorem: Let $G$ be connected graph with $\geq 3$ vertices. Consider the next properties: -1) Every edge in $G$ is incident to a cut vertex. -2) $G$ has all spanning trees independency. -3) The set of all cut vertices in $G$ is dominating. -Then 1)$\Rightarrow$2)$\Rightarrow$3). - -Proof -1)$\Rightarrow$2): Let $e=uv\in E(G)$ and $u$ is a cut vertex in $G$. Assume that $v$ isn't a leaf. Then $G-u$ is disconnected and $d_{G-u}(v)\geq 1$. Therefore $G-\{u,v\}$ is also disconnected. Thus $\{u,v\}$ is a separator in $G$. -2)$\Rightarrow$3): Assume that for some non-cut vertex $u\in V(G)$ the set $N(u)$ consists of non-cut vertices. Then for all $v\in N(u)$ the set $\{u,v\}$ is a separator in $G$. It means that $v$ is a cut vertex in $G-u$. -If there exists a component $A$ of $G-\{u,v\}$ such that $V(A)\cap N(u)=\emptyset$, then $A$ is also a component of $G-v$. It means that $v$ is a cut vertex in $G$ which is a contradiction. Thus every component of $G-\{u,v\}$ contains the vertices from $N(u)$. -Therefore, each vertex $v\in N(u)$ separates the set $N(u)-v$ in $G-u$. But this contradicts the connectedness of $G-u$. $\boxtimes$ -Two counterexamples to the reverse statements of the Theorem. -2)$\nRightarrow$1): - -3)$\nRightarrow$2): $K_{1}+(P_{2}\cup K_{1})$.<|endoftext|> -TITLE: Cobordism and finite sheeted covers of manifolds -QUESTION [6 upvotes]: Let $M$ be an oriented manifold, not necessarily compact. Let $M'$ be a (finite) $k$-sheeted cover and let $\pi:M'\longrightarrow M$ be the covering map. -Question 1 : Is it true that $M'$ is (oriented) cobordant to $k$ disjoint copies of $M$? -Question 2 : If the answer to the above is true then let $W$ be such a cobordism. Is there a map $\widetilde{\pi}:W\longrightarrow M$ such that $\widetilde{\pi}$ restricts to the natural maps on each end? -I would guess that the answers to these questions should be well-known. However, any references, proofs or counter-examples(?) would be helpful. - -EDIT : From the discussion/answers below, it is clear that the answer to (1) is positive when $M$ is a closed, oriented manifold. The characteristic numbers determine the oriented cobordism class of $M$. However, when $M$ is an open manifold, these numbers do not make immediate sense. I'm particularly interested in both questions in this scenario. - -REPLY [4 votes]: The answer is yes to the first question. This follows from the fact that characteristic numbers determine completely cobordism classes. -Actually both questions have two versions: an oriented and an unoriented one. -The fact that the characteristic numbers are multiplied by $k$ under a $k$-fold covering follows from the definition of characteristic numbers through vector fields. (Each Pontrjagin class and Stiefel Whitney class can be obtained by taking some, properly chosen number of generic vector fields and consider the points, where the dimension of the space spaned by them drops by (at least) a given number compered with the possible maximum. (This given number is $1$ for $W$ and 2 for the $P$ classes.) The set of these points is a cycle dual to the corresponding characteristic class. Hence the characteristic numbers can be obtained as the number of intersections of such cycles. Now by the covering map we can pull back the vector fields, we can pull back the cycles, and clearly the number of intersections of the cycles in the covering manifold will be $k$ times that in the base manifold. - -REPLY [3 votes]: Here is Dold's argument showing that if $M' \to M$ is a double cover and an orientation preserving map, than $M'$ is oriented-cobordant to two copies of $M$. -Let $V \subset M$ be a codimension $1$ submanifold representing the $Z_2$-homology class dual to $W_1$ of the double covering $M' \to M.$ Let $N$ be the closed tubular neighbourhood of $V$ in $M.$ Note, that if we take two copies of $M \setminus int N$ and attach them appropriately along the boundaries, then we obtain $M'.$ -Now take two copies of the cylinder $M\times [0,1]$ and attach them along $N \times 1$ to each other so that from the two copies of $M \times {1}$ we obtain $M'$. -We obtain a compact, oriented manifold with boundary, the boundary consisists of two copies of $M \times {0}$ and one copy of $M'$<|endoftext|> -TITLE: Universal covering map from $\mathcal{H}$ to $\mathbb{C}\setminus \mathbb{Z}\oplus i\mathbb{Z}$ (the countably punctured complex plane) -QUESTION [7 upvotes]: It's a consequence of the uniformization theorem for simply connected Riemann surfaces that the universal cover of $\mathbb{C}\setminus(\mathbb{Z}\oplus i\mathbb{Z})$ ($\mathbb{C}$ punctured at all the integral lattice points) is the upper half plane $\mathcal{H}$. -How should I think about this map? How does the map behave near the missing lattice points? -A related question is this: $\mathcal{H}$ is also the universal cover for a punctured torus, whose fundamental group is $F_2$, the free group on two generators. By comparing the punctured torus to the wedge of two circles, I feel like the universal cover for the punctured torus, ie $\mathcal{H}$, ought to be deformation-retractable to an infinite 4-regular tree. Ie, the infinite 4-regular tree ought to be able to be embedded in $\mathcal{H}$ such that the vertices of the tree all lie on the boundary of $\mathcal{H}$. What does this tree look like in $\mathcal{H}$? - -REPLY [11 votes]: On the first question (the universal cover of the complement of a lattice). The missing points are in the image, so it is not the map that "behaves" but the inverse map. -The inverse map behaves in a very simple way: it has infinitely many "logarithmic singularities" over each missing point. -"How to think about the map" is not a well defined question. But the way I think about -it is this. Consider the circular quadrilateral inscribed in the unit disc, say with -vertices 1,-1,i,-i; the sides are arcs of circles orthogonal to the unit circle. All -angles of this quadrilateral are 0. There is a -conformal homeomorphism of this circular quadrilateral onto a (rectilinear) square -with vertices 0,1,1,1+i, sending vertices to vertices. -By Schwarz's symmetry principle, applied very many times, the map extends to a map from the -unit disc to the plane minus the lattice. This is our universal covering map. You can make -a picture. You can express it in terms of special functions (it is a ratio of two solutions -of a very special Heun equation, linear differential equation of second order with regular -singular points at 1,-1,i,-1. -EDIT: I am not sure what exactly you want to know, in your question you mention visualization, rather than computation, but once I computed this map, -arXiv:1110.2696. It can be expressed in terms of hypergeometric functions. -And I also asked a MO question related to it: -Maximum of a function of one variable.<|endoftext|> -TITLE: Is this lemma in elementary linear algebra new? -QUESTION [42 upvotes]: Is anyone familiar with the following, or anything close to it? -Lemma. Suppose $A$, $B$ are nonzero finite-dimensional vector spaces -over an infinite field $k$, and $V$ a subspace of $A\otimes_k B$ -such that -(1) For every nonzero $a\in A$ there exists nonzero $b\in B$ - such that $a\otimes b\in V$, -and likewise, -(2) For every nonzero $b\in B$ there exists nonzero $a\in A$ - such that $a\otimes b\in V$. -Then -(3) $\dim_k(V) \geq \dim_k(A) + \dim_k(B) - 1$. -Remarks: The idea of (1) and (2) is that the spaces $A$ and $B$ -are minimal for "supporting" $V$; that is, if we replace -$A$ or $B$ by any proper homomorphic image, and we map $A\otimes B$ in -the obvious way into the new tensor product, then that map will -not be one-one on $V$. The result is equivalent to saying that if one is -given a finite-dimensional subspace $V$ of a tensor product $A\otimes B$ -of arbitrary vector spaces, then one can replace $A$, $B$ by images -whose dimensions sum to $\leq \dim(V) + 1$ without hurting $V$. -In the lemma as stated, if we take for $A$ a dual space $C^*$, and -interpret $A\otimes B$ as $\mathrm{Hom}(C,B)$, then the hypothesis again -means that $C$ and $B$ are minimal as spaces "supporting" $V$, now as a -subspace of $\mathrm{Hom}(C,B)$; namely, that restricting to any proper -subspace of $C$, or mapping onto any proper homomorphic image of $B$, -will reduce the dimension of $V$. -In the statement of the lemma, where I assumed $k$ infinite, -I really only need its cardinality to be at least the larger -of $\dim_k A$ and $\dim_k B$. -The proof is harder than I would have thought; my write-up is 3.3K. -I will be happy to show it if the result is new. - -REPLY [3 votes]: As suggested by Martin, there is a geometric interpretation of this lemma. Though the proof is probably not shorter than the one proposed by Clément. Nevertheless, this is the kind of very classical reasonning one encouters in the study of secant varieties. -Let us put $a = dim A$ and $b = dim B$. If $a \otimes b \in A \otimes B$, I denote its image in $\mathbb{P}(A \otimes B)$ by $[a \otimes b]$. -I denote by $X_{A,B} = \{(a,b), \textrm{such that} [a \otimes b] \in \mathbb{P}(V) \}$. This is clearly equal to the scheme $(\mathbb{P}(A) \times \mathbb{P}(B)) \cap \mathbb{P}(V)$ (I'll consider only schemes with reduced structure here). -Let us consider the natural projections $p_A : X_{A,B} \longrightarrow \mathbb{P}(A)$ and $p_B : X_{A,B} \longrightarrow B$. The hypothesis given by the OP show that $p_A$ and $p_B$ are surjective. Denote by $\gamma_A$ the dimension of the generic fiber of $p_A$, by $\gamma_B$ the dimension of the generic fiber of $p_B$, by $X_A$ a maximal dimensional irreducible component of the scheme $p_A^{-1}(p_A(X_{A,B}))$ and by $X_B$ a maximal dimensional irreducible component of the scheme $p_B^{-1}(p_B(X_{A,B}))$. -The theorem of the dimension gives $dim \ X_A = a-1 + \gamma_A$ and $dim \ X_B =b-1 + \gamma_B$. -The secant variety $S(X_A,X_B)$ (that is the closure of variety of lines joining a point of $X_A$ and a point of $X_B$) is included in $\mathbb{P}(V)$ and the goal will be to bound below its dimension to get a bound for $dim \ \mathbb{P}(V)$. -The dimension of $S(X_A,X_B)$ is equal to $\dim \ X_A + \dim \ X_B +1 - \delta$, where $\delta$ is the secant defect of $S(X_A,X_B)$. Concretely, if $M$ is a generic point of $S(X_A,X_B)$, then $\delta$ is the dimension of the scheme: -$$\{[a_1 \otimes b_1] \in X_A, \ \textrm{s.t. $\exists [a_2 \otimes b_2] \in X_B$ and $(x,y) \in \mathbb{k}^2$ with $M = x.a_1\otimes b_1 + y.a_2 \otimes b_2$} \}.$$ -It is well known that the secant defect of $S(\mathbb{P}(A) \times \mathbb{P}(B),\mathbb{P}(A) \times \mathbb{P}(B))$ is $2$. Indeed, the parameter family to decompose a rank $2$ matrix as a sum of two rank $1$ matrices is $\mathbb{P}^1 \times \mathbb{P}^1$. (short explanation : as one only needs to construct one of these rank $1$ matrices : choose the image (choice of a $\mathbb{k}^1$ in the image of the rank $2$ matrix, which is isomorphic to $\mathbb{k}^2$) and choose a hyperplane containing the kernel of the rank $2$ matrix). -Assume that $S(X_A,X_B)$ consists only of rank $1$ matrices. Since $X_A$ surjects onto $A$ and $X_B$ surjects onto $B$, we easily deduce that $X_A = \mathbb{P}(A) \times b_0$ and $X_B = a_0 \times \mathbb{P}(B)$ for some $a_0$ and $b_0$ fixed. The dimension of $S(X_A,X_B)$ is then obviously seen to be $a-1+b-1-0 = a+b-2$ and then we have $dim \ \mathbb{P}(V) \geq a+b-2$. -Assume that the $S(X_A,X_B)$ contains a matrix of rank $2$. Then, the generic $M \in S(X_A,X_B)$ has rank $2$. Since $X_A \subset \mathbb{P}(A) \times \mathbb{P}(B)$, $X_B \subset \mathbb{P}(A) \times \mathbb{P}(B)$ and the secant defect of $S(\mathbb{P}(A) \times \mathbb{P}(B),\mathbb{P}(A) \times \mathbb{P}(B))$ is $2$, we deduce that $\delta \leq 2$. As a consequence, -$$dim \ S(X_A,X_B) \geq a-1 + \gamma_A + b-1 + \gamma_B +1 - \delta \geq a+b-3.$$ -If $\delta \leq 1$, then we get in fact: -$$dim \ S(X_A,X_B) \geq a-1 + \gamma_A + b-1 + \gamma_B-1 \geq a+b-2,$$ -and this implies that $dim \ \mathbb{P}(V) \geq a+b-2$, which is what we wanted. -If $\delta = 2$, then the dimension of -$$\{[a_1 \otimes b_1] \in X_A, \ \textrm{s.t. $\exists [a_2 \otimes b_2] \in X_B$ and $(x,y) \in \mathbb{k}^2$ with $M = x.a_1\otimes b_1 + y.a_2 \otimes b_2$} \}$$ is $2$. In view of the explicit decomposition of a rank $2$ matrix as the sum of two rank $1$ matrices, this implies that for every $a$ in $\mathbb{P}(A)$, there is at least a $\mathbb{P}^1$ of $b \in \mathbb{P}(B)$ such that $[a \otimes b] \in X_A$. We deduce that $\gamma_A \geq 1$ and finally: -$$\dim S(X_A,X_B) \geq a-1+1 +b-1 + 1 -2 = a+b-2,$$ -which again implies $dim \ \mathbb{P}(V) \geq a+b-2$.<|endoftext|> -TITLE: Writing a matrix as a sum of two invertible matrices -QUESTION [13 upvotes]: Let $n\geq 2$. Is it true that any $n\times n$ matrix with entries from a given ring (with identity) can be written as a sum of two invertible matrices with entries from the same ring ? - -REPLY [19 votes]: The answer is negative. -There is a nice theorem of M. Henriksen which says that If $n\geq 2$ then every element of $M_n(R)$ is a sum of three units also he proves that there are non-unit matrices in $\bf{M_2(\Bbb{Z}_2[x_1,x_2])}$ that can not be written as a sum of two units. You can find a copy of the article HERE<|endoftext|> -TITLE: Is there a subset of the natural number plane, which doesn't know which of its slices are arithmetic? -QUESTION [20 upvotes]: $\newcommand{\N}{\mathbb{N}}$ -My question, more precisely, is: -Question. Is there a set $B\subset \N\times\N$, such that the set of indices where it is arithmetically definable, that is, $\{ n\in\N \mid B_n\text{ is arithmetic}\}$, is not first-order definable in the structure $\langle\N,{+},{\cdot},0,1,{\lt},B\rangle$? -By $B_n$, I mean the $n^{th}$ slice of $B$, the set $B_n=\{ k\mid (n,k)\in B\}$. And a set is arithmetic, if it is first-order definable in the standard model of arithmetic $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$. -I have need of such a set $B$, with an application ready to go, if -there is such a set $B$. -One might hope to make an example via forcing, say, with -conditions that specify finitely many of the slices. We cannot -allow, however, that arbitrary arithmetic sets appear in the -slices of $B$, because then the $\Sigma_k$ truth predicates would -appear for arbitrarily large $k$, and one can recognize these and -use them to define arithmetic truth from $B$, and thereby tell -which $B_n$ are arithmetic. -So we want to restrict the kinds of arithmetic sets that are -allowed to appear as slices of $B$. Perhaps we want the slices of -$B$ to be themselves somewhat generic, as Cohen reals, say, with -the $B_n$s increasingly $\Sigma^0_k$-generic, and with some of -them fully arithmetically generic (and hence not arithmetic). But -I did not succeed in pushing this idea through. - -Update. The result that Andrew has provided now appears as Lemma 10.1 (credited to him) in my paper with Ruizhi Yang: - -J. D. Hamkins and R. Yang, Satisfaction is not absolute. - -The lemma is used to prove the following, which was the application that I had mentioned in the original question. -Theorem 10. Every countable model of set theory $M$ has elementary -extensions $M_1$ and $M_2$, which agree on the structure of their standard -natural numbers $$\langle \mathbb{N},{+},{\cdot},0,1,\lt\rangle^{M_1}= \langle \mathbb{N},{+},{\cdot},0,1,\lt\rangle^{M_2},$$ and which have -a set $A\subset\mathbb{N}$ in common, extensionally identical in $M_1$ and $M_2$, yet $M_1$ thinks $A$ is first-order definable in $\mathbb{N}$ and $M_2$ thinks it is not. - -REPLY [15 votes]: There are $B$ with this property. -Lets first recall some definitions and notation. Suppose $X \in 2^\omega$ (which we identify with subsets of $\N$ via characteristic functions). Then $X'$ is the Turing jump of $X$, and $X^{(n)}$ is the nth iterate of the Turing jump of $X$. $X$ is said to be $n$-generic (i.e. $\Sigma^0_n$-generic for Cohen forcing) if for every $\Sigma^0_n$ subset $S$ of $2^{< \omega}$ there is a finite initial segment $\sigma$ of $X$ so that either $\sigma \in S$, or there are no extensions of $\sigma$ contained in $S$. $X$ is said to be arithmetically generic if $X$ is $n$-generic for every $n$. It is a standard fact that if $X$ is $1$-generic, then $X' \equiv_T 0' \oplus X$, where $\oplus$ is recursive join. We also have that if $X$ is $n$-generic and $Y$ is $1$-generic relative to $X \oplus 0^{(n-1)}$, then $X \oplus Y$ is $n$-generic. Finally, if $X_0, X_1, \ldots \in 2^\omega$, then $\bigoplus_n X_n = \{\langle n, m \rangle: m \in X_n\}$ notes their recursive join. -Let $A = 0^{(\omega)} = \bigoplus_{n} 0^{(n)}$; the set of true sentences in first order arithmetic. Below, we will construct a $B$ so that: - -If $n \notin A$, then $B_n$ is arithmetically generic, and if $n \in A$, then $B_n$ is $(n+1)$-generic and computable from $0^{(n+1)}$. Hence $\{n \in \mathbb{N} : \text{$B_n$ is arithmetic}\}$ is equal to $A$. -For each $k \in \N$, let $m_i$ be the $i$th element of the set $\{m \in \N : m \notin A \lor m \geq k\}$. Then $C_k = \bigoplus_{i \in \omega} B_{m_i}$ is $(k+1)$-generic. - -A $B$ with the above two properties gives a positive answer to your question by the following reasoning. We can prove by induction that for any $n$, $B^{(n)} \equiv_T 0^{(n)} \oplus C_{n}$. This is trivial when $n = 0$. Now for the inductive case, assume $B^{(n)} \equiv_T 0^{(n)} \oplus C_{n}$. Then we see that $B^{(n+1)} \equiv_T (0^{(n)} \oplus C_n)' \equiv_T 0^{(n+1)} \oplus C_n$ since $C_n$ is $(n+1)$-generic and hence $1$-generic relative to $0^{(n)}$. Finally, $0^{(n+1)} \oplus C_n \equiv_T 0^{(n+1)} \oplus C_{n+1}$ because either $n \notin A$ and so $C_n = C_{n+1}$ or $n \in A$ so $C_{n+1} \equiv_T B_n \oplus C_n$, but then $B_n \leq_T 0^{(n+1)}$. -Hence, $B^{(n)}$ cannot compute $A$ for any $n \in \omega$, since $C_{n}$ is $(n+1)$-generic, and so $0^{(n)} \oplus C_{n} \ngeq_T 0^{(n+1)}$. (Indeed, it's easy to see that $B$ is $GL_n$ for every $n \in \N$). Thus, $A$ is not arithmetically definable relative to $B$. - -So lets turn now to the construction of a $B$ with the required properties (1) and (2) above. We construct $B$ in countably many steps where after step $n$ we will have completely defined $B_0, B_1, \ldots, B_n$ and finitely many other bits of $B$ (i.e. finitely many bits of finitely many $B_i$ where $i > n$). For step $0$, let $B_0$ be an arbitrary real satisfying condition (1). -Given $k \leq n$, let $m_0, \ldots, m_j$ be the elements of $\{m \in \mathbb{N}: m \notin A \lor m \geq k\} \cap \{0, \ldots, n\}$, and define $C_{k,n} = B_{m_0} \oplus \ldots \oplus B_{m_j}$. After each stage $n$, we will have ensured that $C_{n,k}$ is $(k+1)$-generic for every $k \leq n$. -Now at step $n > 0$, for each pair $(i,k)$ where $i,k < n$, do the following. Let $S_{i,k}$ be the $i$th $\Sigma^0_{k+1}$ subset of $2^{< \omega}$. If we can find a finite extension of our approximation of $B$ so that the resulting approximation of $C_k$ extends an element of $S_{i,k}$, then extend our approximation of $B$ in this way. If this is not possible, then since $C_{k,n-1}$ is $(k+1)$-generic, there must be some finite subset of our current approximation to $B$ which cannot be extended to extend an element of $S_{i,k}$. -Next, we finish step $n$ by defining $B_n$. If $n \notin A$ pick $B_n$ to be an element of $2^{\omega}$ extending the finite approximation of $B_n$ that we currently have, which is arithmetically generic relative to $B_0 \oplus \ldots \oplus B_{n-1}$. This clearly ensures that $C_{k,n}$ will be $(k+1)$-generic for each $k < n$, since $C_{k,n-1}$ is $(k+1)$-generic, and $B_n$ is $(k+1)$-generic relative to it. Similarly, $C_{n,n}$ is clearly $(n+1)$-generic. Otherwise, if $n \in A$, let $B_n$ be an arbitrary $(n+1)$-generic computable from $0^{(n+1)}$ and extending the finitely many bits of $B_n$ we have already determined. Now for any $k < n$, let $j_0, \ldots, j_t$ be the elements of $A$ that are $\geq k$ and $< n$. Then since $B_n$ is $1$-generic relative to $0^{(n)}$ which can compute $B_{j_0} \oplus \ldots \oplus B_{j_t}$ we have that $B_{j_0} \oplus \ldots \oplus B_{j_t} \oplus B_n$ is $(k+1)$-generic. Hence $C_{k,n}$ is $(k+1)$-generic, since the remaining elements in the finite join defining $C_{k,n}$ are mutually arithmetically generic (and are hence $(k+1)$-generic relative to $B_{j_0} \oplus \ldots \oplus B_{j_t} \oplus B_n$). Similarly, $C_{n,n}$ is again clearly $n+1$-generic. -To verify that our construction works, note that condition (1) is satisfied by how we pick $B_n$ in the last paragraph. Condition (2) is satisfied by the paragraph before that.<|endoftext|> -TITLE: Toda brackets and factorisation of a sequence of spectra -QUESTION [5 upvotes]: I've found a paper of Spanier's (Higher Order Operations) where he uses the theory of "carriers" to study $n$-th order operations. The set-up is rather general; for example a particular case defines the Massey triple product. Spanier also talks about Toda's 'toral construction', which as far as I can tell is Toda's name for what (everybody else now) calls a Toda bracket. -I am particular interetsed in the results of Section 6 of this paper. Here Spanier sets up an operation associated to a sequence of spaces -$$ -A_0 \xrightarrow{a_1} A_1 \xrightarrow{a_2} A_2 \to \cdots \xrightarrow{a_n}A_n -$$ -In particular he defines an operation $\langle a_1,a_2,\ldots,a_n \rangle$ corresponding to a subset of $[\Sigma^{n-2}A_0,A_n]$, which gives me hope that this is a Toda bracket. -The result I'm particularly interested in is Theorem 6.3: A sequence,as above, is split if there is a commutative diagram of the form -$$ -\begin{array}{cccccccccc} -&&&& B_2 &&&&B_{n-1} \\ -&&&\nearrow_{b_2} & \downarrow_{c_2} &&&& \downarrow_{c_{n-1}}\\ -A_0 & \xrightarrow{a_1} & A_1 & \xrightarrow{a_2} & A_2 & \xrightarrow{a_3} & \cdots & \xrightarrow{} & A_{n-1} \xrightarrow{a_n} & A_n -\end{array} -$$ -such that $b_2a_1,b_3c_2,\cdots,a_nc_{n-1}$ are all null. -Spanier proves that such a splitting exists if and only if the operation$\langle a_0,a_1,\ldots,a_{n-1} \rangle$ is defined and vanishes. (Note that an $n$-th order is defined if and only if all lower order operations are defined and each contain the zero element. An operation vanishes if it is defined and contains the zero element.) -One thing that worries me slightly about the proof is that Spanier uses equalities -$$ -\langle a_1,\cdots,a_{n-2},g'b_{n-1} \rangle = g'_\# \langle a_1,\cdots,a_{n-2},b_{n-1} \rangle -$$ -If I'm not mistaken the shuffling/juggling lemma for Toda brackets is that -$$ -w \circ \langle x,y,z \rangle \subset \langle x,y,wz \rangle -$$ -(and I'm guessing something similar for higher brackets), so I'm not so sure that these are just Toda brackets. -To summarise my questions: - -1.) Is the operation constructed a Toda bracket? (I guess in the sense of Kochman if there is some ambiguity as to what a higher Toda bracket is) -2.) If not, do the obstructions to splitting sequences as above lie in the genuine Toda brackets? I'll be working with sequences of spectra, rather than spaces if it makes a difference - -I would also appreciate if there are any obvious references that I'm missing on this, as the paper by Spanier is the only place I've seen this discussed. - -REPLY [3 votes]: Just to close this off - thanks to Mike-Doherty it appears that the answer is yes (and in fact for spaces this goes back to the paper "The decomposition of stable homotopy" by Joel Cohen.) -Using the terminology of Shipley's paper, given a sequence of maps -$$ -A_{n-1} \xrightarrow{f_{n-1}} A_{n-2} \xrightarrow{f_{n-2}} \cdots \xrightarrow{f_1} A_0 -$$ -then there is an object $X \in \{ f_1,\ldots,f_{n-1} \}$ if and only if $0 \in \langle f_1,\ldots,f_{n-1} \rangle$. The statement that $X \in \{ f_1,\ldots,f_{n-1} \}$ basically corresponds to the existence of the following commutative diagram: -$$ -\begin{array}{cccccccccc} -&&&& \Sigma^{-n+1}F_{n-1}X &&&& \Sigma^{-1} F_1X \\ -&&&\nearrow & \downarrow &&&& \downarrow\\ -\Sigma^{-n} X & \to & A_{n-1} & \xrightarrow{f_{n-1}} & A_2 & \xrightarrow{f_{n-2}} & \cdots & \xrightarrow{} & A_{1} \xrightarrow{f_1} & A_0 -\end{array} -$$ -It is easy to see that the 3-fold bracket is just the usual Toda bracket. In addition the bracket constructed by Kochman is contained in the bracket constructed by Cohen.<|endoftext|> -TITLE: closed subspaces of locally convex inductive limits -QUESTION [7 upvotes]: It's a duplicate of this question, since I really want to get an explanation. -Let $\left(V_{n},\phi_{n,n+1}\right)_{n\in\mathbb{N}}$ be an inductive sequence of LCTV spaces. A locally convex inductive limit of $\left(V_{n},\phi_{n,n+1}\right)_{n\in\mathbb{N}}$ is it's algebraic inductive limit $V=\lim_{\to}V_{n}$, equipped with final locally convex topology w.r.t. $\phi_{n}:V_{n}\to V$. -$V$ can also be described as a quotient of the locally convex direct sum $\bigoplus V_{n}$ by the subspace $D$ , generated by vectors $\left\{ \left(\dots,v_{n},-\phi_{n,n+1}\left(v_{n}\right),\dots\right)|v_{n}\in V_{n}\right\} $. We denote by $\pi:\bigoplus V_{n}\to V\simeq\bigoplus V_{n}/D$ the quotient map. The universal property of V follows from the universal properties of $\bigoplus V_{n}$ and of the quotient. -A map $f:V\to U$ in LCTV is continuous iff maps $f\circ\phi_{n}$ are continuous. Taking $f=\mathrm{id}_V$ one get that if subspace $F\subset V$ is closed then $F_n=\phi_{n}^{-1}\left(F\right)$ is closed. Clearly $F\cong\lim_\to F_n$ as linear space. -The question I am trying to understand is the following: if $F$ is a closed subspace, why it does not have to be the locally convex inductive limit of $F_n$ (why topologies can be different)? -We clearly have $\pi^{-1}(F)=\bigoplus F_n$, which is a closed subspace of $\bigoplus V_n$. Doesn't this force the quotient topology on $\pi^{-1}(F)/\pi^{-1}(F)\cap D$ (which is the locally convex inductive limit topology) be the subspace topology of $F$? -Would it matter if $V_n$ are Banach spaces? - -REPLY [10 votes]: Here is my favorite example which is very natural and simple: Let $\mathscr E (\Omega)$ be the Frechet space of smooth functions on an open set in $\Omega\subseteq\mathbb R^d$ so that it dual $\mathscr E'(\Omega)$ is the space of compactly supported distributions. This is a nice inductive limit of Banach spaces. Hence $V= \mathscr E(\Omega)\times \mathscr E'(\Omega)$ is an inductive limit of Frechet spaces. Now, the "diagonal" $F$ is closed and algebraically isomorphic to the space of test functions $\mathscr D(\Omega)$ but the subspace topology of $V$ on $F$ is much coarser than the topology of $\mathscr D(\Omega)$: the dual of the former is the space of distributions with compact singular support whereas the dual of the latter is the space of all distributions.<|endoftext|> -TITLE: Infinite finitely generated non-amenable groups -QUESTION [10 upvotes]: I am interested if there is an example of an infinite finitely generated non-amenable group that is residually finite but does not contain non-abelian free subgroups. -What examples of infinite finitely gnerated perfect (non-simple) gropus are non-amenable but do not contain free subgroups? -Many thanks, -Elisabeth - -REPLY [9 votes]: Osin and Luck give examples of infinite finitely generated residually finite torsion groups with positive first $\ell^2$-Betti number (and hence non-amenable) in the paper "Approximating the first L2-Betti number of residually finite groups", J. Topol. Anal. 3 (2011), no. 2, 153–160. -http://www.worldscientific.com/doi/abs/10.1142/S1793525311000532<|endoftext|> -TITLE: Isometry group of pseudo Riemannian manifold always a Lie group? (Myers-Steenrod) -QUESTION [9 upvotes]: Myers-Steenrod states that the isometry group of a Riemannian manifold is a Lie group. Is that also true for pseudo Riemannian manifolds? I didn't find anything related to that. -Cheers - -REPLY [12 votes]: Yes. Check out Kobayashi, Transformation Groups in Differential Geometry, theorem 4.1 page 16, and example 2.5 page 8. The automorphisms of a pseudo-Riemannian manifold form a Lie group, as do the automorphisms of a conformal pseudo-Riemannian manifold (in dimension 3 or more), and the automorphisms of a projective connection. -The basic idea of the proof is to show that the bundle of orthonormal frames has a canonical basis of global 1-forms (expressed in terms of the Levi-Civita connection of the pseudo-Riemannian metric). Then you show that no diffeomorphism of a manifold can fix a point and also fix a basis of global 1-forms. You use this to show that the automorphism group actually immerses into the bundle of orthonormal frames, by taking frame $\phi$ and mapping each automorphism $g$ to $g\phi$. The automorphism group orbit equals the orthonormal bundle precisely if the manifold is a homogeneous pseudo-Riemannian space form.<|endoftext|> -TITLE: When are all continuous self-maps of a topological spaces generated by retractions and self-homeomorphisms of prime order? -QUESTION [5 upvotes]: This is probably a very basic question, but I don't know the answer and I also don't see how it might be obvious (which it very well might be). -Given a topological space $X$, when is the set of all its continuous self-maps generated by the subset consisting of - -the self-homeomorphisms of prime order (i.e., there exists some prime number $p$ such that $f^p = id$ while $f^k \neq id$ for all $k \leq p$) and -the continuous retractions (i.e. $f^2 = f$)? - -Is the answer to this a well-known fact? Is there a characterization of the cases where it is true? -I should add that, by generation, I mean everything you can get by applying composition finitely many times. - -Edit: I originally also asked whether the statement is true or not. A counterexample was quickly presented by James Cranch (see below), so I have put more emphasis on a different part of my question, namely if there is a characterization of the cases where the statement is true. - -REPLY [4 votes]: Here is s a different reason why the property is rare: -Consider a topological space $X$ with a continuous surjection $f \colon X \rightarrow X$ that is not a homeomorphism. Assume $f = g_n \circ \ldots \circ g_1$, where each $g_i$ is a -homeomorphism or a continuous retraction. -Since $f$ is surjective, $g_n$ must be surjective and is hence not a nontrivial retraction. It follows that it must be a homeomorphism. We obtain -$g_n^{-1} \circ f = g_{n-1} \circ \ldots \circ g_1$, and we can repeat the arguments to -conclude each $g_i$ is a homeomorphism. Thus $g_n \circ \ldots \circ g_1$ is a homeomorphism, -whereas $f$ is not. Contradiction. -Thus, the statement fails whenever the space has a continuous surjective selfmap that is not a homeomorphism. Thus, it even fails for discrete spaces of infinite cardinality.<|endoftext|> -TITLE: Generalized quasi-perfect numbers -QUESTION [7 upvotes]: A number $n \in \mathbb{N}$ is called quasi-perfect if $\sigma(n) = 2n+1$, where $\sigma$ is the sum of divisors function. It is known that if $n$ is quasi-perfect, then it must be the square of an odd integer (an oddly beautiful Putnam problem from the 1970s). To date no number is known to be quasi-perfect. -My question concerns the existence of 'generalized' quasi-perfect numbers, or rather, let $a,b \in \mathbb{N} \cup \{0\}$ be fixed integers, and call an integer $n$ $(a,b)$ quasi-perfect if it satisfies $\sigma(n) = an+b$. -Are there any known values of $a,b$ for which the number of $(a,b)$ quasi-perfect numbers are known to be infinite? - -REPLY [4 votes]: See the abstract of my PhD thesis, "Generalised quasiperfect numbers", Bulletin Australian Math. Soc., 27 (1983), 153-156, where I consider numbers $n$ with $\sigma(n) = 2n + k^2$, $k$ odd, $(n,k)=1$. -Graeme Cohen<|endoftext|> -TITLE: Sheaf cohomology with compact supports (and Verdier duality?) -QUESTION [8 upvotes]: Consider a manifold and a complex where cochains are sections of vector bundles and coboundary maps are differential operators, which are locally exact except in lowest degree (think de Rham complex). I'd like to know the relationship between the cohomology of this complex and the cohomology of the formal adjoint complex with compact supports (for the de Rham complex, this is again the de Rham complex, but with compact supports, and the relationship is given by Poincaré duality). - -Update: Just added a bounty to raise the question's profile. The biggest obstacle, as came out of the discussion on an unsuccessful previous answer, to a straightforward application of Verdier duality is that it's hard to see how to connect the dual sheaf $\mathcal{V}^\vee$ with the sections of the dual density vector bundles $\Gamma(\tilde{V}^{\bullet*})$. The basic construction of $\mathcal{V}^\vee$ requires, for an open $U\subset M$, the assignment $U\mapsto \mathrm{Hom}_\mathbb{Z}(\Gamma(U,V^\bullet),\mathbb{Z})$, where $\mathrm{Hom}_\mathbb{Z}$ is taken in the category of abelian groups, which is MUCH bigger than $\Gamma(U,V^{\bullet*})$ itself. - -Let me be more explicit, which unfortunately requires some notation. Let $M$ be the manifold, $V^i\to M$ be the vector bundles (non-zero for only finitely many $i$) and $d^i \colon \Gamma(V^i) \to \Gamma(V^{i+1})$ be the coboundary maps. Then $H^i(\Gamma(V^\bullet),d^\bullet) = \ker d^i/\operatorname{im} d_{i-1}$. By local exactness I mean that for every point $x\in M$ there exists an open neighborhood $U_x$ such that $H^i(\Gamma(V^\bullet|_{U_x}), d^\bullet) = 0$ for all except the smallest non-trivial $i$. Now, for each vector bundle $V\to M$, I can define a densitized dual bundle $\tilde{V}^* = V^*\otimes_M \Lambda^{\dim M} T^*M$, which is just the dual bundle $V^*$ tensored with the bundle of volume forms (aka densities). For any differential operator $d\colon \Gamma(V) \to \Gamma(W)$ between vector bundles $V$ and $W$ over $M$, I can define its formal adjoint $d^*\colon \Gamma(\tilde{W}^*) \to \Gamma(\tilde{V}^*)$, locally, by using integration by parts in local coordinates or, globally, by requiring that there exist a bidifferential operator $g$ such that $w\cdot d[v] - d^*[w]\cdot v = \mathrm{d} g[w,v]$. Thus, the formal adjoint complex is defined by the coboundary maps $d^{i*}\colon \Gamma(\tilde{V}^{(i+1)*}) \to \Gamma(\tilde{V}^{i*})$. -There is a natural, non-degenerate, bilinear pairing $\langle u, v \rangle = \int_M u\cdot v$ for $v\in \Gamma(V)$ and $u\in \Gamma_c(\tilde{V}^*)$, where subcript $c$ refers to compactly supported sections. Because $\langle u^{i+1}, d^i v^i \rangle = \langle d^{i*} u^{i+1}, v^i \rangle$ this paring descends to a bilinear pairing in cohomology -$$ - \langle-,-\rangle\colon H^i(\Gamma_c(\tilde{V}^{\bullet*}),d^{(\bullet-1)*}) \times - H^i(\Gamma(V^\bullet),d^i) \to \mathbb{R} . -$$ - -Finally, my question can be boiled down to the following: is this pairing non-degenerate (and if not what is its rank)? - -As I mentioned in my first paragraph, the case $V^i = \Lambda^i T^*M$ with $d^i$ the de Rham differential is well known. Its formal adjoint complex is isomorphic to the de Rham complex itself. Essentially, Poincaré duality states that the natural pairing in cohomology is non-degenerate. I am hoping that a more general result can be deduced from Verdier duality applied to the sheaf $\mathcal{V}$ resolved by the complex $(\Gamma(V^\bullet),d^\bullet)$. I know that the sheaf cohomology $H^i(M,\mathcal{V})$ can be identified with $H^i(\Gamma(V^\bullet),d^\bullet)$. I also know that the abstract form of the duality states that the algebraic dual $H^i(M,\mathcal{V})^*$ is given by the sheaf cohomology with compact supports $H^i_c(M,\mathcal{V}^\vee)$ with coefficients in the "dualizing sheaf" $\mathcal{V}^\vee$. Unfortunately, I'm having trouble extracting the relationship between $\mathcal{V}^\vee$ and my formal adjoint complex $(\Gamma_c(\tilde{V}^{\bullet*}), d^{(\bullet-1)*})$ from standard references (e.g., the books of Iversen or Kashiwara and Schapira). - -REPLY [2 votes]: You will run into some issues with differential equations with singularities. -Consider the differential operator $x\frac{d}{dx} -t $ from the trivial rank $1$ vector bundle on $\mathbb R$ to itself, for some constant $t$. The adjoint map is $-xd/dx -1 - t$, which is another operator in the same class. -If $t$ is not a nonnegative integer, then this complex is locally exact. We have to check that the differential equation $xdy/dx - ty =f(x)$ has solutions for smooth $f$. We will check this for $t<0$, but I think it is also true when $t$ is not a nonnegative integer. A solution is: -$$y = \frac{ f(0) }{t} + x^t \int_{0}^x \left(f(z)-f(0)\right) z^{-1-t} dz $$ -This gives a global solution also, so it is globally exact. Moreover, since any solution to the differential equation is a multiple of $x^{t}$, if $t$ is not a nonnegative integer, there are no nonzero solutions, so the regular cohomology is trivial. -If $t$ is a negative integer, then the dual complex will have nontrivial $H^0$ due to the nonzero solutions, otherwise, say for $-1 -TITLE: Is this a metric on the Grassmannian Manifold? -QUESTION [10 upvotes]: Let $m>n$ and consider the Set -$$S_{m,n}=\{A \in \mathbb{R}^{m \times n}\lvert A^TA=I_n \}.$$ -Does the function $d\colon S_{m,n} \times S_{m,n} \rightarrow \mathbb{R}$ defined by -$$d(A,B)=\sqrt{1-\det(A^TB)}$$ -define a pseudometric on $S_{m,n}$? (A pseudometric satisfies all conditions of a metric except that two elements can also have distance zero.) -Consider the equivalence relation $A \sim B$ if there exist an orthogonal $Q \in \mathbb{R}^{n \times n}$ with $A=BQ$. The set $S_{m,n}$ together with the equivalence relation can be identified with the grassmannian manifold $Gr(n,\mathbb{R}^m)$. Does $d$ define a metric on $Gr(n,\mathbb{R}^m)$? This question interests me because im trying to approximate (interpolate) functions which take values in the grassmanian manifold and the above metric would open up a possibility for approximating such functions. -The difficult part is the triangle-inequality, i.e. for all $A,B,C \in S_{m,n}$ we need to prove that -$$\sqrt{1-\det(A^TC)}\leq \sqrt{1-\det(A^TB)}+\sqrt{1-\det(B^TC)}.$$ -Thanks for any help in advance. - -REPLY [9 votes]: EDIT Actually, Cauchy-Binet suffices as the OP notices in the comments. I'll leave my overkill proof here for your amusement. - -The proof below appeals to a famous result of Schoenberg (I've simplified the statement a bit), and basic linear algebra. - -Schoenberg's theorem (see e.g., [Prop. 3.2, 1]). Let $X$ be a nonempty set and $\psi: X \times X \mapsto \mathbb{R}$ be positive definite kernel. Then, there exists an RKHS $H$ and a map $\varphi : X \to H$ such that - \begin{equation*} - \|\varphi(x)-\varphi(y)\|_H^2 = \frac{1}{2}[\psi(x,x)+\psi(y,y)] - \psi(x,y). -\end{equation*} - -We show that the function $\psi(A,B) = \det(A^TB)$ is positive definite, which as a result of Schoenberg's theorem shows that -\begin{equation*} - 1-\det(A^TB) = \|\varphi(A)-\varphi(B)\|_H^2, -\end{equation*} -from which the triangle inequality is immediate. -To prove the positive definiteness of $\psi$, we show that it is an inner-product by invoking the Cauchy-Binet formula (using Wikipedia's notation, except that for us $A$ is $m \times n$): -\begin{equation*} - \det(A^TB) = \sum_{S \in \binom{[m]}{n}} \det(A^T_{[n],S})\det(B_{S,[n]}) = \sum_{S \in \binom{[m]}{n}} \det(A_{S,[n]})\det(B_{S,[n]}) = \langle \phi(A), \phi(B)\rangle. -\end{equation*} -[1] C. Berg, J. P. R. Christensen, and P. Ressel. Harmonic Analysis on Semigroups: Theory of Positive Definite and Related Functions, Springer GTM 100, 1984.<|endoftext|> -TITLE: A question on some computation of group cohomologies -QUESTION [11 upvotes]: Let $G=H\times J$, where $H\cong J\cong C_2$ (cyclic group of order 2). Let $M \cong \mathbb{Z}$ be a $G$-module via "trivial $H$-action and negation $J$-action". My question is "What are the group cohomologies $H^*(G,M)$?" -I tried to compute them via the Hochschild-Serre spectral sequence $E_2^{p,q}=H^p(J,H^q(H,M))$. The computation of each term is easy, but I don't have any information about arrows and cannot determine $H^*(G,M)$. -More generally, let $G=C_k\rtimes C_2$ be a dihedral group and $M \cong \mathbb{Z}$ a $G$-module via "trivial $C_k$-action and negation $C_2$-action". Can one compute the group cohomologies $H^*(G,M)$? -I would appreciate it if you could provide an explicit computation, or a reference where I can find one. Thank you very much in advance. - -REPLY [6 votes]: See Kuenneth-formula for group cohomology with nontrivial action on the coefficient -Let $C=C_2$ be the cyclic group of order two, $\def\ZZ{\mathbb Z}\ZZ$ the trivial module over $C$ and $S$ the $C$-module which is $\ZZ$ as an abelian group with with the non-trivial action of $C$. You want to compute $H^\bullet(C\times C,M)$ with $M=\ZZ\otimes S$, and the Künneth formula tells us that there is a short exact sequence $$0\to (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^p\to H^p(C\times C,M)\to (Tor^\ZZ_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p+1}\to0$$ -You can easily compute $H^\bullet(C,-)$ with coefficients in every $C$-module and the Tor is also easy to compute. If I am not making too many mistakes, then \begin{gather} -(H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=0 -\end{gather} -for all even $p\geq0$, and -\begin{gather} -(H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^r,\\ -(Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^{r-1} -\end{gather} -for all odd $p=2r-1\geq0$. If follows that in the short exact sequenc above exactly one of the first and third terms are not zero for a given $p$, so we get an isomorphism in all cases. - -For the crossed product $G=C_k\rtimes C_2$ with coefficients $S=\mathbb Z$ with $C_k$ acting trivially and $C_2$ changing sizes. Notice that if $k$ is odd, then the H-L-S spectral sequence has second page $H^p(C_2,H^q(C_k,S))$, and this is zero except when $q=0$ and $p$ is odd, so this gives us the desired result in this case.<|endoftext|> -TITLE: fixed point property for maps of compacts -QUESTION [21 upvotes]: Definition. A topological space $X$ has the Fixed Point Property (FPP) if every continuous self-map $X\to X$ has a fixed point. -Question. If $X$ and $Y$ are homotopy-equivalent compact metrizable spaces and $X$ has the FPP, does it follow that $Y$ also has FPP? Another way to put it: Can one force a fixed point for a self-map of a compact by a "non-homological" argument? -I do not know an answer to this even for finite simplicial complexes, but my primary interest is in locally connected finite-dimensional compacts. - -REPLY [10 votes]: If $X$ is a connected polyhedron, a point $x\in X$ is said to be a global separating point if $X-\{x\}$ is not connected. -Theorem ([1], Theorem 7.1) -In the category of compact connected polyhedra without global separating points, the fixed point property is a homotopy type invariant. -The example by Lopez mentioned in Vidit Nanda's answer shows that the hypothesis about global separating points is fundamental. This theorem is proved using Nielsen theory, which takes into account not only homology but also the fundamental groupoid. -Let $f:X \to X$ be a self map of a compact connected polyhedron. Two fixed points $x_0$, $x_1$ of $f$ are said to be equivalent if there is a path $c$ from $x_0$ to $x_1$ such that $c$ and $f\circ c$ are homotopic. A fixed point class of $f$ is an equivalence class of this relation (there is another definition using the universal cover of $X$ that takes into account also many empty fixed point classes). Each fixed point class has a number associated to it, its index, that measures the number of fixed points in that class. The sum of these indices, taken over the set of fixed point classes is the Lefschetz number of $f$ (Lefschetz-Hopf Theorem). The Nielsen number of $f$, $N(f)$ is the number of fixed point classes having nonzero fixed point index. The number $N(f)$ is a homotopy invariant of $f$ and therefore is a lower bound for the number of fixed points of any map homotopic to $f$. With some local hypotheses (see [1], Main Theorem) there is a map $g$ homotopic to $f$ having exactly $N(f)$ fixed points. To prove this, we can assume that $f$ has isolated fixed points (this goes back to Hopf). The local hypotheses are then used to combine two equivalent fixed points. Some references for this beautiful subject are [2] and [3]. -[1] B. J. Jiang. On the least number of fixed points. Amer. J. Math., 102 (1980), 749-763. -[2] B. J. Jiang. A primer of Nielsen fixed point theory. Handbook of topological fixed point theory, 617--645, Springer, Dordrecht, 2005. -[3] B. J. Jiang. Lectures on Nielsen fixed point theory. Contemporary Mathematics, 14. American Mathematical Society, Providence, R.I., 1983. vii+110 pp.<|endoftext|> -TITLE: Intuition of Kolmogorov-Sinai entropy -QUESTION [6 upvotes]: For a measurable entropy of measurable transformation $T$ from $(X,\mathcal{B},m)$ to itself. -For each finite measurable partition $\mathcal{A}=\{A_i\}_{i=1}^{m}$ of $X$, we can define -$h(\mathcal{A},T,m)$ as $\lim_{n\rightarrow \infty}\frac{1}{n}H(\bigvee_{i=0}^{n-1}T^{-i}\mathcal{A})$. -and measurable h(T,m) is to take sup of all possible finite measurable partition. -It is subtle as least to me to understand $h(T^{n},m)=n h(T,m) (n>0)$. Of course I can prove it almost by definition. However I can not feel the essence of measurable entropy. -For topological entropy, we can understood $h(T^n)=nh(T)$ very well using Bowen balls related to numbers of equivalence of (cut-off) orbits. -I wondered whether there exist similar explanation for measurable entropy. Any reference and commnents will be greatly appreciated. - -REPLY [9 votes]: A good way to understand measurable entropy is via the Shannon-McMillan-Breiman Theorem. Roughly speaking it says that there is a constant $c$ so that most atoms $A$ in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$ have measure $m(A)\approx e^{-cn}$, and the value of $c$ is the measure entropy $h(\mathcal{A},T,m)$. More precisely, given $\epsilon>0$, for all sufficiently large $n$ there is a collection of atoms in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$ whose union has measure a least $1-\epsilon$, and such that each atom in this collection has measure between $e^{-(c-\epsilon)n}$ and $e^{-(c+\epsilon)n}$. Thus measure entropy tells the exponential decay rate of the measure of atoms in $\bigvee_{i=0}^{n-1} T^{-i}\mathcal{A}$. From this it should be clear that $h(T^n,m)=nh(T,m)$. -This interpretation is the starting point for proofs of fundamental theorems such as Ornstein's result that Bernoulli shifts with the same entropy are measurably isomorphic. With good control of the exponential sizes of atoms, a partial isomorphism can be built using Rohlin towers, and then successively refined with combinatorial input from the Marriage Lemma to converge to an isomorphism. A lucid account of this proof, starting from the basics, is in Paul Shields' book "The Theory of Bernoulli Shifts", now available for free and easy to find via web search.<|endoftext|> -TITLE: Diffeomorphisms of the sphere conjugate to a rotation -QUESTION [9 upvotes]: What are sufficient condition on a given diffeomorphism of the sphere (say, given explicitly with formulas) that can ensure that it is conjugate to a rotation, in the group of diffeomorphism of the sphere? -Also, if I take a given rotation of some fixed angle, is there a neigbourhood of it which consists only of elements conjugate to a rotation? - -REPLY [8 votes]: An irational rotation $f$ is arbitrarily distorted in $\text{Diff}^{\infty}S^2$, this was proved by Calegari and Freedman http://arxiv.org/pdf/math/0509701.pdf and in a more general setting by Militon, see http://arxiv.org/abs/1005.1765. This means that for any diverging sequence $g(n)$(diverging as slow as one wants), there is a subsequence $f^{n_i}$ and a finite set $S$ in $Diff(S^2)$ generating a group $G = $ such that $\{f^{n_i}\} \in G$ for all $i$ and that $l_S(f^{n_i}) \leq g(n_i)$. ($l_S(g)$ is the word length of $g$ in the generating set $S$). To make it a little bit more clear, let's consider the following example: Let's take the translation $a: x \to x+1$ in $Diff(R)$. If we consider the subgroup $a:x \to x+1$, and $b : x \to 2x$ in $Diff(\mathbb{R})$, observe that $b^{n}ab^{-n} = a^{2^n}$, this implies that $a$ is at least exponentially distorted in $Diff(\mathbb{R})$, because we can write $a^{2^n}$ as a product of $O(n)$ elements in $a$ and $b$. -One condition that ensures that a given $f: S^2 \to S^2$ is conjugate to a rotation is that the sequence of derivatives of powers of $f$, $(Df^k)_{k \in \mathbb{z}}$, is a bounded sequence. The idea is that in that case, one can take an arbitrary riemannian metric $g$, take and averaging sequence $g_n = \frac{1}{n}\sum_{i=1}^n ({f}^n)^*g$, and observe that the bounds in derivatives imply that $g_n$ has a subsequence converging to a metric $g_{av}$ invariant by $f$. If $g_{av}$ is smooth, by the uniformization theorem, $g$ is conformally equivalent to the standard $S^2$, where $f$ is acting by rotations. -Using these two ideas, one can get a group theoretical way of characterizing any $f: S^2 \to S^2$ that is conjugate to a rotation: $f$ is conjugate to a rotation if and only if for any subsequence $(f^{n_i})_i$ of powers of $f$ there exist a subsequence $m_k$ of $n_i$ and a finite set $S$ of diffeomorphisms generating a group $G = $ such that $\{f^{m_k}\} \subset G$ and $l_S(f^{m_k}) \leq k$. For more details, see the introduction of my paper http://arxiv.org/abs/1307.4447. -As rpotrie mentioned, it might be of interest to you some diffeomorphisms called "pseudorotations", that can be defined as diffeomorphisms of $S^2$ having exactly two periodic points. Birkhoff conjectured that any real analytic measure preserving pseudorotations is an actual rotation. There are many counterexamples in the $C^{\infty}$ category, in the case that the rotation angle is not diophantine, all of them as far as I know constructed by the same method. See http://www.math.psu.edu/katok_a/pub/Herman-survey-horo-corrected.pdf for a survey on this kind of constructions. There is a very positive result known as Herman's last geometric theorem in the case that $\alpha$ is diophantine. See http://www.math.jussieu.fr/~bassam/herman_ann_ens_second_revision.pdf. Using this method one can also construct diffeomoprhisms of $S^2$ that are arbitrarily distorted but not conjugate to rotations.<|endoftext|> -TITLE: On the local structure of Deligne-Mumford stacks -QUESTION [6 upvotes]: Is it true that for any DM stack $\mathcal{X}$ (quasicompact, separated, of finite type over a field) there is a Zariski covering $\mathcal{U}_i \to \mathcal{X}$ by open substacks, such that for all $i$ there is an etale finite surjective morphism $\phi_i: Z_i \to \mathcal{U}_i$, where $Z_i$ is a scheme? (Of course it will be true if we remove the condition that the morphisms $\phi_i$ must be finite.) -Upd. Oh, I think by Theorem 6.1 of Laumon, Moret-Bailly "Champs algebriques" it is equivalent to saying that $\mathcal{X}$ admits a Zariski covering by quotient stacks. - -REPLY [2 votes]: Yes! A nice reference for this is Lurie's new book Theorem 1.2.5.9. where he began (more or less) with what you said as a definition.<|endoftext|> -TITLE: A property that forces the NORM to be induced by an INNER PRODUCT -QUESTION [16 upvotes]: Let $(E, \|\cdot\|)$ be a real normed vector space such that for any $a,b\in E$, -$$ \|x +y\|^2 + \|x-y\|^2 \geq 4 \|x\|\cdot \|y\| $$ -I want to show that the norm is induced by an inner product. Any suggestion or references would be helpful. - -REPLY [23 votes]: If $E$ is to be a Hilbert space, a proof must establish more or less directly that the inequality implies the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 2\lVert x\rVert^2 + 2\lVert y\rVert^2$ for all $x,y \in E$. Since both, your hypothesis and the parallelogram law, are conditions on all $2$-dimensional subspaces of $E$, one can assume that $\dim E = 2$. -I. J. Schoenberg proves the following stronger result as Theorem 2 in A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961-964: - -If the inequality $\lVert x + y \rVert^2 + \lVert x-y\rVert^2 \geq 4$ holds for all unit vectors $x,y \in E$ then $E$ is an inner-product space. - -After reducing to the case $\dim E = 2$, the proof proceeds by showing that the inequality forces the curve $\Gamma$ described by $\lVert x \rVert = 1$ to be an ellipse. To achieve this, Schoenberg shows that $\Gamma$ must coincide with the boundary of the John ellipse of the unit ball by a nice geometric argument. The proof is then completed by appealing to Day's theorem that the parallelogram law $\lVert x + y\rVert^2 + \lVert x-y\rVert^2 = 4$ for unit vectors characterizes inner-product spaces, see Theorem 2.1 in Some characterizations of inner-product spaces, Trans. Amer. Math. Soc. 62 (1947), 320-337. - -Added: Your hypothesis appears as condition $(7)$ in Schoenberg's article. It is pointed out in footnote 5 that the parallelogram law and your condition are equivalent.<|endoftext|> -TITLE: Image of abelian varieties -QUESTION [6 upvotes]: Let $k$ be an arbitrary field, and let $\varphi:A\to B$ be a morphism of abelian varieties over $k$. -If $k$ has characteristic zero, then $\varphi(A)$ has the structure of an abelian subvariety of $B$ which is defined over $k$. -Question: For an arbitrary field $k$, has $\varphi(A)$ the structure of an abelian subvariety of $B$ which is defined over $k$? - -REPLY [10 votes]: In general, if $f:G \rightarrow H$ is any homomorphism between smooth group schemes of finite type over a field $k$, the image $f(G)$ is always a smooth closed $k$-subgroup of $H$. (This is a special case of general results in SGA3, but it seems more instructive to give the direct argument in this case rather than wade through SGA3 for this purpose. The argument below has all of the same ideas as in the affine case treated in every book on linear algebraic groups, except that the argument is given in geometric terms without reference to coordinate rings so that it works without an affineness hypothesis. In the argument below we work with schemes throughout.) -Give the closure $Z$ of $f(G)$ the "schematic image" structure; this corresponds to the kernel of $O_H \rightarrow f_{\ast}(O_G)$. We want to show that $Z$ is a smooth closed $k$-subgroup of $H$ and $f:G \rightarrow Z$ is surjective. That is, we want (i) $Z$ is equi-dimensional with the coherent $\Omega^1_{Z/k}$ flat over $O_Z$ of constant rank equal to $\dim(Z)$, (ii) $m:H \times H \rightarrow H$ carries $Z \times Z$ into $Z$, (iii) $f:G \rightarrow Z$ is surjective. (It is obvious that $Z$ contains the identity point and is carried into itself under inversion since $f$ is a continuous map that is a $k$-homomorphism, so for the group aspect it is multiplication that we may focus upon.) -Since $f(G)$ is a constructible set whose formation commutes with extension of the ground field (in the sense that for an extension field $K/k$, the preimage of $f(G)$ under $H_K \rightarrow H$ is $f_K(G_K)$) and the formation of "schematic image" of $f$ commutes with extension of the ground field too (as the formation of $f_{\ast}(O_G)$ commutes with flat base change, due to quasi-compactness of $f$), we may apply extension of the ground field up to an algebraic closure to reduce to the case when $k$ is algebraically closed. In this case since $Z$ inherits reducedness from $G$, it is generically smooth (as $k$ is algebraically closed), and thus $Z$ is smooth once it is a subgroup of $H$ (by using $k$-point translations). -Since $k$ is algebraically closed and $Z$ is reduced (so $Z \times Z$ is reduced), $Z$ is a $k$-subgroup of $H$ provided that $Z(k)$ is a subgroup of $H(k)$ in the ordinary sense. For any $z \in Z(k)$, it suffices to show that left multiplication $\ell_z$ on $H$ carries $Z$ into itself. But $Z$ is the closure of $f(G)$, so it suffices to show that $\ell_z(f(G)) \subset Z$. In other words, we want $\ell_z \circ f:G \rightarrow H$ to land inside $Z$. Since $k = \overline{k}$ and $Z$ is closed in $H$, it is equivalent to show that $\ell_z \circ f$ carries $G(k)$ into $Z$. That is, for $g \in G(k)$ we want $z f(g) \in Z$. Thus, we are reduced to checking that for any $g \in G(k)$, right multiplication $\rho_{f(g)}$ carries $Z$ into itself. It suffices to show that it carries $f(G)$ into $Z$, and obviously it even carries $f(G)$ into $f(G)$ (since $f$ is a $k$-homomorphism). -That completes the proof that $Z$ is a smooth closed $k$-subgroup of $H$, and it remains to check (still with $k = \overline{k}$) that $f:G \rightarrow Z$ is surjective. The dense image $Y$ is constructible by Chevalley's theorem, and equality of constructible subsets of a finite type scheme over an algebraically closed field can be checked on rational points. It is easy to check that $Y(k) = f(G(k))$, and this is a subgroup of $Z(k)$. By constructibility, there is a dense open subset $U \subset Z$ contained in $Y$, so $U(k) \cdot U(k) \subset Y(k)$. -Thus, to conclude that $Y = Z$ it suffices to check that if $Z$ is any smooth finite type group over an algebraically closed field $k$ and if $U \subset Z$ is a dense open subset then $m(U \times U) = Z$. By constructibility of $m(U \times U)$, it is equivalent to check that its set of $k$-points $U(k) \cdot U(k)$ exhausts $Z(k)$. For any $z \in Z(k)$, the translate $z U^{-1}$ is a dense open subset of $Z$, so it meets $U$. The non-empty overlap must contain a $k$-point since $k = \overline{k}$, so $zu^{-1} = u'$ for some $u, u' \in U(k)$. Hence, $z = u' u \in U(k) \cdot U(k)$ as desired. -Remark: If you apply generic flatness and translation arguments over $\overline{k}$, it also follows that the natural map $G \rightarrow f(G)$ is flat and so $f(G)$ represents the fppf quotient group sheaf $G/(\ker f)$, thereby making contact with the quotient formulation in Cesnavicius' answer.<|endoftext|> -TITLE: What is the definition of being smooth for a function from a Lie group to a Fréchet space? -QUESTION [5 upvotes]: In representation theory of real groups, one is confronted with the notion of smoothness for functions defined on a Lie group with values in a Fréchet space (e.g. see Wallach's Real Reductive Groups I, 1.1.1). What is the precise definition of this? Presumably, the definition should be given in terms of local coordinates, so I am actually asking what smoothness means for a function defined on an open subset of $\mathbb{R}^n$ with values in a Fréchet space $V$. (If $V$ is a Banach space or $n = 1,$ I have no problem.) Proceeding in the usual manner, at some point one has to see $\mathrm{Hom}(\mathbb{R}^n,V)$ as a topological vector space of the same kind as $V$ in order to be able to define the second derivative and so on. On p. 52 of his Representation Theory of Semisimple Groups, Knapp mentions that there exists a canonical topological vector space structure on $\mathrm{Hom}(\mathbb{R}^n,V).$ What is this canonical structure? Does it turn $\mathrm{Hom}(\mathbb{R}^n,V)$ into a topological vector space of the same kind as $V$? - -REPLY [3 votes]: Since the above answers and comments refer to the concept of differentiability for functions not just with values IN an infinitely dimensional space, but also defined ON one, despite the fact that your question addresses functions defined on a Lie group and so a finite dimensional space and you explicitly state that this reduce to the case of functions on euclidean space by localisation, let me take the liberty of adding some information on this situation. This was considered in some detail in the article "Espaces de fonctions différentiables a valeurs vectorielles", Jour. d'Anal. 4 (1954-56) 88-148 by Laurent Schwartz which is probably still the definitive statement on the subject. The definition is exactly as in the scalar case using limits of difference quotients. Schwartz uses the theory of nuclear spaces and tensor products which had just been invented by -Grothendieck for the somewhat analogous case of holomorphic functions and the fact that the smooth functions form a nuclear space makes the theory somewhat simpler.<|endoftext|> -TITLE: Periods of Twists of Modular Forms -QUESTION [5 upvotes]: Let $f \in S_2(\Gamma_1(N))$ be an eigenform. By a theorem of Shimura, there are associated "periods" $\Omega_f^\pm$ such that, after normalizing by these periods, the L-function associated to $f$ takes algebraic values. -If $\chi$ is a Dirichlet character, one can form the twist $f_\chi$ of $f$. -How are the periods $\Omega^\pm_f$ and $\Omega^\pm_{f_\chi}$ related? In particular, are they algebraic multiples of each other? - -REPLY [7 votes]: By a famous theorem of Manin, one can define $\Omega^{\pm}$ such that $L(f\otimes\chi,j)\in \Omega^{\epsilon}_{f}\mathbb Q$ with $\chi(-1)(-1)^{j}=\epsilon$. So the period depends on $\chi$ only insofar as you need to know $\chi(-1)$ to determine if you should choose $\Omega_f^{+}$ or $\Omega_f^{-}$. -This result is the key property allowing one to construct the $p$-adic $L$-function of a modular form.<|endoftext|> -TITLE: A well-behaved $A$ that is almost contained in every element of some filter for a countable arithmetically closed family $\mathfrak X$ -QUESTION [8 upvotes]: The question has relevance for constructing Scott sets with certain extra desirable properties. -Suppose that $\mathfrak X$ is a countable arithmetically closed family of subsets of $\mathbb N$: whenever $B\in \mathfrak X$ and $C$ is definable in $\langle \mathbb N,+,\cdot, <,B, 0,1\rangle$, then $C\in\mathfrak X$. -Let us fix some $C\notin\mathfrak X$. I would like to find a set $A$ that is almost contained in either $B$ or $B^c$ for every $B\in\mathfrak X$, which has the property that $C$ is not in the arithmetic closure of $\mathfrak X$ with $A$. Thus, closing $\mathfrak X$ and $A$ under definability does not introduce $C$. I am willing to add such an $A$ by forcing (for a ground model set $C$) but I am inclining more and more towards a counterexample. Is there an arithmetically closed family $\mathfrak X$ and a set $C\notin\mathfrak X$ so that an $A$ almost contained in every element of $\mathfrak X$ or its complement always "codes in" $C$? - -REPLY [6 votes]: In my paper, A variant of Mathias forcing that preserves $\mathsf{ACA}_0$ [Archive for Mathematical Logic 51 (2012), 751–780; arXiv:1110.6559, doi:10.1007/s00153-012-0297-4], I show that $F_\sigma$-Mathias forcing preserves $\mathsf{ACA}_0$. When restricted to $\omega$-models the main preservation theorem gives: -Theorem. If $\mathfrak{X}$ is an arithmetically closed Turing ideal and $G$ is $F_\sigma$-Mathias generic over $\mathfrak{X}$, then the Turing ideal generated by $\mathfrak{X}\cup\{G\}$ is also arithmetically closed. -The meaning of "$F_\sigma$-Mathias generic over $\mathfrak{X}$" is that the forcing consists $F_\sigma$-Mathias conditions coded in $\mathfrak{X}$ and that the generic $G$ should meet all of the dense sets that are definable over $\mathfrak{X}$ in the language of second-order arithmetic. (As usual, there is some flexibility in the second requirement.) -To derive the above from Theorem 4.3 of my paper, observe that the names I used are a way of formalizing Turing functionals with built-in oracles from $\mathfrak{X}$. Therefore, the evaluation of a $G$-total name is a total computable function with respect to some oracle $G \oplus X$ where $X \in \mathfrak{X}$, and all such functions can be represented by a $G$-total name in this way. It follows that the evaluation of all $G$-total names, which is how I define the generic extension, is simply the Turing ideal generated by $\mathfrak{X}\cup\{G\}$. -The result you want then follows from the following: -Lemma. For every partial name $F$ coded in $\mathfrak{X}$ the collection of $\mathfrak{X}$-coded conditions $(a,A,\mu)$ such that, for some $x$, $(a,A,\mu)$ forces that $F(x) \neq C(x)$ is dense. -Indeed, we may assume that $(a,A,\mu)$ forces that $F$ is total, otherwise some extension forces that $F(x)$ is undefined for some $x$. Then, if $(a,A,\mu)$ does not decide all values of $F$, then some extension will decide a value $F(x)$ to be different from $C(x)$. Finally, if $(a,A,\mu)$ decides all values of $F$ then the resulting evaluation of $F$ is computable from $F$ and $(a,A,\mu)$. Therefore, $F \neq C$ since $C \notin \mathfrak{X}$. -Note that the above is really a meta-lemma since $F(x) \neq C(x)$ is expressible in $\mathfrak{X}$ for fixed input $x$ but not for all inputs at once. Nevertheless, the lemma gives a family of open dense sets for $F_\sigma$-Mathias forcing over $\mathfrak{X}$ and any generic $G$ that meets all of these dense sets (and all those that are definable from $\mathfrak{X}$) will be as required by the theorem: the arithmetic closure of $\mathfrak{X}\cup\{G\}$ will not contain $C$. -Note that the same result cannot be achieved with ordinary Mathias forcing since Blass has shown that every Mathias generic set computes all hyperarithmetic reals. - -An earlier version of this answer outlined a tweaking of the cone avoiding result at the end of my paper to achieve the same goal. While that solution did work, it introduced some unnecessary complexity since the point of that result is that one can still arrange that the generic $G$ does not compute $C$ even if $C \in \mathfrak{X}$, provided that $C$ is not computable.<|endoftext|> -TITLE: Hodge Theory (Voisin) -QUESTION [8 upvotes]: I have a strong understanding of Representation Theory but am interested in learning from -Voisin, Hodge Theory and Complex Algebraic Geometry. -What are the prerequisites to learning from this textbook? I haven't studied differential geometry in a while but am decently comfortable with the basics of the subject. -Any textbook recommendations for a sort of roadmap for this textbook? - -REPLY [13 votes]: Voisin's book is quite (but not always) self-contained and well-written. For the supplementary references, you may use the first chapters of Griffiths-Harris "principles of algebraic geometry" which can help you to understand and motivate the complex backgrounds of complex Hodge theory. The book by Carlson, Müller-Stach, Peters, "Period mappings and Period domains" is more readable and self-contained than Voisin's book. The book by Bertin, Demailley, Illusie and Peters, "Introduction to Hodge theory" is less famous but a very good reference specially if you are interested in interactions between complex Hodge theory and Hodge theory in characterstic $p$. Note that nowadays, there are a lot of online lecture notes that are simplified and can help you to understand the content of Voisin's book much more easily. For example this or this which is more detailed. By googling you can find even more references. Also, your background of representation theory can help you a lot in Hodge theory, as it constantly appears in studying Hodge theory and in Voisin's book (for example the very important notion of local systems is nothing but studying representations of the fundamental group).<|endoftext|> -TITLE: Rational points on circular spirals -QUESTION [9 upvotes]: Is it the case that every unit-radius circular spiral, -$$x = \cos(t)$$ -$$y = \sin(t)$$ -$$z = c \cdot t$$ -for $c \in \mathbb{R}^+$ -is dense in rational-coordinate points -(i.e., all three coordinates rational)? -Perhaps not? In which case: -For which $c>0$ is the spiral dense in rational points? - -          - - -Related MO question: -Rational points on a sphere in $\mathbb{R}^d$. - -Answered (by Noam Elkies, Mark Sapir, Will Jagy, Robert Israel). Remarkably to me, for no $c \neq 0$ is the spiral dense in rational -points, but for $c=0$, the unit circle, it is dense in rational points. - -REPLY [6 votes]: Suppose $\cos(t) $ and $\sin(t) $ are rational but not in $\{-1,0,1\}$. Since the Gaussian integers are a UFD, we can write $z = \exp(it) = \cos(t) + i \sin(t) = \prod_j p_j^{d_j}$ where $p_j$ are Gaussian primes and $d_j$ are integers, with finitely many (and at least one) $d_j$ nonzero. -Similarly, if $s = mt/n$ for coprime integers $m,n$ and $\cos(s)$ and $\sin(s)$ are rational, write -$w = \exp(is) = u\prod_j p_j^{e_j}$ where $u \in \{\pm 1, \pm i\}$. Now $w^n = z^m$, so $d_j m = e_j n$ for all $j$. In particular, $m/n = e_j/d_j \in {\mathbb Z}/d_j$. We conclude that for all nonzero $c$, there is at most a discrete set of rational points on the spiral. -EDIT: In fact, we don't need the $y$ component: for all nonzero $c$, $\{(ct, \cos(t)): t \in {\mathbb R}\}$ has only a discrete set of rational points. -Note that $\cos(nt) = T_n(\cos(t))$ where $T_n$ is the $n$'th Chebyshev polynomial of the first kind. These satisfy the recursion -$T_n(x) = 2 x T_{n-1}(x) - T_{n-2}(x)$ with $T_0(x) = 1$ and $T_1(x) = x$. -It's easy to prove by induction that for any odd prime $p$, if the $p$-adic -norm $|x|_p = q > 1$, then $|T_n(x)|_p = q^n$, while if $q \le 1$, $|T_n(x)|_p \le 1$. On the other hand, for $p=2$, if $|x|_2 = q > 2$ then $|T_n(x)|_2 = 2 (q/2)^n$ while if $|x|_2 \le 2$ then $|T_n(x)|_2 \le 2$. -Now suppose $\cos(t) = x$ and $\cos(s) = w$ are nonzero and rational and $s/t = m/n$ is rational, where $m$ and $n$ are positive integers. -We have $T_n(w) = \cos(ns) = \cos(mt) = T_m(x)$. Fix $t$ and take $0 < |m/n-1| < \epsilon$ where $\epsilon$ is small. If $\epsilon$ is sufficiently small, there is some prime $p$ such that $|w|_p \ge \max(p,|x|_p^2)$ (if $p \ne 2$) or -$\max(4, |x|_2^2)$ (if $p = 2$). - But then (in the case $p > 2$) -$|T_n(w)|_p = |w|_p^n \ge \max(p^n,|x|_p^{2n}) > |T_m(x)|_p$ -or (in the case $p=2$) -$|T_n(w)|_2 = 2 (|w|_2/2)^n \ge \max(2^{n+1}, 2 (|x|_2^2/2)^n) > |T_m(x)|_2$. -The contradiction shows that $s$ can't be taken too close to $t$.<|endoftext|> -TITLE: Probability that a convex shape contains the unit ball -QUESTION [21 upvotes]: This probability problem seems interesting and I don't know if it has been solved before. -If you pick $n$ points uniformly at random from the surface of a $d$ dimensional sphere of radius $r>1$ with center at the origin, what is the probability that the convex hull of these points contains the unit ball (of dimension $d$) centered at the origin? - -REPLY [5 votes]: I misread the problem as that you just wanted the origin in the convex hull. This is a nice problem whose solution deserves to be better known, and this provides an upper bound for the probability that the convex hull contains the unit ball, and the limiting behavior as $r \to \infty$. -J. G. Wendel; - "A Problem in Geometric Probability," - Mathematica Scandinavica 11 (1962) 109-111. - Zbl 108.31603 -http://www.oeis.org/A008949 -A special case ($4$ points on the $2$-sphere) was A6 on the 1992 Putnam exam. -You say both $d$-dimensional sphere and $d$-dimensional ball, though I would say the unit $3$-dimensional ball is bounded by a $2$-sphere. I'll assume you mean the ball in $\mathbb{R}^d$. -Consider the kernel of the map $\mathbb{R}^n \to \mathbb{R}^{d}$ of linear combinations of the points. Generically, this kernel has dimension $n-d$. The origin is contained in the convex hull precisely when this kernel intersects the positive orthant, since then some linear combination can be rescaled to have total weight $1$. -By symmetry, all orthants are hit equally often. (Negating a point reflects the kernel, and the action this generates is transitive on the orthants.) A generic $a$-dimensional subspace of $\mathbb{R}^{n}$ hits $2\sum_{i=0}^{a-1} {n-1 \choose i}$ out of $2^{n}$ orthants, so the probability that the convex hull of $n$ random points contains the origin is -$$2^{-n+1}\sum_{i=0}^{n-d-1} {n-1\choose i}. $$ -This is an upper bound for the probability that the unit ball is contained in the convex hull. -The origin is contained in the convex hull with probability $1/2$ when the kernel and its orthogonal complement have the same dimension, when $n=2d$.<|endoftext|> -TITLE: A non-hyperfinite type III factor from an action of the free group on the circle -QUESTION [8 upvotes]: We define below a von Neumann algebra $\mathcal{M}$ from an action of the free group on the circle, and we prove that $\mathcal{M}$ is a non-hyperfinite type ${\rm III}$ factor. - -Question : Is $\mathcal{M}$ of type ${\rm III}_{0}$, ${\rm III}_{\lambda}$ or ${\rm III}_{1}$ ? - - -Definition : Let $s, r_{\theta}: \mathbb{R} / \mathbb{Z} \to \mathbb{R} / \mathbb{Z} $, defined by $s( x) = x^{2}$ (choosing representatives in $[0,1[$) and $r_{\theta} (x) = x+\theta$. -Now, identifying $ \mathbb{R} / \mathbb{Z}$ and $\mathbb{S}^{1}$, we define the action $\alpha$ of $\mathbb{F}_{2} = \langle a, b \vert \ \rangle$, generated by $\alpha (a) = s$ and $\alpha (b) = r_{\theta}$ in Homeo($\mathbb{S}^{1}$). -Lemma: If $\theta$ is transcendental, the action $\alpha$ is faithful. -Proof: A relation $s^{n_{1}}r_{\theta}^{m_{1}}...s^{n_{k}}r_{\theta}^{m_{k}} = e $ can be translated into an algebraic equation in $x$ and $\theta$, where $\theta$ has to be a root $\forall x$. Then, if $\theta$ is transcendental, we are sure that there is no relation. $\square$ -Remark: For a fixed transcendental $\theta$, each non-trivial relations can be realized for at most finitely many $x \in \mathbb{R} / \mathbb{Z}$, i.e. roots of the related algebraic equation. -Theorem: $\mathcal{M} = L^{\infty}(\mathbb{S}^{1}, Leb) \rtimes_{\alpha} \mathbb{F}_{2} $ is a non-hyperfinite type III factor. -Proof : The action $\alpha$ of $\mathbb{F}_{2}$ on $\mathbb{S}^{1}$ is: - -(a) Measure class preserving: the set of null measure subspaces is invariant. -(b) Essentially free: a fixed point set for $\gamma \ne e$ is at most finite, so with null measure. -(c) Properly ergodic: ergodicity comes from irrational rotation, next, every $\mathbb{F}_{2}$-orbit have null measure. -(d) Non-amenable (Edit, Aug. 2018): for any $\eta > 0$, there is $n \in \mathbb{Z}_{>0}$ such that $\lfloor n\theta \rfloor < \eta$. Now, for $\lambda=Leb$ and $g = r_{\theta}^n$, $\partial (\lambda g)/\partial \lambda = 1$ because $\lambda$ is $g$-invariant. It follows that the action $\alpha$ is indiscrete, and then by the proposition below, it is non-amenable. -(e) Non equivalent measure preserving: by ergodicity, an equivalent invariant measure $m$ is proportional to $Leb$. Then $m([1/4 , 1/2]) = 2m([1/16 , 1/4])$, and by $\alpha(a)$ invariance, $m([1/4 , 1/2]) = m([1/16 , 1/4])$. In fact, the only invariant measure are $0$ or $\infty$. -(a), (b), (c) give a factor, (d) gives non-hyperfinite, (e) gives a type ${\rm III}$. $\square$ - - -Here are two extracts of the following recent paper (April 2018): -Bartholdi, Laurent. Amenability of groups and $G$-sets. Sequences, groups, and number theory, 433--544, Trends Math., Birkhäuser/Springer, Cham, 2018. - -REPLY [2 votes]: Theorem: $\mathcal{M}$ is of type ${\rm III}_{1}$. -Proof: By Corollaire 3.3.4 below, we need to show that the ratio set of the action $\alpha$ is $[0, \infty)$. By the classification below, it is enough to prove that the interval $[0,2)$ is included in the ratio set. -Consider $r_{\theta}$ and $s$ on $[0,1)$ directly, i.e., $r_{\theta}(x) = \lfloor x+\theta \rfloor$ and $s(x) = x^2$. -Let $r \in [0,2)$, $\epsilon > 0$ and $A \subseteq [0,1)$ be a Borel set with $\mu(A)>0$. -Let $x_0 \in A$ such that for any neighborhood $V$ of $x_0$ in $[0,1)$ we have $\mu(V \cap A)>0$. -Consider integers $n,m$ and let $f:=r^n_{\theta} \circ s \circ r^m_{\theta}$. -Then $f(x) = \lfloor \lfloor x + m\theta \rfloor^2 + n\theta \rfloor$ and $f'(x) = 2\lfloor x + m\theta \rfloor$. Note that $f$ depends on the integers $n$ and $m$, whereas $f'$ depends on $m$ only (and is well-defined up to a finite set). -By ergodicity, we can choose $m$ such that $|2r^m_{\theta}(x_0)-r|<\epsilon$, so that for a sufficiently small neighborhood $V$ of $x_0$, $|f'(x)-r|<\epsilon$ for all $x \in V$. Let $C:=V \cap A$ and $D:=s \circ r^m_{\theta}(C)$. Then $\mu(C)>0$ and $\mu(D)>0$, and by definition of ergodicity, we can choose $n$ such that $\mu(C \cap r^n_{\theta}(D))>0$. -Thus, for $B= f^{-1}(C) \cap C = f^{-1}(C \cap r^n_{\theta}(D))$ we have: - -$\mu(B)>0$, -$B \cup f(B) \subset A$, -$|f'(x)-r|<\epsilon$, for all $x \in B$. - -The result follows. $\square$ -Remark: If we replace $s:x \mapsto x^2$ by $s_n: x \mapsto x^n$ with $n \ge 2$, we generate a von Neumann algebra $\mathcal{M}_n$ which is also a non-hyperfinite ${\rm III}_1$ factor (the proof is similar). Are they isomorphic? -It should work as well if we replace $s$ by any bijection of $[0,1)$ polynomial of degree $\ge 2$. - -An extract of the following paper of Alain Connes (page 88): -Connes, A. Structure theory for Type III factors. Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 2, pp. 87--91. Canad. Math. Congress, Montreal, Que., 1975. - -An extract of the following paper of Alain Connes (page 195): -Connes, Alain. Une classification des facteurs de type ${\rm III}$. (French) Ann. Sci. École Norm. Sup. (4) 6 (1973), 133--252. - - -Conclusion: this action $\alpha$ of $\mathbb{F}_2$ on $\mathbb{S}^1$ generates a non-hyperfinite ${\rm III}_1$ factor $\mathcal{M}$. -Let $\Omega$ be a cyclic-separating vector (i.e. $\mathcal{M}\Omega$ and $\mathcal{M}'\Omega$ dense in $H$) and $\sigma=(\sigma_t^{\Omega})$ be the one-parameter modular group. Then $\widetilde{\mathcal{M}}:=\mathcal{M} \rtimes_{\sigma} \mathbb{R} $ is a ${\rm II}_{\infty}$ factor called the core of $\mathcal{M}$. -It is independent (up to isomorphism) of the choice of $\Omega$, and is isomorphic to $\mathcal{N} \otimes B(H)$, with $\mathcal{N}$ a non-hyperfinite ${\rm II}_1$ factor of fundamental group $\mathbb{R}_{+}^*$. -By Theorem 2.23 below, $$\widetilde{\mathcal{M}} = L^{\infty}(\mathbb{S}^1 \times \mathbb{R}_{+}^*, Leb) \rtimes_{\widetilde{\alpha}} \mathbb{F}_2$$ where the action $\widetilde{\alpha}$ of $\mathbb{F}_2$ on $\mathbb{S}^1 \times \mathbb{R}_{+}^*$ is given by $$\gamma(x,\lambda) = (\gamma(x), [\gamma'(x)]^{-1}\lambda),$$ -with $x \in \mathbb{S}^1$, $\lambda \in \mathbb{R}_{+}^*$ and $\gamma \in \mathbb{F}_2$ identified with $\alpha(\gamma)$ or $\widetilde{\alpha}(\gamma)$ when appropriate. -Remarks and questions: -Note that $\widetilde{\mathcal{M}}$ has a Cartan subalgebra. Can we deduce that $\mathcal{N}$ has also a Cartan subalgebra? - If so, $\mathcal{N}$ cannot be isomorphic to $L(\mathbb{F}_2)$, because this last has no Cartan subalgebra, by this paper of Dan Voiculescu. Anyway, can we deduce that $L(\mathbb{F}_2)$ has also for fundamental group $\mathbb{R}_{+}^*$? -If so, it would solve the free group factors isomorphism problem, according to this paper of Florin Rădulescu. - -Extracts of the following book of Masamichi Takesaki (pages 28 and 17): -Takesaki, M. Theory of operator algebras. III. Encyclopaedia of Mathematical Sciences, 127. Operator Algebras and Non-commutative Geometry, 8. Springer-Verlag, Berlin, 2003. xxii+548 pp. -The result below provides the core $\widetilde{\mathcal{R}}$ of the crossed-product $\mathcal{R}$. - -From Corollary 2.5 below, $\delta(x,sx) = \delta(sx,x)^{-1}$, whereas Lemma 2.4 provides $\delta(sx,x)$.<|endoftext|> -TITLE: Injective simplicial maps between Arc complexes -QUESTION [5 upvotes]: Let $A(S)$ denotes the Arc complex of a finite type hyperbolic surface $S$ with nonempty boundary. Let $\lambda:A(S)\rightarrow A(S)$ be a map such that on triangulations of $S$ i.e. on the top dimensional simplices of $A(S)$ the map $\lambda$ is induced by a homeomorphism $\Phi$ of $S$. -Question: How to show that $\lambda$ is induced by a homeomorphism of $S$ in the whole of $A(S)$? -P.S: In the context I have heard about it the speaker said it followed from "Mosher's connectivity argument." I searched about it but I didn't find any reference or answer. - -REPLY [3 votes]: I'm not as familiar with this area as I should be, but it seems as though the desired result follows from the main theorem of the following paper: -Irmak, McCarthy. Injective simplicial maps of the arc complex, Turk J Math, 34 (2010) , 339 – 354. -Here is their Theorem: - -Let $S$ be a compact, connected, orientable surface of genus $g$ with $b \geq 1$ boundary components. If $\lambda : A(S) \to A(S)$ is an injective simplicial map, then $\lambda$ is induced by a homeomorphism $H : S\to S$. - -If $\lambda$ is induced by a homeomorphism on top simplices of $A(S)$, then it is at least (I hope) injective. I see that Mosher's connectivity argument is mentioned and used in this paper as "Theorem 3.9", so presumably this argument is exactly what you're looking for. -Finally, in case you are hunting for more references: I saw all of this in the lovely monograph by McCarthy and Papadopolous called Simplicial actions of mapping class groups. It is available here.<|endoftext|> -TITLE: Uncountable model of bounded arithmetic with an elementary end extension -QUESTION [9 upvotes]: Theorem 1.53 (3) in page 227 of Hajek and Pudlak's book, Metamathematics of First-Order Arithmetic, says: -Theorem. If $M$ is a countable model of $I\Delta_{0}$ such that $M$ has a proper elementary end extension, then $M\models PA$. -Is the above theorem still true if we drop the countability assumption of $M$? - -REPLY [10 votes]: Having seen the other answers, let me point out yet another quick argument using just Lowenheim-Skolem. -Let $M$ be an uncountable model and $N$ a proper elementary end extension of it. Expand $N$ by adding an extra unary predicate naming the subset $M$. Call the resulting structure $(N,M)$. Take a countable elementary substructure of it, say $(N',M')$. Then $N'$ is a proper elementary end extension of $M'$ and $M'$ is countable, so the theorem applies. Hence $M'$ is a model of PA and so is $M$ since they are elementarily equivalent.<|endoftext|> -TITLE: Abscissa of convergence of Dirichlet series -QUESTION [7 upvotes]: Let $D(s)$ be a Dirichlet series with abscissa of convergence $\sigma_c=\sigma_a$. Does it follow that the Dirichlet series defined by $P(s)=D(s)\bar D(\bar s)$ has the same abscissa of convergence? -In light of Noam D. Elkies answer below, I would like to know about conditions on the coefficients that also lead to the divergence of the square? What if the coefficients are completely multiplicative? - -REPLY [11 votes]: That's not true even for power series with real coefficients. -Let -$$ -D(s) = \sqrt{1-2^{-s}} - = 1 - \frac12 2^{-s} - \frac18 4^{-s} - \frac1{16} 8^{-s} - \frac5{128} 16^{-s} -- \cdots . -$$ -Then $P(s) = 1 - 2^{-s}$, so -$\sigma_c(D) = \sigma_a(D) = 0$ but -$\sigma_c(P) = \sigma_a(P) = -\infty$.<|endoftext|> -TITLE: Is it true that if $M$ is injective then $S^{-1}M$ is also injective -QUESTION [11 upvotes]: Let $R$ be a commutative ring with identity and let $S$ be a multiplicative subset of $R$. Is it true that for any injective $R$-module like $M$, $S^{-1}M$ (as the $S^{-1}R$-module) is also injective ? - -REPLY [13 votes]: When $R$ is noetherian, yes: By Baer's criterion it suffices to prove that the map -$\hom_{S^{-1} R}(S^{-1} R,S^{-1} M) \to \hom_{S^{-1} R}(J,S^{-1} M)$ -is surjective for every ideal $J \subseteq S^{-1} R$. Write $J = S^{-1} I$ for some ideal $I \subseteq R$. Since $I$ (and of course $R$) are of finite presentation, this map is isomorphic to -$S^{-1} \hom_R(R,M) \to S^{-1} \hom_R(I,M),$ -which is surjective since $\hom_R(R,M) \to \hom_R(I,M)$ is surjective. -The same proof also shows the converse: If $M$ is injective locally (either in the sense that all $M_{\mathfrak{p}}$ are injective over $R_{\mathfrak{p}}$, where $\mathfrak{p}$ runs through all prime ideals, or if there is a basic open cover $\mathrm{Spec}(R) = \cup_i D(f_i)$ such that each $M_{f_i}$ is an injective $R_{f_i}$-module), then $M$ is injective. See also here. -You can also see this as a consequence of the fact that the Ext functor, when restricted to f.g. modules over a noetherian ring, commutes with localization.<|endoftext|> -TITLE: Why does $H^i(X_{ét},\mathbb{Q}_p)$ have a Hodge-Tate structure? -QUESTION [18 upvotes]: Let $X$ be a variety over a $p$-adic field $K$. -Is there a simple or intuitive explanation of why the $G_K$ representation $H^i(X_{ét},\mathbb{Q}_p)$ is Hodge-Tate? More precisely, why do the powers of the cyclotomic character appear and not other characters? - -REPLY [2 votes]: You need to assume that your variety is proper and smooth. -The properness implies that the vector space $H^{1}_{et}(X_{\overline{K}}, \mathbb{Q}_p)$ is of finite dimension. And the smoothness implies the following decomposition which is proved by Falting: -$(\mathbb{C}_p \otimes H^{i}(X_{\overline{K}}, \mathbb{Q}_p)=\oplus_{j \in \mathbb{Z}} \mathbb{C}_p(-j) \otimes H^{i-j}(X,\Omega^{j}_{X})$<|endoftext|> -TITLE: Dualizing sheaf of an associated bundle -QUESTION [8 upvotes]: Let $G$ be a reductive algebraic group defined over an algebraically closed field $k$ of characteristic p, let assume p is good prime for simplicity. Fix $B$ a Borel subgroup of $G$. Then for every $B$-variety $X$, we can define an associated bundle $G\times^B X$. Suppose $X$ has the dualizing sheaf $\omega_X$. My question is whether there exists a formula for computing dualizing sheaf of $G\times^B X$ in terms of $\omega_X$. -Any reference is appreciated. Thanks in advance! - -REPLY [6 votes]: Yes, there is a formula for this due to Brion (it's Lemma 2 in his paper Multiplicity-Free Subvarieties of Flag Varieties). First, for any $B$-equivariant coherent sheaf $\mathcal F$ on $X$, there is a natural $G$-equivariant coherent sheaf $ G \times^B \mathcal F $ on $G \times^B X$. This assignment in fact is an equivalence of categories. (See Brion's paper for details on the construction.) Now assume that $X$ is Cohen-Macaulay. Then we have -$$ \omega_{G \times^B X} = G \times^B \big( \omega_X( 2\rho )\big), $$ -where by $ \omega_X(2\rho) $ we mean the bundle $\omega_X$ on $X$ twisted by the trivial line bundle on $X$ with $B$-weight $2\rho$. (The twist by $2\rho$ comes from the fact that $\omega_{G/B}$ is the $G$-equivariant bundle on $G/B$ whose fiber is the 1-dimensional $B$-module with weight $2\rho$. Here I'm using the convention that $B$ corresponds to the positive roots.)<|endoftext|> -TITLE: Dimensions and number of complex irreducible representations for SL3(Z/pZ) -QUESTION [12 upvotes]: This is my first post here :) -I have the following two related questions. While looking in Conway's ATLAS (the 1985 one) for $SL_3(Z/pZ)$, where he does the cases $p=2,3,5,7$, I saw that the dimension of the biggest irreducible representation is ~$p^3$. More precise, as the order of the group is $p^3(p^3-1)(p^2-1)$, it looks like the dimension of the biggest irreducible representation is a polynomial of degree 3 in p, that "obviously" dividers the order. -For p=2, the biggest one has dimension $8=2^3$ -For p=3, the biggest one has dimension $39 = 3(3^2+3+1)$ -For p=5, the biggest one has dimension $186=(5+1)(5^2+5+1)$ -For p=7, the biggest one has dimension $456=(7+1)(7^2+7+1)$ -Is this true in general? Are there infinitely many primes for which the biggest irr rep of $SL_3(Z/pZ)$ has dimension $O(p^3)$ (or even weaker bounds, like $o(p^4)$). -The second question is about the number of irreducible representations. While looking in the ATLAS, looks like there are quite few of them, relative to the dimension of the group. Is there a result that computes the growth rate of the number of conjugacy classes/number of irr rep, as $p\to\infty$? -I am interested in a result that somehow combines both. Something in the spirit "small", but "few" representations. - -REPLY [6 votes]: The question of the maximal dimension of irreducible representations of simple finite groups (including those of Lie type) is considered in "The largest irreducible representations of simple groups" by Larsen, Malle and Tiep (arXiv:1010.5628); for finite groups of Lie type (not necessarily simple) over large enough field, it is a theorem of Seitz ("Cross-characteristic embeddings of finite groups of Lie type", Proc. LMS 1990, 166-200, section 2) that the maximal dimension is at most $|G:T|_{p'}$, where $T$ is a maximail torus of smallest possible order (i.e., maximally split). In particular, if there is a split maximal torus, this shows that the principal series representations are indeed those with maximal dimension.<|endoftext|> -TITLE: Largest convex hull of a unit length path -QUESTION [8 upvotes]: What is the largest area possible for the convex hull of a path of unit length lying on a plane? For what paths is that largest area attained? - -REPLY [15 votes]: The answer seems to be $\frac{1}{2\pi}$, using a semi circle. See -Moran, P. A. P. "On a problem of S. Ulam." Journal of the London Mathematical Society 1.3 (1946): 175-179.<|endoftext|> -TITLE: Continuous functions as uniformly continuous function -QUESTION [8 upvotes]: Three question concerninng metrics on the real line: -Is there a metric $d$ on $\Bbb{R}$ such that a function $f : (\Bbb{R},d) \longrightarrow (\Bbb{R},d)$ ( or $f : \Bbb{R} \longrightarrow (\Bbb{R},d)$ or $f : (\Bbb{R},d) \longrightarrow \Bbb{R}$) is continuous if and only if $f : \Bbb{R} \longrightarrow \Bbb{R}$ is uniformly continuous ? - -REPLY [9 votes]: EDIT: The answer now applies to arbitrary topologies, using an idea by Pietro Majer from the comments. -Proposition: There are no topologies $\tau_0,\tau_1$ on $\mathbb R$ such that $f\colon\mathbb R\to\mathbb R$ is uniformly continuous in the Euclidean metric iff $f\colon(\mathbb R,\tau_0)\to(\mathbb R,\tau_1)$ is continuous. -Proof: -$\tau_1$ cannot be indiscrete (lest all functions are uniformly continuous), hence we can fix a $\tau_1$-closed set $F$ and points $a\in F$, $b\notin F$. For every Euclidean closed set $A$ and $c>0$, let $f_c(x)=a+c\operatorname{dist}(x,A)$. Then $f_c$ is uniformly continuous, hence continuous from $(\mathbb R,\tau_0)$ to $(\mathbb R,\tau_1)$, hence the $\tau_0$-closed set $f_c^{-1}(F)$ includes $A$ and excludes all points of Euclidean distance $(b-a)/c$ from $A$. The intersection of such sets for all $c$ is just $A$. This shows that $A$ is $\tau_0$-closed, i.e., $\tau_0$ refines the Euclidean topology. -Let $f\colon\mathbb R\to\mathbb R$ be a Euclidean-continuous but not uniformly continuous function, such as $f(x)=x^2$. For every $n>0$, $f_n=f\restriction[-n,n]$ can be extended to a uniformly continuous function on $\mathbb R$. By assumption, this function is continuous from $(\mathbb R,\tau_0)$ to $(\mathbb R,\tau_1)$, hence $f_n$ is continuous from $([-n,n],\tau_0)$ to $(\mathbb R,\tau_1)$. Since $\tau_0$ refines the Euclidean topology, every point has a $\tau_0$-open neighbourhood included in some $[-n,n]$, thus $f=\bigcup_nf_n$ is continuous from $(\mathbb R,\tau_0)$ to $(\mathbb R,\tau_1)$. However, it is not uniformly continuous in the Euclidean metric, a contradiction.<|endoftext|> -TITLE: GOE Version of Longest Increasing Subsequence -QUESTION [12 upvotes]: Let $S_n$ be the symmetric group equipped with uniform measure. For any $\pi\in S_n$, let $L_n=L_n(\pi)$ denote the longest increasing subsequence. A celebrated result of Baik, Deift and Johansson states that -$$P\left(\frac{L_n-2\sqrt{n}}{n^{1/6}}\leq s\right)\rightarrow F_2(s),$$ -where $F_2(s)$ is the Tracy-Widom distribution, describing the largest eigenvalue of a GUE matrix. The connection between longest increasing subsequence, representation theory and random matrix theory is now well established. - -Question: is there another natural statistic on $S_n$ resembling something akin to $L_n$ which follows a GOE distribution? Specifically, calling such a statistic $M_n$, I'm interested in a similar result of the form -$$P\left(\frac{M_n-2\sqrt{n}}{n^{1/6}}\leq s\right)\rightarrow F_1(s),$$ -where $F_1(s)$ is now the Tracy Widom distribution of the largest eigenvalue of a GOE. Ideally, the scaling $n^{1/6}$ should be the same. I would also prefer to retain uniform measure on $S_n$ but, of course I would be very interested to hear of other examples. As well, the shift can be different as well. - -I'm aware of numerous results in combinatorics where $F_1$ comes up in. For example, the fluctations of random aztec diamond tilings follow an Airy process and therefore have GOE behavior. There the scaling is $n^{1/3}$. As well, it comes up in viscous walkers and nonintersecting brownian motion. I mention these examples because there are definitely tangentially linked to $S_n$ through representation theory and growth processes. However, as I said I would ideally like an example of something that naturally occurs on $S_n$. - -REPLY [12 votes]: Involutions $s=s^{-1}$ in $S_n$ are modeled by the Tracy-Widom distribution $F_1$ for real symmetric matrices (GOE): -Take as $S_n^\ast$ the subset of involutions in $S_n$, and let -$M_n$ be the corresponding random variable for the longest increasing subsequence. Then the limit distribution is -$$P\left(\frac{M_n − 2\sqrt n}{n^{1/6}}\leq s\right)\rightarrow F_1(s).$$ -J. Baik and E.M. Rains, Symmetrized random permutations (1999). -The same paper also gives a generalization that is modeled by the Tracy-Widom distribution $F_4$ for the GSE (Gaussian symplectic ensemble). For a recent review article, see: -G. Olshanski, Random permutations and related topics (2011).<|endoftext|> -TITLE: A model category which is an additive category -QUESTION [5 upvotes]: Let $\cal C$ be a model category which is also additive. Suppose that the homotopy category $\operatorname{Ho}\mathcal C$ is additive, for example this is true when the weak equivalences in $\cal C$ is closed under biproducts (see this question). -If we take a cofibrant object $X$ and a fibrant object $Y$ then there is a natural isomorphism -$$ - \operatorname{Ho}\mathcal C(X,Y) \cong \mathcal C(X,Y)/\sim -$$ -where $\sim$ is the homotopy relation. Is this always a group isomorphism? - -REPLY [8 votes]: Yes, because the projection functor from the model category to the homotopy category preserves coproducts of cofibrant objects. That is actually the way of showing that the homotopy category has coproducts. - -REPLY [4 votes]: If by "additive" you mean an $\mathbf{Ab}$-enriched category with a zero object and biproducts, then yes. Let $\mathcal{M}$ be model category that is additive in this sense, let $\mathcal{M}_c$ be the full subcategory of cofibrant objects, let $\mathcal{M}_f$ be the full subcategory of fibrant objects, and let $\mathcal{M}_{cf} = \mathcal{M}_c \cap \mathcal{M}_f$. Here are the relevant facts: -$\DeclareMathOperator{\Ho}{Ho}$ - -The coproduct of a family of cofibrant objects is automatically a homotopy coproduct, so the localising functor $\mathcal{M}_c \to \Ho \mathcal{M}$ preserves coproducts. Dually, the localising functor $\mathcal{M}_f \to \Ho \mathcal{M}$ preserves products. -Hence, the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ preserves the zero object and biproducts. (Note that $\mathcal{M}_{cf}$ is an additive subcategory of $\mathcal{M}$.) -A category with a zero object and biproducts is automatically enriched over commutative monoids in a unique way, and a functor that preserves zero objects and biproducts is similarly enriched. Thus, there is a unique enrichment of $\Ho \mathcal{M}$ over commutative monoids that makes the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ an enriched functor. -Since $\mathcal{M}_{cf}$ is actually $\mathbf{Ab}$-enriched and the localising functor $\mathcal{M}_{cf} \to \Ho \mathcal{M}$ is full and essentially surjective on objects, $\Ho \mathcal{M}$ is also $\mathbf{Ab}$-enriched. - -Now, let $X$ and $Y$ be any two objects in $\mathcal{M}$. In order for the hom-set map -$$\mathcal{M}(X, Y) \to \Ho \mathcal{M}(X, Y)$$ -to be a group homomorphism, it is sufficient that the localising functor $\mathcal{M} \to \Ho \mathcal{M}$ preserve either the coproduct $X + X$ or the product $Y \times Y$. (We already know that it preserves initial and terminal objects.) Thus it suffices to take either $X$ cofibrant or $Y$ fibrant.<|endoftext|> -TITLE: Are there insane families in $L$? -QUESTION [12 upvotes]: Let $A,B\subseteq\omega$. We write $A\subseteq^*B$ if $A\setminus B$ is finite, if additionally $B\setminus A$ is infinite then we write $A\subsetneq^*B$, otherwise we write $A=^*B$. -We say that a $\cal A\subseteq P(\omega)$ is almost disjoint if for every two distinct $A,B\in\cal A$ we have $A\cap B=^*\varnothing$. We say that $\cal A$ is maximal almost disjoint, or MAD, if there is no $\cal B$ strictly containing $\cal A$ which is almost disjoint. -At the other end of the spectrum we say that $\cal A\subseteq P(\omega)$ is a tower if $\cal A$ is well-ordered by $\subsetneq^*$. -Finally, we define $\mathcal B=\{B_\alpha\mid\alpha<\kappa\}$ to be insane if it is MAD, and there exists a tower $\mathcal A=\{A_\alpha\mid\alpha<\kappa\}$ with the following property: $$\beta<\alpha\implies B_\beta\subseteq^*A_\alpha\\ \beta\geq\alpha\implies B_\beta\cap A_\alpha=^*\varnothing.$$ -In that case we say that $\cal A$ is an associated tower for $\cal B$. -Note, for example, that if $\cal B$ is insane and $\cal A$ is an associated tower then $A_{\alpha+1}\setminus A_\alpha=^*B_\alpha$. - - -Questions. - -Is the existence of insane families consistent with $\sf ZFC$? -If the answer is yes to the previous question, is there an insane family in $L$? -If the answer is yes to the previous question, can this notion be extended to every regular cardinal $\kappa$? (replacing "finite" by ${<}\kappa$ in the definition of $\subseteq^*$ and so on.) - -REPLY [12 votes]: Your requirements are inconsistent; there is no insane family. -Suppose towards contradiction that we have an insane family -$\mathcal{B}=\{B_\alpha\mid\alpha\lt\kappa\}$, witnessed by tower -$\langle A_\alpha\mid\alpha\lt\kappa\rangle$. For finite $k$, let $b_k$ be any -element in $[(A_\omega-A_k)\cap B_k]-\bigcup_{j\lt k}B_j$. There are such elements, since $B_k$ is almost disjoint from $A_k$ and from the earlier $B_j$ for $j\lt k$, and $B_k$ is almost contained in $A_\omega$. Note that the $b_k$ are -distinct, and so $B=\{b_k\mid k\lt\omega\}$ is infinite. By -maximality, $B$ must have infinite intersection with some -$B_\beta$. Note that $B$ has exactly one element from each $B_k$ -for $k\lt\omega$. So it must be that $\beta\geq\omega$. But in -this case, since $B\subset A_\omega$, we have infinitely many -elements in $B_\beta\cap A_\omega$, which violates the second insanity clause. -So there is no insane family.<|endoftext|> -TITLE: Non-commutative reduced rings of order $p^2$ -QUESTION [10 upvotes]: Let $p$ be a prime number. Is there a non-commutative reduced ring of order $p^2$? (Note that any ring of order $p^2$ with identity is commutative). - -REPLY [3 votes]: A more conceptual path to essentially the same solution: -Lam, a first course in noncommutative rings, pag. 207: (12.7) Theorem. A nonzero ring $R$ is reduced iff $R$ is a subdirect product of domains. [Yes, it works also for unitless rings] -Finite subdirect product of rings are direct products: Chinese remainder theorem. [This is an optional step, not needed to obtain only commutativity but used to obtain the full structure of the ring] -Finite domains are division rings (since injective endofunctions of a finite set are surjective) -Now apply Wedderburn (finite division rings are Galois fields) [Kaplansky, Jacobson, Herstein et. al. have many generalizations of Wedderburn's theorem (whose proof, however, uses Wedderburn's result, plus steps related to the first stem in this proof)]. -With the same proof (without Wedderburn): -reduced (right) artinian rings are exactly the finite direct products of (possibly skew) fields.<|endoftext|> -TITLE: What is known about dynamics on Grassmannians? -QUESTION [12 upvotes]: I have found myself becoming interested in dynamical systems given by homeomorphisms acting on $G(r,d)$, the space of $r$-dimensional subspaces of $\mathbb{R}^d$. I tried to do a literature search and failed to turn up any useful references or papers. I'd like to know what is currently known and/or where I can learn more about such things. -My motivation for studying such dynamical systems is the following: in general, a one-dimensional quasi-periodic family of self-adjoint operators can be thought of as one which fibers over an irrational rotation $ \omega \mapsto \omega + \alpha $ of the circle $\mathbb{T} := \mathbb{R} / \mathbb{Z}$. This notion and its generalizations to operator families fibering over minimal translations of $ \mathbb{T}^d = \mathbb{R}^d / \mathbb{Z}^d $ are extremely well-studied, with many lovely results. -Now, instead of $\mathbb{R} / \mathbb{Z}$, it is reasonably natural to think of the circle as the real projective line $ \mathbb{R} \mathbb{P}^1 = G(1,2)$, and, in this case, the rotations mentioned above now correspond to the natural action of $ \mathrm{SO}(2,\mathbb{R}) $ on $G(1,2)$. In this case, it is well-known that the corresponding action is minimal if and only if it is uniquely ergodic if and only if the corresponding angle of rotation is an irrational multiple of $\pi$. I'd like to know to what extent one can generalize dynamical statements like this to the action of $\mathrm{SO}(d,\mathbb{R})$ on $G(r,d)$. -More precisely, given $R \in \mathrm{SO}(d)$, $R$ naturally preserves a version of normalized Lebesgue measure on $G(r,d)$, which one could realize, for example, by (a constant multiple times) the pushforward of $d-1$ dimensional Lebesgue measure on $S^{d-1} \subset \mathbb{R}^d$ under the quotient map $S^{d-1} \to G(r,d)$. It is then natural to ask when Lebesgue measure is ergodic with respect to the action of $R$, or when this action is uniquely ergodic (in which case Lebesgue measure is necessarily the unique preserved measure), or minimal. -Still more generally, are there interesting homeomorphisms on $G(r,d)$ which do not arise from the action of a matrix? -Obviously, there have to be some difficulties in higher dimensions. For example, it is easy to see that any $R \in \mathrm{SO}(2n+1)$ necessarily has a fixed point on $S^{2n}$, so the induced action on $G(r,2n+1)$ can never be minimal or uniquely ergodic. - -REPLY [3 votes]: I suggest that you also look at the related frame flows, say on a compact smooth Riemannian manifold with negative sectional curvatures. The action (on the set of positively oriented orthonormal frames) moves the first vector of the frame according to the geodesic flows and the remaining ones by parallel transport. It was shown by Brin and Pesin that the time-1 map of the frame flow is a partially hyperbolic diffeomorphism, and later was shown by Brin that the flow is ergodic with respect to the product $\mu\times m$ of the Liouville measure $\mu$ on the unit tangent bundle and the Haar measure $m$ on $SO(n-1)$ ($n$ being the dimension of the manifold) for a set of metrics (with negative curvature) that is open and dense in the $C^3$ topology. Curiously, the dimensions $7$ and $8$ are apparently more complicated than all others.<|endoftext|> -TITLE: Knotted projective planes and fake complex projective space -QUESTION [6 upvotes]: Paul Melvin gave a talk at Knots in Washington last year in which he asked whether the connected sum of an odd twist-spin of a classical knot and a standard cross-cap embedding of ${\mathbb R}P^2$ is a standard cross-cap. I believe this question to be a standard one, and I have two questions about it. -First, I think that Paul stated that if the result is knotted, then there is a non-standard ${\mathbb C}P^2$. Please may I have an outline of the argument that gives the construction of the non-standard ${\mathbb C}P^2?$ -Second, for aesthetic reasons, I imagine that there is a difference in using a cross-cap with normal Euler class $2$ and using one with normal Euler class $-2$. Won't the construction give a $\pm {\mathbb C}P^2$ (or better ${\mathbb C}P^2$ or $\overline{{\mathbb C}P}^2$) depending on the normal Euler class of the cross-cap? -Any other folk-lore would be appreciated. - -REPLY [2 votes]: I can't think of another way to get $CP^2$ from an embedded $RP^2$ in the 4-sphere, so I would guess you are right. But why would the knottedness of the $RP^2$ imply that the $CP^2$ is exotic? After all, there are plenty of knotted 2-spheres in $S^4$ whose double branched covers are $S^4$. In other words, a knotted $RP^2$ might produce an exotic involution on the genuine (I don't want to say real, for obvious reasons!) $CP^2$ rather than an exotic $CP^2$.<|endoftext|> -TITLE: Is there a "by hand" proof on the symmetry of the Atiyah class of $TX$? -QUESTION [6 upvotes]: Let X be a complex manifold and $TX$ its tangent bundle. The Atiyah class $\alpha(E)\in \text{Ext}^1(E\otimes TX, E)$ for a vector bundle $E$ is defined to be the obstruction of the global existence of holomorphic connections on $E$. We can refer to M, Kapranov's paper "Rozansky–Witten invariants via Atiyah classes" or Sasha's answer to this mathoverflow question Atiyah class for non-locally free sheaf. -Now if we take $E=TX$ to be the tangent bundle of $X$ itself, we can prove that the Atiyah class $\alpha(TX)\in \text{Ext}^1(TX\otimes TX, TX)$ is symmetry, which means, $\alpha(TX)$ is in fact in $\text{Ext}^1(S^2(TX), TX)$. -The proof in Kapranov's paper is given in the language of torsor. In J. Roberts and S. Willerton's paper "On the Rozansky–Witten weight systems" (p. 29) they explain Kapranov's proof and mentioned that this symmetry "corresponds in differential geometry to the vanishing of the torsion." -This comment remind me about the similar result in Riemannian geometry that the curvature tensor is symmetric, which relies on the fact that the connection is torsion-free. -$\textbf{My question}$ is: Is there a direct "by hand" proof of the symmetry of the Atiyah class $\alpha(TX)$ without using torsors? - -REPLY [2 votes]: Here is a different answer, which is nevertheless the same as Damien's second proof written in a different language, and that can be found in Markarian's famous paper "The Atihay class, Hochschild cohomology, and the Riemann-Roch theorem". Look at Proposition 1 (i).<|endoftext|> -TITLE: Ring $R$ such that $R^n$ contains unimodular elements that are not part of a free basis for all $n \geq 2$ -QUESTION [5 upvotes]: Let $R$ be a commutative ring. A vector $(c_1,\ldots,c_n) \in R^n$ is unimodular if $Rc_1 + \cdots + Rc_n = R$. Say that a vector $\vec{v} \in R^n$ is a basis element if there exists a free basis for $R^n$ containing $\vec{v}$. It is clear that all basis elements of $R^n$ are unimodular. Moreover, if $\vec{v} \in R^n$ is unimodular and $V = R \cdot \vec{v}$, then there exists an $R$-submodule $W \subset R^n$ such that $R^n = V \oplus W$ (proof : if $\vec{v} = (c_1,\ldots,c_n)$ and $1 = a_1 c_1 + \cdots + a_n c_n$ with $a_i \in R$, then we can define a surjection $\phi : R^n \rightarrow R$ via the formula $\phi(x_1,\ldots,x_n) = a_1 x_1 + \cdots + a_n x_n$; the map $\phi$ is split via the inclusion $R \hookrightarrow R^n$ that takes $1$ to $\vec{v}$). Clearly $V \cong R^1$, but it is not necessarily true that $W$ is a free $R$-module, so it does not follow that $\vec{v}$ is a basis element. -It is clear that unimodular vectors in $R^1$ are basis elements. -Question : Can someone give me an example of a ring $R$ such that for all $n \geq 2$, there exist unimodular vectors in $R^n$ that are not basis elements? -Such rings have to be pretty weird; for instance, it is standard that if a ring satisfies Bass's stable range condition $SR_{d+2}$, then for $n \geq d+2$ all unimodular vectors in $R^n$ are basis elements. This means that rings $R$ as in our question must either be non-Noetherian or have infinite Krull dimension. - -REPLY [4 votes]: Start with the integers. Adjoin variables $X_{in}$ and $Y_{in}$ for all $1\le i \le n$. Mod out by all relations of the form -$$\sum_{i=1}^nX_{in}Y_{in}=1$$ -Call the resulting ring $R$. -Then, by construction, any $(X_{1n},X_{2n},\ldots,X_{nn})$ is a unimodular row over $R$. I claim it's not a basis element. Equivalently, I claim that the complement of this row (i.e. the cokernel of the linear map $R\rightarrow R^n$ that it defines) is not free. -To see this, let $S$ be any ring over which there exists a unimodular row $(t_i,\ldots,t_n)$ which is not a basis element. (Such rings are known to exist by results of Raynaud, or, alternatively, of Mohan Kumar and Nori). Write $\sum_{i=1}^nt_iu_i=1$. Map $R$ to $S$ by -$$X_{in}\mapsto t_i$$ -$$Y_{in}\mapsto u_i$$ -and, for each $m\neq n$, -$$X_{1m}\mapsto 1$$ -$$Y_{1m}\mapsto 1$$ -$$X_{jm}\mapsto 0 \quad (j\neq 1)$$ -$$Y_{jm}\mapsto 0 \quad (j\neq 1)$$ -Now let $P_n$ be the $R$-module that is the complement of $(X_{i1},\ldots, X_{in})$. Then $P\otimes_RS$ is the complement of $(t_1,\ldots,t_n)$, and hence not free. Therefore $P$ cannot be free, so $(X_{i1},\ldots,X_{in})$ cannot be a basis element.<|endoftext|> -TITLE: Kaplansky's conjecture and Martin's axiom -QUESTION [17 upvotes]: Recall Kaplansky's conjecture which states that every algebra homomorphism from the Banach algebra C(X) (where X is a compact Hausdorff topological space) into any other Banach algebra, is necessarily continuous. The conjecture is equivalent to the statement that every algebra norm on C(X) is equivalent to the usual uniform norm. -By work of Dales and Esterle, CH refutes the conjecture. On the other hand Solovay (building on work of H. Woodin) proved the consistency of Kaplansky's conjecture with ZFC. -As far as I know, the models constructed for the consistency of Kaplansky's conjecture also satisfy the Martin's axiom (MA). So my question is: -Question. Is Kaplansky's conjecture + the negation of MA consistent with ZFC? -Is this question open? - -REPLY [11 votes]: The answer is yes. -Recall that if $S$ is a Suslin tree, then PFA($S$) denotes the forcing axiom for the class of all proper posets which preserve $S$. PFA($S$) is consistent with ZFC relative to a supercompact cardinal, like PFA, and shares many of the consequences of PFA, but a model of PFA($S$) naturally does not satisfy MA (since $S$ is Suslin). -Claim: PFA($S$) implies Kaplansky's conjecture. -To prove this it suffices to show that each of the forcings used in Todorcevic's PFA proof preserve the Suslin tree $S$. I won't go through all of the details of the PFA proof; let me just point to the appropriate references. -Woodin proved (see Dales-Woodin, chapter 3) that if Kaplansky's conjecture fails, then there is a nonprincipal ultrafilter $U$ and an order-embedding -$$ (\langle \mathrm{id} \rangle/U, <_U) \to (\omega^\omega, <^*) $$ -where $\langle \mathrm{id} \rangle$ is the set of all $f\in\omega^\omega$ with $\lim f(n) - 1 = \lim n - f(n) = \infty$, and $<^*$ is eventual dominance. Assume PFA($S$) and suppose such an embedding exists. -MA implies that for any nonprincipal ultrafilter $U$, there is an order-embedding $(2^{\omega_1}, <_{lex}) \to (\omega^\omega/U, <_U)$. ($<_{lex}$ is the lexicographical ordering.) Todorcevic's proof of this (see Partition Problems in Topology, Theorem 7.7) uses a poset that is not only ccc but Knaster, i.e. every uncountable sequence of conditions has an uncountable subsequence which are pairwise compatible. It's well known that such posets preserve Suslin trees; so PFA($S$) also gives us such an embedding. It's not hard to show that this embedding can be modified to map into $\langle\mathrm{id}\rangle$, so we get an embedding -$$ (2^{\omega_1}, <_{lex})\to (\omega^\omega, <^*) $$ -Now Todorcevic shows (PPIT Theorem 8.8) that there can't be such an embedding in a model of PFA. The poset he uses is -$$ \mathrm{Col}(\omega_1, 2^\omega) * \mathbb{P}_{\mathcal{A},\mathcal{B}} $$ -where $\mathrm{Col}(\omega_1, 2^\omega)$ is the usual countably-closed collapse of $2^\omega$ to $\omega_1$, $(\mathcal{A},\mathcal{B})$ is an $(\omega_1,\omega_1)$-gap found in the extension, and $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is the ccc poset which freezes that gap. It's again well-known that countably-closed forcings preserve Suslin trees, so we just need to show that the same holds for $\mathbb{P}_{\mathcal{A},\mathcal{B}}$, i.e. after forcing with $\mathbb{P}_{\mathcal{A},\mathcal{B}}$, $S$ is still ccc. For this it suffices to show that after forcing with $S$, $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is still ccc. But $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is ccc whenever $(\mathcal{A},\mathcal{B})$ is truly a gap, i.e. it's not split by some real $a\subseteq\omega$, and since $S$ adds no reals it follows that $\mathbb{P}_{\mathcal{A},\mathcal{B}}$ is still ccc in the extension by $S$. This completes the proof.<|endoftext|> -TITLE: What is the maximum-entropy distribution given mean, variance, skewness, and kurtosis? -QUESTION [9 upvotes]: $X\in \mathbb{R}$. Which distribution $P(X)$ has the highest possible entropy given its expected value, variance, skewness, and kurtosis? Is it an exponential family distribution of the form $P(X) \propto \exp(a\cdot x +b\cdot x^2 + c\cdot x^3 +d\cdot x^4)$ in analogy to the normal distribution being the maximum-entropy distribution given mean and variance? If so, where can I read about it? -I am interested in this question because I want to model data with high skewness and high kurtosis, but I am still looking for an appropriate distribution. - -REPLY [9 votes]: Yes, your $P(X) \propto \exp(a\cdot x +b\cdot x^2 + c\cdot x^3 +d\cdot x^4)$ maximises the entropy $-\int P(X){\rm log} P(X)dX$ for prescribed first four moments, if the skewness and curtosis lie in a certain range: -M. Rockinger and E. Jondau, Entropy densities with an application to autoregressive conditional skewness and kurtosis (2002). -This simple generalization of the normal distribution holds also if higher moments are prescribed, provided that the highest prescribed moment is even. If, for example, only mean, variance, and skewness, are prescribed, then $d$ would be zero and the distribution fails to normalize. In this case the maximum entropy distribution exists, but it has a more complicated form, as discussed here. -You'll still have to determine the coefficients $a,b,c,d$ from the given first four moments. The cited 2002 paper gives a description of an efficient method. -There may be no solution (typically, if the skewness is too large relative to the kurtosis), meaning that a maximum entropy solution of this simple form does not exist (see Figure 1 of the paper). The combination of a prescribed skewness and kurtosis typically leads to a bimodal distribution (see Figures 2 and 3), which may or may not be desirable for your application.<|endoftext|> -TITLE: Is There a maximal space that is a P-space? -QUESTION [5 upvotes]: we guess there is no maximal space which is also a P-space. Am I right? Do u know a counter example? -clarifications: -Maximal space is that space with topology $\tau$ which is maximal crowded topology on X. -crowded: a topology with no isolated point = dense in itself. -P-space:every $G_\delta$ set is open = every prime ideal is a Z-ideal = every prime ideal is a maximal ideal. (for others ref 4M of Gillman , Jerison) - -REPLY [6 votes]: Given a space $(X,\tau)$, the collection of all $G_\delta$-subsets of $X$ form a base for a stronger topology $\tau_\omega$ on $X$. It is easy to see that $(X,\tau_\omega)$ is a $P$-space (sometimes called the $G_\delta$ modification of $X$). If the original space $(X, \tau)$ is maximal, then there are two possibilities: -1) $(X, \tau_\omega)$ has isolated points. This will happen if and only if the original space has a point of countable pseudocharacter (i.e. a point which is a $G_\delta$). -2) $(X, \tau_\omega)$ is crowded. Then by maximality $\tau_\omega= \tau$ and therefore the original space is already a $P$-space. -It is known that if there are no measurable cardinals then every maximal space has countable pseudocharacter, so it cannot be a $P$-space. -I think it is still open whether there can be a [homogeneous] maximal space with uncountable pseudocharacter (see Question 168 [169] in Open Problems in Topology II). Note that by the remarks above, a maximal space in which every point has uncountable pseudocharacter must be a $P$-space.<|endoftext|> -TITLE: fiber bundle and free action -QUESTION [7 upvotes]: In Spanier's book " Algebraic topology" a fiber bundle is defined as follows: -A fiber bundle $\xi=(E,B,F,p)$ consists of a total space $E$, a base space $B$ and a fiber $F$ and a bundle projection $p:E\rightarrow B$ such that there is exists an open covering $\{U\}$ of $B$ and for each $U\in \{U\}$ a homeomorphism $\phi_U: U\times F\rightarrow p^{-1}(U)$ such that the composite $p\circ \phi_U: U\times F\rightarrow U$ is projection map. -Question: If a compact lie group G (e.g circle group) acts freely on a paracompact space X then does $(X,X/G,G,\pi)$ (where $X/G$ is orbit space and $\pi$ is orbit map) forms a fiber bundle. -If $\pi$ is covering projection then it seems the above is true but what I am not able to verify this is in general case. - -REPLY [9 votes]: The answer is yes if you assume paracompact spaces are Hausdorff, and no if not. -This is because in Hausdorff spaces, paracompactness implies complete regularity, and it is a theorem of Gleason that if compact Lie group $G$ acts on a completely regular space $X$, then every orbit $Gx$ admits a tube neighbourhood. This implies that a free action is a principal $G$-bundle in this case. See also Bredon's "Introduction to compact transformation groups", Theorem II.5.8. -On the other hand, its easy to cook up counter-examples for non-Hausdorff paracompact spaces. In fact, just take your favourite free $G$-space $X$, and give $X$ the indiscrete topology.<|endoftext|> -TITLE: Efficiently determine if convex hull contains the unit ball -QUESTION [14 upvotes]: Given a set of $n$ points in $\mathbb{R}^d$, is there an algorithm to determine if the convex hull contains the unit ball centered at the origin in polynomial time? The convex hull itself might have an exponential number of facets so we cannot afford explicitly to compute it. -My main interest is not in computer precision so we can make whatever assumptions help avoid that in relation to the points themselves (for example they only have integer coordinates). - -REPLY [14 votes]: The problem is NP hard. Here is a proof sketch. -The problem is to determine if there is a point $y$ with $\|y\|=1$ outside of the convex hull of given points $x_1,\dots, x_n$. Note that such point exists if and only if there is hyperplane at distance less than $1$ from the origin such that all points $x_1, \dots, x_n$ and $0$ lie on one side of the hyperplane (consider a hyperplane $\pi$ separating $x_1,\dots, x_n, 0$ and $y$). -So the problem can be stated as follows: is there $a\in {\mathbb R}^d$ s.t. - -$\langle a, x_i\rangle \leq 1$ (that is, all $x_i$ lie in the half-space $\{x: \langle a, x\rangle\leq 1\}$); -$\|a\| > 1$. - -Note that this problem is equivalent to the following well-known problem: - -We are given a convex polytope $\cal P$, a positive definite matrix $A$ and a number $t$, find $x\in \cal P$ such that $x^T A x > t$. ($\cal P$ is described by a system of linear equations). - -The optimization version of this problem is: - -Quadratic Programming (Non-convex Linearly Constrained Quadratic Programming with Positive Definite Matrix). We are given a convex polytope $\cal P$ and and a positive definite matrix $A$, find $x\in \cal P$ that maximizes $x^T A x$. - -This problem is known to be NP hard. -It is even NP-hard to optimize $x^T A x$ when $\cal P$ is the unit cube $\{(b_1, \dots, b_d): -1\leq b_i \leq 1\}$ (this problem is known as Integer Quadratic Programming with Positive Definite Matrix). In particular, the MAX CUT problem is a special case of this problem. Let $G$ be a graph on $n$ vertices and $L$ be its Laplacian. Then $\max_{x\in\{\pm 1\}^n} x^T L x$ is equal to the size of the maximum cut in $G$ ($L$ is positive semi-definite, not positive definite, but this difference is not important; e.g. we can consider $L'=L+\varepsilon I$ with very small $\varepsilon$). The NP-hardness of MAX CUT was proved by Karp in -Richard M. Karp (1972). Reducibility Among Combinatorial Problems. In R. E. Miller and J. W. Thatcher (editors). Complexity of Computer Computations. New York: Plenum. pp. 85–103.<|endoftext|> -TITLE: Dualizing Complexes -QUESTION [5 upvotes]: Let $R$ be a dualizing complex of a Noetherian graded algebra $A$ (not necessary commutative). For any $M\in D_c^b(A)$, there is a natural morphism -$$ -\theta: R\Gamma_m(M) \to RHom_{A^o}(RHom_A(M,R),R\Gamma_m(R)) -$$ -in $D^+(A)$. -Notations and terminologies we refer to the paper [Dualizing complexex over noncommutative graded algebras, A. Yekutieli], in which the author show that $\theta$ is an isomorphism whenever $R$ is balanced. Nevertheless I wonder whether or not $\theta$ is an isomorphism for arbitrary $R$. -Concerning the natural morphism, one can take $M$ to be a left bounded complex of injective modules, and take $R$ to be a complex of bimodules which are injective over $A$ and $A^{op}$. Thus $\Gamma_m(R)$ is also a complex of bimodules which are injective ofer $A$ and $A^{op}$ because $A$ is Noetherian. The natural morphism $\theta$ is given by the following morphism: -$$ -\Gamma_m(M) \to Hom_{A^{op}}(Hom_A(M,R),\Gamma_m(R)). -$$ - -REPLY [2 votes]: For an arbitray dualizing complex (not necessary balanced), I am going to prove the following version of local duality theorem. I will be grateful for any comments and corrections. Notations and concepts are refered to Here. - -Theorem: Let $A$ be a connected graded Noetherian algebra, and let $R\in D(A^e)$ be a dualizing complex (not necessary balanced). Then for any $X\in D^b_{fg}(A)$, the natural morphism - $$ -\theta: R\Gamma_{\mathfrak{m}} (X) \to R\text{Hom}_{A^o}(R\text{Hom}_A(X,R),R\Gamma_{\mathfrak{m}}(R)) -$$ - is an isomorphism in $D(A)$. - -Proof of the Theorem: Firestly choose an injective resolution $X\to I$ in $K^+(A)$, and also choose an injective resolution $R\to E$ in $K^+(A^e)$. Then one knows that $A^o$-modules occures in $\text{Hom}_A(A/A_{\geq n},E)$ and $\Gamma_\mathfrak{m}(E)$ are all injective. -So for any $n\geq 1$ we get a commutative diagram in $K(A)$: -$$ -\begin{array}{ccc} -\text{Hom}_A(A/A_{\geq n},I) & \xrightarrow{f_n} & \text{Hom}_{A^o}(\text{Hom}_A(I,E),\text{Hom}_A(A/A_{\geq n},E))\\ -\downarrow_{p_n} & & \downarrow_{q_n}\\ -\Gamma_\mathfrak{m}(I) & \xrightarrow{g} & \text{Hom}_{A^o}(\text{Hom}_A(I,E),\Gamma_\mathfrak{m}(E)), -\end{array} -$$ -where all $p_n,q_n,f_n,g$ are the canonical jomomorphisms of complexes. -Note that $\lim_{n\to \infty} p_n$ and $\lim_{n\to \infty} f_n$ are all quasi-isomorphisms. So to prove that $g$ is a quasi-isomorphism it quivalent to show that $\lim_{n\to\infty}q_n$ is a quasi-isomorphism. -Since $H^i(\text{Hom}_A(I,E))$ is finitely generated and vanishes when $|i|\gg0$, one can choose a resolution $P\xrightarrow{\sim} \text{Hom}_A(I,E)$ in $D(A^o)$ such that $P$ is bounded above and all $P_i$ are finitely generated pojective $A^o$-modules. Again we have a commutative diagram in $K(A)$: -\begin{array}{ccc} -\text{Hom}_{A^o}(\text{Hom}_A(I,E),\text{Hom}_A(A/A_{\geq n},E)) &\xrightarrow{u_n} & \text{Hom}_{A^o}(P,\text{Hom}_A(A/A_{\geq n},E))\\ -\downarrow_{q_n} & & \downarrow_{r_n}\\ -\text{Hom}_{A^o}(\text{Hom}_A(I,E),\Gamma_\mathfrak{m}(E)) & \xrightarrow{v} & \text{Hom}_{A^o}(P,\Gamma_\mathfrak{m}(E)), -\end{array} -Where $r_n,u_n,v$ are the natural homomorphism of complexes. -Now note that $\lim_{n\to \infty} u_n$ and $v$ are quasi-isomorphisms. From the fact that $\text{Hom}_{A^o}(P,-)$ commutes with direct limits of complexes with a common lower lound, one obtains that $\lim_{n\to \infty} r_n$ is a quasi-isomorphism, and so is $\lim_{n\to\infty}q_n$ too. Therefore the cononical homomorphism $g$ is a quasi-isomorphism. -At last, the construction of $\theta$ tells that it is an isomorphism in $D^+(A)$.<|endoftext|> -TITLE: Diffeomorphism group of the 2-sphere with $C^0$ topology -QUESTION [9 upvotes]: What is known about the homeomorphism (or homotopy) type of the group of $C^\infty$ diffeomorphisms of $S^2$ equipped with $C^0$ topology? -The group is the image of the inclusion $\mathrm{Diff}(S^2)\to \mathrm{Homeo}(S^2)$, where $\mathrm{Diff}(S^2)$ has $C^\infty$ topology. The inclusion is (I think) a homotopy equivalence, yet it is not a homeomorphism onto its image, so it does not seem to help answer the question. - -REPLY [3 votes]: The homotopy type seems to be independent of considering this group with $C^0$ or $C^{\infty}$ topology. As far as I know, the group of oriented diffeomorphisms of $S^2$ is homotopy equivalent to the group $PSL(2, C)$ (or $SO(3, R))$. I do not know direct reference, but maybe I can give a sketch of the proof here: -We are going to prove that the stabilizer of the triple of points $(0, 1, \infty)$ is homotopy equaivalent to a point. First of all, we need to establish the isomorphism between this stabilizer and the space of all smooth almost complex (<=> complex, in 2 dimensions these notions are the same) structures on the sphere $S^2$. In order to do it we use the Riemann's existence theorem, which states, that any complex curve of genus $0$ is isomorphic to $CP_1$: -Step 1: We fix the standard complex structure $I$ on the sphere $S^2$. -Step 2: We map the diffeomorphism $f$ to the structure $f^*I$ -Step 3: For $f$ and $g$ if $f^*I = g^*I$ then $g^{-1}f$ preserves $I$ and also $(0, 1, \infty)$ and hence it's identity. So $f = g$. Our mapping is injective. -Step 4: By the Riemann's existence theorem our mapping is surjective. -The space of all almost complex structures on a 2-dimensional sphere is contractable, because it is just a quotient space of the space of all riemannian metrics modulo conformal factors. Only thing we need to check is that nearby almost complex tensors will correspond to the nearby diffeomorphisms, and it can be done using smth like infinite-dimensional inverse function theorem... -And then, weakening the topology to the $C^0$ obviously won't make this space uncontractable. -And on a homeomorphism type... Maybe it makes sense to ask what is the corresponding to the $C^0$ topology on the space of complex structures. I suspect that it's topology given by the norms of quasiconformal mappings, but actually, I dunno.<|endoftext|> -TITLE: How strong is "all sets are Lebesgue Measurable" in weaker contexts than ZF? -QUESTION [21 upvotes]: Famously, Solovay showed that, if $\textrm{ZFC}$ plus $\textrm{IC}$ (the existence of an inaccessible cardinal) is consistent, then so is $\textrm{ZF}$ plus $\textrm{DC}$ (dependent choice) plus $\textrm{LM}$ (all subsets of $\mathbb{R}$ are Lebesgue measurable). And Shelah showed that, conversely, the consistency of $\textrm{ZF}+\textrm{DC}+\textrm{LM}$ implies the consistency of $\textrm{ZFC}+\textrm{IC}$. -Since $\textrm{LM}$ is a statement of third-order arithmetic, it makes sense to consider it in theories weaker than $\textrm{ZF}$ set theory. I am curious what is known about the consistency strength of $\textrm{DC} + \textrm{LM}$ in such weaker contexts. Of specific interest to me is the consistency strength of $\textrm{DC}+\textrm{LM}$ over the base logic of a boolean topos (= higher-order classical logic). - -REPLY [17 votes]: The consistency of ZFC + IC is perhaps a little bit too much to ask, but I believe the next best thing is true: -Conjecture. Every boolean topos1 with dependent choice in which every set of reals is Lebesgue measurable contains a well-founded model of ZFC. In fact, every real is contained in a well-founded model of ZFC. -The proof that Con(ZF + DC + LM) implies Con(ZFC + IC) shows that in any model $V$ of ZF + DC + LM, $\aleph_1$ (of $V$) is an inaccessible cardinal in the constructible universe $L$ and therefore the inner model $L$ satisfies ZFC + IC. In fact, it shows that $\aleph_1$ is inaccessible in $L[a]$ for every real $a$ and therefore $L_{\aleph_1}[a]$ is a well-founded model of ZFC that contains $a$. I will now argue that we can still make sense of "$L_{\aleph_1}[a]$ is a well-founded model of ZFC that contains $a$" in a boolean topos with dependent choice and I will explain why I believe why this is true in a boolean topos with dependent choice in which all sets of reals are Lebesgue measurable. -Since there are no actual material sets around, we must first find a substitute. A boolean topos can still make sense of HC, the collection of all hereditarily countable sets, by amalgamating all countable well-founded extensional structures $(\mathbb{N},E)$. More precisely, one can show that any two such structures have at most one transitive embedding between them and that any two of them can be amalgamated into a third. (All that is needed for this is arithmetic transfinite recursion.) Then the limit of this directed system of countable well-founded extensional structures under transitive embeddings is the required HC. -Once we have HC, we can do a bit of set theory in there, thinking of its elements as actual material sets. In fact, assuming dependent choice, HC is a very nice model: it satisfies comprehension, replacement, choice and the basic combinatorial axioms but it doesn't have powersets, of course, since every set is countable. The ordinals of HC form a well-ordering that we will call $\aleph_1$. Working in HC, one can construct $L_\eta[a]$ for every $\eta \in \aleph_1$ and every $a \subseteq \omega$. Therefore, one can make sense of the substructure $L_{\aleph_1}[a]$ of HC, which is a well-founded model of a fragment of ZFC + $V = L[a]$ and the question whether $L_{\aleph_1}[a]$ is a model of ZFC makes sense. -The key ingredient that connects these ideas with Lebesgue measurability is the following theorem of Raisonnier: -Theorem. Assume ZF + DC. If there is an uncountable well-orderable subset of $\mathbb{R}$ then there is a non-measurable set of reals. -I don't know whether this goes through in a boolean topos with dependent choice. However, since this is a theorem of "ordinary mathematics," I conjecture that it does! The theorem does use some "fancy objects" such as rapid ultrafilters but these do make sense in a boolean topos and, in the presence of dependent choice, so does Lebesgue measure. I could be wrong, but I can't see any immediate obstructions to Raisonnier's Theorem in a boolean topos with dependent choice. -Now, the $a$-constructible reals $\mathbb{R}^{L[a]} = \mathbb{R}\cap L_{\aleph_1}[a]$ form a well-orderable set of reals since $L_{\aleph_1}[a]$ has a definable wellordering. Therefore, assuming that Raisonnier's Theorem goes through, in any boolean topos with dependent choice where all sets of reals are Lebesgue measurable, it must be the case that $\mathbb{R}^{L[a]}$ is countable for every $a \subseteq \omega$. Then, the usual argument that shows that $\aleph_1$ is inaccessible in $L[a]$ goes through to show that $L_{\aleph_1}[a]$ is a model of ZFC. -To summarize, at Alex's request, the above sketches a proof of the following: -Theorem. In a boolean topos with dependent choice where all well-orderable sets of reals are countable, every real is contained in a well-founded model of ZFC. -So the truth of the initial conjecture rests only on the truth of Raisonnier's Theorem in a boolean topos with dependent choice. Note that the conclusion is rather strong. Since the models are well-founded, all $\Sigma^1_2$ statements that are true in such models are also true in the ambient topos. In particular, every such topos proves that ZFC is not only consistent but also $\Sigma^1_2$-sound. -To get the relative consistency of ZFC+IC as in Shelah's Theorem, we would need to continue the construction of $L$ past $\aleph_1$ and show that this leads to a model of ZFC in the limit. It is unlikely that this is possible without some kind of additional completeness assumptions on the topos. - -1For the sake of brevity, by "topos" I always mean "topos with a natural number object."<|endoftext|> -TITLE: Using the decomposition $641 = 5^4 + 2^4$ to factor $F_5$ -QUESTION [6 upvotes]: The question in the title arises from a problem in Stewart's "Galois Theory, Third Edition" (and possibly elsewhere) which has been bugging me for a few days since reading it: -Problem 19.5 (p. 224) asks: -Use the equations - -$641 = 5^4+2^4 = 5\cdot 2^7+1$ - -to show that 641 divides $F_5$. - -Now the latter expression is related to Euler's proof of his Theorem 8 in E134 and the ideas contained in the proof of that theorem is simple enough to lead to 641 being a candidate divisor of $F_5$ which can then be easily checked by hand/calculator. -However, the fact that Stewart includes the other expression as well intrigues me; I have been trying to use factorizations via sums of two squares to see how this expression might arise; for example since -\[ -F_5 = 65536^2+1^2 = 62264^2+20449^2 -\] -one can find this factor as -\[ -641 = \gcd(65536*62264-1*20449, 65536*20449+1*62264) -\] -But this approach is unsatisfactory since -1). Stewart does not mention the latter decomposition of $F_5$ as a sum of two squares -2). This approach makes no use of the decomposition of the potential factor as a sum of two squares/fourth powers -So, does anyone else have a clue as to what theorem/approach Stewart may have intended by including this decomposition. In particular, are there other less well-known theorems dealing with factoring Fermat numbers by expressing potential factors as Generalized Fermat numbers or some other similarly out of the hat approach? Or did Stewart include this expression for no good reason (which seems doubtful given the clarity of the approaches he takes throughout the rest of the book)? -BTW: As to why this is posted here, although the question asked in the book is certainly not research level, the intricacy of the other methods used to demonstrate 641 is a candidate factor indicate that if there is indeed a way to use the decomposition $5^4 + 2^4$ to factor $F_5$, such a method likely involves some deeper mathematics that is/was research-level. - -REPLY [12 votes]: There is no deep mathematics involved here. The proof goes as follows (see, for example, W.A. Coppel, Number Theory: An Introduction to Mathematics, Springer, 2009, p. 160). Since $641=5\cdot2^7+1$ $=5^4+2^4$, we have $5\cdot 2^7\equiv -1 \;(\mathrm{mod}\; 641)$ and $2^4\equiv -5^4 \;(\mathrm{mod}\; 641)$. Thus -$$2^{32}=2^4\cdot 2^{28}\equiv -5^4\cdot 2^{28}=-(5\cdot 2^7)^4\equiv -(-1)^4=-1 \;(\mathrm{mod}\; 641).$$<|endoftext|> -TITLE: calculating Littlewood-Richardson coefficients -QUESTION [12 upvotes]: It is known that if $\alpha,\beta,\gamma$ are three partitions then the Littlewood-Richardson coefficient $c_{\alpha \beta}^{\gamma}$ is positive when the triple ($\alpha,\beta,\gamma$ -) occurs as eigenvalues of Hermitian $n \times n$ -matrices $A, B, C$ with $C = A + B$ which can be seen from the following paper. -http://www.ams.org/journals/bull/2000-37-03/S0273-0979-00-00865-X/S0273-0979-00-00865-X.pdf -Is there a way to calculate the exact value of $c_{\alpha \beta}^{\gamma}$ by using this Hermitian matrices and their eigenvalues? - -REPLY [10 votes]: A sensible "yes" to your question would imply that the computational complexity of the calculation of Littlewood-Richardson coefficients is the same that of a calculation of eigenvalues of Hermitian matrices, so that they can be calculated in polynomial time. This seems to be impossible, see H. Narayanan, On the complexity of computing Kostka numbers and Littlewood-Richardson coefficients. - -REPLY [10 votes]: It's hard to prove that there isn't a way to do something, but I think the answer is no. -The saturation conjecture, now a theorem of Knutson and Tao, says that $c_{(N \alpha) (N \beta)}^{N \gamma} >0$ implies $c_{\alpha \beta}^{\gamma} >0$ for any positive integer $N$ and any partitions $\alpha$, $\beta$ and $\gamma$. Note that the corresponding statement for eigenvalues of Hermitian matrices is obvious. I suspect that any simple answer to your question would lead to a simple proof of the saturation conjecture and, while several proofs are now known, I would describe none of them as simple. -There is a relationship in the other direction. If $\alpha$, $\beta$ and $\gamma$ are partitions with $d$ parts then $c_{(N \alpha) (N \beta)}^{N \gamma}$ is a polynomial of degree $\binom{d-1}{2}$ in $N$ and the leading term (up to constants I'm not going to remember) is the volume of the space of triples $(A,B,C)$ of Hermitian matrices with spectra $(\alpha, \beta, \gamma)$ and $A+B=C$. So computing $c_{(N \alpha) (N \beta)}^{N \gamma}$ for enough values of $N$ allows you to compute the volume of a space of Hermitian matrices. -As a heurisitic, Hermitian matrix questions are about computing volumes of regions in $\mathbb{R}^M$; LR questions are about counting lattice points in those regions. Either one approximates the other, and the asymptotic behavior of lattice point counts tells you about volumes, but just knowing volumes will never tell you lattice point counts. -My favorite survey on the relation between the two problems is this one.<|endoftext|> -TITLE: Elementary end extensions of models of Peano Arithmetic in uncountable languages -QUESTION [9 upvotes]: A well-known theorem of Mills asserts that there is a model of Peano Arithmetic $M$ in an uncountable language such that $M$ has no elementary end extension (e.e.e.). I ask whether every complete extension of $PA$ in an uncountable language can have models of every cardinality with arbitrary large e.e.e.'s. To put it in more precise form, suppose $L$ is an uncountable language expanding the language of arithmetic $\{+,.,0,1,<\}$. By $PA(L)$ we mean $PA^{-}$ plus the induction axiom for all $L$-formulas $\phi(x,\bar{y})$. Let $T$ be a complete $L$-theory extending $PA(L)$ and let $\kappa$ be a cardinal $\geq |L|$. -Question. Does $T$ have a model $N$ such that $|N|=\kappa$ and $N$ has e.e.e.'s of every cardinality $\geq\kappa$? - -REPLY [6 votes]: It may be of interest to note that Jerabek and Dorais's positive solution to the above question and their use of the MacDowell-Specker theorem motivated me to take a closer look at the following paper of myself: -Sh.Mohsenipour, On Keisler Singular-Like Models, Math. Logic Quart. 54 (2008), 324-330, -in order to obtain another proof and then I realized that the following model theoretic result can be easily derived from the main result of the paper: -Theorem. Suppose $L=\{<,\dots\}$ is a first-order language where $<$ is a linear order and $T$ is an -$L$-theory such that for any finite $T^{'}\subset T$, there is a strong limit cardinal $\theta$ so that $T^{'}$ has a -$\theta-$like model $M$ and also $M$ can be represented as the union of an e.e.e. chain of its submodels. Let $\lambda$ be a cardinal $\geq|L|$, then: -(i) $T$ has a model $N$ of cardinality $\lambda$ such that $N$ has e.e.e.'s of every cardinality $\geq\lambda$. -(ii) If $\lambda$ is a singular cardinal, then the above $N$ can be a $\lambda-$like model with the additional property that for any singular cardinal $\lambda^{'}$ with $cf(\lambda^{'})<\lambda$ and $\lambda^{'}>\lambda$, $N$ has a $\lambda^{'}-$like e.e.e. -Now turning to the above question it is easily seen that the MacDowelـ‎Specker theorem implies that every completion $T$ of $PA(L)$ satisfies the hypothesis of the above theorem and therefore by part (i) we get another solution to the question. -This is a magic property of MO that makes you understand even your own work better!<|endoftext|> -TITLE: Partition $\Bbb{R}$ into a family of sets each one homeomorphic to the Cantor set -QUESTION [9 upvotes]: It is known that there is no (nontrivial) partition of $\Bbb{R}$ into a countable number of closed set. But is there a partition of $\Bbb{R}$ into sets, each one homeomorphic to the cantor ternary set? - -REPLY [18 votes]: Let $f$ be the x-coordinate of Hilbert's space-filling curve, -whose graph is shown here: - -Then the sets $\{f^{-1}(t)\}_{t\in [0,1]}$ -form a partition of the interval [0,1] into Cantor sets. -An easy variation of the above construction produces a partition of the reals (take the point-preimages of $F$, where $F:\mathbb R\to \mathbb R$ is the periodic extension given by $F(t):=\lfloor t\rfloor+f(t-\lfloor t\rfloor)$).<|endoftext|> -TITLE: Are larger large cardinals less expressible? -QUESTION [7 upvotes]: First note to the following well known theorems:‎‎ -Theorem (1): ‎The ‎notion ‎of ‎"‎$‎‎x$ ‎is a strongly inaccessible cardinal‎" ‎is ‎first ‎order ‎expressible ‎and ‎‎$‎‎\Pi_{1}$‎. -Theorem (2):‎‎ ‎The ‎notion ‎of ‎"‎$‎‎x$ is a measurable cardinal‎" ‎is ‎first ‎order ‎expressible ‎but ‎not ‎‎$‎‎‎\Pi‎_{1}$‎.‎ -Theorem (3): ‎The ‎notion ‎of ‎"‎$‎‎x$ is a Reinhardt cardinal‎" ‎is ‎not ‎first order expressible.‎ -‎ -Now there are some questions here:‎ -‎ -Question (1): ‎Are ‎larger ‎large ‎cardinals ‎more ‎complicated ‎in ‎first ‎order ‎expressibilty? ‎Is ‎there ‎any ‎exception?‎ -‎ -Question (2): ‎Is ‎there a‎ ‎non ‎first ‎orde‎r expressible large cardinal weaker than Reindhardt cardinal?‎ -‎ -Question (3): ‎What ‎is ‎the ‎largest $‎‎\Pi_{1}$ - ‎expressible ‎large ‎cardinal? ‎For ‎example the notions of being a ‎Mahlo ‎or ‎weakly ‎compact ‎cardinal ‎are ‎first ‎order ‎expressible ‎and‎ $‎‎\Pi_{1}$. - -REPLY [9 votes]: For question (1), there are many exceptions. For example, being superstrong is $\Sigma_2$ expressible, since it is witnessed inside a sufficiently large $V_\theta$, but this is stronger than strong in consistency strength, and being strong is $\Pi_3$. Similarly, being supercompact up to an inaccessible is $\Sigma_2$, but stronger than supercompact in strength, while being supercompact is $\Pi_3$. One can make many similar examples of very strong $\Sigma_2$ properties, by asserting that a strong $\Pi_3$ property holds up to an inaccessible cardinal. This reduces the complexity of the assertion, but is stronger consistency-wise. -For question (2), I would suggest Vopěnka's principle as a commonly considered large cardinal concept that, if consistent, is not first-order expressible as a single assertion in the language of set theory. This axiom is typically formulated in second-order theories such as GBC.<|endoftext|> -TITLE: Is there a truth approximation on a‎ cumulative hierarchy‎‎‎‎? -QUESTION [5 upvotes]: ‎‎‎Note ‎to ‎the ‎following ‎well known theorem:‎ -Theorem (1): ‎If ‎‎$‎‎\kappa‎‎$ ‎be a ‎‎"measurable" ‎cardinal ‎and ‎‎$‎‎‎\mathcal{F}‎$ be a‎ ‎"non-principal ‎$‎‎‎\kappa‎$-complete normal‎" ‎ultrafilter ‎on ‎it ‎then:‎‎ ‎‎‎‎$‎‎‎\langle V_{‎‎\kappa ‎+1‎},\in ‎‎\rangle ‎‎\cong ‎\prod_{‎\mathcal{F}‎}‎‎\lbrace ‎‎‎\langle ‎‎V_{‎\alpha +1‎}, \in \rangle‎~|~‎\alpha‎\in ‎\kappa‎ ‎\rbrace‎‎‎‎‎$ ‎‎‎‎ -Proof: ‎Chang and‎ ‎Kiesler, ‎Model ‎Theory, ‎Page ‎241.‎ -‎ -‎In ‎the ‎other ‎words ‎the ‎above ‎theorem ‎says ‎that ‎the ‎"truth" of a particular sentence ‎in ‎‎$‎‎‎\kappa +1‎$ th ‎level ‎of ‎von ‎Neumann's ‎cumulative hierarchy ‎is "‎dependent" ‎on ‎the ‎"truth" ‎of ‎that ‎sentence ‎in ‎lower ‎levels. ‎So ‎if a ‎sentence ‎be ‎"almost everywhere" ‎true ‎below ‎‎$‎‎‎\kappa ‎+1‎$ th ‎level, ‎it ‎cannot ‎be ‎false ‎in ‎‎it. In fact the lower stages give us a "truth approximation" for the ‎$‎‎‎\kappa +1‎$ th ‎stage. ‎Now ‎consider ‎the ‎following ‎definition: ‎‎ -‎ -Definition (1): ‎We ‎say ‎that a‎ ‎cumulative ‎hierarchy ‎$‎W‎$‎ ‎has a ‎‎‎truth approximation property at ‎$‎‎‎\delta‎$ ‎th level (‎$‎‎tap(W,‎\delta‎)$‎) ‎iff ‎there exists an ultrafilter ‎$‎\mathcal{F}‎‎‎$ ‎on ‎$‎‎\delta‎‎$ ‎such that:‎‎ ‎‎‎‎‎‎‎$‎‎‎\langle W_{‎‎\delta +1‎‎},\in ‎‎\rangle ‎‎\cong ‎\prod_{‎\mathcal{F}‎}‎‎\lbrace ‎‎‎\langle W_{‎\alpha ‎+1‎}, \in \rangle‎~|~‎\alpha‎\in ‎\delta‎ ‎\rbrace‎‎‎‎‎$ ‎‎ -‎ -Corollary (1): ‎$‎‎ZFC‎\Longrightarrow‎ \forall \kappa \in measurable~cardinal~~~tap(V,‎\kappa‎)$‎‎ -‎ -Corollary (2): ‎$‎‎ZFC + ‎\exists~a~measurable~cardinal ‎\Longrightarrow‎ ‎\exists ‎‎\delta>‎\omega‎~~~tap(V,‎\delta‎)‎‎‎‎$‎ -‎ -Now ‎there ‎are ‎some ‎natural ‎questions:‎ -‎‎‎ -Question (1): ‎Is ‎the ‎use ‎of non-trivial ‎‎$‎‎‎\kappa‎$-additive‎ normal measure ‎in ‎proof of ‎theorem ‎(1) ‎‎essential? ‎In ‎other ‎words ‎can ‎one ‎find a‎ ‎weaker ‎large ‎cardinal ‎axiom than existence of a measurable cardinal, ‎like ‎‎$‎A‎$ ‎such ‎tha‎t:‎ -‎ -(a) ‎$‎‎ZFC+A‎\Longrightarrow ‎‎\exists ‎‎\delta‎>‎\omega‎~~~tap(V,‎\delta‎)‎‎$‎‎ -‎ -Moreover can ‎‎‎$‎‎ZFC$ alone ‎prove ‎that ‎there is an ordinal ‎$‎\delta‎$‎ such that "‎$‎V‎$ ‎has a ‎truth approximation property at level ‎$‎\delta‎$‎‎‎"? Precisely ‎is ‎the ‎following ‎statement ‎true?‎ -‎ -(b) ‎‎$‎‎Con(ZFC)‎\Longrightarrow ‎Con(ZFC+‎\forall ‎‎\delta‎>‎\omega‎~~~\neg tap(V,‎\delta‎)‎)‎$‎‎ -‎Question (2): Is the inverse of corollary (2) true? In the other words does truth approximation property of von Neumann's cumulative hierarchy in a certain stage imply the existence of a measurable or weaker large cardinal? Precisely which one of these statements are true? -‎‎‎(a) $‎‎ZFC+‎\exists ‎‎\delta>‎\omega‎~~tap(V,‎\delta‎) ‎\Longrightarrow ‎‎\exists~‎a~strongly~inaccessible~cardinal‎‎$‎‎ -(b) ‎$‎‎ZFC+‎\exists ‎‎\delta>‎\omega‎~~~tap(V,‎\delta‎) ‎\Longrightarrow ‎‎\exists~‎a~measurable~cardinal‎‎$ -‎ -Question (3): ‎Are ‎there ‎any ‎known ‎"truth approximation properties" ‎for ‎other ‎famous ‎cumulative hierarchies ‎like ‎‎$‎‎L$ ‎and ‎‎$‎J‎$‎? Precisely is there any large cardinal axiom like ‎$‎‎A$ and ‎$‎B‎$‎ ‎which the following statements be true:‎ -‎ -(a) ‎‎$‎‎ZFC+A‎\Longrightarrow ‎‎\exists‎‎\delta>‎\omega‎~~~tap(L,‎\delta‎)‎‎‎$‎ -‎ -(b) ‎‎$‎‎ZFC+B‎\Longrightarrow ‎‎\exists‎‎\delta>‎\omega‎~~~tap(J,‎\delta‎)‎‎‎$‎ ‎‎‎‎ -‎ -More simplified, are there any large cardinals ‎$‎‎‎\kappa‎$ ‎and ‎‎$‎‎\lambda‎‎$ and ultrafilters ‎$‎‎\mathcal{F}‎$ ‎and ‎$‎‎‎\mathcal{G}‎$ ‎on them ‎‎‎‎such ‎that the following statements be true?‎ -‎ -(c)‎‎ ‎‎‎‎‎‎‎$‎‎‎\langle L_{‎‎‎\kappa‎‎ ‎+1‎},\in ‎‎\rangle ‎‎\cong ‎\prod_{‎\mathcal{F}‎}‎‎\lbrace ‎‎‎\langle L_{‎\alpha +1‎}, \in \rangle‎~|~‎\alpha‎\in ‎\kappa‎ ‎\rbrace‎‎‎‎‎$ ‎‎ -‎ -(d) ‎‎‎‎‎‎‎$‎‎‎\langle J_{‎‎‎\lambda‎‎ ‎+1‎},\in ‎‎\rangle ‎‎\cong ‎\prod_{‎\mathcal{G}‎}‎‎\lbrace ‎‎‎\langle J_{‎\alpha +1‎}, \in \rangle‎~|~‎\alpha‎\in ‎‎\lambda‎‎ ‎\rbrace‎‎‎‎‎$ ‎‎ -‎ - -REPLY [4 votes]: For question (1), the answer is the truth approximation property at $\delta$ implies the existence of a measurable cardinal. This is simply because the filter $\mathcal{F}$ witnessing your isomorphism must be countably complete, or else the ultraproduct on the right hand side will have an ill-founded $\omega$, preventing the isomorphism. But if an ultrafilter is countably complete, then the $\kappa$ for which it is $\kappa$-complete but not $\kappa^+$-complete is a measurable cardinal. That is, the degree of completeness of any countably complete ultrafilter is a measurable cardinal. -Thus, the converse of corollary 2 is true, and this answers question 2. -For question 3, you can take A and B to be "there is a measurable cardinal". The way I think about it is this. If $\delta$ is a measurable cardinal, with normal measure $\mathcal{F}$, then $\kappa$ is represented in the ultrapower $j:V\to M$ by the function $\alpha\mapsto\alpha$, and so $\kappa+1$ is represented by the function $\alpha\mapsto\alpha+1$. Thus, the $L_{\delta+1}$ of $M$, which is the real $L_{\delta+1}$, is represented by the function $\alpha\mapsto L_{\alpha+1}$, which gives your property (c). And similarly with $J_{\delta+1}$ and $\alpha\mapsto J_{\alpha+1}$, which gives (d).<|endoftext|> -TITLE: Variety $X$ such that $TX$ is ample on any curve in $X$ -QUESTION [9 upvotes]: Let $X$ be a smooth complex projective variety such that the restriction of $TX$ on any curve $C$ in $X$ is ample. Is true in this case that $X$ is isomorphic to $\mathbb CP^n$? -I guess the above condition implies that that $TX$ is nef (i.e. $O(1)$ is nef on $\mathbb P(TX)$), but it is not clear for me that this condition implies that $TX$ is ample (which then implies that $X\cong \mathbb CP^n$ according a theorem of Mori). - -REPLY [17 votes]: In this very famous paper: -Mori, Shigefumi, -Projective manifolds with ample tangent bundles. -Ann. of Math. (2) 110 (1979), no. 3, 593–606. -it is proven that over an algebraically closed field of characteristic 0 $\mathbb P^n(K)$ is the only manifold $X$ with ample tangent bundle. -In the introduction the author points out that the statement is true under the weaker conditions: 1) $-K_X$ is ample; 2) for any non constant map $\mathbb P^1\to X$ the pull back of $T_X$ is ample on $\mathbb P^1$.<|endoftext|> -TITLE: Reduction mod $p$ of units in a ring of integers -QUESTION [6 upvotes]: Let $\mathcal{O}_k$ be the ring of integers in an algebraic number field $k$ and let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_k$. I'm looking for conditions on $k$ and $\mathfrak{p}$ which will ensure that the image of the group of units $(\mathcal{O}_k)^{\ast}$ in $\mathcal{O}_k/\mathfrak{p}$ is all of $(\mathcal{O}_k/\mathfrak{p})^{\ast}$. More generally, what is known about this image? -My area of research is geometry and not number theory (the above problem arose while studying some geometric problems), so I apologize if the answer to the above is standard stuff. - -REPLY [6 votes]: I don't know of any result that specifically looks at units but there are a lot of results on looking at the image of a fixed finitely generated subgroup $G$ of $k^*$ in the units of the residue fields. The granddad of these questions is the Artin conjecture for primitive roots. There are results of Gupta and Murty (which probably have been improved) and guarantee the image is everything for infinitely many primes provided that the rank of $G$ is big enough. You'd have to look at the papers to see if "big enough" applies when $G$ is group of units. If you are willing to assume GRH then much stronger results are known.<|endoftext|> -TITLE: Consistency results using nonstandard models -QUESTION [7 upvotes]: Are there any consistency results in set theory (or in mathematics) that can be proved using nonstandard models of ZFC but not using transitive models of ZFC? - -REPLY [8 votes]: Yes. Harvey Friedman has identified several (natural) combinatorial statements that are equivalent to the $1$-consistency of $\mathsf{ZFC}$ or strengthenings of it via large cardinals (here, $1$-$\mathrm{Con}(T)$ is the assertion that all $\Sigma^0_1$ consequences of $T$ are true). In particular, this means that the statements in question hold in the standard model, and any model of their negation is necessarily non-standard (in fact, it is not an $\omega$-model). -See for example - -Harvey M. Friedman. Finite functions and the necessary use of large cardinals, Ann. of Math. (2), 148 (3), (1998), 803–893. MR1670057 (2002b:03108). - -Harvey's research has actually produced a nice variety of such statements. -If one relaxes your requirements that these be consistency results, and that any proof must necessarily involve non-standard models, a different example appears in the work of Greg Hjorth on descriptive set theory of equivalence relations, see - -Greg Hjorth. Thin equivalence relations and effective decompositions, J. Symbolic Logic, 58 (4), (1993), 1153–1164. MR1253912 (95c:03119). - -In his argument, Greg must analyze non-standard Ehrenfeucht-Mostowski models (coming from sharps). There is actually a different version of the argument, due to Sy Friedman and Boban Velickovic. Their argument also uses non-standard models, but their methods are different. I do not know, however, that this use is essential.<|endoftext|> -TITLE: Counting representations of $k[x,y]$ when $k$ is finite -QUESTION [13 upvotes]: $\newcommand{\GFq}{\mathbf F_q}$ -Let $r_n(q)$ denote the number of isomorphism classes of $n$-dimensional modules of the $\GFq$-algebra $\GFq[x,y]$. Is it known whether there exists a polynomial $p_n(t)$ (with say, integer coefficients) such that $r_n(q) = p_n(q)$ for all prime powers $q$? -Here are the values that I know (see arXiv:1212.6157v2): - -$r_1(q) = q^2$ -$r_2(q) = q^4 + q^3 + q^2$ -$r_3(q) = q^6 + q^5 + 2q^4 + q^3 + 2q^2$ -$r_4(q) = q^8 + q^7 + 3q^6 + 3q^5 + 5q^4 + 3q^3 + 3q^2$ - -Note: for the free algebra $\GFq\langle x, y\rangle$ this is well known, since it is the number of $n$-dimensional representations of the quiver with one vertex and two loops. In fact, it is also known now that this polynomial has non-negative integer coefficients (this follows from the work of Mozgovoy or Hausel, Letellier and Rodriguez-Villegas). - -REPLY [14 votes]: The number you seek is closely related to the number of commuting pairs of $n\times n$ matrices over $\mathbf{F}_q$. This is a polynomial and the formula was first given by Feit and Fine in a 1960 Duke paper. The stack of $\mathbf{F}_q[x,y]$ modules is the stack quotient of the commuting variety -$$C(n) = \{(A,B)\in End(\mathbf{F}_q^n)^2 : [A,B]=0\}$$ -by $Gl(n)$ acting by conjugation. The number you have asked for is the number of points in the coarse space of this stack. I don't know much about that number, but the "stacky" number of points is much better behaved --- in other words, instead of counting modules, you should be counting modules by 1 over the number of automorphisms of the module. That number is equal to the number of commuting pairs of matrices over $\mathbf{F}_q$ divided by the number of elements in $Gl(n,\mathbf{F}_q)$. The generating function of that number has a nice product expansion: -$$\sum_{n=0}^{\infty} \frac{|C(n)|}{|Gl(n)|} t^n = \prod_{k=1}^\infty \prod_{m=1}^\infty (1-q^{2-m}t^k)^{-1}$$ -This is in Feit and Fine's paper or you can find a more modern motivic derivation of this in my paper with Morrison. http://arxiv.org/abs/1206.5864<|endoftext|> -TITLE: inverse problem for ergodic measures -QUESTION [6 upvotes]: It is a basic fact in the weak-* topology, the set of invariant measures for a dynamical system is closed, compact, and convex in the weak-* topology. Furthermore, the set of ergodic measures is equal to the set of extremal points of the the set of invariant measure. -In the symbolic space, the set of ergodic measures are dense in the set of invariant measures. However, for the general dynamic model, there still lacks a clear picture of a characterization of ergodic measure. -According to my knowledge, there are few results to give more topological characterizations of ergodic measure. There is a very natural question to ask: for a given convex set $A$, can we realize it into a set of invariant measures for some dynamical system? To be precise, I am asking whether there exists a dynamical system whose invariant measure set is $A$. -It is obvious for a convex set in a finite dimension. an we say something about infinite dimensional convex sets? -Any reference and comments will be appreciated in advance. - -REPLY [12 votes]: Let $X$ be a compact metric space and $T\colon X\to X$ be a continuous map. The set of $T$-invariant Borel probability measures $\mathcal{M}_T(X)$ is well known to be non-empty, convex, compact, and metrizable. Moreover, its extreme points $\mathcal{M}^e_T(X)$ coincide with ergodic invariant measures and every invariant measure can be represented as a barycenter of a unique measure supported on $\mathcal{M}^e_T(X)$ by the ergodic decomposition theorem. This can be found in Walter's Introduction to Ergodic Theory [Theorem 6.10 and Remark (2) page 153] or in Downarowicz's Entropy in Dynamical Systems (Cambridge University Press 2011). Hence, $\mathcal{M}_T(X)$ is a Choquet simplex for every compact dynamical system. -It turns out that any abstract Choquet simplex is affinely homeomorphic to the set of invariant measures of some minimal dynamical system. -This is a result of Downarowicz [The Choquet simplex of invariant measures for minimal flows. Israel J. Math. 74 (1991), no. 2-3, 241--256], which completes the line of research begun in 50's (see references therein).<|endoftext|> -TITLE: Are these rings of functions isomorphic? -QUESTION [10 upvotes]: Let $R$ be the ring of all functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ which are continuous outside $(-1,1)$ and let $S$ be the ring of all functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ which are continuous outside a bounded open interval containing zero (depended on $f$). Is it true that $R \cong S$? - -REPLY [8 votes]: An idempotent $u$ is called primitive if the equation of idempotents $$(1-u)x=0$$ has a unique non-zero solution $x=u$. The primitive idempotents in $R$ are $\chi_{\{a\}}$ for $a \in (-1,1)$ and $\chi_{(-\infty,-1]}$ and $\chi_{[1,+\infty)}$. For ease of notation give two last mentioned primitive idempotents index $-1$ and $1$ respectively. The primitive idempotents in $S$ are indexed by $a \in \mathbb R$. -A ring isomorphism $\phi \colon R \rightarrow S$ would then define a bijection $\tilde \phi \colon [-1,1] \rightarrow \mathbb R$. Let $A=\tilde \phi^{-1}(\mathbb Q) \subseteq [-1,1]$, where $\mathbb Q$ denotes the rational numbers. Then $\chi_A$ is an idempotent in $R$ with the property that $\chi_A \cdot \chi_{\{a\}} \neq 0$ for all $a \in A$ and $\chi_A \cdot \chi_{\{a\}}= 0$ for all $a \in [-1,1] \smallsetminus A$. -The idempotent $\phi(\chi_A) \in S$ must be of the form $\chi_B$ for some $B \subseteq \mathbb R$. We must have $$\chi_B \cdot \chi_{\{\tilde \phi(a)\}}=\phi(\chi_A \cdot \chi_{\{a\}}) \neq 0$$ for all $a \in A$ and $$\chi_B \cdot \chi_{\{\tilde \phi(a)\}}=\phi(\chi_A \cdot \chi_{\{a\}}) = 0$$ for all $a \in [-1,1] \smallsetminus A$. Therefore $B$ contains all the rational numbers and no numbers that are not rational. But $\chi_{\mathbb Q}$ is not an element of $S$ and we reach a contradiction. Therefore there is no ring isomorphism $\phi \colon R \rightarrow S$.<|endoftext|> -TITLE: Integral with Dirac delta (me or wolfram mathematica?) -QUESTION [6 upvotes]: I asked the question on math.stackexchange but didn't get an answer so I came here. -I tried to compute with Wolfram Mathematica the following integral -$$I=\int_0^\pi\int_{-\infty}^\infty x e^{-\mathrm jx\cos(\theta-\varphi)}f(\alpha\cos(\theta-\psi))\mathrm \, \mathrm dx\mathrm \, \mathrm d\theta$$ -where $-\pi\leq\varphi,\psi\leq\pi$. -Assuming that the inner intergral is by definition the derivative of the Dirac delta function I get: -$$I=2\pi \mathrm j\int_0^\pi\delta'(\cos(\theta-\varphi))f(\alpha\cos(\theta-\psi)) \, \mathrm \, \mathrm d\theta$$ -Wolfram Mathematica tells me that -$$I_\mathrm{wolfram}=0$$ -Then I tried to do it by hand. If I use the definition of a delta function: -$$ -\begin{eqnarray} -\delta^{'} (\cos(\theta-\varphi))&=&\left[ \sum_i \frac{\delta(\theta-\varphi-\frac{\pi}{2}-i\pi)}{|\sin(\frac{\pi}{2}+i\pi)|}\right]^{'}=\sum_i \delta^{'}\left(\theta-\left(\varphi+\frac{\pi}{2}+i\pi\right)\right) -\end{eqnarray} -$$ -$$ -\begin{eqnarray} -I_\mathrm{me}&=&2\pi \mathrm j\sum_i\int_0^\pi\delta'\left(\theta-\left(\varphi+\frac{\pi}{2}+i\pi\right)\right)f(\alpha\cos(\theta-\psi)) \, \mathrm d\theta=\\ -&=&-2\pi \mathrm j \sum_i f'(\alpha\cos(\varphi+\frac{\pi}{2}+i\pi-\psi))u(\pi -2 \varphi ) u(2 \varphi +\pi ) -\end{eqnarray} -$$ -where $u(\cdot)$ is a Heaviside function. Here I used the fact that $$\int f(x)\delta'(x-a)\mathrm dx=-\int f'(x)\delta(x-a)\mathrm dx=-f'(a)$$ -And $I_\mathrm{me}$ is actually not $0$. -So where's the catch? - -REPLY [5 votes]: Since the solutions to this query and its twin on MSE are riddled with errors, I have decided to bite the bullet and give a rigorous treatment. Perhaps I can begin with the the general remark that most of the inquiries about distributions here and on its two siblings are about the Dirac distribution and its derivatives, usually in connection with their composition with a smooth function. As far as I can tell, none of the treatments that one finds employ a rigorous definition of this concept -(although this has been known for over 50 years) but rely on formal manipulations under the motto---if it works for smooth functions, then it will work for distributions. While this might be acceptable,even desirable, for a theoretical physicist, one expects more from a forum for research mathematicians. And it can lead to disaster, as the present case shows. -And so to the definitions. Here there is another pitfall. There are two such, one -as in Schwartz which uses duality and does not correspond to the usage of physicists---it generalises the notion of composition for measures. -The one we give here is simpler and does correspond to that of the physicists. It uses the fact that a distribution $T$ has the form $D^n F$ for some continuous function $F$. (This is true locally for every distribution and globally for the ones in question here). One then defines $T\circ \phi$ to be $(\frac 1{\phi'} D)^n F\circ \phi$ for suitable smooth functions $\phi$---more on this later. We remark that there is an issue here---one has to show that this is independent of the partular representation of the distribution as a (higher) derivative of a continous function. This is covered in the reference below. -If one applies this to $\delta'$, which is, up to a factor, the third derivative of $|x|$, one gets the formula -$$\delta'\circ \phi=\pm \frac 1{\phi'(0)^2}\delta',$$ -the sign being that of $\phi'(0)$. -This is valid for smooth diffeomorphisms with zero at the origin and whose derivative there is non zero. One can extend this to functions such as are required here (the cosine function) using standard partition of unity methods (called "recollement des morceaux" by Schwartz). The basic idea is to cover the real line with open intervals on each of which $\phi$is a diffeomorphism whose derivative is non-zero at any zero and then to sum the corresponding expressions over the zeros. -The integral in the OP can then be obtained in one line and takes on a simple and elegant form as an infinite sum over the zeros of the cosine terms---since I am typing this on my iPad I would rather leave the details to the poser. -Final remark: The above method is due to J. Sebastião e Silva and a detailed treatment can be found in "An Introduction to the Theory of Distributions" by J. Campos Ferreira, which is based on lectures given by the former in Lisbon in the late 60's.<|endoftext|> -TITLE: Applications of forcing in model theory -QUESTION [10 upvotes]: What are the major applications of (set theoretic) forcing in model theory? - -REPLY [7 votes]: There is some overload of the word forcing' in these replies.Model theoretic forcing' is a specific technique of Abraham Robinson, named in analogy with set theoretic forcing, but aimed primarily at least at first towards the construction of existentially closed models. The name has been used for numerous extensions since. But it should not be confused with applications of set theoretic forcing in model theory.<|endoftext|> -TITLE: Do all varieties have only finitely many etale covers of fixed degree -QUESTION [7 upvotes]: I've been wondering about the following "finiteness statement" concerning etale covers for a while. -Let $K$ be a field of characteristic zero, not necessarily algebraically closed. A variety over $K$ is a smooth quasi-projective geometrically connected scheme over $K$. -Let $X$ be a variety, and let $d$ be an integer. -Question. Are there only finitely many varieties $Y$ over $K$ such that there exists a finite etale morphism $Y\to X$ of degree $d$ (up to $K$-isomorphism)? -This is true if $K$ is algebraically closed. In fact, by a standard Lefschetz principle argument, we may and do assume $K$ is the field of complex numbers. Then, the statement follows from GAGA and the fact that the topological fundamental group of $X$ is finitely generated. -In general, there might be problems with twists, but I'm not sure. -Note that if $\dim X = 0$ then the only variety $Y$ satisfying the above hypotheses is $Y= $ Spec $K$, because we require our varieties to be geometrically connected (over $K$). -If it helps, in my set-up, we may assume $K$ is finitely generated over $\mathbf Q$, or that $K$ is even a number field. - -REPLY [6 votes]: Just to add one word to ulrich's answer: for a geometrically irreducible $K$-variety $X$, there is a short exact sequence of etale fundamental groups. -$$ -0 \to \pi_1^{\text{et}}(X\otimes_K \overline{K}) \to \pi_1^{\text{et}}(X) \to \text{Gal}(\overline{K}/K) \to 0. -$$ -An étale degree $d$ cover is equivalent to a homomorphism $\phi:\pi_1^{\text{et}}(X)\to \mathfrak{S}_d$. This homomorphism gives an irreducible cover of $X\otimes_K \overline{K}$ if the image of $\pi_1^{\text{et}}(X\otimes_K \overline{K})$ is a transtive subgroup of $\mathfrak{S}_d$. But for a given homomorphism $\phi_{\overline{K}}:\pi_1^{\text{et}}(X\otimes_K \overline{K}) \to \mathfrak{S}_d$ with transitive image, there may be many extensions of this to a homomorphism $\phi$. That does not in itself prove that there are infinitely many étale covers of, say $\mathbb{G}_{m,K}$. But it does help organize the infinitely many covers that ulrich listed.<|endoftext|> -TITLE: Solving $x^k+(x+1)^k+(x+2)^k+\cdots+(x+k-1)^k=(x+k)^k$ for $k\in\mathbb N$ -QUESTION [5 upvotes]: This question has been asked previously on math.SE without receiving any answers. -https://math.stackexchange.com/questions/479740/solving-xkx1kx2k-cdotsxk-1k-xkk-for-k-in-mathbb-n -Letting $k$ be a natural number, can we solve the following $k$-th degree equation ? -$$x^k+(x+1)^k+(x+2)^k+\cdots+(x+k-1)^k=(x+k)^k\ \ \ \ \cdots(\star).$$ -The following two are famous: -$$3^2+4^2=5^2, 3^3+4^3+5^3=6^3.$$ -I've tried to find the other integers which satisfy $(\star)$, but I can't find any nontrivial solution. Then, I suspect that the following might be proven true. -My expectation: There is no integer which satisfies $(\star)$ except $(k,x)=(2,-1),(2,3),(3,3)$. -The followings are what I found: -1. In $k=4$ case, there is no integer which satisfies $(\star)$. -Proof: Suppose that there exists an integer $x$ which satisfies $(\star)$ -Then, considering in mod $4$, we reach a contradiction in each remainder, so the proof is completed. -2. Supposing that the one's digit of $k$ is $1$, then there is no integer which satisfies $(\star)$. -Proof: In $k=1$ case, it's obvious. Letting $k=10n+1$ ($n$ is a natural number), let's consider in mod $5$. Let $a_l=l^k$ (mod $5$) ($l$ is an integer). -(i) The $n=1,3,5,\cdots$ cases : -$$a_{5m}\equiv 0, a_{5m+1}\equiv 1, a_{5m+2}\equiv 3, a_{5m+3}\equiv 2, a_{5m+4}\equiv 4.$$ -(ii) The $n=2,4,6,\cdots$ cases : -$$a_{5m}\equiv 0, a_{5m+1}\equiv 1, a_{5m+2}\equiv 2, a_{5m+3}\equiv 3, a_{5m+4}\equiv 4.$$ -Suppose that there exists an integer $x$ which satisfies $(\star)$. Letting $l=x+k-1$, we get -$$(0+1+2+3+4)\times 2n+a_l \equiv a_{l+1}\ \Rightarrow\ a_l\equiv a_{l+1}.$$ -in both (i) and (ii). However, this doesn't happen because of the above. Hence, the proof is completed. -3. Suppose that $k+1$ is a prime number. If there exists an integer $x$ which satisfies $(\star)$, then $k=2$. -Proof: Let's consider when $k=p-1$ ($p$ is a prime number which is more than or equal to $5$). We are going to consider in mod $p$. By Fermat's little theorem, note that -$$a^{p-1}\equiv 0 (a\equiv0), 1(a\not\equiv 0).$$ -Suppose that there exists an integer $x$ which satisfies $(\star)$. -(i) The $x\not\equiv1$ case (the multiples of $p$ exists in the integers from $x$ to $x+p-2$): we get -$$0+1\times (p-2)\equiv 1.$$ -However, this doesn't happen because $p\ge5$. -(ii) The $x\equiv1$ case (there is no multiples of $p$ in the integers from $x$ to $x+p-2$): we get -$$1\times (p-1)\equiv 0.$$ -However, this doesn't happen. Hence, the proof is completed. - -REPLY [9 votes]: Here are a few partial results (the most interesting is probably (3)) towards this problem, but I don't have any hope that these will lead to a general solution. Your problem is a little similar to the famous Erdos-Moser question (which still remains unsolved): does there exist a solution to $1^n+2^n+\ldots+k^n= (k+1)^n$ apart from $1+2=3$? -$1$. If $k$ is even and there is a solution to the equation, then $k \equiv 2\pmod{16}$. -Proof: We may suppose that $k\ge 4$, and let $2^a$ exactly divide $k$. Then $x^k+ \ldots+(x+k-1)^k \equiv k/2 \pmod{2^{a+2}}$, while $(x+k)^k$ is either $0$ or $1 \pmod{2^{a+2}}$. It follows that $a=1$ and that $k/2\equiv 1\pmod 8$, or $k\equiv 2\pmod {16}$ as claimed. -$2$. If $k\equiv 1\pmod 4$ then the equation has no solutions. For $(x+1)^k+\ldots+(x+k-1)^k$ must be even, whereas $(x+k)^k-x^k$ is odd. -$3$. If $k\ge 5$ is odd and squarefree then the equation has no solutions. -Proof: Let $p$ be a prime dividing $k$. Note that $(p-1)\nmid k$. We make use of the fact that if $p$ is odd and $(p-1)\nmid k$ then $1^k+2^k+\ldots+(p-1)^k +p^k\equiv 0\pmod {p}$; this follows easily by picking a primitive root $\pmod p$ and summing the resulting geometric series. Therefore $x^k+\ldots+(x+k-1)^k \equiv 0\pmod p$, and so if this equals $(x+k)^k$ then $p$ must divide $x$. Since $k$ was assumed to be square-free, we conclude that $k|x$. But now note that $0^k+1^1+\ldots+(k-1)^k < k^k$ (for $k\ge 5$), and that -$(\ell k)^k+\ldots + (\ell k +k-1)^k > ((\ell+1)k)^k$ for all $\ell \ge 1$ and $k\ge 5$. -$4$. If $k$ is odd, and $p$ is an odd prime dividing $k-1$ such that $(k,p-1)=1$ then the equation has no solutions. In particular if $k \equiv 1\pmod{6}$, or $1\pmod {10}$ or $1\pmod{p(p-1)}$ for any odd prime $p$, there are no solutions. -Proof: If $p|(k-1)$ and $(p-1)\nmid k$ then we find from the observation used in (3) that -$(x+1)^k+\ldots+(x+k-1)^k \equiv 0 \pmod {p}$. Thus $(x+k)^k \equiv x^k \pmod p$. Since $(k,p-1)=1$ by assumption the map $x\to x^k \pmod {p}$ permutes all the residue classes $\pmod p$. Thus it follows that $(x+k)\equiv x \pmod p$, but this is impossible since $x+k \equiv x+1 \pmod p$.<|endoftext|> -TITLE: How fast can we numerically calculate Kloosterman sums? -QUESTION [15 upvotes]: Define the usual Kloosterman sum by $$S(m,n;c) = \sum_{\substack{x \pmod{c} \\ (x,c) = 1}} e\Big(\frac{mx + n\overline{x}}{c}\Big),$$ -where $x \overline{x} \equiv 1 \pmod{c}$, and $e(x) = e^{2 \pi i x}$. I'm interested in knowing if there is a faster way to calculate values of this rather than just adding up the $\varphi(c) \leq c$ values of the sum. By calculate here, I just mean to find a good numerical approximation; the definition already gives an exact expression for it as an element of $\mathbb{Z}[e^{2\pi i/c}]$ which probably can't be improved by much in general. I'm interested in $m$ and $n$ fixed, and how fast it can be computed in terms of $c$. The case $c=$prime is probably the most interesting. I'd also be interested to know if people have considered the speed of evaluation of the full list of values of $S(m,n;c)$ for $c \leq C$. -For the case of a Gauss sum to modulus $q$, I have heard that it is "well-known" how to calculate it in around $\sqrt{q}$ steps as follows. The main point is to use the fact that the completed Dirichlet $L$-function has the approximate functional equation (See Theorem 5.3 of Iwaniec and Kowalski's book) -$$L(1/2, \chi) = \sum_{n} \frac{\chi(n)}{\sqrt{n}} V(\frac{n}{X \sqrt{q}}) + \epsilon(\chi) \sum_{n} \frac{\overline{\chi}(n)}{\sqrt{n}} V(nX/\sqrt{q}).$$ -Here $V$ is a special function that in practice can be calculated (e.g. in Mike Rubinstein's Lcalc package), $\epsilon(\chi) = i^{-a} \tau(\chi) q^{-1/2}$ where $a$ is $0$ or $1$ if $\chi$ is even or odd, $\tau(\chi)$ is the Gauss sum, and $X> 0$ is a parameter to choose. Write this in the form $L(1/2, \chi) = A(X) + \epsilon(\chi)B(X)$. -Taking say $X=1$ and another value of $X$ close to $1$ will give different values for $A(X)$ and $B(X)$, and so one can solve for $\epsilon(\chi)$. The function $V(x)$ has rapid decay for $x \gg 1$, so it takes around $\sqrt{q}$ terms in the sum to evaluate $A(X)$ and $B(X)$, far fewer than the full $q$ terms needed to evaluate the Gauss sum by definition. - -REPLY [8 votes]: This is not quite an answer to your interesting question, but rather a couple -of related comments. First your nice "well known" observation on the Gauss sum -can perhaps be expressed more simply by noting that if $\chi(-1)=1$ then Poisson summation gives -$$ -\tau(\overline{\chi}) \sum_{-\infty}^{\infty} \chi(n) e^{-\pi n^2/q} = -\sqrt{q} \sum_{-\infty}^{\infty} \overline{\chi}(n) e^{-\pi n^2/q}, -$$ -so that about $\sqrt{q}$ terms are needed to approximate $\tau(\chi)$. A similar argument of course holds if $\chi(-1)=-1$, and in any case this is closely related to what you wrote. -Secondly, one can consider a different related problem: namely rapid algorithms for counting points on curves over finite fields. For elliptic curves there is a rapid algorithm of Schoof which allows this in polynomial time, but for general curves I don't think an algorithm faster than the naive one is known. One survey is Elkies http://www.math.harvard.edu/~elkies/modular.pdf but this is now twenty years old so perhaps -much more is known. Schoof's algorithm is useful in computing Kloosterman sums over -large finite fields of characteristic two or three: see work of Lisonek and -others on this topic http://link.springer.com/chapter/10.1007%2F978-3-540-85912-3_17 -The above is somewhat orthogonal to the problem you have in mind. My only other thought -was to try to use divisor function in arithmetic progressions to compute Kloosterman sums, but I did not obtain anything nontrivial.<|endoftext|> -TITLE: Extending a partial order while preserving an automorphism -QUESTION [11 upvotes]: It is well known that if $(P, \leq)$ is a partial order then $\leq$ can always be extended to a linear order. This is sometimes called Szpilrajn´s theorem although it had been previously proved by Banach, Kuratowski and Tarski. -Now suppose that $f$ is an automorphism of $(P, \leq)$ and we want to extend $\leq$ to a linear order in such a way that $f$ remains an automorphism. Of course, this is not always possible since $f$ could have a finite orbit and automorphisms of linear orders can´t have finite orbits; but I wonder if this is the only obstruction. So let $A$ be the collection of all those $f$´s for which it is possible (for instance $Id_P \in A$); here are my questions: -1) Is it true that If $f$ has no finite orbits then $f \in A$? -2) Is $A$ a subgroup of $Aut(P, \leq)$? -and the vaguer -3) If the answer to 1 is negative. Can we somehow characterize the elements of $A$? -Perhaps this is all well known and studied but I couldn´t find anything at all in the literature, so references are also appreciated. - -REPLY [5 votes]: I couldn't find anything in the literature either, but the answer to the first question is positive. Let $G$ be a group acting on a space $X$. Say that $G$'s action on $X$ has the invariant order-extension property provided that every $G$-invariant partial order on $X$ (i.e., partial order $\le$ such that $x\le y$ iff $gx\le gy$, for each $g\in G$) extends to a $G$-invariant linear order on $X$. -I managed to show: -Theorem. The following are equivalent for an abelian group $G$: - -$G$'s action on $X$ has the invariant order extension property. -$X$ has a linear $G$-invariant order. -No element of $G$ has an orbit of finite size greater than one. - -Here's the proof. -The answer to Question 1 follows by letting $G$ be the subgroup of permutations generated by $f$. -(By the way, I assume that your no-finite-orbits condition really means: no finite orbits of size greater than one. Fixed points aren't a problem.)<|endoftext|> -TITLE: How to efficiently sample uniformly from the set of p-partitions of an n-set? -QUESTION [11 upvotes]: Let $n,p \in \mathbb{N}_+$ with $p \leq n.$ Let $\mathcal{P}$ denote the set of partitions of $\{1, \ldots, n\}$ into $p$ nonempty sets. How can I efficiently sample uniformly from $\mathcal{P}$? - -REPLY [9 votes]: The sets you're interested in are counted by Stirling Numbers of the Second Kind, which satisfy the recursion -$$\left\{{n \atop k}\right \}=\left\{{n-1 \atop k-1}\right \}+k \left\{{n-1 \atop k}\right \}$$ -Here the first term represents those partitions where $n$ is its own set, and the remaining term represents inserting $n$ into a partition of $\{1, \dots, n-1\}$. This recursion can also be used to generate a set partition recursively: -With probability $\left\{{n-1 \atop k-1}\right \}/\left\{{n \atop k}\right \}$ put $n$ in its own set, and make the rest a partition of $n-1$ elements into $k-1$ sets chosen uniformly at random. Otherwise generate a uniform random partition of $n-1$ elements into $k$ sets, and insert $n$ into a set uniformly chosen from those sets. -The algorithm would run in time on the order of $nk$, with the main overhead being computing (or looking up) all of the Stirling Numbers up to $\left\{{n \atop k}\right \}$ at the start of the algorithm.<|endoftext|> -TITLE: redundant relations -QUESTION [10 upvotes]: The situation is the following: I have a group of matrices (given by generators), and I am trying to find a presentation. Now, this is in general very hard (undecidable?) but here I want to know if people know heuristics for pruning the set of relators -- in other words, by computing further generations of the group, I get a number of words $w_1, \dots, w_k$ which represent the identity. No doubt some of these relations are consequences of others, so is there some nice way to remove the "dependent relations"? - -REPLY [11 votes]: This doesn't answer the main question, but it does address the parenthetical question of the decidability of computing presentations in matrix groups. In so doing, I hope it helps to clarify Joel and Benjamin's answers. -Bridson and I showed that the problem of computing presentations in matrix groups over $\mathbb{Z}$ is undecidable, in this paper. More precisely, we produce a recursive sequence of subsets $S_n\subseteq SL_{m_n}(\mathbb{Z})$ and an r.e. sequence of integers $r_n$ such that: - -for each $n$, $\langle S_n\rangle$ is finitely presentable (in principle!), but -the set -$\{n\in\mathbb{N}\mid b_1(\langle S_n\rangle)=r_n\}$ -is r.e. but not recursive. - -So $b_1(\langle S_n\rangle)$ (the first Betti number) is not computable. Since it's straightforward to compute $b_1$ from a presentation, it follows that presentations are not computable. -Unfortunately, in our examples, $m_n\to\infty$ as $n\to\infty$. It would be great to have examples in which $m_n$ was bounded (and even better to have examples in which $m_n=3$). - -REPLY [7 votes]: Let's ignore the source of the group, and concentrate on the question you seem to be asking. Let $G = \langle x_1,x_2,\ldots,x_r \mid r_1,r_2,\ldots,r_s \rangle$ be a finite group presentation, where the $r_i$ are defining relators. Can we decide whether $r_s$ is redundant; that is, whether $G = \langle x_1,x_2,\ldots,x_r \mid r_1,r_2,\ldots,r_{s-1} \rangle$. -As others have pointed out, the theoretical answer is no, but the problem is a semi-decidable probelm in the sense that, if the answer is yes, then you can prove constructively that it is yes. In practice there are two well studied and implemented algorithms that you can use to try and do this. -The first is Todd-Coxeter coset enumeration, which is most useful when the groups are finite, so that is probably not the best choice here. -The second is Knuth-Bendix completion, and that is the most likely to be useful here. You run it on the presentation $\langle x_1,x_2,\ldots,x_r \mid r_1,r_2,\ldots,r_{s-1} \rangle$. It will probably not halt, but that doesn't matter. Just run it for a few minutes or a few days, or whatever. Then interrupt it (or wait until it exceeds some bound - I usually tell it stop when it has produced a million rewrite rules or something). Then you can quickly check whether $r_s$ reduces to the identity under the rewrite rules that you have generated. If $r_s$ really is a consequence of the other relators, then you know that this method will work if you run the program for long eneough but, except in the rare cases when the process completes with a confluent presentation, you cannot use this method to prove the answer is no. -The only practical ways of showing that $r_s$ is not a consequence of the other relators is to compute some of the quotients of $G$ and see if you get different results by leaving out $r_s$. There are algorithms available for computing finite quotients, virtually abelian quotients, virtually nilpotent quoteints, etc. -Knuth-Bendix completion and the quotient algorithms are available in both GAP and Magma.<|endoftext|> -TITLE: Monoidal transformations are isomorphisms at dualizable objects -QUESTION [7 upvotes]: Here is a cute observation: Let $F,G : \mathcal{C} \to \mathcal{D}$ be a symmetric monoidal functors between symmetric monoidal categories, and let $\eta : F \to G$ be a monoidal transformation. Then for every dualizable object $V \in \mathcal{C}$ the morphism $\eta_V : F(V) \to G(V)$ is actually an isomorphism! The inverse is given by -$G(V) \cong G(V^*)^* \xrightarrow{\eta_{V^*}*} F(V^*)^* \cong F(V).$ -Of course one has to check that this is, in fact, inverse to $\eta_V$. A nice corollary: If $F,G$ are cocontinuous, and every object of $\mathcal{C}$ is a colimit of dualizable objects (which happens quite often), then every monoidal transformation $F \to G$ is an isomorphism, i.e. $\mathrm{Hom}_{c\otimes}(F,G)$ is a groupoid. -Is this well-known? Does somebody know a reference for this observation? The proof is not so hard, but this is one of dozens of lemmas which I need in my thesis and with which I don't want to waste time / place with the proof. In an example this appears in Remark 2.4.7 in Lurie's On the classification of topological field theories. - -REPLY [9 votes]: This goes back at least to Saavedra-Rivano "Categories Tannakiennes." In fact, this has a generalisation to Frobenius functors, in Day-Pastro "Note on Frobenius monoidal functors," and to more general settings, as monoidal bicategories: Lopez Franco, Street, Wood; "Duals Invert," (App. Categor. Str. Vol 19).<|endoftext|> -TITLE: Is compass and straight edge geometry complete? -QUESTION [8 upvotes]: Euclid's first three postulates are the basis of compass and straight edge constructions which are as complex as arithmetic. -The constructions themselves may be expressed as a formula with each of the six operations of addition, subtraction, multiplication, division, complex conjugate, and square root corresponding to a simple compass and straight edge construction. - -Q1 Do Gödel's incompleteness theorems apply in the feild of Compass-and-straight edge-constructions ? -Q2 Do we have any problems in compass and straight edge constructions which have neither been constructed nor proved impossible, and may just be un-decidable ? - -The question asked nearly two months back here on maths.SE got many opinions but no answers.> -The opinions ranged from addition of new postulates to Tarski's Elementary Geometry (in order to make it inconsistent) to finding a (weak) subset of Peano Arithmetic which can be encoded as a statement in Euclidean Geometry. -EDIT -We talk of statements such as - - - -$\mathcal construct$ regular $(2^{2^n}+1)$-gon : -undecided; maybe undecidable ( taken from a comment by @AngelaRichardson, based on Gauss–Wantzel theorem) -$\mathcal construct$ regular n-gon : decidable; True -$\mathcal construct$ trisection of angle $\theta$: decidable; -False(Is this how you classify impossible constructions) - - Motivation was to introduce undecideability without a direct referance to arithmatic not withstanding the fact that each of the above bullets would have an equivalent arithmetical statement with the same decideability status; -constructability here may take into account the decideability of the equivalent arithmetical statement or the Construction itself based, maybe , on a decision routine for Tarski's Elementary geometry]. -Besides geometry, dynamical systems is another field which can be used to state decidability problems ; of course the essential language remains that of arithmetic but the context changes. ( Decidability in dynamical systems ) - -REPLY [7 votes]: Here is a precise formulation, which shows the problem to be open. -Let $L$ be the language of Tarski-style geometry, with the two relations $Collinear(p,q,r)$ and $Equidistant(p,q,r,s)$. -Let $E$ be the constructible numbers, and $R$ the real numbers. Then $E$ and $R$ are structures for $(+,\times)$, and $E^2$ and $R^2$ are $L$-structures. $Th(E^2,L)$ is the first-order theory of $E^2$ in the language $L$. -Then I believe: $Th(E^2,L)$ is decidable iff $Th(E,+,\times)$ is decidable. Whether that decidability holds is an open question; if so, there is a complete recursive axiomatization of that theory. -I can envision a decision procedure for simple sentences about $E^2$, which would rely heavily on the algorithm for deciding sentences about $R^2$. We can enumerate constructions and algebraic symmetries, enough to either verify constructibility, or show an element of odd order in the Galois group that prevents constructibility. Given a potential construction or symmetry, we can check if it has the right properties using the Tarski procedure. I've never seen this all written down, but in principle someone could have done it all in the 50's. -For the particular sentences you mention, the sentences and decision procedure would look like this. - -For trisectability, we have the sentence "For any angle $PQR$, there are congruent angles $PQX$, $XQY$, $YQR$ inside it." We decide that this is false in $E^2$ by finding the element of odd order in the appropriate Galois group. -For constructiblity, for each $n$, we have the sentence that "there exist $P_1\ldots P_n$ forming a regular $n$-gon". We decide this either by finding the construction, if $n$ is of the appropriate form, or by finding the element of the Galois group, as before. - -This account of the question isolates whether constructibility adds any incompleteness or undecidability to geometry. It avoids immediately dragging in the Godel phenomena of the integers. Whether we can define the integers in $E$ or $E^2$ is the content of the open question.<|endoftext|> -TITLE: What is the relationship between the bramble number and the strict bramble number of a graph? -QUESTION [10 upvotes]: A bramble in a graph $G$ is a set of connected subgraphs $H_1, \dots, H_m$ such that for every $i, j$, either $H_i$ intersects $H_j$ in a vertex, or there exists an edge of $G$ with one end in $V(H_i)$ and one end in $V(H_j)$. The order of the bramble is the minimum $|X|$ such that $X \subseteq V(G)$ and $X \cap V(H_i) \neq \emptyset$ for all $i$. The bramble number of $G$ is the maximum order of a bramble in $G$. Denote it $Br(G)$. -Say a strict bramble in a graph $G$ is a set of connected subgraphs $H_1, \dots, H_m$ such that for all $i, j$, $H_i$ intersects $H_j$ in a vertex. Definite analogously the order of a strict bramble and the strict bramble number of a graph, and denote the strict bramble number by $sBr(G)$. -Clearly, the bramble number is at least the strict bramble number. They also won't be equal in general. For example, in $K_t$, the bramble number is $t$, but the strict bramble number will be essentially $t/2$. -Is this tight? Is it true that $sBr(G) \ge Br(G)/2$? The grid theorem implies that there is a function $f$ such that the $sBr(G) \ge f(Br(G))$, but it might be true as well with a linear function. - -REPLY [8 votes]: Yes, $sBr(G) \geq Br(G)/2$ for all graphs $G$. -To see this, let $\mathcal{Y}:=Y_1, \dots, Y_m$ be a bramble of $G$ with a minimum hitting set $X$ such that $|X|=br(G)$. It will be slightly more convenient to let $Y_i$ be sets of vertices which induce connected subgraphs. For each $x \in X$, let $\mathcal{Y}_x$ be the sets of $Y_1, \dots, Y_m$ that $x$ belongs to. For each subset of $X'$ of $X$, let $\mathcal{F}(X')$ be the family of sets of the form $\bigcup_{x \in X'} Y_x$, where $Y_x \in \mathcal{Y}_x$. Finally, let $\mathcal{B}$ be the family of sets obtained by taking the union of $\mathcal{F}(X')$ over all subsets $X'$ of $X$ of size $k:=\lfloor Br(G)/2\rfloor+1$. -Note that since $\mathcal{Y}$ is a bramble, each set in $\mathcal{B}$ does induce a connected subgraph. Moreover, $\mathcal{B}$ is a strict bramble since any two sets in $\mathcal{B}$ meet in a vertex of $X$. Finally, I claim that every hitting set of $\mathcal{B}$ has size at least $Br(G)/2$. If not, then there is a hitting set $Z$ for $\mathcal{B}$ of size $k' -TITLE: Character of normal ultrafilters -QUESTION [7 upvotes]: The character of an ultrafilter $U$, denoted $\chi(U)$, is the minimal size of an $A \subseteq U$ such that $(\forall x \in U ) (\exists y \in A) y \subseteq x$. This cardinal characteristic has been studied for ultrafilters on $\omega$. For all nonprincipal $U$, $\omega_1 \leq \chi(U)$, and it is known that, consistently, there exists nonprincipal $U$ with $\chi(U) < 2^\omega$. My question is, for normal ultrafilters $U$ on a measurable cardinal $\kappa$, is it possible that $\chi(U) < 2^\kappa$? -Here are some relevant papers about ultrafilters on $\omega$, by Shelah, Brendle, and Hart: -MR1686797, MR0987317, MR2365799, MR2847327 -(available through http://www.ams.org/mathscinet/) - -REPLY [6 votes]: To get the ball rolling... -One can show easily that $\chi(U)$ must be at least $\kappa^+$, since otherwise one can take the diagonal intersection of a $\kappa$-sized family and find a single set that supposedly generates $U$, which is impossible. -Thus, your situation would require that $\kappa$ is measurable and $2^\kappa\gt\kappa^+$, a situation already whose consistency strength strictly exceeds that of a measurable cardinal. Silver showed that this is consistent relative to a $\kappa^+$-supercompact cardinal, but it is now known to be equiconsistent with a cardinal $\kappa$ that is $\kappa^{++}$-tall. But it follows from this that one cannot prove that your situation is consistent if one starts only from the assumption that measurable cardinals are consistent (unless that assumption is inconsistent). We need to use a stronger hypothesis. -But it is conceivable that we might hope to mimic the methods on $\omega$ higher up, if we start with a supercompactness assumption on $\kappa$....<|endoftext|> -TITLE: Singularizing forcing of "small" cardinality? -QUESTION [16 upvotes]: Can there be a large cardinal $\kappa$ and a forcing of size $\kappa$ that makes $\kappa$ a singular cardinal? The motivation is that the standard Prikry forcing does not have a dense set of size $\kappa$. -Edit: -In response to some attempts at a positive answer, let me explain something that does not work. If $\mathbb{P}$ is the Prikry forcing and $\mathbb{Q}$ is something like $Coll(\kappa,2^\kappa)$, one may expect under suitable indestructibility hypotheses, $\mathbb{P}$ works in $V^\mathbb{Q}$. But this never works. -The following lemma is based on an exercise in Kunen's book: Suppose $\kappa$ is a singular cardinal and $\mathbb{R} = \{ f : f$ is a partial function from $\kappa$ to $2$ with domain bounded below $\kappa \}$, ordered by extension. Then $\mathbb{R}$ collapses $\kappa$ to $cf(\kappa)$. -Proof: Suppose for simplicity $cf(\kappa) = \omega$, and let $\langle \kappa_n : n \in \omega \rangle$ be an increasing cofinal sequence. If $G \subseteq \mathbb{R}$ is generic, define in $V[G]$ the function $f : \omega \to \kappa$ by $f(n) = \beta$ where $\beta < \kappa_n$ and for some $\delta$, the ordinal $\kappa_n \cdot \delta + \beta$ is the $\kappa_n$-th element of $\{ \alpha : \bigcup G(\alpha) = 1 \}$. A simple density argument shows that $f$ is surjective. -Now let $\kappa$ be our large cardinal. It follows from a general folklore fact that there is a dense embedding $e : Add(\kappa,1) \times Coll(\kappa,2^\kappa) \to Coll(\kappa,2^\kappa)$. After forcing with $\mathbb{P}$, the $Add(\kappa,1)$ of the ground model becomes the forcing with bounded functions from the lemma, and the map $e$ is still a dense embedding. So if $G \times H$ is $\mathbb{P} \times \mathbb{Q}$-generic, then by the lemma, $\kappa$ is collapsed to $\omega$. Therefore in $V^\mathbb{Q}$, $\mathbb{P}$ collapses $\kappa$ to $\omega$. -I suspect that if a positive answer is possible, the forcing must be significantly different from the standard Prikry forcing or some combination of it with simple forcings. - -REPLY [11 votes]: The answer to your question is no. We have the following theorem. - -Theorem. Suppose $\kappa$ is a regular uncountable cardinal and $|P|=\kappa.$ Then $\Vdash_P cf(\kappa)=|\kappa|.$ - -Proof. Let $\tau$ be a name of an unbounded subset of $\kappa.$ We show there is $f\in V, f:\kappa\to\kappa$ such that $\Vdash_P f''[\tau]=\kappa.$ The result will follow immediately. -Let $(p_i: i<\kappa)$ enumerate $P$. Define by induction, for each pair $(i, j)\in \kappa\times \kappa$ an ordinal $\alpha_{ij}<\kappa$ and a condition $q_{ij}$ such that: -(1) $(i', j') < (i,j) \implies \alpha_{i'j'} < \alpha_{ij},$ (where $<$ denote any well ordering of $\kappa\times\kappa$ of order type $\kappa,$ say like Godel ordering) -(2) $q_{ij}\leq p_i$ and $q_{ij}\Vdash \alpha_{ij}\in \tau.$ -Then $f$ defined by $f(\alpha_{ij})=j$ is as required as can be proved easily. --- -Note on 23/07/2015: I realized that the result with the same proof has appeared in Hiroshi Sakai's paper ``Semiproper ideals'' as Fact 2.2.<|endoftext|> -TITLE: Does a section of a morphism of schemes give a subscheme? -QUESTION [6 upvotes]: Let $f:X\rightarrow Y$ be a morphism of schemes (or algebraic spaces), and $s:Y\rightarrow X$ is a section to $f$, i.e. $fs=1_{Y}$. -Question: Is $s$ an (closed) immersion? - -REPLY [4 votes]: Prop. If $f\colon X\longrightarrow Y$ is a separated morphism of schemes, then every section $\sigma\colon Y\longrightarrow X$ is a closed immersion. -Proof. Let $\sigma\in X(Y)$ be a section of the $Y$-scheme $X$. Let's take the fibered product $\left(X\times_X Y,p_X,q_X\right)$, where we are considering $Y$ as a $X$-scheme endowing it with the structural morphism $\sigma\colon Y\longrightarrow X$. The projections $p_X$ and $q_X$ induce a closed immersion of $Y$-schemes $X\times_X Y\longrightarrow X\times_Y Y$, since $\ast$ $f\colon X\longrightarrow Y$ is separated. In a similar way, the identity map of $Y$ and the section $\sigma$ induce a morphism $Y\longrightarrow X\times_X X$ of $X$-schemes. In other words we have the following commutative diagram. - -At this point the solution of the exercise follows by the fact that a composition of closed immersions is a closed immersion and by the fact that the morphisms $q^{-1}_X$ and $p^{}_Y$ are isomorphisms, then, in particular, closed immersions. -$\ast$ Let $Y$ be a separated $Z$-scheme. Then for any $Y$-schemes $X_1,X_2$, the canonical morphism $X_1\times_Y X_2\longrightarrow X_1\times_Z X_2$ is a closed immersion. -The claim above is the Proposition 3.9 (f) at pag.101 of Algebraic Geometry and Aritmetic Curves.<|endoftext|> -TITLE: Proving ZFC results using large cardinals -QUESTION [9 upvotes]: There are many $ZFC$ results that their proof uses forcing: The idea is that we force the statement to be true, and then using absoluteness (or other reasons) we conclude that the result is true in $ZFC$. -Question. Are there any $ZFC$ results such that in their proof we use large cardinals (and then remove the use of that large cardinal by some arguments like absoluteness)? - -REPLY [4 votes]: I am not sure if this is a sort of example you asked for but it seems interesting so I will mention it. There are several ZFC results that are proved in the following way: You assume the result does not hold and that gives you a highly saturated ideal (over a small large cardinal which Fremlin calls quasi measurable). Then you force with this ideal, form the well founded generic ultrapower in the extension and argue towards a contradiction. An example of such a theorem is: For every sequence $\{X_n : n < \omega\}$ of subsets of $[0, 1]$ there is a disjoint refinement of full outer measure, i.e., there is a sequence $\{Y_n : n < \omega\}$ of pairwise disjoint sets such that $Y_n \subseteq X_n$ and $X_n$ and $Y_n$ have same (Lebesgue) outer measure. This is due to Gitik and Shelah. Another such application is: For every $X \subseteq [0, 1]$, there is a $Y \subseteq X$, such that $X$ and $Y$ have same outer measure and the distance between any two points of $Y$ is irrational - So $Y$ is Vitali like inside $X$ and it has full outer measure.<|endoftext|> -TITLE: What are some characterizations of the strong and total variation convergence topologies on measures? -QUESTION [9 upvotes]: I asked this question on StackExchange a few days ago but didn't get any response, so I thought I would try here. -The Wikipedia article on convergence of measures defines three kinds of convergence: total variation, strong, and weak. For weak convergence, a number of equivalent formulations are given, but not for strong or total variation convergence. I have two questions. - -Are there some equivalent formulations for those notions of convergence too? In particular, is it true that $\mu_n$ strongly converges to $\mu$ iff $E_{\mu_n} f\to E_\mu f$ for every bounded measurable function $f$? According to an answer to this question, the left-to-right direction holds if strong convergence is strengthened to total variation convergence. -Are there some nice characterizations of these topologies that are not stated in terms of convergent sequences? - -REPLY [9 votes]: To summarize the situation. Let $(X,\mathcal{F})$ a measurable space. -The space $M(X,\mathcal{F})$ of all real-valued signed measures on $(X,\mathcal{F})$ is a Banach space wrto the total variation norm $\|\mu\|:=|\mu|(X)$ where the (non-negative) measure $|\mu|:= \mu_+ +\mu_-$ is the variation of $\mu$. The space $M(X,\mathcal{F})$may be isometrically embedded as a norm-closed subspace of the dual space $B(X,\mathcal{F})^*$ of the Banach space $B(X,\mathcal{F})$ of all bounded measurable functions on $X$, $B(X,\mathcal{F})$ with the uniform norm $\|\cdot\|_\infty$ ( that dual is in general much larger, since it also contains all additive measures). Note that the space of simple functions $S(X,\mathcal{F})$, linear span of characteristic functions of measurable sets, is norm dense in $B(X,\mathcal{F})$ via the usual approximation $f_n(x):= \lfloor nf(x)\rfloor/n$. -So the above "strong convergence of measures", that is with test functions in $B(X,\mathcal{F})$, is the weak* convergence of $B(X,\mathcal{F})^*$, in the particular cas eof sequences in $M(X,\mathcal{F})$. In particular any such convergent sequence of measures is norm bounded (as any w* convergent sequence of elements in a dual space), and it is "weakly convergent" in the above sense, that is with characteristic functions, hence also with simple functions as test. It is not norm convergent in general (as an example, take as said e.g. $\mu_n$ absolutely continuous w.r.to the Lebesgue measure on $X:=[0,1]$ -and with densities $g_n\in L^1$ weakly convergent but not norm convergent). -Conversely, a norm bounded sequence of measures $\mu_n$ that weakly converges to $\mu$ wrto test functions $g$ in $S(X,\mathcal{F})$ also converges with test functions $f$ in $B(X,\mathcal{F})$, again an elementary and general fact. You can see it writing -$$\langle \mu_n, f\rangle - \langle\mu, f\rangle=\langle\mu_n-\mu, g\rangle+ \langle\mu_n-\mu, g -f\rangle $$ -with a simple function $g$, so that -$$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|\le \limsup_{n\to\infty}|\langle\mu_n-\mu, g\rangle|+ \big(\sup_n \|\mu_n\|+\|\mu\|\big)\|g -f\|_\infty $$ -whence -$$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|=0,$$ -since simple functions are $\|\cdot\|_\infty$ dense.<|endoftext|> -TITLE: Signature of compact oriented 4-manifold -QUESTION [7 upvotes]: I was told that the signature of $S_1\times F_3$ is zero, where $F_3$ is a compact oriented 3-manifold. Let $M_4$ be a fibre bundle with $S_1$ as a the base manifold and $F_3$ as the fibre. Assume $M_4$ is oriented. Can one show that the signature of $M_4$ is zero? - -REPLY [14 votes]: Yes, in several ways. It can be proven by cohomological methods (Meyer, ''Die Signatur von Faserbündeln'', PhD thesis in Bonn, early 1970s), L-theory (Lück-Ranicki ''Surgery obstructions if fibre bundles'') or index theory (a footnote in Atiyah ''The signature of fibre-bundles'', the details worked out by myself (arXiv:0902.4719)). -The result is that the signature is multiplicative in oriented fibre bundles of odd fibre dimension. When the base is $S^1$, there is a shorter argument: -Consider the rational Leray-Serre spectral sequence, which collapses for degree reasons. The terms that contribute to the middle dimensional cohomology of $M_4$ are $E_{2}^{0,2}=H^0 (S^1; H^2 (F))$ and $E_{2}^{1,1}=H^1 (S^1 ;H^1 (F))$ (with local coefficients, of course). Now observe that both are dual to each other by Poincare-duality, so in particular have the same dimension. The term $E_{2}^{1,1}=H^1 (S^1 ;H^1 (F))$ is a subgroup of $H^2 (M_4)$. The point now is that the cup product is trivial on this subspace, by the multiplicativity of the spectral sequence. -By the above argument, $dim (E_{2}^{1,1})=1/2 dim H^2 (M_4)$, and so we have found a Lagrangian subspace, in particular, the signature is null.<|endoftext|> -TITLE: Uniformly distributed sequence in $\mathbb{R}$ -QUESTION [6 upvotes]: We say that a sequence $(x_n)_{n=1}^\infty \subseteq \mathbb{R}$ is "uniformly distributed in $[a,b]$", with $a < b$, if $(x_n)_{n=1}^\infty \cap [a,b] \neq \varnothing$ and -$$\lim_{N \to \infty} \frac{\#\{n \leq N : x_n \in [c,d]\}}{\#\{n \leq N : x_n \in [a,b]\}} = \frac{d-c}{b-a}$$ -for all $[c,d] \subseteq [a,b]$, with $c < d$, (here $\#S$ is the number of elements of a finite set $S$). -We say that a sequence $(x_n)_{n=1}^\infty \subseteq \mathbb{R}$ is "uniformly distributed in $\mathbb{R}$" if $(x_n)_{n=1}^\infty$ is uniformly distributed in any $[a,b]$, with $a < b$. -I have a construction for a sequence uniformly distributed in $\mathbb{R}$. Are there in the literature examples of uniformly distributed sequence in $\mathbb{R}$? References? -Thank you all for your help. - -REPLY [3 votes]: Yes definitely. -You can generate equidistributed sequences on $\bf R$ in the same fashion that you would do on $[0,1) \simeq {\bf R}/{\bf Z}$: by using an ergodic transformation preserving the Lebesgue measure. Note that the Lebesgue measure on $\bf R$ has infinite mass and thus the situation is slightly more subtle on $\bf R$ than on $[0,1)$. -The study of transformations preserving a measure of infinite mass is the topic of infinite measure theory. The standard reference is the book of J. Aaronson, an introduction to infinite measure theory edited by the AMS. The first example of a conservative ergodic transformation from $\bf R$ to $\bf R$ preserving the Lebesgue measure given in the book (and perhaps the simplest example) is the Boole transformation $$x \mapsto x - {1\over x}.$$ -Applying the Hopf ratio ergodic theorem, we deduce that for almost all $x\in \bf R$, the sequence given by $x_{n+1} = x_n -{1\over x_n}$, $x_0=x$ is equidistributed according to your definition (which is almost the standard one). -Another way to generate an equidistributed sequence is to use a random walk on $\bf R$. So let $X_1$,...$X_k$... be an iid sequence of real valued random variables defined on some space $(\Omega, {\cal F}, P)$ which are both integrable and with zero expectation. We need a "non-arithmetic" assumption on the law of the $X_i$ because we don't want that the variables take values in a discrete subgroup of $\bf R$. $P_X = {1\over 1+\alpha}(\delta_{-\alpha}+\alpha \delta_1)$ with $\alpha$ positive irrational is ok. -Theorem the sequence $(S_n(\omega))$ is equidistributed on $\bf R$ for almost all $\omega$. -I think that this result is stronger than the CLT. Maybe this can be deduced from the local CLT. Anyway, this follows from the conservativity and ergodicity of the lift of the shift on the skew-product $\Omega\times \bf R$, the same argument as above applies. In addition to the previous reference, there are probably a few probability books where this is discussed but I can't remember any from the top of my head.<|endoftext|> -TITLE: Where to break paragraphs in a proof? -QUESTION [18 upvotes]: I have some rules of thumb about writing research papers that I can actually articulate. For example, leave all definitions as late as possible (but not later!), so the reader won't fear "Do I need to understand this definition for this theorem?" nor wonder "Did I understand that theorem, insofar as it didn't seem to use the previous definition?" -One I don't have, though, is where to include paragraph breaks in proofs. If I follow the usual maxim that each paragraph should be about a single idea, then interpreted variously, this leads to sentence-long or page-long paragraphs. -More specifically, do you think of "It remains only to show some big thing" as an announcement with which to begin a new paragraph, or as a capstone with which to end the previous one? - -Do you have a maxim describing where to put paragraph breaks in proofs, that doesn't lead to ones that are too short or too long? - -REPLY [15 votes]: Here is a useful quote from Zinsser's "On writing well", a timeless guide to all such matters: - -Keep your paragraphs short. Writing is visual—it catches the eye - before it has a chance to catch the brain. Short paragraphs put air - around what you write and make it look inviting, whereas a long chunk - of type can discourage a reader from even starting to read. -But don't go berserk. A succession of tiny paragraphs is as annoying - as a paragraph that's too long. I'm thinking of all those midget - paragraphs—verbless wonders—written by modern journalists trying to - make their articles quick 'n' easy. Actually they make the reader's - job harder by chopping up a natural train of thought. - -REPLY [3 votes]: I'm not sure what type of answer to give here, except don't be tasteless, and do what looks right. The obvious thing to do, if an idea is spanning multiple paragraphs, is to make it a separate proposition and cite it from within the larger argument. I've found how-to-write-readable-computer-code type guides rather useful in this regard. -The harder problem appears to be when to group separate ideas into a larger paragraph, for which it seems difficult to consistently defend any universal prescriptivist stance. Here's an example (yes, I understand that it is a bit of a low-blow to bring up something you wrote a long time ago): it has often confounded me why the proof of Lemma 7 on page 18 of your honeycomb paper spans three paragraphs. On the other hand, I was really glad that Lemma 8 was broken into various pieces, otherwise it would have taken me much longer to parse. -Maybe the golden rule should apply: would I want to read this in someone else's paper? Personally, I find it much more psychologically satisfying to see "only one more thing remains" at the end of a paragraph rather than at the beginning of the next one. - -REPLY [3 votes]: While writing, I usually read aloud the text: I like that what I write be not very different from what I would say. Since paragraphs are separated by pauses, which are usually quite obvious when talking, a bad paragraph break or a missing one tends to make itself quite obvious in this way. -This does not constitute a rule, of course, for paragraph breaking, but is a useful way to probe them. -Good paragraph breaking is like pornography, in a way. -Regarding your specific example: without context, I don't think we can decide if a «It remains only to show some big thing» is better at the end of a paragraph or at the beginning. If you have a paragraph whose point is to explain why to obtain the desired result A it is enough to prove B, then it may well end with a «It remains only to show that B holds.» On the other hand, if that point was done earlier in the argument, and since then time has been spent on doing some other menial task, it would be nice to start the paragraph that begins the remaining task with an announcement that «It remains only to show that B holds» if only to be nice to the reader who might have by now forgotten the big lines of our reasoning.<|endoftext|> -TITLE: Separation of almost disjoint families by ground model almost disjoint families -QUESTION [6 upvotes]: Suppose that $V$ is a model of $\sf ZFC$, and for concreteness I should point that at this point I am interested in $V=L$ as a ground model. -Suppose that $V[c]$ is a Cohen extension of $V$ where $c$ is a real number, and $\{A_n\mid n\in\omega\}$ is an almost disjoint family in $V[c]$ of subsets of $\omega$. Can we always find an almost disjoint family $\{A'_n\mid n\in\omega\}$ such that $A_n\subseteq A'_n$? -What about larger cardinals? Suppose that $\{A_\alpha\mid\alpha<\kappa\}$ is an almost disjoint family of subsets of a regular $\kappa$ (almost disjoint means that the intersection of any distinct two is of size ${<}\kappa$). Can we find in $V$ a separation family? - -While this is not the case I care about, I am curious about the case where the almost disjoint family is maximal, or at least of size $\kappa^+$. Can we still find such family? - -REPLY [9 votes]: Here's a possibly simpler example than Joel's, in the case of $\omega$. Let $A_0$ be $c$ (where I think of the Cohen real as a subset of $\omega$), and let the rest of the $A_n$'s be any partition of $\omega-c$. (In fact, you could let $A_1=\omega-c$ and let all the later $A_n$'s be empty, if you're willing to let the empty set into your almost disjoint family.) An easy genericity argument shows that the only ground-model subsets $A_0'$ of $\omega$ that satisfy $c\subseteq A_0'$ are cofinite. So $A_0'$ would have to meet $A_1'$ infinitely.<|endoftext|> -TITLE: Can one show the equivalence of the abstract and classical Jordan decompositions for simple Lie algebras without complete reducibility? -QUESTION [21 upvotes]: The following fact is basic in the theory of complex Lie algebras: - -Theorem. Let ${\mathfrak g} \subset {\mathfrak gl}_n({\bf C})$ be a simple Lie algebra, and let $x \in {\mathfrak g}$. Let $x = x_s + x_n$ be the Jordan decomposition of $x$, thus $x_s, x_n \in {\mathfrak gl}_n({\bf C})$ are commuting elements that are semisimple and nilpotent respectively. Then $x_s,x_n$ also lie in ${\mathfrak g}$. - -The standard proof of this theorem (e.g. Proposition C.17 of Fulton-Harris) proceeds, roughly speaking, by using complete reducibility of ${\mathfrak g}$-modules to reduce to the case when ${\bf C}^n$ is irreducible, at which point Schur's lemma may be applied. This fact allows one to show that the abstract Jordan decomposition for simple (or semisimple) Lie algebras is preserved under linear representations. -My (somewhat informal) question is whether the above theorem can be established without appeal to the complete reducibility of ${\mathfrak g}$-modules. I would also like to exclude any use of Casimir elements or the Weyl unitarian trick (since these give completely reducibility fairly quickly), and also wish to avoid using deeper structural facts about Lie algebras, such as the theory of Cartan subalgebras or root systems. I am happy to use the Killing form and to assume that it is non-degenerate on simple Lie algebras (and more generally, to assume Cartan's criteria for solvability or semisimplicity); I am also happy to use the closely related fact that all derivations on a simple Lie algebra are inner. -The best I can do without complete reducibility is to show that $x_s = x'_s + h$, $x_n = x'_n - h$, where $x'_s \in {\mathfrak g}$ is ad-semisimple, $x'_n \in {\mathfrak g}$ is nilpotent, and $h \in {\mathfrak gl}_n({\bf C})$ is nilpotent and commutes with every element of ${\mathfrak g}$; one can also show that $h$ vanishes when restricted to ${\mathfrak gl}(W)$ for every ${\mathfrak g}$-irreducible module $W$ of ${\bf C}^n$ (this is basically a Schur's lemma argument after observing that $h$ has trace zero on $W$), which shows that $h$ vanishes if one assumes complete reducibility. But it seems remarkably difficult to conclude the argument without complete reducibility. -Alternatively, I would be happy to see a proof of complete reducibility that did not use any of the other ingredients listed above (Casimirs, the unitary trick, or root systems). (My motivation for this is that I have been trying to arrange the foundational theory of finite-dimensional complex Lie algebras in a way that moves all the "elementary" theory to the front and the "advanced" structural theory to the back, at least according to my own subjective impressions of the elementary/advanced distinction. I had thought I had achieved this to my satisfaction in these notes, until someone recently pointed out a gap in my proof of the above theorem, which I have thus far been unable to bridge without using complete reducibility.) - -REPLY [7 votes]: This is how I do this in my third year course on Lie algebras: Since we may assume that the Killing form $\kappa$ of $\mathfrak g$ is is non-degenerate, we can make use of the direct sum decomposition ${\rm End}({\mathfrak g})= -{\rm ad}({\mathfrak g})\oplus M$, where $M=({\rm ad}({\mathfrak g}))^\perp$. The subspace $M$ has the property that $[{\rm ad}({\mathfrak g}),M]\subseteq M$, where the commutator brackets are taken in ${\rm End}({\mathfrak g})=\mathfrak{gl}({\mathfrak g})$. Hence we can write any $D\in{\rm Der}({\mathfrak g})\subset {\rm End}({\mathfrak g})$ as $D={\rm ad}\,x+m$ for some $x\in {\mathfrak g}$ and $m\in M$. For any $y\in{\mathfrak g}$ we then have $[D,{\rm ad}\,y]=[{\rm ad}\,x,{\rm ad}\,y]+[m,{\rm ad}\,y]$. Since $[D,{\rm ad}\,y]={\rm ad}(Dy)\in {\rm ad}({\mathfrak g})$ and $[m,{\rm ad}\,y]\in M$, it follows -that $D={\rm ad}\,x$ (one should keep in mind here that the centre of $\mathfrak g$ is trivial). We thus conclude that all derivations of $\mathfrak g$ are inner, and this does not rely on Weyl's theorem on complete reducibility. -At this point one can use the fact that the semisimple and nilpotent parts, $D_s$ and $D_n$, of the Jordan decomposition of the endomorphism $D\in\mathfrak{gl}(\mathfrak{g})$ are again derivations of $\mathfrak g$. This is a nice (and elementary) exercise which can be found in Jacobson's book on Lie algebras, and one can replace $\mathfrak g$ by any finite dimensional algebra $A$, not necessarily associative or Lie. By the above, $D_s={\rm ad}\,x_s$ and $D_n={\rm ad}\,x_n$ for some $x_s,x_n\in {\mathfrak g}$ which are uniquely determined since $\mathfrak{z}(\mathfrak{g})=0$. The elements $x_s$ and $x_n$ commute as so do $D_s$ and $D_n$ in $\mathfrak{gl}(\mathfrak{g})$. The fact that the Jordan decomposition exists in $\mathfrak{gl}(V)$ has to be proven, of course.<|endoftext|> -TITLE: Second Stiefel Whitney class of quotients of odd spheres -QUESTION [9 upvotes]: I don't know much of algebraic topology so the following question could be very silly. Let $G$ a finite subgroup of $U(n)$ that acts linearly (the action induced by the action of $U(n)$ on $\mathbb{C}^{n}$) and freely on the unit sphere $S^{2n-1}\subset \mathbb{C}^{n}$. Let $X$ be the quotient manifold $$X:=S^{2n-1}/G$$ -I want to compute $w_{2}(X)\in H^{2}(X,\mathbb{Z}_{2})$ i.e. the second Stiefel-Whitney class of the tangent bundle of $X$. More precisely i'd like to know if it is $0$ or not and under which hypotheses on $G$ it vanishes. Is this a standard problem? Is there some reference for this type of calculation or something similar? -Thank you in advance! - -REPLY [3 votes]: Because $G \subset U(n)$, the stabilised tangent bundle of $X = S^{2n-1}/G$ has an obvious complex structure. The determinant bundle $\lambda$ is the quotient of $S^{2n-1} \times \mathbb{C}$ by $G$ acting as $(p, z) \mapsto (g(p), (\det g) z)$. Now $\det : G \to \mathbb{R}/\mathbb{Z}$ descends to a homomorphism $H_1(X) \to \mathbb{R}/\mathbb{Z}$, so defines an element $\alpha \in H^1(X; \mathbb{R}/\mathbb{Z})$. Then $c_1(\lambda) \in H^2(X; \mathbb{Z})$ is the image of $\alpha$ under the Bockstein map associated to $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$. Its mod 2 reduction equals $w_2(X) \in H^2(X; \mathbb{Z}/2\mathbb{Z})$. -Equivalently, $w_2(X)$ is the image of $\alpha$ under the Bockstein of $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{R}/2\mathbb{Z} \to \mathbb{R}/\mathbb{Z} \to 0$. Whether $\alpha$ is in the image of $H^1(X; \mathbb{R}/2\mathbb{Z})$ depends on whether one can make a consistent choice of square roots of $\det g$ for $g \in G$ (essentially the same lifting problem that appears in José's answer). I believe this is possible if and only if any $\bar g$ in the abelianisation of $G$ with $\det \bar g = -1$ has order divisible by 4 (or if you like: $g^2$ is not a commutator for any $g \in G$ with $\det g = -1$). That is then a criterion for $w_2(X)$ to vanish.<|endoftext|> -TITLE: Barr's theorem and constructivity? -QUESTION [10 upvotes]: Barr's covering theorem assert that any Gorthendieck topos can be covered by a Grothendieck topos (even a locale) satisfying the axiom of choice (and hence also the law of excluded middle). Its corrolary is that if one can deduce from geometrical hypothesis $H$ a geometric conclusion $C$ using (AC) and classical logic, then one can deduce $C$ from $H$ in any Grothendieck topos. -The proof of this theorem rely heavily on the axiom of choice (actually on the fact that the topos of sheaf over a boolean locale satisfy the axiom of choice) and hence is not valid in a general elementary topos. -My question is : Is there any know examples of something that can be proved using Barr's theorem (a deduction of geometric conclusion from geometric hypothesis which can be proved using (AC) ) and which is false in some elementary topos (preferably with a natural number object) ? -Also, could you confirm me that the weaker form of barr's theorem "every topos can be covered by a boolean topos" is actualy true also for elementary topos ? (it seems that one can consider the topos of double negation sheaf over the internal frame of nuclei of $\Omega$ in every elementary topos) -Edit : Thinking about Zhen Lin answer and comment I realize that my question was unclear : what I am looking for is an exemple of a geometric theory $H$ possibly stated in the language of some elementary topos, which have a geometric consequence $C$ deducible using the axiom of choice such that $C$ cannot be deduce from $H$ constructively. ie, I'm essentially looking for a proof that barr's theorem can't be constructive. In this situation, the completeness theorem for geometric logic (mentioned in Zhen Lin's comment) does not apply because as the theory $H$ can be stated in an elementary topos it does not mean anything to say that $C$ can be deduced from $H$ in any Grothendieck toposes. - -REPLY [5 votes]: In every Grothendieck topos, the following sequent is valid for the natural numbers object $N$, -$$x : N \vdash \bigvee_{n : \mathbb{N}} x = s^n (z)$$ -where $\mathbb{N}$ is the set of natural numbers, $s : N \to N$ is the internal successor operation, and $z : 1 \to N$ is the zero element. In words, "every natural number is standard". This can be proved directly without appealing to any kind of boolean completeness theorem. However, it is false in any non-standard topos, provided one makes the appropriate allowances to deal with the fact that such toposes do not have all infinite colimits: note that in every cocomplete topos, the indicated sequent is valid if and only if the collection $\{ s^n \circ z : 1 \to N \}$ is a jointly epimorphic family. It goes without saying that non-standard toposes exist: for instance, take the elementary topos associated with any non-trivial ultrapower of $V_{\omega + \omega}$. -Every elementary topos can be covered by a boolean topos: see Proposition 4.5.23 in [Sketches of an elephant, Part A]. Apparently this is a theorem of Freyd in [1972, Aspects of topoi], but I have not found the precise statement there. The construction is as follows: for an elementary topos $\mathcal{E}$, let $q$ be the local operator on $\mathcal{E}_{/ \Omega}$ defined by the formula $q (x) = ((x \Rightarrow t) \Rightarrow t)$, where $\Rightarrow$ is the Heyting implication and $t$ is the generic subobject $1 \rightarrowtail \Omega$ in $\mathcal{E}$ considered as a truth-value in $\mathcal{E}_{/ \Omega}$; then the obvious geometric morphism $\mathbf{sh}_{q} (\mathcal{E}_{/ \Omega}) \to \mathcal{E}$ is a geometric surjection.<|endoftext|> -TITLE: Fermat's proof for $x^3-y^2=2$ -QUESTION [29 upvotes]: Fermat proved that $x^3-y^2=2$ has only one solution $(x,y)=(3,5)$. -After some search, I only found proofs using factorization over the ring $Z[\sqrt{-2}]$. -My question is: -Is this Fermat's original proof? If not, where can I find it? -Thank you for viewing. -Note: I am not expecting to find Fermat's handwritings because they may not exist. -I was hoping to find a proof that would look more ''Fermatian''. - -REPLY [6 votes]: Lemma. -Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $n=t^2+2u^2$ divides $bt-au$. -Proof. -Assume the theorem is false, and let $m$ be a minimal counterexample. Evidently $m > 1$ since the theorem is trivially true for $m=1$. -Note that $b$ is coprime to $m$. Let $A$ be an integer such that $Ab \equiv a\!\pmod{m}$, chosen so that $\tfrac{-m}{2} < A \le \tfrac{m}{2}$. Then $A^2+2 = lm$ for some positive integer $l < m$. Clearly $l$ cannot be a smaller counterexample than $m$, and so there exist coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. -Let $t = \tfrac{ar+2bs}{m}$ and $u=\tfrac{br-as}{m}$. Direct calculation confirms the equations for $a$, $b$, and $n$. From $n=t^2+2u^2$, we deduce that $t$ is an integer because $u$ is an integer, and $t$ and $u$ are coprime because $\gcd(t,u)$ divides both $a$ and $b$. Finally, note that $n$ divides $bt-au=sn$. -Hence $m$ is not a counterexample, contradicting the original assumption. $\blacksquare$ -Corollary. -Let $a$ and $b$ be coprime integers with $m$ an integer such that $m^3=a^2+2b^2$. Then there are coprime integers $r$ and $s$ such that $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$. -Proof. -Evidently $m$ is odd since $a^2+2b^2$ is at most singly even. And $a$ and $m$ must be coprime. Using the theorem, we have $m=r^2+2s^2$ and $m^2=t^2+2u^2$. Then $m$ divides $a(ur-ts)=t(br-as)-r(bt-au)$, and therefore $m \mid (ur-ts)$. The lemma can then be reapplied with $a$ and $b$ replaced by $t$ and $u$. Repeating the process, we eventually obtain integers $p$ and $q$ such that $p^2+2q^2=1$. The only solution is $q=0$ and $p=\pm1$. Ascending the path back to $a$ and $b$ (reversing signs along the way, if necessary) yields $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$, as claimed. $\blacksquare$ -Theorem. -The Diophantine equation $X^3 = Y^2+2$ has only one integer solution, namely $(x,y) = (3, \pm 5)$. -Proof. -Evidently $y$ and $2$ are coprime. By the corollary, we must have $b=1=s(3r^2-2s^2)$ for integers $r$ and $s$. The only solutions are $(r,s)=(\pm 1,1)$. Hence $a=y=r(r^2-6s^2)=\pm 5$, so $(x,y)=(3,\pm 5)$. $\blacksquare$<|endoftext|> -TITLE: realizing fusion categories as subfactors of the hyperfinite -QUESTION [8 upvotes]: Let R be the hyperfinite II_1 or the hyperfinite III_1 factor (pick which ever one you prefer), and let Bim(R) denote the tensor category of R-R-bimodules. - -This question is inspired by the recent article -The classification of subfactors of index at most five -by Jones, Morrison, and Snyder. -More precisely, I am interested in Thm 1.1 and Thm 2.10 of the above paper (both of them are older results). -Given the above, -I expect the following to be true: - -(1) Any unitary fusion category can be embedded in Bim(R). - (2) Any two embeddings are conjugated by an automorphism of R. - -however, I am not sure if things have ever been formulated in this way. -My questions are: -• What is the closest result to (1) and (2) available in the literature? -• is it easy to adapt/use some existing proofs to get (1) and (2), and how? - -REPLY [6 votes]: 1, and indeed its generalization to the amenable case, is in "Amenable tensor categories and their realizations as AFD bimodules" by Hayashi and Yamagami, see Section 7. I don't think 2 has appeared in the literature, though I'd expect that it's true. -In the fusion case, if the universal grading group is trivial, I think you can just look at the algebra object $\oplus_x x \otimes x^*$ in $\mathcal{C}$ and use the usual reconstruction and uniqueness for subfactors. It's possible I'm missing a technicality, but I think this should work (in particular, Popa doesn't assume irreducibility). If the grading group is non-trivial there still may be some trick that reduces it to the known cases, but it's not clear to me.<|endoftext|> -TITLE: Is there a universal property for graded localization? -QUESTION [8 upvotes]: Question: Let $S$ be a graded ring and $ f \in S_+$. Does the ring $ S_{(f)}$ which consists of degree $ 0$ elements of $ S_f$ represent a nice functor? - - Motivation: Let $ X = {\rm Spec} A$. Assume that $D(f) \subseteq D(g)$. It is messy to think about the restriction map $ A_g \to A_f $ using formulas. Things are much more transparent in the language of representable functors. Indeed, $A_g$ represents the functor -$$ R \mapsto \{ \text{maps $ A \to R$ which send $ g $ to a unit} \} $$ -Since $ D(f) \subseteq D(g) $ we have a natural transformaion -$$ \{ \text{maps $ A \to R$ which send $ f $ to a unit} \} \subseteq \{ \text{maps $ A \to R$ which send $ g $ to a unit} \} $$ -The yoneda lemma gives us a ring homomorphism $ A_g \to A_f $ which is the restriction map in the structure sheaf of the affine scheme. I just want to emphasize that I am not saying the yoneda lemma construction is better, it is just easier for me to keep in my head than a bunch of formulas. -Now let $S$ be a graded ring. We construct $ {\rm Proj} S $ by gluing together the affine schemes $ {\rm Spec} S_{(f)}$ where $ f \in S_{+}$ is homogeneous. In order to glue we need ring isomorphisms -$$ S_{(fg)} \cong (S_{f})_{\frac{g^{\deg f}}{f^{\deg g}}} $$ -If $ S_{(f)}$ represented a nice functor, we could try and mimic the construction of the restriction maps in the structure sheaf of an affine scheme. As it stands, the standard construction of this isomorphism is messy and very hard to remember (for me). - -REPLY [4 votes]: It seems that Martin has provided the answer which Daniel sought. But the question in the title doesn't appear to be answered yet: Is there a universal property for graded localization? More precisely, for the degree-0 part of the localization? -I believe there is one. Let $S$ be an $\mathbb{N}$-graded ring. Let $M \subseteq S$ be a multiplicatively closed set consisting only of homogenous elements. Assume that $M$ contains an element of degree $1$. Let $R$ be an arbitrary ring. Then -$$ \mathrm{Hom}_{\mathrm{Ring}}(S[M^{-1}]_0, R) \cong \mathrm{Hom}_{\mathrm{grRing}}(S, R[X])_M / R^ \times. $$ -An element of the right hand side is an equivalence class (modulo rescaling) of a graded ring homomorphism $S \to R[X]$ which maps the degree-$n$ elements of $M$ to polynomials whose coefficient of $X^n$ is a unit in $R$. -A ring homomorphism $\varphi : S[M^{-1}]_0 \to R$ gives rise to such an equivalence class $[\psi]$ by picking an element $f \in M \cap S_1$ and defining $\psi(x) := \varphi(x/f^n) X^n$ for homogeneous elements $x$ of degree $n$. -Conversely, such an equivalence class $[\psi]$ gives rise to a ring homomorphism $\varphi$ by setting $\varphi(x/s) := \psi(x)|_{X=1} \cdot \psi(s)|_{X=1}^{-1}$. -I sincerely hope that there is some better way to express this universal property.<|endoftext|> -TITLE: Can one measure the infeasibility of four color proofs? -QUESTION [24 upvotes]: Terms like "impractical" and "unfeasible" are used to say the Robertson, Sanders, Seymour, and Thomas proof of the four color theorem needs computer assistance. Obviously no precise measure is possible, for many reasons. -But is there an informed rough estimate what a graph theorist would need to verify the 633 reducible configurations in that proof? $10^4$ hours? $10^8$ years? -I am not asking if other proofs are known. I want to know if graph theorists have an idea what scale of practicality we are talking about when we say the Robertson, Sanders, Seymour, and Thomas proof is impractical without machine assistance. - -REPLY [8 votes]: As explained in Tony's answer, in order to answer this question you need to separately answer how long it would take to prove reducibility of the configurations (Theorem 2) and how long it would take to prove unavoidability of the configurations (Theorem 3). -There's an interesting alternate proof of the 4-color theorem due to John Steinberger, which differs from RSST in that the proof of reducibility only uses the easier notion of "D-reducibility" rather than the more elaborate "C-reducibility." The cost is that the unavoidable set is much longer and the proof of unavoidability is also longer. As Tony explained, unavoidability was the "easy" part, so it's possible that for a human Steinberger's proof would be easier to verify. Even if it is not easier, he provides some additional detail of the estimate of "a few months" from RSST. -In discussing the proof of unavoidability, Steinberger discusses the files which serve as certificates of unavoidability. That is, there's a file which tells the computer how to prove this particular case. Of these files, in verbose human readable form, Steinberger writes: - -While the resulting output may be readable at a normal pace it is also -quite large: over 3’000’000 lines for the Robertson et al. proof, over -13’000’000 lines for our proof. A mathematician checking these proofs -at the rate of one line per second and working 9 hours a day would -take over 3 months to read the Robertson et al. proof and over a year -to read ours. - -This makes much more precise the "few months" estimate of RSST. -Unfortunately he does not give a human estimate for the reducibility portion. Instead he says it takes a 2010 personal computer "around 10 hours." If I understand things correctly, the problem here is that in order to check D-reducibility, you need to check something for every choice of 4-coloring on the boundary of your configuration. For large configurations this is an enormous number of cases. The computer time here large, which suggests that a human might prefer to do the RSST proof. -As discussed in the comments, a plausible estimate is that the Steinberger proof of reducibility would likely take somewhere between $10^4$ and $10^6$ hours (i.e. somewhere between a year and a lifetime).<|endoftext|> -TITLE: a question on lattice invariants -QUESTION [5 upvotes]: Let $V$ be an $n$-dimensional vector space and $L$ be a lattice in $V$, i.e. a free $\mathbb{Z}$-module of full rank. Define the following two numbers. Let $\lambda$ be the minimal positive real number such that the ball centered at the origin of radius $\lambda$ contains a basis of the lattice. Let $\mu$ be the minimal positive real number such that the ball centered at the origin of radius $\mu$ contains a basis of a sublattice of full rank. Apparantly $\lambda \geq \mu$. But it seems to me that (at least when $n=2$), these two numbers are always the same. Can one construct a example with $\lambda > \mu$? - -REPLY [7 votes]: Yes, as long as $n \geq 5$. The standard example is the lattice generated by -${\bf Z}^n$ and $(\frac12,\frac12,\ldots,\frac12)$. Then $\mu=1$ and -$\lambda = n/4$.<|endoftext|> -TITLE: Omitting types and Baire category -QUESTION [6 upvotes]: What is the relation between omitting types theorems in model theory and the baire category theorem? - -REPLY [5 votes]: OK, found the thesis. -What you can prove is that the omitting types theorem follows from the Baire category theorem. I haven't thought about whether the other direction holds in the sense of reverse mathematics. -One starts by fixing a signature $\tau$, and assigning an appropriate topology to the space $\mathcal E_\tau$ of $\tau$-structures. This is a proper class, but the topology has a set-sized basis, so arguments can be formalized in a straightforward fashion. -The topology of elementary classes is defined by picking as a basis the classes of the form $M_\tau(\phi)$ for $\phi$ a $\tau$-sentence. Here, $M_\tau(\phi)$ simply denotes the class of $\tau$-structures that satisfy $\phi$. Similarly, if $\Phi$ is a set of sentences (a theory), then $M_\tau(\Phi)=\bigcap_{\phi\in\Phi}M_\tau(\phi)$ is the class of models of $\Phi$. -This topology is not quite metrizable, but admits a uniform structure: A basis of uniformity is attained by considering the relations $\equiv_\Phi$, where $\Phi$ varies over the finite sets of $\tau$-sentences, and two $\tau$-structures $\mathcal M$ and $\mathcal N$ are $\Phi$-equivalent, $\mathcal M\equiv_\Phi\mathcal N$ iff for each $\phi\in\Phi$, we have that $\mathcal M\models\phi$ iff $\mathcal N\models\phi$. -As a consequence, $\mathcal E_\tau$ is regular and (if $\tau$ is countable) it is pseudo-metrizable, with pseudo-metric $$d(\mathcal M,\mathcal N)=\inf\{1/n\mid\mathcal M\models_{\Phi_n}\mathcal N\},$$ -where $\phi_1,\phi_2,\dots$ enumerates the $\tau$-sentences and $\Phi_n=\{\phi_i\mid i< n\}$. -Notoriously, the space is complete (every Cauchy net converges to an ultraproduct of the members of the net) and compact (and logical compactness can be seen as a consequence of this). -The topological version of the omitting types theorem then takes the following form: - -Theorem. Let $\tau$ be a countable signature, and let $T$ be a consistent theory in this language. Let $C$ be a countable set if constant symbols, disjoint from $\tau$. Let $\Sigma(x)$ be a $1$-type in language $\tau$. If for all $c\in C$ we have that - $$ M_{\tau\cup C}(T\cup\Sigma(c))$$ - is closed nowhere dense in $M_{\tau\cup C}(T)$, then there is a (countable) model of $T$ that omits $\Sigma(x)$. - -The point is that $T$ locally omits $\Sigma(x)$ iff for any $c\in C$, the open class $M_{\tau\cup C}(T\cup\Sigma(c))$ has empty interior in $M_{\tau\cup C}(T)$. The key to this argument is that if $\tau'$ expands $\tau$, then the projections - $$ P_{\tau'\tau}:\mathcal E_{\tau'}\to\mathcal E_\tau $$ -(that map $\tau'$-structures to their restrictions to $\tau$) are uniformly continuous. -This can be used to give an essentially topological proof of the omitting types theorem, as a consequence of the Baire category theorem, as follows: Letting $\phi(x)$ vary over all $\tau\cup C$-formulas in one free variable $x$, define - $$ X=\bigcap_{\phi(x)}\bigcup_{i\in\mathbb N}M_{\tau\cup C}(\exists x\phi(x)\to\phi(c_i)). $$ -The important properties of $X$ are that it is a dense $G_\delta$ subset of $\mathcal E_{\tau\cup C}$, the projection $P_{\tau\cup C,\tau}\upharpoonright X:X\to\mathcal E_\tau$ is continuous, open, and onto, and - $$ X=\{\mathcal M\in\mathcal E_{\tau\cup C}\mid\{c^\mathcal M\mid c\in C\}\mbox{ is the universe of an elementary substructure of }\mathcal M\}.$$ -Letting $X(T)=X\cap M_{\tau\cup C}(T)$, we have that $X(T)$ is dense in $M_{\tau\cup C}(T)$, so each $M_{\tau\cup C}(\Sigma(c_i))\cap X(T)$ is closed nowhere dense in $X(T)$, and therefore a $G_\delta$ in the compact regular space $M_{\tau\cup C}(T)$. Therefore $X(T)\cap\bigcup_{i\in\mathbb N}M_{\tau\cup C}(\Sigma(c_i))$ has empty interior and its complement in $X(T)$ is dense (so, relevant for us, non-empty). If $\mathcal M$ is any structure in this complement, then for each $i$ there is a $\sigma_i(x)\in\Sigma(x)$ such that $\mathcal M\models\lnot\sigma_i(c_i)$. So, using the key property of the space $X$, the elementary substructure of $\mathcal M$ with universe the $c_i^\mathcal M$ is countable, and omits $\Sigma(x)$, completing the proof. -The pretty idea of considering $\mathcal E_\tau$ as a topological space is classical. One can see some of it in the book on Models and ultrapowers by Bell and Slomson. Another early reference is - -Andrzej Ehrenfeucht, and Andrzej Mostowski. A compact space of models of first order theories, Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys., 9, (1961), 369–373. MR0148536 (26 #6043). - -Modern attention to this topological approach comes from the work of Xavier Caicedo, who saw its usefulness in the setting of abstract logics. See for example his three papers - -Compactness and normality in abstract logics, Ann. Pure Appl. Logic, 59 (1), (1993), 33–43. MR1197204 (93m:03062), -Continuous Operations on Spaces of Structures. In Quantifiers: Logics, Models and Computation, pp. 263-296, Synthese Lib., 248, Kluwer Acad. Publ., Dordrecht, 1995, -and -The abstract compactness theorem revisited. In Logic and foundations of mathematics (Florence, 1995), pp. 131–141, Synthese Lib., 280, Kluwer Acad. Publ., Dordrecht, 1999. MR1739865 (2001b:03039). - -The specific presentation I sketched above comes from Julián's undergraduate thesis, written under the supervision of Xavier Caicedo: - -Julián Mariño Von Hildebrand. Topología en los espacios de models de la lógica de primer orden, Undergraduate thesis, Dept. of Mathematics, Universidad de los Andes, Bogotá, 1995. - -(Julián is a friend from many years ago, and I haven't talked to him in ages. Thanks, I think I'll use this question as an excuse to give him a call.)<|endoftext|> -TITLE: Serre duality for compactly supported sheaves -QUESTION [10 upvotes]: Given a smooth quasi-projective variety $X$ over $\mathbb{C}$ and bounded complexes of vector bundles $(P,d)$ and $(P',d')$ with compactly supported cohomology. It is well-known that such complexes satisfy Serre-duality. The standard proof that I have heard is to complete $X$ to a projective variety $\bar{X}$. -For me the intuition behind Serre duality is integration, and the intuition for the above result is that one can always integrate compactly supported differential forms. Unfortunately, I've never seen a place where this result is actually proven in this way. -Question: Is there a reference which proves Serre duality using compactly supported Dolbeault cohomology? -The proof I have in mind is the standard proof of Serre duality for projective varieties, but I remember in Serre's original paper there are some tricky points of topology on Frechet spaces which are complicated. I haven't actually seen the compactly supported Dolbeault theory used in any other papers since then, so I'm wondering whether there is any newer reference with the above statement and which covers its properties more systematically. - -REPLY [6 votes]: Looks like this question fell in between the cracks. But just in case anyone is still interested in an answer, a fairly detailed proof of Serre's duality theorem, which applies to cohomologies with compact supports (including the Dolbeault complex, and other complexes of differential operators) can be found in Sec.5.1 of - -Tarkhanov, N. N., Complexes of differential operators. Revised and updated translation from the Russian by P. M. Gauthier, Mathematics and its Applications (Dordrecht). 340. Dordrecht: Kluwer Academic Publishers. xviii, 396 p. (1995). ZBL0852.58076.<|endoftext|> -TITLE: The origins of forcing in mathematical logic and other branches of mathematics -QUESTION [15 upvotes]: As everyone knows, forcing was created by Cohen to answer questions in set theory. -Question 1. What are the first applications of set theoretic forcing in other branches of mathematical logic, like number theory, computability theory, complexity theory and model theory. -Question 2. What are the first applications of set theoretic forcing in other branches of mathematics like topology, algebra, analysis, .... -Update. Here I will collect the answers and will add a few that I am aware: -1) Scott, "A proof of the independence of the continuum hypothesis": models of higher order theories of the Real numbers. -2) Feferman, "Some applications of the notions of forcing and generic sets": Number theory. -3) Fernando Tohmè, Gianluca Caterina, Rocco Gangle, "Forcing Iterated Admissibility in Strategic -Belief Models": Game theory (in particular epistemic game theory). -4) Solovay, Tennenbaum, "Iterated Cohen extensions and Souslin's problem" : Analysis-Topology. -5) Shelah, "Infinite abelian groups, Whitehead problem and some constructions" : Algebra. --- - -Edits: - -1) I think, the work of Silver on the independence of gap two cardinal transfer principle, and Chang's conjecture is essentially the first application of forcing in model theory. -2) Are there any applications of forcing in dynamical systems? -3) What about "Recursion theory" and "Complexity theory"? -4) What about other branches of mathematics not mentioned above or in the answers? - -REPLY [7 votes]: Boris Tsirelson constructed the first Banach space that did not have a subspace isomorphic to some $\ell_p$, $1\le p < \infty$ or $c_0$. This eventually led to the Gowers-Maurey construction of hereditarily indecomposable Banach spaces. Tsirelson writes that his forcing construction was motivated by Cohen's forcing arguments: -http://www.tau.ac.il/~tsirel/Research/myspace/remins.html<|endoftext|> -TITLE: Embedding of flat surfaces -QUESTION [7 upvotes]: Let $S$ be a orientable compact surface with a flat euclidean structure with conical singularities (cf. [T] for instance). Let also $\mathcal P$ be a polyhedral euclidean decomposition of $S$ (with vertices at the singular points of the euclidean structure of $S$). -Question 1: can $S$ be realized (as a polyhedral surface) in an euclidean space? If yes, what is known about the set of such `euclidean polyhedral realizations'? -Question 2: same questions than above, but for $\mathcal P$ (i.e. an euclidean polyhedral decomposition of $S$ is fixed). --- -Remarks: - -there is no assumption on the genus of $S$ (when $S$ has genus 0, an answer to the (first part of) Question 1 is given by Alexandrov theorem) -A similar question (but for Riemann surfaces) already appears in [B] (page 9). I guess that a lot has been done on this problem since. Any relevant reference would constitute an interesting answer. - -Thanks for any help! - -[B] Bers L., Riemann surfaces (1958) -[T] Troyanov M. -- Les surfaces a singularités coniques. Enseign. Math. 32 (1986), 79–94. - -REPLY [8 votes]: It may be that this theorem of Burago & Zalgaller (partially) answers your question? - -Theorem (Burago-Zalgaller 1.7). - Every polyhedron $M$ admits an isometric piecewise-linear $C^0$ - immersion into $\mathbb{R}^3$. - If $M$ is orientable or has a nonempty - boundary, then $M$ admits an isometric piecewise-linear $C^0$ - embedding into $\mathbb{R}^3$. - -This is from: - -Yu. D. Burago and V. A. Zalgaller. -"Isometric piecewise linear immersions of two-dimensional manifolds with polyhedral metrics into $\mathbb{R}^3$. -St. Petersburg Math. J., 7(3):369--385, 1996. Translated by S. G. Ivanov. -English translation: -Scanned PDF (15MB). - -Here is their definition of a "polyhedron": - -By a two-dimensional manifold with polyhedral metric (in brief, - a polyhedron), we mean a metric space endowed with the structure - of a connected compact two-dimensional manifold (possibly with - boundary) every point $x$ of which has a neighborhood isometric to - the vertex of a cone. ... The metric is locally - flat everywhere except - for a finite collection of points; these points are the `true' vertices." - -A nice phrase they use in Lemma 2.2 to describe the mapping is that each -triangle becomes a pleated surface in $\mathbb{R}^3$.<|endoftext|> -TITLE: Can Tarski decide constructibility in elementary geometry? -QUESTION [10 upvotes]: Can the decision routine for Tarski's Elementary geometry be extended to decide when an existence claim in that theory can be instantiated by a compass and straightedge construction? -The answer does not follow from mere existence of Tarski's decision routine since the natural definition of constructible quantifies over "any finite number" of steps. But can the routine in fact be refined to decide constructibility? -I got this question from thinking about Mazur's comments on geometry versus arithmetic for Euclid in http://www.math.harvard.edu/~mazur/preprints/meaning_error.pdf, and I see the question occurs in discussion of Is compass and straight edge geometry complete? mixed with many more or less precise variants and calls for more precision. It is not answered there. -In case it helps I specify that I mean Tarski's first order theory of elementary geometry, not supplemented by a predicate for integer lengths or any other predicates. Since each construction is expressible in the first order theory, a construction that works in one model works in all. But is there a decision routine to tell if such a construction exists? - -REPLY [6 votes]: Here is a partial affirmative answer, where we consider existence-and-uniqueness assertions rather than just existence assertions. -Consider the collection of algebraic reals, which form a real-closed field, and give each such algebraic real a finite name, of the form, "the $k^{th}$ solution of $p(x)=0$", for a specific integer polynomial $p$ and specific number $k$. These descriptions are expressible in the language of ordered fields, and so because of Tarski's decidability result, we can decide all questions in this language about these reals under these descriptions. Thus, we have a computable presentation of the algebraic reals, in which we can decide all questions of geometry for algebraic points in Euclidean space using Tarski's algorithm. -Consider now the reals that are constructible by straightedge and compass. These are exactly those algebraic reals in the quadratic closure of $\mathbb{Q}$. I claim that we can decide which of our algebraic reals is constructible, since when we are given an algebraic real $r$, we have a description of a polynomial realizing $r$ as algebraic. Using the computable splitting algorithms (e.g. this or this), we can find a minimal polynomial for $r$ and thereby tell if $r$ is constructible or not. -Now, consider a weakening of your question, where we have not just an existence claim, but an existence-and-uniqueness claim. That is, suppose that we prove there is a unique $r$ for which $\varphi(r)$ in the theory of real-closed fields. Now, we may search in our computable model until we find the instantiating instance using our finite descriptions of the algebraic reals. And given this $r$, we can as explained above decide whether or not this $r$ is constructible. Thus, we have an algorithm to decide all such existence-and-uniqueness instances of your phenomenon, whether the instantiating point is constructible or not.<|endoftext|> -TITLE: What is the max of $n$ such that $\sum_{i=1}^n\frac{1}{a_i}=1$ where $2\le a_1\lt a_2\lt \cdots\lt a_n\le m$? -QUESTION [14 upvotes]: Let $N(m)$ be the max of $n$ such that $\sum_{i=1}^n\frac{1}{a_i}=1$ -where $a_i \ (i=1,2,\cdots,n)$ are integers which satisfy $2\le a_1\lt a_2\lt\cdots\lt a_n\le m$. -Question 1 : What is $N(99)$? -Question 2 : What is $N(m)$? -Examples : I'm going to represent $\sum_{i=1}^n\frac{1}{a_i}$ as $(a_1,a_2,\cdots,a_n).$ -The followings are two examples for $(m,n)=(99,42)$. -$(15,17,20,21,22,26,27,30,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60, 63,66,70,75,76,77,78,80,84,85,88,90,91,95,96,99)$ -$(17,18,20,21,22,24,26,27,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60,63,66,70,72,75,76,77,78,80,84,85,88,91,95,96,99)$ -Remark : -Question 1 has been asked on math.SE. -https://math.stackexchange.com/questions/488173/what-is-the-max-of-n-such-that-sum-i-1n-frac1a-i-1-where-2-le-a-1-l -$99$ has no special meaning except that $99$ is not too small and not too large. -Question 2 might be somewhat ambiguous. The best answer would be to represent $N(m)$ by $m$ if it is possible. Also, finding both the max of $m$ and the min of $m$ would be needed. -Motivation : The beginning was the following: -"$\sum_{k=2}^n \frac 1k$ is not an integer for any $n$." -(the proof and the other related facts can be seen at https://math.stackexchange.com/questions/494174/proving-that-the-finite-sum-of-the-each-reciprocal-of-any-sequence-of-integers-w). - -REPLY [17 votes]: Note that $N(m)$ cannot be significantly larger than $m(1-1/e)$, since the sum of the reciprocals of any $m(1-1/e)$ integers not exceeding $m$ is at least $\sum_{m/e < k < m} 1/k \sim 1$. -I've proved that in fact, $N(m)$ is asymptotic to $m(1-1/e)$; in other words, you can have essentially as many summands as size considerations allow you to have. See Theorem 1 and equation (5) of my paper "Denser Eygptian fractions". (There I consider the equivalent dual problem - instead of fixing $m$ and trying to maximize the number of summands, I fix the number of summands and try to minimize $m$.)<|endoftext|> -TITLE: A problem on a specific integer partition -QUESTION [11 upvotes]: Let $n$ be a positive integer, we consider partitions of the following form : -$$n = d^{2}_{1} + d^{2}_{2} + ... + d^{2}_{r}$$ such that : - -$d_{i}\vert n$ -$1=d_{1} -TITLE: How Random is Cohen? -QUESTION [7 upvotes]: Suppose that $V$ is a universe of $\sf ZFC$, and $c$ is a Cohen generic real over $V$. Is it possible that $c$ is also generic in other senses? I know that it can't be random or Sacks or whatnot because those reals have different properties with regards to preservation of measure and category, or minimality properties. -So to put my question in precise terms: - -Suppose that $V$ is a model of $\sf ZFC$ and $c$ is Cohen generic over $V$. Is there a complete Boolean algebra $B$ in $V$ which does not embed the Cohen algebra [densely, or at least unboundedly] such that $c$ is definable from the generic filter for $B$ as well? - -If this is a factor we can assume $V=L$, but a general answer is interesting nonetheless. -From this there is a second question, if $M$ is an intermediate model between $V$ and $V[c]$, such that $c$ is not Cohen generic over $M$. Is it possible for it to still be a generic set for another partial order in $M$? - -REPLY [8 votes]: The central facts governing the relation between generic filters arising from different forcing notions are the following: -Theorem. Suppose that $G\subset\mathbb{P}$ and $H\subset\mathbb{Q}$ are $V$-generic. - -If $V[G]=V[H]$, then there are conditions $p\in G$ and $h\in H$ such that $\text{RO}(\mathbb{P}\upharpoonright p)\cong\text{RO}(\mathbb{Q}\upharpoonright h)$. -If $V[G]\subset V[H]$, then there are conditions $p\in G$ and $h\in H$ such that $\text{RO}(\mathbb{P}\upharpoonright p)$ is isomorphic to a complete subalgebra of $\text{RO}(\mathbb{Q}\upharpoonright h)$. - -It follows that if $c$ is a Cohen real, then it isn't $V$-generic for any forcing notion, except those that look just like Cohen forcing below a condition. Other forcing notions, of course, can add Cohen reals as a by-product, and this often happens at limit stages in finite support iterations or products of forcing. In these cases, these other forcing notions have Cohen forcing as a complete subalgebra, and they can be viewed as adding a Cohen real, followed by the corresponding quotient forcing. -I take the first part of the theorem to say basically that a generic filter determines in a sense the forcing notion that gave rise to it, and that a generic filter can't by accident be generic for a totally different forcing notion. -I can post a proof sketch of the theorem, if you like. For the first claim, let $\dot H$ be a $\mathbb{P}$-name, and then let $p\in G$ force that $\dot H$ is $\check V$-generic and that $\check V[\dot G]=\check V[\dot H]$. One then looks at the map $q\mapsto [\![q\in\dot H]\!]^{\mathbb{P}\upharpoonright p}$. -One must restrict to the forcing below a condittion in this theorem, because otherwise one can do silly sneaky stuff like this: let $\mathbb{B}$ be Cohen forcing, and let $\mathbb{C}$ be the lottery sum of trivial forcing with Cohen forcing. So if $c$ is $V$-generic for $\mathbb{B}$, then it can also be added by the second, by opting for the part of the forcing the selects Cohen forcing; but $\mathbb{B}$ does not embed densely into $\mathbb{C}$, because $\mathbb{C}$ has an atom, the part that opted instead for trivial forcing, and there is no way to embed anything below that atom. So this technically counts as an affirmative answer to your highlighted question. But what it shows is that really one wants instead to consider cones in the forcing.<|endoftext|> -TITLE: Laws of Iterated Logarithm for Random Matrices and Random Permutation -QUESTION [22 upvotes]: The law of iterated logarithm asserts that if $x_1,x_2,\dots$ are i.i.d $\cal N(0,1)$ random variables and $S_n=x_1+x_2+\cdots+x_n$, then -$$\limsup_{n \to \infty} S_n/\sqrt {n \log \log n} = \sqrt 2, $$ almost surely. -1. Random matrices -Let $(x_{ij})_{1 \le i < \infty, 1 \le j \le i }$ be an array of i.i.d random $\cal N(0,1)$ variables, define $x_{ji}=x_{ij}$ and let $M_n$ be the $n$ by $n$ matrix obtained by restricting the array to $1 \le i \le n, 1 \le j \le n$. -Let $\lambda(n)$ denote the maximum eigenvalue of $M_n$. The variance and limiting distribution of the maximum eigenvalue (for these and related classes of matrices) were discovered by Tracy and Widom. -Question 1: What is the law of iterated logarithm for this scenario? Namely, can one identifies functions $F(n)$ and $G(n)$ such that -$\limsup_{n \to \infty} (\lambda (n)- {\bf E}(\lambda(n))/ F(n) = 1,$ almost surely, and -$\liminf_{n \to \infty} (\lambda (n)- {\bf E}(\lambda(n))/ G(n) = -1$ almost surely. -Update (May, 24, 2015): A full answer for the limsup (including constants) and a partial answer for the liminf was achieved by Elliot Paquette and Ofer Zeitouni: arxiv.org/abs/1505.05627 ! -2. Random permutations -Choose a sequence of real numbers in the unit interval at random independently. -Use the first $n$ numbers in the sequence to describe a random permutation $\pi$ in $S_n$. Look at the length $L_n$ of the maximum increasing sub-sequence of $\pi$. The expectation, variance and limiting distributions of the maximal increasing subsequence are known by the works of Logan-Shepp, Kerov-Vershik, (expectation) Baik, Deift & Johansson (variance and limiting distribution) and many subsequent works. -Question 2: What is the law of iterated logarithm for this scenario? -3. Motivation. -My initial motivation is to try to understand the law of iterated logarithm itself, and more precisely, to understand why do we have the $\sqrt {\log \log n}$ factor. (I cannot rationally expect to understand the $\sqrt 2$.) - -REPLY [15 votes]: My initial answer was wrong. Instead of completely deleting it, I left it at the bottom of the post. I use notation now that slightly differ from my original post. -As before, I give only a partial answer related to the max eigenvalue question and the limsup. Recall that the tail estimates for the max eigenvalue is -$$(P(\lambda_1(n) >2\sqrt{n}+tn^{-1/6})\sim e^{-Ct^{3/2}}$$ -(see e.g. Ledoux's reviews or -http://math.univ-lyon1.fr/~aubrun/recherche/smalldeviations/smalldeviations.pdf). -Now, the question is what is the scale in which the recentered and rescaled maximal eigenvalues start behaving like independent random variables. -In my original post I essentially claimed this starts happening at scale roughly $n$. This is wrong: the correct scale is given by the so-called GUE minor process (described by Baryshnikov in 2001). Relevant to Gil's problem is the scaling limit computed by Forrester and Nagao: http://arxiv.org/pdf/0801.0100.pdf. They show that the correct scaling in $n^{2/3}$, that is -that the (centered and rescaled) maximal eigenvalues of the GUE(n) and GUE(m) decorelate if $m-n>>n^{2/3}$ and are strongly correlated in $m-n<C'\log n$, i.e. -$t_{a_n}\sim C'' (\log n)^{2/3}$. Unravelling the definitions, -this will give you an upper bound on the limsup that is $t_n\sim -(\log n)^{2/3}$; that is, this will show that the function $F(n)$ -is at most $C (\log n)^{2/3}/n^{1/6}$. The lower bound on the limsup will follows a similar pattern, using the asymptotic independence stated above, and will give the same function $F(n)$. -One can also compute constants (for the a.s. convergence) in the same manner. -Filling in the details in the above is beyond a mathoverflow answer, I may return to it in detail later. -I speculate that the liminf will use the lower tail estimates, which are of the -form $e^{-Ct^{3})}$. Working with the same argument would then give -$(\log n)^{1/3}/n^{1/6}$ as the correct scaling. -OLD (WRONG) POST: -The following is a partial answer with some hints on how to complete it; I may revisit it later with more details. The answer below refers to the max eigenvalue question, to the limsup, and does not attempt to get the constants, only the scale. -First, the upper bound: recall that the tail estimates for the max eigenvalue is -$$(P(\lambda_1(n) >2\sqrt{n}+t\sqrt{n})\sim e^{-Cnt^{3/2}}$$ -(see e.g. Ledoux's reviews or -http://math.univ-lyon1.fr/~aubrun/recherche/smalldeviations/smalldeviations.pdf). Also recall (a kindly neighbor next door, A. Dembo, pointed out the usefulness of this observation in the current context) that the sequence $\lambda_1(n)$ is monotone in $n$. So, the limsup can be taken over a sequence $a_n=(1+\epsilon)^n$ instead of $n$. If you go over that sequence, you get from Borel-Cantelli that you cannot exceed $t_{a_n}\sqrt{a_n}$ infinitely often as long as $$\sum_n e^{-C t_{a_n}^{3/2} a_n}<\infty.$$ This will happen as soon as $a_nt_{a_n}^{3/2}>C'\log n$, i.e. -$t_{a_n}\sim C'' (\log n/a_n)^{3/2}$. Unravelling the definitions, -this will give you an upper bound on the limsup that is $t_n\sim -(\log \log n /n)^{2/3}$; that is, this will show that the function $F(n)$ -is at most $C (\log \log n)^{2/3}/n^{1/6}$. -(Remark to @Carlo Beenakker: the difference with TSAW is that the tail estimates there are proportional to $t^3$, not $t^{3/2}$; This will be in line with the liminf computation in the RM case, in which case I expect the same discrepency in exponents of the $\log \log n$ as you noted, see below.) -For the lower bound (on the limsup), you need to take a sequence that -is sparse enough so that you have approximate independence. -Here is a leap of faith, that would require a separate proof: if -you take an $N\times N$ GUE and resample the top $\epsilon N\times \epsilon N$ -entries, the maximum eigenvalue does not change much if $\epsilon$ is small. -I believe this follows from delocalization results for the eigenvectors -of GUE, but have not checked details. If that is true, then for an exponentially -growing sequence $a_n=C^n$, the sequence $\lambda_1(a_n)$ - will behave like an i.i.d. sequence, to which -you now can again apply the independent Borel-Cantelli to match the previous upper bound. -I speculate that the liminf will use the lower tail estimates, which are of the -form $e^{-C(nt^{3/2})^2}$. Working with the same argument would then give -$(\log \log n/n^2)^{1/3}$ as the correct scaling.<|endoftext|> -TITLE: Overlapping sets -QUESTION [6 upvotes]: Consider the following problem: -Let $F \subseteq 2^{I}$ be a finite family of finite subsets of some index set $I.$ -Let $F_x$ be defined as the number of elements of $F$ that contains $x.$ -Assume that for each $x \in I,$ $F_x$ is an even number. -Under what conditions on $F$ can we find a subset $F'$ of $F$ such that -$2F'_x = F_x$ for all $x$? -Note that if $I$ is the set of natural numbers, then for any family $F$ which consists of intervals (satisfying the extra property that $F_x$ is always even), this problem is solvable, and it is easy to prove. -(However, it does not generalize to rectangles, where there are counter examples) -Now, there are several ways to generalize/specialize this problem in various directions, for example, this question is such a generalization. -I am curios if there is any research done on problems that is similar to this type of problem above. - -REPLY [4 votes]: Your question is equivalent to whether the combinatorial discrepancy of the dual hypergraph is zero. (In case you are unfamiliar with the notion, start here: http://en.wikipedia.org/wiki/Discrepancy_of_hypergraphs) This problem is well-known to be NP-complete, you can easily reduce e.g., 1-in-3 SAT to it.<|endoftext|> -TITLE: Quotient of Coxeter group -QUESTION [6 upvotes]: Since the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$ seems to have resisted attacks of some powerful programs, I will turn to a group that seems to be a bit easier to analyse. The group -$$H := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19} \rangle$$ -only has one of the simple groups that are a quotient of $G$, $J_1$. Therefore it seems it will be easier to analyse. -Adding the relation $(abcbc)^i$ gives the trivial group for all $i$ less than 25 except for 15, where it gives $J_1$. What is the group -$$I := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^3, (bc)^7, (abc)^{19}, (abcbc)^{25} \rangle?$$ -I have checked on magma using knuth bendix, and it takes too long, so magma stops the calculation. - -REPLY [13 votes]: As you may know, in the family of groups -$H^{m,n,p} := \langle a, b, c \ | \ a^2, b^2, c^2, (ab)^2, (ac)^m, (bc)^n, (abc)^p \rangle$ -considered by Coxeter, there is only one remaining case for which finiteness has not been decided, $H^{3,7,19}$, which is your group $H$. It is possible that $H \cong J_1 \times {\rm PSL}(2,113)$, since only two finite simple quotients are known. The kernel of the homomorphism onto $J_1$ is perfect. I haven't quite proved it, but I think the kernel onto ${\rm PSL}(2,113)$ is also perfect. -A homomorphism onto ${\rm PSL}(2,113)$ is induced by the map onto ${\rm SL}(2,113)$ -$$a \mapsto \left(\begin{array}{rr}58& 57\\ 50& 55\end{array}\right), -b \mapsto \left(\begin{array}{rr}55& 69\\ 2& 58\end{array}\right), -c \mapsto \left(\begin{array}{rr}100& 67\\ 43& 13\end{array}\right).$$ -The image of $abcbc$ under this map has order $57$. -With the ACE coset enumerator, I have been able to show that adding the extra relation $(abcbc)^n$ gives the trivial group for $n=25,26,27,28,29$, but I've got stuck on $n=30$.<|endoftext|> -TITLE: Local polynomial form of holomorphic functions -QUESTION [8 upvotes]: It is well-known that a germ of a holomorphic function of $n\geq 2$ variables, with at most an isolated singularity (i.e. the singularity of the analytic variety defined by the zero locus of the function), is locally conjugate (by right composition with a local biholomorphism) to a polynomial (in fact, a finite jet of the function). -This statement is not true anymore when $n\geq 3$ if the singularity is not isolated, although I can't find a reference containning examples of germs not locally conjugate to polynomials. Yet if I remember correctly, it is true when $n=2$. -Also I would be interested in knowing what happens in the meromorphic case, which I believe is still open. -I would be grateful for any pointer in the litterature towards such results! -PS: I hope this question is not a duplicate. I couldn't possibly browse through all the search results for requests like «polynomial holomorphic function», and was unable to think about sharper keywords… - -REPLY [11 votes]: The following example of non-algebraic germ is due to Whitney, see -H. Whitney: Local properties of analytic varieties, Differential and Combinatorial Topology 205–244, Princeton University Press (1965). -Take the germ of singularity in $\mathbb{C}^3$ given by $f(x,y,z)=0$, where $$f=xy(x+y)(x- zy)(x-e^zy).$$ -The set $\{f=0\} \cap \{z= \lambda \}$ is given by five distinct lines through the origin. -If $f$ would be conjugate to a polynomial, the cross-ratios of any four of these lines would depend algebraically on $\lambda$. But this is impossible, because we have $\lambda$ and $e^{\lambda}$. -So this germ is not conjugate to a polynomial (or algebraic) germ.<|endoftext|> -TITLE: Is the field of constructible numbers known to be decidable? -QUESTION [30 upvotes]: By the field of constructible numbers I mean the union of all finite towers of real quadratic extensions beginning with $\mathbb{Q}$. By decidable I mean the set of first order truths in this field, in the language of 0,1, + and $\times$, is recursive. Is this field either known to be decidable, or known not to be? -As of 1963 Tarski's question of whether this field is decidable was open -- so i doubt any simple adaptation of his result on real closed fields can settle this question. He conjectured that the only decidable fields were finite, real closed, or algebraically closed. See Julia Robinson, The decision problem for fields, Theory of Models (Proc. 1963 Internat. Sympos. Berkeley), North-Holland, Amsterdam, 1965, pp. 299–311. especially pages 302 and 305. -Much has gone on since 1963, and Tarski's general conjecture is well refuted, but I do not find a solution to this problem. - -REPLY [14 votes]: Ziegler's theorem, a special case of which is that the theory of Euclidean fields (ordered fields in which every positive has a square root) is undecidable, already shows that ancient -Greek geometry was creative, not mechanical. In English, see -http://www.michaelbeeson.com/research/papers/Ziegler.pdf -The constructible field is just the smallest Euclidean field (but of course we don't yet know that it is undecidable).<|endoftext|> -TITLE: Is there any monoid in which the product of two non-invertible elements could be invertible? -QUESTION [9 upvotes]: I think the title speaks for itself. Thus I just explain the story behind the question. Of course, you may want to skip the story. -Story: Currently, I teach a course in linear algebra and matrices with a mathematician colleague of mine. In preparing for the class, we discussed together about one of the standard theorems of matrices saying the product of two invertible matrices is invertible. To understand the theorem and its converse, I naturally came to the question asked in the title. To my surprise, neither the colleague I discussed with, nor four other mathematician colleagues of mine could come up with an answer. But the question seems very natural and it is very surprising if it has been remained unnoticed. That is why I came to MO to find the answer. -PS. None of the colleagues I asked the question is an expert in algebra, and I am a mathematics educator. -PPS. You may replace "monoid" with whatever you wish, providing that you keep the rest of the title intact! Please keep your example natural (if there is one)! - -REPLY [5 votes]: Since the question came from an undergraduate level perspective, here's a slight variant on the answers given thus far that students might appreciate. -Take the vector space P of real polynomials. There are two operators on P, D=differentiation and I=integration (to make integration a linear operator, we "set c=0"). Now DI is the identity map, but I is not onto (as the constant polynomials are not in its image) and D is not injective (as D(c) = 0 for any constant polynomial c).<|endoftext|> -TITLE: Is a given point in the interior of the convex hull of a given finite collection of points? -QUESTION [7 upvotes]: Suppose I have the convex hull $P$ of a finite collection of points in $\mathbb{R}^d,$ and I want to see whether a point $p$ is contained in $P.$ This is a standard (some would say the standard linear programming problem: we are determining whether $p = \sum_{i=1}^{V(p)} \lambda_i v_i,$ with $\lambda_i \geq 0,$ $\sum_i \lambda_i = 1.$ (sum being over the vertices of $P.$ -But now, suppose I want to know whether $p$ is strictly contained in $P.$ (that is, whether $p$ is in the interior of $P$). Of course, this already assumes that $P$ has nonempty interior, but let's suppose we have some reason to know that. How do we check that? One way is to maximize $\lambda_i$ for $i=1, \dots, V(P)$ -- all the maxima should be positive. This is rather inefficient ($V(P)$ could be large). Another approach is to shrink $P$ slightly (that is, find a point in the interior, say the barycenter of the vertices, call it the origin, shrink the vertices by $1-\epsilon$, etc). This works, except that it is not clear what the right value of $\epsilon$ is. -I am probably missing something obvious. - -REPLY [13 votes]: Take your linear program and add the objective function max $x$, and the inequalities $\lambda_i - x \geq 0$. If the point is on the exterior, the optimum solution has $x=0$. Otherwise, there is a solution with $x > 0$.<|endoftext|> -TITLE: Comparison of two traces -QUESTION [5 upvotes]: Suppose $X$ is a smooth quasi-projective variety over $\mathbb{C}$ and $Z$ a proper subscheme, there is a formal duality isomorphism (here we consider the Zariski topology) due to Hartshorne: -$$ tr: R^n\Gamma_{Z}(\Omega_X^n) \cong \mathbb{C} $$ -given in proposition 5.2 of his paper,"On the de Rham cohomology of algebraic varieties." In our setting there is a really simple trace in the analytic topology given by integration. -$$ R^n\Gamma_{cs}(\Omega_X^n) \cong \mathbb{C} $$ -where we are considering cohomology with compact support. -Question: Do these traces satisfy compatibility under GAGA and the universal $\delta$ map -$$ R^n \Gamma _{Z_{an}}(\Omega_{X_{an}}^n) \to R^n \Gamma_{cs}(\Omega_X^n) $$ I'd draw the diagram I want but I'm too incompetent. On the other hand I expect it's clear from the question. - -REPLY [5 votes]: The issue of equality of maps in the derived category is always a delicate issue. In the algebraic setting the commutativity is explained in Lipman's Asterisque 117. When you get into the analytic setting, then there is a comparison between the algebraic map and the one coming from De Rham theory that involves a period, namely $\frac{1}{2 \pi i}$. -As a matter of fact, when comparing the algebraic and the analytic traces, a sign arises that depends on the dimension of $X$. This is explained carefully in: -Sastry, Pramathanath; Tong, Yue Lin L.: -The Grothendieck trace and the de Rham integral. -Canad. Math. Bull. 46 (2003), no. 3, 429–440.<|endoftext|> -TITLE: Does $\text{Mat}_n(k)$ have some universal properties similar to its universal enveloping algebra? -QUESTION [5 upvotes]: Let $k$ be a field and $\text{Mat}_n(k)$ be $n \times n$ matrices over $k$. Let's consider $\text{Mat}_n(k)$ as an associative algebra and denote $gl_n(k)$ be the same $k$-linear space as $\text{Mat}_n(k)$ but considered as a Lie algebra. We can form the universal enveloping algebra $U(gl_n(k))$ of $gl_n(k)$, which satisfies the universal property, $\forall$ unital associative $k$-algebra $A$, we have -$$ -Hom_{k-alg}(U(gl_n(k)),A)\cong Hom_{Lie}(gl_n(k),F(A)), -$$ -where $F$ is the forgetful functor from $k$-alg to Lie. -Now we know that the associative algebra $\text{Mat}_n(k)$ is not the universal enveloping algebra $U(gl_n(k))$. However, by the universal property, we have -$$ -Hom_{k-alg}(U(gl_n(k)),\text{Mat}_n(k))\cong Hom_{Lie}(gl_n(k),gl_n(k)), -$$ -which makes $\text{Mat}_n(k)$ "special" in some sense. -$\textbf{My question}$ is: Could we make the above observation more precise? Does $\text{Mat}_n(k)$ have some universal properties similar to $U(gl_n(k))$? - -REPLY [8 votes]: Of course, what's a universal property is open to interpretation, but $\mathrm{Mat}_n(k)$ corepresents something very nice: $\mathrm{Hom}_{k-alg}(A,\mathrm{Mat}_n(k))$ is the space of representations of $A$ on $k^n$ (not up to any kind of equivalence, of course). This can also be interpreted as the $k$ points of an affine algebraic $k$-variety $\mathrm{Rep}_n(A)$, as long as $A$ is finitely generated: if $a_1,\dots, a_m$ are generators of $A$, and $r_1(\mathbf{a}), \dots, r_\ell(\mathbf{a})$ are the non-commutative polynomials giving relations, then $\mathrm{Rep}_n(A)$ is defined by taking $n\times n$ matrices $M_1,\dots, M_m$ with each matrix coefficient a variable, and letting the relations be $r_k(\mathbf{M})=0$. -In fancier language, we can define its points over any $k$-algebra $R$ by $$\mathrm{Rep}_n(A,R)=\mathrm{Hom}_{R-alg}(A\otimes_k R,\mathrm{Mat}_n(R)).$$ -You can interpret this as saying that the functor from commutative to non-commutive $k$-algebras given by $\mathrm{Mat}_n$ is the right adjoint to the functor sending $A$ to $k[\mathrm{Rep}_n(A)]$, much like the universal enveloping algebra is left adjoint to the forgetful functor from algebras to Lie algebras that just remembers the bracket.<|endoftext|> -TITLE: Diophantine equation - $a^4+b^4=c^4+d^4$ ($a,b,c,d > 0$) -QUESTION [7 upvotes]: How can I find the general solution of $a^4+b^4=c^4+d^4$ ($a,b,c,d > 0$)? -And how did Euler find the solution $158^4+59^4=133^4+134^4$? - -REPLY [2 votes]: some material related to your question can be found here : https://sites.google.com/site/tpiezas/013<|endoftext|> -TITLE: On permuted sum of squares of primes in a list -QUESTION [6 upvotes]: We want to pick a set of distinct primes (if not possible, then just positive numbers) $p_1,p_2,\dots,p_k$ such that there exists $t$ permutations, $\sigma_1(\cdot)$,$\sigma_2(\cdot),\dots,\sigma_t(\cdot)$, of the primes such that the sum of vectors $$(p_1^2,p_2^2,\dots,p_k^2)+(p_{\sigma_1(1)}^2,p_{\sigma_1(2)}^2,\dots,p_{\sigma_1(k)}^2)+\dots+(p_{\sigma_t(1)}^2,p_{\sigma_t(2)}^2,\dots,p_{\sigma_t(k)}^2)=(T,T,T,....,T)$$ where $T=O(k^c)$ for some constant $c > 0$. -Is this possible and how do you do this? Given a $k$, can $t$ be as small as $O(\log(k))$? -For every $k$ is there a polynomially big $T$ and a $t$ that is logarithmic in $k$? -From Gerry's comments: let us fix $t=2$, for every $k$, is there a $T$ that is polynomial in $k$ that satisfies the above relations? His comments provide existence of values of $k$ such that $T=k$ but does not cover all $k$. - -REPLY [6 votes]: For large $k$ it is always possible to do this with $30$ permutations. Write $k$ as $30a+6b+5c$ where $0\le b\le 4$ and $0\le c\le 5$. Then the permutations will be a product of $a$ disjoint $30$-cycles, $b$ disjoint $6$-cycles, and $c$ disjoint $5$ cycles (take a cycle consisting of the first thirty primes, then the next thirty primes etc until we get to the last five primes). The resulting permutation has order thirty and its powers are the thirty permutations we want. (The basic idea here is what was expressed in Gerry Myerson's comments to the question. Myerson's observation is that when $k$ is a multiple of $3$, one can use products of three cycles, and three permutations suffice. In general this would allow one to take the smallest prime dividing $k$ as a possible answer. What is covered here is all the cases when $k$ does not have a small prime factor.) -For the sums to add up to the same number, we must find a number $N$ such that $30N$ is expressible as the sum of thirty squares of distinct primes in at least $a$ ways; $6N$ is expressible as the sum of six squares of distinct primes in at least $b$ ways; and $5N$ is expressible as the sum of five squares of distinct primes in at least $5$ ways. This can be arranged thanks to the Hardy-Littlewood circle method. -It may not be easy to find an exact reference that does so however. So here's a little context. It is an old conjecture that every integer that is $4 \pmod {24}$ is the sum of four squares of primes. This remains difficult to prove. However Hua showed in 1938 that every large integer that is $5 \pmod{24}$ may be written as a sum of five squares of primes. In his argument, which is based on the circle method, it would be possible to arrange for the primes to be distinct, and to guarantee many solutions. In fact for five squares of primes (assuming the congruence condition $\pmod {24}$) there will be about $n^{3/2-\epsilon}$ representations of a large number $n$. Note that representations where a prime is repeated, or representations with a given prime will be at most $O(n^{1+\epsilon})$ in number. So one can guarantee many representations ($n^{1/2-\epsilon}$ at least) as a sum of five squares of distinct primes, and no primes shared among two such solutions. Once one has the result for five squares, of course every integer that is $j \pmod{24}$ for $j\ge 5$ may be expressed in many ways as a sum of $j$ squares of distinct primes. That finishes the job. -This problem seems a little strange to me; I would appreciate some motivation of where it came from. -Update: Actually I can give a reference that will work. It is work of Harman and Kumchev: see http://arxiv.org/pdf/0902.4190.pdf . Harman and Kumchev show that with very few exceptions, any number that is $3\pmod {24}$ and not a multiple of $5$ can be expressed as a sum of three squares of primes in $\gg n^{1/2-\epsilon}$ ways. Note that one doesn't have to worry about Siegel zeros, as this is an almost all result. From this and the argument above one can get that $12$ permutations suffice (writing $k=12 a +4b+3c$).<|endoftext|> -TITLE: Multiplying by irrational numbers in combinatorial problems -QUESTION [28 upvotes]: This is getting no attention on stackexchange. -Everybody knows that the number of derangements of a set of size $n$ is the nearest integer to $n!/e$. -It had escaped my attention until last week, when I wrote this answer, that the number of sequences of distinct elements of a set of size $n$ (including sequences of length $0$) is the nearest integer to $n!e$, provided $n\ge2$. The sequence whose $n$th term is the nearest integer to $n!e$ satifies the recurrence $x_{n+1} = (n+1)x_n + 1$. -How widespread is this operation of mulitplying by an irrational number and then rounding, in combinatorial problems? Are there other standard examples? Is there some general theory accounting for this? And, while I'm at it, where is this in "the literature"? -(I'm not sure whether we should include things like Fibonacci numbers or solutions of Pell's equation as examples of the same thing.) - -REPLY [2 votes]: A remarkable paper by Colin Defant, Troupes, Cumulants, and Stack-Sorting, uncovers another surprising combinatorial appearance of $e$. Let $s$ denote the stack-sorting map (for the definition, see Defant's paper) and let $\mathrm{des}(\pi)$ denote the number of descents of a permutation $\pi$. If $\sigma$ is a permutation of $\{1, 2,\ldots, n-1\}$ chosen uniformly at random, then the expected value of $\mathrm{des}(s(\sigma))+1$ is -$$\biggl(3 - \sum_{i=0}^n {1\over i!}\biggr)n.$$ -Multiplying the above number by $n!$ yields the integer $\lceil n!(3 - e)n\rceil$. You might expect that there is some simple connection to derangements, but no combinatorial connection to derangements is known (as of 2021).<|endoftext|> -TITLE: An analysis proof of the Hall marriage theorem -QUESTION [14 upvotes]: The Hall marriage theorem has several relatively easy combinatorial proofs. Are there short analytical or topological reformulations and proofs of that theorem? - -REPLY [9 votes]: Here is an only slightly more direct proof from convex duality and total unimodularity of the constraint matrix. -Let $G = (L \cup R, E)$ be a bipartite graph satisfying Hall's condition, where $|L| = |R| = n$. The matching polytope $P_G$ for this graph is the convex hull of the indicator vectors $x \in \mathbb{R}^E$ of all matchings, and can be written as: -$$ -\forall v \in L \cup R: \sum_{e \ni v}{x_{e}} \leq 1\\ -\forall e \in E: x_e \geq 0 -$$ -I.e., if $B$ is the incidence matrix of $G$, then $P_G = \{x: Bx \leq 1, x\geq 0\}$. Let us verify this is the matching polytope. It is clear that all matchings satisfy these constraints, and that all integral points in $P_G$ are matchings. Since the incidence matrix of a bipartite graph is easily seen to be totally unimodular, all vertices of $P_G$ are integral, and we are done. -A maximum cardinality matching in $G$ is naturally formulated as the linear optimization problem $\max\{1^Tx: x\in P_G\}$, whose dual is -$$ -\min 1^Ty\\ -\text{ subject to}\\ -\\ -\forall e=(u,v) \in E: y_u + y_v \geq 1\\ -\forall u \in L \cup R: y_u\geq 0 -$$ -(Notice the solution to the above linear optimization problem is just the size of the minimum vertex cover of $G$.) By (strong) duality, which is a consequence of the Hahn-Banach theorem, $G$ has a matching of size $n$ if and only if the optimum of the above program is at least $n$. To prove Hall's theorem we must show that if the above optimization problem has optimal value strictly less than $n$, then Hall's condition is violated. Again by total unimodularity, there exists an integral optimal solution; let $y$ be such a solution and take $S = \{u: y_u = 1\}$ and $S_L = S \cap L$, $S_R = S \cap R$. Since $y$ is feasible, $N_G(L \setminus S_L) = S_R$, and we have -$$ -|L \setminus S_L| = n - |S_L| = n - |S| + |S_R| > |S_R| = N_G(L \setminus S_L), -$$ -contradicting Hall's condition. -This approach gives König's theorem immediately, and you get Egerváry's theorem in the weighted case.<|endoftext|> -TITLE: Is there an underlying topological space for ind-schemes? -QUESTION [7 upvotes]: An ind-scheme over a base scheme $S$ can be defined in several ways. For simplicity, lets assume that $S$ is the spectrum of an algebraically closed field $k$. We can define a $k$-ind-scheme as a functor from $k$-algebras to sets, such that there exists a sequence of $k$-schemes with closed embeddings -$X_1 \to X_2 \to ...\to X_n \to ...$ -and a natural isomorphism of $X$ with the functor -$R \mapsto \lim_{\rightarrow}(X_n(R))$. -Such a functor is a sheaf in the Zariski topology and can be uniquely extended as a sheaf to operate on all $k$-schemes. -Is there a sensible way to define the "underlying topological space" of an ind-scheme $X$? In other words, is it true that the direct limit of the topological spaces underlying the $X_n$ (an increasing union really) is independent of the presentation? -If we can take all $X_n$ to be quasi-compact, it seems to me that the answer is positive. since for different (quasi-compact) presentations, we will have closed embeddings $f_n:X_n\to X'_{m(n)}$. Am I right? -Edit: -Even though I wrote "any sensible way to define the underlying topological space", I now realize that I am actually only interested in the "direct limit" topology. The main question is the independence of presentation in the general or some special cases. - -REPLY [4 votes]: The underlying set of a functor $X : \mathsf{Alg}(k) \to \mathsf{Set}$ is given by $\mathrm{colim}_{K/k} X(K)$, where $K/k$ runs through all field extensions. The topology is defined using open subfunctors. A reference is the book by Demazure and Gabriel on algebraic groups. I have learned it from a script by Marc Nieper-Wißkirchen. I'm not sure what happens for Ind-schemes, though.<|endoftext|> -TITLE: Does the notion of graphs with vertex multiplicity exist? -QUESTION [11 upvotes]: I need to use graphs where each vertex gets a natural number, $b(v)$, its multiplicity. These numbers indicate how many 'replications' of the vertex we have. -It is actually a way to write in a compact way a graph which has a lot of twin vertices: two vertices $u$ and $v$ are twin if $N(u)=N(v)$ (they are not adjacent). -Since the twin relation is an equivalence relation, there is a unique way to write any graph in a reduced form as a graph with vertex multiplicities. -Here is an example: - -This representation is particularly useful when vertices represent some items with large quantity. -Does someone know if such a concept already exists ? What's the usual name ? Do you have some references ? - -REPLY [3 votes]: This notion is complementary to the notion of vertex multiplication, in which every vertex is replaced with a homogeneous clique. This goes back to Lovasz' proof of the Weak Perfect Graph Theorem, and probably even earlier. - -REPLY [2 votes]: As well as the use of this by Lovász for perfect graphs (mentioned by earlier answers), this has been used by Häggkvist to find high-degree four-chromatic triangle-free graphs. See -Häggkvist, R. (1981), "Odd cycles of specified length in nonbipartite graphs", Graph Theory (Cambridge, 1981), pp. 89–99, MR0671908.<|endoftext|> -TITLE: On triangulated categories of pro-objects -QUESTION [5 upvotes]: Which term is used for model categories whose homotopy categories are triangulated? Stable proper model categories? -I want $Ho(Pro-M)$ to be triangulated ($Pro-M$ is the category of pro-objects of M) and the functor $Ho(M)\to Ho(Pro-M)$ to be an exact full embedding. Which restrictions on M are needed to this end? - -I looked several papers on homotopy categories of pro-objects, yet I was not able to find a clear answer to this question. In particular, is it possible to take an $M$ such that $Ho(M)$ is the motivic stable homotopy category here? -P.S. I would like to understand the relation between the approaches of: "t-model structures" (Fausk, Isaksen), "Duality and pro-spectra" ( Christensen, Isaksen), "Model structures for pro-simplicial presheaves" (Jardine), "Strict model structures for pro-categories" (Isaksen), and "Stability in pro-homotopy theory" (Seymour). -Upd. Moreover, I would like $SH$ (or its compact objects) to become cocompact in the corresponding $Ho(Pro-M)$. Does this mean that the objects of $M$ should be fibrant in $Pro-M$ (or only the fibrant ones?), and that all the objects of $Pro-M$ should be cofibrant? - -REPLY [5 votes]: Take this answer with a grain of salt since I can only provide vague references. Nevertheless, I claim that the properness of $M$ should be sufficient. If $C$ is a stable $\infty$-category, then $Pro(C)$ is stable as well, because the suspensions and loops can be computed "levelwise". For any $\infty$-category $C$, the Yoneda embedding $C\to Pro(C)$ is fully faithful and preserves colimits and finite limits, so in particular it is exact. Now if $M$ is a proper model category with underlying $\infty$-category $\tilde M$ and $Pro(M)$ is equipped with Isaksen's strict model structure, then the underlying $\infty$-category of $Pro(M)$ is $Pro(\tilde M)$, so in particular $Pro(M)$ is a stable model category and $Ho(M)\to Ho(Pro(M))$ is fully faithful and triangulated. -Pro-objects in accessible $\infty$-categories are discussed briefly in J. Lurie, Derived Algebraic Geometry XIII, §3.1, and the fact that the Yoneda embedding is fully faithful and exact is proved in Higher Topos Theory, §5.3, in the dual setting of Ind-objects, but I'm afraid that more precise references for the claims I've made may not exist. However, for accessible $\infty$-categories such as those of motivic spectra, I think the above claims can be justified rigorously enough using those references.<|endoftext|> -TITLE: quotient by finite group actions that are smooth -QUESTION [5 upvotes]: Let $X$ an affine normal scheme of finite type over a field $k$ of characteristic zero. -Let $G$ a finite group acting on $X$ and $Y=X/G=Spec(K[X]^{G})$. -We assume that $Y=\mathbb{A}^{n}=k[f_{1},\dots,f_{n}]$ with $f_{i}\in K[X]^{G}$. -Then we have a finite flat surjective morphism $\pi:X\rightarrow Y$ generically étale of group $G$. -Do we have that $\pi$ is a complete intersection? - -REPLY [15 votes]: No, there is no reason that $X$ should be a complete intersection. For instance, begin with $Z=\mathbb{A}^2_k$ with coordinates $(z_0,z_1)$. Consider the action on $Z$ of the group of $n^{\text{th}}$ roots of unity, $\mu_n$, by $\zeta\cdot(z_0,z_1) = (\zeta z_0,\zeta z_1)$. The ring of invariants, $k[z_0,z_1]^{\mu_n}$, is generated by the monomials $y_{a,b} = z_0^az_1^b$ for all pairs $(a,b)$ of nonnegative integers with $a+b=n$. So the finitely generated $k$-algebra $k[z_0,z_1]^{\mu_n}$ has Krull dimension $2$ and has $n+1$ elements $y_{a,b}$ in a minimal set of generators. Yet, already the vector space of quadratic relations among these generators has dimension roughly $n^2/2$. Indeed, it is generated by the binomials, -$$ Q_{(a_0,b_0),(a_1,b_1),(a_2,b_2),(a_3,b_3)} = y_{a_0,b_0}y_{a_1,b_1}-y_{a_2,b_2}y_{a_3,b_3},$$ -where $a_0+a_1=a_2+a_3$ (which then forces the same identity for the $b$ indices). Of course these generators are not linearly independent. However, using a monomial ordering, there are roughly $n^2/2$ leading monomials that occur among these binomials. The dimension of the vector space is at least as large as the number of leading monomials, thus roughly $n^2/2$. When $n$ is large, $k[z_0,z_1]^{\mu_n}$ cannot possibly be a complete intersection of dimension $2$. -Let $Y$ be the quotient of $Z$ by this action of $\mu_n$. The point is that there is a larger group, $\mu_n\times\mu_n$, that acts on $Z$, e.g., $(\zeta_0,\zeta_1)\cdot(z_0,z_1) = (\zeta_0 z_0,\zeta_1 z_1).$ Of course the original action comes from the diagonal subgroup $\Delta(\mu_n) \leq \mu_n \times \mu_n$. Since this group is Abelian, there is an induced action of the quotient group $G=(\mu_n\times \mu_n)/\Delta(\mu_n)$ on the quotient variety $Y = Z/\Delta(\mu_n)$. The quotient of $Y$ by this induced action is the same as the quotient of $Z$ by the full action of $\mu_n\times \mu_n$. But, of course, $k[z_0,z_1]^{\mu_n\times \mu_n}$ equals $k[z_0^n,z_1^n]$, which is a polynomial ring. -Edit. Of course you can compute the precise dimension of the vector space of quadratic relations is $n(n-1)/2$. Thus, already for $n=3$, $Y$ is not a complete intersection.<|endoftext|> -TITLE: $(\varphi, \Gamma)$-modules of finite height -QUESTION [7 upvotes]: Maybe the answer to my question is obvious. -Let $p$ be a prime $\geq 3$. Let $D$ be an étale $(\varphi, \Gamma)$-module over $A_{\mathbb{Q}_p} = \{ \sum_{n \in \mathbb{Z}} a_n X^n \, \vert \, a_n \in \mathbb{Z}_p, a_n \to 0 \mbox{ as } n\to -\infty \}$ of finite rank. -Recall that the action of $\varphi$ is given by $\varphi(X) = (1+X)^p -1$ and that of $\gamma \in \Gamma$ by $\gamma(X) = (1+X)^{\omega(\gamma)} -1$ where $\omega$ is the $p$-adic cyclotomic character. -Denote by $A^+$ the subring of $A_{\mathbb{Q}_p}$ of power series without denominators, that is $A^+ = \{ \sum_{n \in \mathbb{N}} a_n X^n \, \vert \, a_n \in \mathbb{Z}_p \}$. -Say that $D$ is of finite height if $D= D^+ \otimes_{A^+} A_{\mathbb{Q}_p}$ where $D^+$ is an étale $(\varphi, \Gamma)$-module over $A^+$. -Is there an étale $(\varphi, \Gamma)$-module $D$ over $A_{\mathbb{Q}_p}$ (of finite rank) which is not of finite height and such that for all étale $(\varphi, \Gamma)$-module of rank $1$ $\delta$ over $A_{\mathbb{Q}_p}$, $D \otimes \delta$ is not of finite height ? -Obviously if such an object exists, it should be of rank $\geq 2$. - -REPLY [9 votes]: Proposition: If $D$ is not of finite height, then nor is $D \otimes \delta$ for any $\delta$ of rank 1. -Proof: It suffices to show the contrapositive: if $D$ is of finite height so is $D \otimes \delta$ for any rank 1 $\delta$. This follows easily from the fact that every continuous character of $G_{\mathbf{Q}_p}$ (thus any etale rank 1 $(\varphi, \Gamma)$-module) is the product of an unramified character and a character factoring through $\Gamma$; twisting by a character of either type will just change the actions of $\varphi$ and of a generator of $\Gamma$ on $D$ by elements of $\mathbf{Z}_p^\times$, and thus sends a $(\varphi, \Gamma)$-stable $\mathbf{A}^+_{\mathbf{Q}_p}$-sublattice in $D$ to one in $D(\delta)$. -Since representations that aren't of finite height do exist, the answer to your question is thus "yes" -- any non-finite-height $D$ will do.<|endoftext|> -TITLE: rings in which every element is a sum of two commuting idempotents -QUESTION [10 upvotes]: Is there a known characterization of the rings $R$ (containing $1$) with this property: every element of $R$ is a sum of two commuting idempotents. - -REPLY [6 votes]: The trivial/elementary proof (in particular, it does not use the axiom choice). -A ring $R$ satisfies your condition iff it satisfies the identity $x^3=x$. -Pf. -Will shows above with a short computation that if $R$ satisfies the condition, then the characteristic divides $6$. So if $z=e+f$ with $e,f$ commuting idempotents, then $z^3=(e+f)^3=e+6ef+f = e+f=z$. -Suppose $z^3=z$ for all elements of $R$. Now $R$ has characteristic dividing $6$ since $2=(1+1)^3=8$. Thus we have $R$ is a product $R_1\times R_2$ of a ring $R_1$ of characteristic $2$ and $R_2$ of characteristic $3$. The ring $R_1$ is boolean (using $z-1=(z-1)^3=z^3-3z^2+3z-1=z^2-1$, whence $z^2=z$) and so all elements are idempotent and we are done. So it suffices to handle $R_2$, i.e., assume the characteristic is $3$. -Note $z^2$ is idempotent and $z=z^2+(z-z^2)$. We claim the latter is idempotent. Then $(z-z^2)^2= z^2-2z^3+z^4=2z^2-2z=z-z^2$. Done. -Answer. -I believe that the answer is that the ring be a subdirect product of copies of $Z/2$ and $Z/3$, which in this case is equivalent to being in the variety of rings generated by $Z/6$. -My original revised answer below says that a ring satisfying the desired condition is a subdirect product of copies of $Z/2$ and $Z/3$ and hence in the variety generated by $Z/6$. -Suppose $R$ is a ring in the variety generated by $Z/6$. Since $R$ is a direct limit of finitely generated rings and your property is closed under direct limit, we may assume that $R$ is finitely generated. But then $R$ is finite because a variety generated by a finite ring is locally finite. -But if $R$ is finite and in this variety, then $R$ is a reduced finite commutative ring and hence a finite direct product of domains belonging to the variety generated by $Z/6$. But $Z/2$ and $Z/3$ are the only domains in this variety. -Original Revised Answer. -A necessary condition is that $R$ is a subdirect product of copies of $Z/2$ and $Z/3$. -Claim 1: The class of rings satisfying your property is closed under direct product and homomorphic images. -Pf. Exercise. -Claim 2: If the ring is indecomposable (not a direct product), then the ring either is a boolean algebra or satisfies $z^3=z$. -Pf. See Will's answer and the comments. -Claim 3: the ring $R$ is reduced (i.e., has no non-zero nilpotents). -Pf. Clear since it is a product of a ring satisfying $z^2=z$ and one satisfying $z^3=z$. -Theorem 12.7 in Lam's book on non-commutative rings shows a subdirectly irreducible reduced ring is an integral domain. Since 0,1 are the unique idempotents of an integral domain, we conclude that $R$ is subdirect product of $Z/2$ or $Z/3$. -Now boolean algebras (=subdirect products of $Z/2$) all have the desired property so it remains to see which subdirect products of $Z/3$ have the property.<|endoftext|> -TITLE: Embedding points in 2D based on distance estimates? -QUESTION [6 upvotes]: Suppose we have a collection of exactly $N$ points (say $N=1000$), with each point belonging to 2-dimensional Euclidean space $\mathbb{R}^2$, but we don't know the coordinates of the points. Suppose that we instead have, for some pairs of points, an approximation for the Euclidean distance between them. -Question: How can we find an approximation of the coordinates of these points (up to flip/rotations of the plane) in the plane? -I.e., we want to embed the points in the plane in a way that is as consistent as possible with the estimated distances. (Note: These distances might turn out to be inconsistent, since they're approximations.) -I need something that I can actually implement on a computer. -Motivation: This question arises from studying geometric graphs in the plane: we can approximate the Euclidean distance between vertices by studying the number of common neighbours. More common neighbours imply the vertices are more likely to be closer together. But not all pairs of vertices have common neighbours, in which case we don't have an estimate. - -REPLY [6 votes]: One standard thing is to simulate a network of springs, one per data that you have, such that each spring wants to have the length corresponding to your estimate. In other words, if $l_{ij}$ is your estimate, you can look at the energy $$H((x_i)) = \sum [l_{ij} - \|x_j-x_i\|]^2$$ or something similar. Of course the sum runs over all pairs for which you have an estimate. Then minimize H, which can be done numerically in a reasonably efficient way because of the shape of $H$. -Software packages like graphviz do that, and so does Mathematica IIRC. - -REPLY [5 votes]: For complete distance information, this is known as "multidimensional scaling" and heavily used in social sciences or psychology. Actually, there is a very simple approach in this case based on the singular value decomposition and this goes back to Households and Young in the paper "Discussion of a set of points in terms of their mutual distances". -In the case of limited data, one can approach this via matrix completion, i.e. try to find a full distance matrix, i.e. a matrix $D$, such that $D_{i,j} = \|x_i-x_j\|$ for some points $x_i$ which has the known entries (up to an error) in the right places. See, e.g. "Solving Euclidean Distance Matrix Completion -Problems Via Semidefinite Programming" by Alfakih, Khandani and Wolkowicz or "Low-rank optimization for distance matrix completion" by Mishra, Meyer and Sepulchre.<|endoftext|> -TITLE: What are the essential properties of algebraic closure on an arbitrary structure? -QUESTION [7 upvotes]: Define the "model theoretic" notion of a closure function as follows: -Definition (1): Let $D$ be a non-empty set. A function $cl:P(D)\longrightarrow P(D)$ called a closure function iff it has the following properties: -$(1)~\forall A\subseteq D~~~~A\subseteq cl(A)$ -$(2)~\forall A,B\subseteq D~~~~A\subseteq B\Longrightarrow cl(A)\subseteq cl(B)$ -$(3)~cl(cl(A))=cl(A)$ -We say that $cl$ is a "good" closure on $D$ if it has the following property too: -$(4)~\forall A\subseteq D~~~~cl(A)=\bigcup_{B\in P_{<\omega}(A)}cl(B)$ -and a "pregeometry" if we have: -$(5)~\forall A\subseteq D~~~~\forall a,b\in D~~~~~~a\in cl(A\cup \lbrace b\rbrace)\setminus cl(A)\Longrightarrow b\in cl(A\cup \lbrace a\rbrace)$ -There are many trivial good closures on a given set which are not related to any structure, but even there are some non-trivial natural good closures on the domain of an arbitrary $\mathcal{L}$-structure $\mathcal{M}$ like well known "algebraic closure" ($acl_{\mathcal{M}}$), "definable closure" ($dcl_{\mathcal{M}}$) and "structural closure" ($scl_{\mathcal{M}}$) which is defined as follows: -$\forall A\subseteq Dom(\mathcal{M})~~~~~scl_{\mathcal{M}}(A):=Dom(\langle A\rangle_{\mathcal{M}})$ -and $\langle A\rangle_{\mathcal{M}}$ is the substructure of $\mathcal{M}$ generated by the set $A$. -Now the main question is about the "essential" properties of these closure functions on an arbitrary structure. In the other words is "goodness" the unique essential property of $acl_{\mathcal{M}}$, $dcl_{\mathcal{M}}$ and $scl_{\mathcal{M}}$ on an arbitrary $\mathcal{L}$ - structure $\mathcal{M}$? Precisely: -Question (1): Let $D$ be an arbitrary non-empty set, and $cl:P(D)\longrightarrow P(D)$ be a good closure function on $D$, is there a first order language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathcal{M}$ such that: $Dom(\mathcal{M})=D$ and $scl_{\mathcal{M}}=cl$? -Question (2): What is the answer of question (1) for $acl_{\mathcal{M}}$ and $dcl_{\mathcal{M}}$? -Remark (1): Note that producing a negative answer for the above questions needs finding a property $P$ different from "being a good closure" and proving that the functions $acl$, $dcl$ and $scl$ have the property $P$ on "any" structure and so there is no such language and structure for an arbitrary function $cl:P(D)\longrightarrow P(D)$ because it is possible that such function have the property $\neg P$.So one can re ask the above questions by the following re stating: -"Let $D$ be an arbitrary non-empty set, and $cl:P(D)\longrightarrow P(D)$ be a good closure function with property $P$ on $D$, is there a first order language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathcal{M}$ such that: $Dom(\mathcal{M})=D$ and $scl_{\mathcal{M}}=cl$ ($acl_{\mathcal{M}}=cl$ or $dcl_{\mathcal{M}}=cl$)? -So we can go further and try to characterize "all" essential properties of $scl, acl, dcl$ on an arbitrary structure and ask the following question: -Question (3): What is the property $P$ such that for any non-empty set $D$ and any function $cl:P(D)\longrightarrow P(D)$ which is a good closure function with property $P$, the answer of the question (1) or (2) be true? -In the other direction It is well known that $acl_{\mathcal{M}}$ is a pregeometry on "strongly minimal" structures. So: -Question (4): Is there a known type of $\mathcal{L}$-structures which the functions $scl$ or $dcl$ be a pregeometry on them? - -REPLY [3 votes]: $\textbf{Question 1}$ The answer to this question is affirmative. The closure operators that you call good closure operators are normally called algebraic closure operators. Furthermore, each algebraic closure operator is of the from $\mathrm{scl}_{\mathcal{A}}$ for some algebra $\mathcal{A}$. A proof of this fact is given here on Theorem 3.5 in Chapter 2. Suppose that $C$ is an algebraic closure operator on some set $A$. -Let $\mathcal{F}_{n}$ be the set of all $n$-ary functions $f:A^{n}\rightarrow A$ such that $f(a_{1},...,a_{n})\in C(a_{1},...,a_{n})$ whenever $a_{1},...,a_{n}\in A$. Let $\mathcal{A}=(A,\bigcup_{n}\mathcal{F}_{n})$. Then it is easy to see that $C=\mathrm{scl}_{\mathcal{A}}$.<|endoftext|> -TITLE: Is there a 3d equivalent of this picture? -QUESTION [7 upvotes]: This question arises apropos of an earlier question I asked that was (VERY!!!) helpfully answered by Anton Petrunin: -Fitting a mesh to a density function -The picture below is the image of a regular equilateral triangular lattice in the complex plane under the map $z \mapsto z^2$: - -(for whatever reason, I seem to be missing a few links in the picture above, but I hope this is sufficient). This picture has two nice visual properties, namely that the points are distributed in a "regular" or "predictable" way, and that the density of points becomes more concentrated as we move towards the origin. My question is, are there configurations of points in 3 dimensions that also have these two visual properties? Owing to Liouville's theorem, I realize that I won't be nearly as fortunate in constructing a nice and easy conformal mapping, so I'm just looking for any possible way to place points in $\mathbb{R}^3$ where I could get some kind of behavior similar to that pictured above. The best I can think of would be some sort of concentric geodesic buckyball-type structures, or possibly the cartesian image of a regular lattice in spherical coordinates, but I'm wondering if there's anything that's more interesting. -I should also mention that I don't necessary need the configuration to be "completely regular". That is, I'm happy to tolerate occasional points that are inconsistent with the rest of the configuration; there should just be some kind of visual consistency that one would readily observe by inspection. - -REPLY [7 votes]: The restriction to conformal maps is a natural one, as it means that there is no affine distortion in the neighbourhood of a point. Specifically, the Voronoi cells of the points will not be oblated or prolated, which is not the case for non-conformal maps. Once we insist on that restriction, Liouville's theorem insists that the only possibilities are Möbius transformations (compositions of Euclidean transformations and geometric inversions). -It seems like a good idea to invert a lattice, and the dense face-centred cubic lattice is more aesthetically pleasing than the ordinary cubic lattice. Geometric inversion of the face-centred cubic lattice ($A_3 \cong D_3$ in Sphere Packings, Lattices and Groups) yields an arrangement of points, all of which are contained within the unit ball, and the density diverges to infinity as you approach the origin. Specifically, we take the set of points: -$$\{ \dfrac{(x,y,z)}{x^2+y^2+z^2} : x,y,z \in \mathbb{Z}, \enspace x+y+z \equiv 1 \mod 2\}$$ -Here's a three-dimensional rotating view of the configuration, where I've made the points semitransparent: - -Is this what you're looking for? Unlike your two-dimensional configuration, this is bounded and infinitely dense in any open ball containing the origin.<|endoftext|> -TITLE: A problem on infinite dimensional metric space -QUESTION [7 upvotes]: Let $(X_{n},d_{n})_{n \in \mathbb{N}}$ be a sequence of complete geodesic metric spaces such that: -$X_{n}$ is a regular$^1$ CW-complex of constant local dimension$^3$ $n$, it is of finite type$^4$, boundaryless$^2$, unbounded, uniform$^5$, and it is the $n$-skeleton of $X_{n+1}$, which is n-connected. Moreover, the distances $d_{n}$ , $d_{n+1}$ generate the same topology on $X_{n}$ and $\forall x,y \in X_{n} \ d_{n+1}(x,y) \le d_{n}(x,y)$. -Finally $(X_{n},d_{n})$ is quasi-isometric to $(X_{n+1},d_{n+1})$, through the inclusion map $X_{n} \subset X_{n+1}$, and a distance $d$ on $ \bigcup{X_{n}}$ is defined (for $x, y \in X_{n_0}$) by $d(x,y) := lim_{n (\ge n_0) \to \infty} d_{n}(x,y)$. -Definition : Let $X:=\overline{\bigcup{X_{n}}}$ be the completion of the metric space $\bigcup{X_{n}}$ with $d$. -Question : Is $X$ weakly contractible ? -Remark: Some of these conditions could be useless for a proof, and others, highly generalized. -Motivation: See here for applications to geometric group theory and noncommutative geometry. - -$^1$Regular (for a CW complex) : the attaching maps are homeomorphism (see this post). -$^2$Boundaryless (for a regular CW complex) : the boundary of each closed cell is contained is the union of the boundaries of other closed cells. -$^3$Constant local dimension : the topological dimension of all neighborhood of all point, is constant. -$^4$Finite type : finitely many $r$-cells ending in a fixed $(r-1)$-cell. -$^5$Uniform : For all $r$-cell $c_{1}$ and $c_{2}$, there is a neighborhood $n_{1}$ of $c_{1}$ and $n_{2}$ of $c_{1}$, such that $n_{1}$ is homeomorphic to $n_{2}$. - -REPLY [12 votes]: This is not true even in finite dimensions. There exists a decreasing sequence of complete Riemannian metrics on the plane, pairwise Lipschitz equivalent, such that the pointwise limit is isometric to the standard sphere without one point. Then the completion is the sphere. -To construct such a sequence, consider the metric of the punctured sphere in geodesic polar coordinates: $ds^2= dr^2+\sin^2 r\,d\varphi^2$ and add a term like $2^{-n}f(r)dr^2$, where $f(r)=1/r$ for $r$ near 0. This makes the distance to the origin infinite, so the metric is complete. But the additional term goes to zero as $n\to\infty$, so the limit is the standard metric of the punctured sphere. -To make an infinite-dimensional example, take a metric product with your favorite contractible infinite-dimensional cell complex.<|endoftext|> -TITLE: What's the link between topological spaces as locales and topological spaces as infinity-groupoids? -QUESTION [10 upvotes]: I've seen texts that talk about topological spaces being essentially locales, like Topology via Logic by Vickers, and texts related to homotopy theory that talk about topological spaces being essentially infinity-groupoids. -However, I've never seen any treatment of topology that combines these perspectives or talks about the relationship between locales and infinity-groupoids. Where should I be looking? - -REPLY [7 votes]: I don't think it's a good idea to mix the slogans "topological spaces are locales" and "topological spaces are $\infty$-groupoids." I think the former slogan encapsulates what we ended up defining as topological spaces while the latter slogan encapsulates what we should've defined as topological spaces, at least if we're algebraic topologists (recall that $\infty$-groupoids are only supposed to capture topological spaces up to, say, weak homotopy equivalence). -But let me make a guess anyway: the answer should be "higher sheaf theory." The story for $1$-types should be that the category of locally constant sheaves on a locale is equivalent to the category of functors from a (pro?-)groupoid to $\text{Set}$. This groupoid can be recovered by a suitable version of Tannaka reconstruction and should be regarded as the fundamental groupoid of the locale (although this won't agree with the fundamental groupoid in the ordinary sense without some local connectivity assumptions). To get the fundamental $2$-groupoid we need to consider locally constant $2$-sheaves (stacks?), whatever those are, and that should be equivalent to the category of $2$-functors from a $2$-groupoid to $\text{Gpd}$. And so forth. There are some nLab articles suggesting that someone has worked all this out.<|endoftext|> -TITLE: $A \otimes^L_B C$ computing the derived fiber product of schemes -QUESTION [5 upvotes]: Let $A \rightarrow B$ and $C \rightarrow B$ be two maps of schemes. How can I compute the derived fiber product $A \otimes^L_B C$? I'm guessing this is a dg-scheme. -For instance - let $B=\mathbb{A}^1$ and $A = 0 \in \mathbb{A}^1$, and $C \in \mathbb{A}^1$ some point (either $0$ or $1$). -Question: The example I'm really interested in is $\tilde{\mathfrak{g}} \otimes^L_{\mathfrak{g}} 0$, where $\mathfrak{g}=\mathfrak{sl}_2$ and $\tilde{\mathfrak{g}}=\{ X,(0 \subset V \subset \mathbb{C}^2) | X \in \mathfrak{sl}_2, X V \subset V \}$ is the Grothendieck-Springer resolution. - -REPLY [7 votes]: As usual, you replace $A$ (or $C$) with a (sheaf of) DG-algebras which is flat over $B$ and compute the usual tensor product of it with $O_A$ over $O_B$. In case when $A$ is a complete intersection subscheme a good choice for such DG-algebra is the Koszul complex. Then the derived fiber product is given by the pullback to $C$ of the Koszul resolution of $O_A$. -For example, if $A = 0 \subset \mathbf{A}^1$ then $O_A \cong \{k[t] \stackrel{t}\to k[t]\}$, so the derived fiber product is empty if $C = 1$ and $\{k \stackrel{0}\to k\} = k[\epsilon]/\epsilon^2$ with $\deg\epsilon = -1$, $d\epsilon = 0$ if $C = 0$. The same computation works as well in your second example.<|endoftext|> -TITLE: How did Riemann calculate the first few non-trivial zeros of the zeta-function? -QUESTION [18 upvotes]: Does anyone know how Riemann calculated the first few non-trivial zeros of the Zeta function? I am wondering if he approximated the integral, $\frac{1}{2 \pi i} \int_{R} \frac{{\xi}^\prime(z)}{\xi (z)} dz$ over appropriate rectangle(s) in the critical strip. This still seems difficult, however, without a computer. - -REPLY [13 votes]: In searching through the Riemann Nachlass in Gottingen (including those -folders not listed as connected with $\zeta(s)) $ there is no -evidence -- at least that has been saved -- that Riemann computed -anything more than the first few zeros (I think up to ordinate about 80). -The method he used was the expansion that is now called the Riemann-Siegel -formula. I did not see any use, e.g., of an approach based on -Euler-Maclaurin. The limited accuracy Riemann obtained reflects that of -the error term in the R-S formula.<|endoftext|> -TITLE: Sum over integer compositions -QUESTION [5 upvotes]: Sorry if the question is trivial - are there closed form expressions or good approximations for the sum of a symmetric function taken over all integer compositions (into given number of parts) of a number? -More precisely, I'm interested in: -$$ -S(n,k) = \sum_{a1+ \cdots +a_k = n, \ \ a_i \geq 1} \phi_k(a_1,\dots,a_k) -$$ -where $\phi_k = \prod_i a_i^p$, e.g. for $p=-2$, but I'm curious even about $p=1$ -I realize that I can bound this (using AM-GM ineq.) by replacing all terms by the most (un)balanced composition, but this seems quite weak as a bound. -EDITED: partition composition - -REPLY [3 votes]: Yes, there are closed-form expressions. The number $S(n,k)$ you are looking for is the number of weighted integer compositions with weighting function $f(a)=a^p$. Many recursions and other representations for this number exist. -For example, the number $S(n,k)$ from above is precisely the number $d_{S,f}(n,k)$ defined in Eger (2013), Restricted weighted integer compositions and extended binomial coefficients (for $S=\{1,2,3,…,\}$ and $f(a)=a^p$). This paper says that $S(n,k)$ is an extended binomial coefficient, and gives various representations of the extended binomial coefficients. Other relevant literature would be Fahssi (2012), The polynomial triangles revisited, and Shapcott (2013), C-color compositions and palindromes. More relevant literature can be found in the references of these works. Other work of C. Shapcott also addresses part-products of integer compositions, which is related to your case $p=1$.<|endoftext|> -TITLE: Question about Godel's Proof book (Ernest Nagel / James R. Newman) -QUESTION [5 upvotes]: I am seeking some assistance on a Mathematics/logic problem which I am having trouble with. -My problem is understanding how a number is calculated in a book called Gödel's Proof (about Kurt Gödel's Incompleteness Theorems)...page 97. Basically, the problem is regards to the (G) formula, whose meta-mathematical statement refers to itself as being not 'demonstrable'. -My problem is that I can't work out how Gödel number (g) could be sub (n, 17, n): -(1) ~ (∃x) Dem (x, Sub (y, 17, y)) Gödel number = n -(G) ~ (∃x) Dem (x, Sub (n, 17, n)) Gödel number (g) = sub (n, 17, n) -(G) is derived from (1), due to specialization of the 'y' variable - replacing y with the Gödel number for (1), which is n. -The sub (n, 17, n) function is really a shorthand for a formula of a formal system (which has the Gödel number 'n'), where any instance of the variable which has the Gödel number '17' (say 'y') is replaced by the Gödel number for 'n' itself.....kind of self-swallowing. I assume that a fellow reader of the book could put it all in context, someone with stronger mathematical/logic ability than my own. -The way I interpret the formula (G) is that it states that (1) is not a theorem (although it is as there world be a number x that fits the criteria), and since (G) is a specialization of (1), I can see how it would be a reference to 'itself', however, I cannot workout how (G)'s Gödel number would be sub (n, 17, n). -I would like to understand more clearly how the number sub (n, 17, n) could be the Gödel number for the formula (G). This is the only difficulty I have with the book..just want to understand the nuts and bolts of (G). -I would be immensely grateful for any help with this and thank you for reading my post. - - -I've drawn up this table to illustrate my interpretation of (G). - -REPLY [8 votes]: I don't have the book in front of me right now, so the following may not exactly match what it says, but it should be close enough to give you the right idea. The function sub is defined so that, if $a$ is the Gödel number of a formula $\alpha$, if $b$ is the Gödel number of a variable $v$, and if $c$ is any natural number, then $\text{sub}(a,b,c)$ is the Gödel number of the formula obtained by substituting the numeral for $c$ in place of the variable $v$ in the formula $\alpha$. So, in the case at hand, since $n$ is the Gödel number of formula (1) and 17 is the Gödel number of the variable $y$, $\text{sub}(n,17,n)$ is the Gödel number of the formula you get by substituting the numeral for $n$ in place of $y$ in (1). But the result of that substitution is (G), so $\text{sub}(n,17,n)$ is the Gödel number of (G).<|endoftext|> -TITLE: Topological Hochschild cohomology? -QUESTION [8 upvotes]: Let $A$ be a $E_\infty$-ring spectrum. By EKMM, it may be treated as a commutative algebra in the appropriate category. In particular, one may define topological Hochschild homology as $A\wedge_{A\wedge A} A$, like for usual commutative algebra. -My question is: can one say something about $F_{A\wedge A}(A,A)$ (function spectum), that is about topological Hochschild cohomology? Does the Gerstenhaber bracket make sense in this context? If $A$ is, say, the K-theory, does homotopy groups of its Hochschild cohomology contain some interesting elements? -Of course, it is enough to have $A_\infty$-ring structure on $A$ for this questions, but I am interested only in $E_\infty$. Besides, I am interested in spectra $F_{A\otimes S^n}(A, A)$, where $S^n$ is the $n$-sphere. - -REPLY [11 votes]: The topological Hochschild cohomology (that I'll denote now THC) makes sense whenever $A$ is at least an $E_1$-algebra. In particular, you can construct THC of an $E_\infty$-algebra. There is a result called Deligne's conjecture but which is now a theorem stating that THC of an $E_1$-algebra is an $E_2$-algebra. In particular, if you take the homology of THC of something, the resulting graded abelian group has a Gerstenhaber algebra structure. If you take homotopy groups, you get a commutative algebra with a degree 1 bracket but I don't think it's going to satisfy the axioms of a Gerstenhaber algebra in general. -Taking the endomorphisms over $A\otimes S^{n-1}$ is a perfectly fine construction called higher THC. It can be defined as soon as $A$ is an $E_{n}$-algebra although the definition is slightly more involved (a good reference is http://www.math.northwestern.edu/~jnkf/writ/cotangentcomplex.pdf). Higher Deligne's conjecture tells you that this higer Hoschild cohomology is an $E_{n+1}$-algebra. In particular taking homology, you get a Gerstenhaber algebra with a bracket of degree $n$. -Note that in the case where $A$ is $E_\infty$, there is a nice construction of higher THC in the following paper of Ginot Tradler and Zeinalian (they restrict to $E_\infty$-algebras in chain complexes but the case of spectra is similar) -http://arxiv.org/abs/1205.7056 -Edit: I just noticed that you were asking more specifically what THC of $KU$ is. It turns out that the unit map $KU\to F_{KU\wedge KU}(KU,KU)$ is an equivalence. The same is true if you replace $KU$ by $E_n$ (the height $n$ Lubin-Tate spectrum). This remains true for the higher dimensional versions of THC. The unit map $E_n\to F_{S^d\otimes E_n}(E_n,E_n)$ is an equivalence. The reason for this is essentially the fact that $E_n$ is étale aver the $K(n)$-local sphere. You can look at http://geoffroy.horel.org/HHC%20of%20the%20LT%20ring%20spectrum.pdf for more details.<|endoftext|> -TITLE: Linear Algebra without Choice -QUESTION [23 upvotes]: We consider the field of "usual" linear algebra. -Q. Which aspects of it can be carried out without the Axiom of Choice? -Q. Do interesting "exotic" phenomena appear in presence of (some instance of) the negation of the Axiom of Choice? -Without Choice, vector spaces may have a basis (hence, in particular, be dimensional) or not and hence be adimensional. [As Andreas Blass observes in the comments, the terminology "dimensional/adimensional" should rather be used to denote the property of having all bases of the same cardinality, rather than just having a basis, as there are vector spaces with two bases of different cardinality] -Q. Could the following property of a vector space $V$ -Property ($\star$) Every injective endomorphism of $V$ is an automorphism. -be a valid substitute for finite-dimensionality for the class of not-necessarily-dimensional vector spaces over a field? -Would the linear algebra of vector spaces verifying ($\star$) be reasonably similar to the usual one for finite dimensional spaces? - -REPLY [36 votes]: Some things about vector spaces which are consistent with the failure of choice: - -Vector spaces may have bases of different cardinality. In particular, this means that the notion of "dimension" is not well-defined. It follows from the Boolean Prime Ideal theorem (which is strictly weaker than $\sf AC$ itself) that if there is a basis, then its cardinality is unique. See Sizes of bases of vector spaces without the axiom of choice for more details. -The existence of a basis is no longer hereditary. That is, it is consistent that there is a vector space which has a basis, but it has a subspace which doesn't have a basis. You can find the example in Goldstern's answer If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?, and what is even more interesting is the fact that the vector space without a basis has a direct complement which has a well-ordered basis. -It is consistent that there is a vector space, that all its endomorphisms are scalar multiplications (which is not $(0)$ or the field itself). In particular every non-zero endomorphism is an automorphism, and this answers yours final question. Indeed every non-zero endomorphism is an injective endomorphism and an automorphism. These spaces were the main topic of my masters thesis, where I somewhat extended Lauchli's original result (and construction) of such spaces. You can find somewhat of an outline of the general result here: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice? -It is consistent that there is a vector space, which is not finitely generated, which is (naturally) isomorphic to its algebraic double dual. In particular this can be $\ell_2$. See my answer at Does the fact that this vector space is not isomorphic to its double-dual require choice? for details. - -There are other properties which fail for non-finitely generated vector spaces in $\sf ZFC$ which are consistent with the failure of choice. The list is long, and these just a few I could write about from the top of my head. -Whether or not any of them is equivalent to the axiom of choice is usually an open (and a difficult) question. But it is usually the case that if a property requires some form of choice (like a basis, or extension of functinoals, etc.) then it can fail in suitable models of $\sf ZF+\lnot AC$.<|endoftext|> -TITLE: Hyperbolic Manifolds which fiber over the circle -QUESTION [14 upvotes]: If $N^2$ is a closed, orientable surface of genus at least $2$, and if $\phi$ is an (orientation-preserving) pseudo-Anosov mapping on $N$, then one can form the closed orientable 3-manifold $M^3$ by gluing the boundary components of $N^2\times [0,1]$ via $\phi$. In other words, $M^3$ is fibered over $S^1$, with fiber $N^2$. $M^3$ can also be given a hyperbolic structure (much like $N^2$ can). So my question is: - -Does there exist a closed orientable hyperbolic manifold $M^k$, fibered over $S^1$ with a fiber $N^{k-1}$, when $k\neq 3$? - -No such $M$ exists for $k=2$. It is also known (via Mostow rigidity) that for such an $M^k$ to exist for $k>3$, we cannot have the fiber $N^{k-1}$ hyperbolizable as well. But I don't know anything further than that. - -REPLY [8 votes]: I make a remark here that is well-known to experts, that $\pi_1(M)$ is an extension of $\pi_1(N)$ by an infinite-order outer automorphism, explaining partly Misha's comment. -Consider the sequence $\pi_1(N^{k-1}) \to \pi_1(M^k) \overset{\varphi}{\to} \mathbb{Z} $. Choose $\alpha\in \pi_1(M)$ such that $\varphi(\alpha)=1\in \mathbb{Z}$. Then conjugation $g\mapsto \alpha^k g\alpha^{-k}, g\in \pi_1(N)$ gives an automorphism of $\pi_1(N)$. If this automorphism is conjugation by an element of $\pi_1(N)$ for some $k$, then there exists $h\in \pi_1(N)$ such that $\alpha^k g \alpha^{-k}=hgh^{-1}$ for all $g\in \pi_1(N)$. Let $\alpha'=h^{-1}\alpha^k$, then $\alpha' g= g \alpha'$ for all $g\in \pi_1(N)$. Consider the maximal cyclic subgroup $C<\pi_1(M)$ containing $\alpha'$ (there is a unique such group since $\pi_1(M)$ is the fundamental group of a closed hyperbolic manifold, so $C$ is the stabilizer of the axis of $\alpha'$ by the proof of Preissman's theorem). Also by Preissman's theorem, the subgroup generated by $\langle \alpha', g \rangle$ is abelian and therefore cyclic, and therefore $g\in C$. But this implies that $\pi_1(N)\leq C$, so $\pi_1(M)\leq C$, a contradiction. -Thus, the conjugation by $\alpha$ maps to a infinite order element of $Out(\pi_1(N))$. -Then if $\pi_1(N)$ is hyperbolic, one can deduce that $\pi_1(N)$ splits over $\mathbb{Z}$ by Rips' theory. However, an aspherical closed $k-1$-manifold group cannot split over $\mathbb{Z}$ for $k>2$.<|endoftext|> -TITLE: When to use more exciting function spaces than ordinary Sobolev spaces? -QUESTION [20 upvotes]: In which kinds of PDEs are the more interesting function spaces required? I am thinking of spaces such as Besov and Triebel spaces, and their weighted versions. -For example, Sobolev spaces $L^2(0,T;H^1)$ can be used in linear parabolic PDEs. If we weaken this to consider certain parabolic PDEs with monotone operators then we can need to use spaces like $L^p(0,T;W^{1,q}).$ For degenerate problems, one can use weighted Sobolev spaces. -Where next? And what to read to get a good understand of the next steps? I would be grateful for any answers. - -REPLY [18 votes]: Spatial weights would be relevant in non-homogeneous settings in which one expects the behaviour at different regions of space to be different. For instance, if there is an obstacle or a boundary, a weight that depends on the distance to the boundary would be natural in order to capture boundary effects. If the initial data is originally assumed to be concentrated at the origin, then weights involving the distance $|x|$ to the origin are also natural. Similarly, weights involving time $t$ are sometimes natural in evolution equations, particularly if one is trying to describe decay or blowup in time. -More generally, if there is a natural singular set in physical space or frequency space, then it is natural to weight one's spaces around that set. The $X^{s.b}$ spaces mentioned in Willie's answer are a good example of this in the frequency domain (and Sobolev spaces themselves reflect the privileged nature of the frequency origin for many PDE, as the zero set for the symbol of the underlying linear operator (e.g. the Laplacian)). -If one needs to prevent the solution from concentrating all its mass or energy into a ball, then Morrey or Campanato spaces are occasionally useful. -As for the frequency-based refinements to Sobolev spaces (e.g. Besov and Triebel-Lizorkin, but also Hardy spaces, BMO, BV, etc.), these are "within logarithms" of Sobolev spaces, in the sense that if the ratio between the finest and coarsest spatial scale of interest (or equivalently, the ratio between the highest and lowest frequency scale of interest) is comparable to $N$, then the ratio between a Besov or Triebel-Lizorkin norm and its Sobolev counterpart (as plotted for instance on this type diagram: http://terrytao.wordpress.com/2010/03/11/a-type-diagram-for-function-spaces/ ) is at most a power of $\log N$. Because of this, Sobolev spaces generally suffice for all "subcritical" or "non-endpoint" situations in which one does not have to contend with a logarithmic pileup of contributions from each scale. If one is working in a critical setting (which is more or less the same thing as a scale-invariant or a dimensionless setting), these refinements can often be necessary to stop the logarithmic divergences caused by such things as the failure of the endpoint Sobolev inequality, e.g. $H^{n/2}({\bf R}^n) \not \subset L^\infty({\bf R}^n)$. (In this particular case, one can sometimes replace the Sobolev space $H^{n/2}$ with the smaller Besov space $B^{n/2}_1$ to recover the endpoint embedding, although there is no free lunch here and this will likely make some other estimate in one's analysis harder to prove.) -In general, unless one is perturbing off of an existing method, one does not proceed by randomly picking function spaces and hoping that one's argument closes. Often the function spaces one ends up using are dictated by trying to directly estimate solutions (or approximations to solutions). For instance, if one is trying to establish a local well-posedness result for a semilinear evolution equation in some standard space, e.g. $H^s({\bf R}^n)$, one can try to expand the $H^s$ norm of that solution using a basic formula such as the Duhamel formula or the energy inequality. In trying to estimate the terms arising from that formula by harmonic analysis methods (e.g. Holder inequality, Sobolev embedding, etc.), one is naturally led to the need to control the solution in other norms as well. If all goes well, all the norms on the right-hand side can be controlled by what already has on the left-hand side plus the initial data, and then one has a good chance of closing an argument; if not, one often has to tweak the argument by either strengthening or weakening the norms one is trying to control, as dictated by what the harmonic analysis is telling you. The final norms one uses to close the argument often arise from a lengthy iteration of this procedure (which unfortunately is often hidden from view in the published version of the paper, which usually focuses on the final choice of spaces that worked, rather than the initial guesses which didn't quite work but needed to be perturbed into the final choice). -Ultimately, in PDE one is usually more interested in the functions themselves, rather than the function spaces (though there are exceptions, e.g. if one is taking a dynamical systems perspective, or is relying on a fixed-point theorem exploiting the global topology of the function space). The reason that function spaces appear so prominently in PDE arguments is that functions have an infinite number of degrees of freedom, and the basic physical features of such functions (e.g. amplitude, frequency, location) are not easy to define directly in a precise and rigorous fashion. Function space norms serve as mathematically rigorous proxies for these physical statistics, but in the end they are only formal tools (with the exception of some physically natural norms or norm-like quantities, such as the mass or energy) and one should really be thinking about the physical features of the solution to the PDE directly. I discuss this point at http://terrytao.wordpress.com/2010/04/02/amplitude-frequency-dynamics-for-semilinear-dispersive-equations/ in the setting of semilinear dispersive equations (but there are similar perspectives for other PDE also).<|endoftext|> -TITLE: Has the conformally field theoretic metric on a Calabi-Yau variety been proved to exist? -QUESTION [20 upvotes]: Let $(X,g)$ be a compact Kähler manifold. Physics allows us to consider a supersymmetric sigma model with target $(X,g)$, which is a N=2 two-dimensional field theory. -From the two-dimensional point of view, the metric $g$ can be thougt as a coupling constant. -At the classical level, this theory is conformally invariant. But it is well known that in a quantum theory, coupling constant are in general renormalized: there is only a effective coupling constant which depends on an energy scale $\mu$. Here, we have an effective metric $g(\mu)$. The dependence on $\mu$ of $g(\mu)$ is controlled by a differential equation of the form $dg/d(log \mu)= \beta(g)$ where the beta function $\beta$ can be computed order by order in perturbation theory. By definition, the quantum theory will be conformal iff the beta function vanishes. At the first order in perturbation theory, $\beta$ is proportional to the Ricci curvature and so one can solve $\beta=0$ if $X$ is Calabi-Yau ($c_{1}(X)=0$) and there is a unique up to scaling such solution in each Kähler class $[\omega]$ (by Yau's proof of the Calabi conjecture). But in general, there are higher order corrections. One can show that the Kähler class remains unchanged but this is not the case for the metric in this class. If one could solve the equation $\beta = 0$ order by order in perturbation theory and prove that the solutions converge, then this should define in each Kähler class of $X$ a unique metric $g_{CFT}$. -All this is rather well known from the mid 80's. At this time, there was an article of Nemeschansky and Sen, "Conformal invariance of supersymmetric $\sigma$-models on Calabi-Yau manifolds", which, I have the impression, has closed the question among physicists. In this article, it is proved that one can solve $\beta =0$ order by order -but there is no result of convergence (as far as I understand). -So my question is: -Has $g_{CFT}$ been mathematically proved to exist? -As the perturbative expansion is only valid for large Kähler class, it is likely that such a restriction is necessary. Maybe a better question is: is the perturbative expansion -expected to converge for a large Kähler class or is it just some asymptotic expansion -(as it is often the case for such question in perturbative quantum field theory)? -In any case, the question of the existence of $g_{CFT}$ is rather well posed mathematically and I would like to know if it has been studied/solved. Any reference will be appreciated. - -REPLY [15 votes]: To the best of my knowledge of the literature on this topic, the answer is: not really (a few exceptions appear below). Let me first give a rigorous statement of the problem: the physical equation $\beta = 0$ can be expressed as -$\sum_{k=0}^{\infty} \epsilon^k \beta_k = 0$, -where $\epsilon$ is a physical parameter one imagines is "small," and where $\beta_k$ -is a symmetric two tensor depending on a Riemannian metric which arises as a certain universal expression in the curvature tensor and its derivatives, and which moreover obeys a natural scaling law and a natural "homogeneity" in terms of the number of derivatives of $g$ which are required to express it. These terms $\beta_k$ are computable in theory, although my understanding is that this has only been done on a piecemeal basis for the first few terms. There are many references one can easily find which do this for this and other sigma models. The first two terms are $\beta_0 = \mbox{Vol}(M) g$, $\beta_1 = \mbox{Rc}$, the term $\beta_2$ is a quadratic expression in the curvature tensor. Considering the equation up to order $k$ in the $\epsilon$ power expansion is referred to in physics literature as "up to $k$-loops." -So, with this background, there are at least two interesting questions one can ask: -1) Given $N \in \mathbb N \cup \{\infty\}$, can one construct on a given manifold a solution to $\sum_{k=0}^N \epsilon^k \beta_k = 0$? -2) Can one construct on a given manifold a one-parameter family of Riemannian metrics satisfying $\frac{\partial g}{\partial t} + \sum_{k=0}^N \epsilon^k \beta_k = 0$? -The original question was related to 1) above, but 2) is as important in the physics literature (AFAIK). Of course, the case $N=1$ of question 1) corresponds to solving the usual Riemannian Einstein equation (although one may need to justify ignoring the term $k=0$ in some cases, which physicists have ways of doing). In the Kähler setting the metric can come from the Calabi-Yau theorem as the poster stated. For question 2), the case $N=1$ corresponds to the (properly normalized) Ricci flow equation. -Now that I have properly stated the question, let me just say that I focus on studying elliptic/parabolic equations on manifolds, and for a while got interested in exactly this question. After much literature digging, I found only a few rigorous mathematical results: -http://arxiv.org/abs/0904.1241 -http://arxiv.org/abs/1108.0526 -http://arxiv.org/abs/1205.6507 -The first paper considers the case $N=2$ of question 2). In this case the operator is still second-order, and so in special settings the equation can be rendered parabolic, and then known techniques can be applied. This paper is certainly interesting, although the techniques cannot really be extended to $N > 2$. The last two papers address the homogeneous setting. -Moving beyond $N = 2$ is, IMHO, extremely difficult from a PDE point of view, since one has very little information on the form of the terms $\beta_k$ beyond the general qualitative statements I made above. For instance, an interesting side question is: do there exist arbitrarily large values of $k$ for which $\beta_k$ corresponds to an elliptic operator of a Riemannian metric? This is true for $k=1$, but I believe it is very unlikely to be true for higher $k$ since I believe it is "known" in the physics sense that all of the terms arising in $\beta_k$ for $k > 1$ are at least quadratic combinations of curvature and derivatives. I.e., one cannot expect a term like $\Delta \mbox{Rc}$ to show up, which would be elliptic. -One can imagine considering the case $N > 2$ on homogeneous spaces, where in principle the equations reduce to a system of ODE, but again one does not have much in the way of understanding the form of the general term $\beta_k$. -It is also not inconceivable that if one starts with say a Ricci-flat metric, and is allowed to choose $\epsilon$ very small with respect to this metric, that one can perturb it to a solution to the higher order equations, but again without at least a little more hard data on the form of the $k$-th term this seems tricky. -EDIT: In response to Robert's comments, yes, I meant $\mbox{Vol}(M) g$ (fixed above). Also, let me give a reference: -http://www.sciencedirect.com/science/article/pii/0550321389904227 -This paper computes the terms $\beta_3$ and $\beta_4$ (The formula for $\beta_4$ covers half a page of terms, and each of these is shorthand for a complicated curvature expression already. Written out fully would take 2 pages at least!). As far as whether one can expect the $\beta$ functions to be simpler on a Kähler manifold, I can't speak to the physics well enough to really say. From a mathematical point of view, I don't really see any advantage beyond the fact that of course each term could be expressed in terms of a Kähler potential. -The case of symmetric spaces is probably more tractable. Looking for solutions within the class $\nabla \mbox{Rm} \equiv 0$ could certainly simplify the terms $\beta_k$ greatly. My understanding of how these terms are derived is extremely fuzzy, but let me speak anyways: I believe the different summands in $\beta_k$ correspond in some sense to different sized "loop diagrams" (see the above reference for some examples), and moreover the qualitative behavior of such terms (i.e. number of derivatives of $\mbox{Rm}$ which appear is (somehow) related to the topology of the diagram. One main difficulty in performing these calculations to derive the terms $\beta_k$ is the sheer number of diagrams which must be considered, which grows wildly with $k$. If, on the other hand, one could a priori rule out a large number of such diagrams (by assuming $\nabla \mbox{Rm} \equiv 0$), perhaps these calculations become more tractable, or at least some stronger qualitative statements could be made. -Lastly, the quantity $\epsilon$ (which, also note is usually $\alpha'$ in physics literature) is, I believe, meant to be small but fixed. Physically it apparently is meant to represent the "string tension."<|endoftext|> -TITLE: Pseudo-differential operators with compactly supported symbols -QUESTION [7 upvotes]: If the symbol $p(x,\xi)$ of a pseudodifferential operator $P$ has compact $x$-support, then for any Schwartz function $f$, $Pf$ has compact $x$-support. -Is the reverse true? Namely that if some PDO $P$ with a symbol $p(x,\xi)$ from some reasonable symbol class has a property that $Pf$ has compact $x$-support for any Schwartz $f$, then does it imply that $p(x,\xi)$ has compact $x$-support? - -REPLY [12 votes]: Yes. Denote by $S_K$, $K$ compact, the closed subspace of Schwartz space $S$ consisting of all $f$ such that the support of $Pf$ is contained in $K$. The assumption and the Baire category theorem imply that $S=S_K$ for some $K$. It follows that the $x$-support of the Schwartz kernel $k(x,y)$ of $P$ is contained in $K$, hence also the $x$-support of the symbol $p(x,\xi)=\int e^{i\xi y} k(x,x-y)\,\mathrm dy$.<|endoftext|> -TITLE: On integers as sums of three integer cubes revisited -QUESTION [6 upvotes]: It is easy to find binary quadratic form parameterizations $F(x,y)$ to, -$$a^3+b^3+c^3+d^3 = 0\tag{1}$$ -(See the identity (5) described in this MSE post.) To solve, -$$x_1^3+x_2^3+x_3^3 = 1\tag{2}$$ -in the integers, all one has to do is to check if one term of $(1)$ can be solved as a Pell-like equation $F_i(x,y) = 1$. For example, starting with the cubes of the taxicab number 1728 as $a,b,c,d = 1,-9,12,-10$, we get, -$$a,b = x^2-11xy+9y^2,\;-9x^2+11xy-y^2$$ -$$c,d = 12x^2-20xy+10y^2,\;-10x^2+20xy-12y^2$$ -We can then solve $a = x^2-11xy+9y^2 = \pm 1$ since it can be transformed to the Pell equation $p^2-85q^2 = \pm 1$, thus giving an infinite number of integer solutions to $(2)$. -Question: How easy is it to find a quadratic form parameterization to, -$$x_1^3+x_2^3+x_3^3 = Nx_4^3\tag{3}$$ -for $N$ a non-cube integer? I'm sure one can see where I'm getting at. If one can solve, -$$x_4 = c_1x^2+c_2xy+c_3y^2 = \pm 1\tag{4}$$ -as a Pell-like equation, then that would prove that, -$$x_1^3+x_2^3+x_3^3 = N\tag{5}$$ -is solvable in the integers in an infinite number of ways. (So far, this has only been shown for $N = 1,2$). The closest I've found is a cubic identity for $N = 3$ in a 2010 paper by Choudhry, -$$\begin{aligned} -x_1 &= 2x^3+3x^2y+3xy^2\\ -x_2 &= 3x^2y+3xy^2+2y^3\\ -x_3 &= -2x^3-3x^2y-3xy^2-2y^3\\ -x_4 &= xy(x+y) -\end{aligned}$$ -which is a special case of eq.58 in the paper. -Anybody knows how to find a quadratic form parametrization to $(3)$? (If one can be found, hopefully $(4)$ can also be solved.) - -REPLY [3 votes]: Perhaps if you start with my three-rational-cubes identity -$$ - ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3} - - \biggl(\frac{(a^2+3b^2)^2-(6ab)^2}{6a(a^2+3b^2)}\biggr)^{\!3} - - \biggl(\frac{(a^2-3b^2)6ab^2}{(a^2+3b^2)^2}\biggr)^{\!3} -$$ -you might be able to find something?<|endoftext|> -TITLE: Illustrating Edward Nelson's Worldview with Nonstandard Models of Arithmetic -QUESTION [36 upvotes]: Mathematician Edward Nelson is known for his extreme views on the foundations of mathematics, variously described as "ultrafintism" or "strict finitism" (Nelson's preferred term), which came into the split light recently because of his claim, quickly recanted, of proving $PA$ inconsistent (but he's still working on it). He believes in Robinson Arithmetic, but not the induction schema of Peano Arithmetic (although he does accept weak forms of induction interpretable in Robinson's $Q$, like induction for bounded formulas; see his book Predicative Arithmetic). He even believes that exponentiation is not total! Here is a quote explaining his viewpoint: -“The intuition that the set of all subsets of a finite set is finite—or more generally, that if $A$ and $B$ are finite sets, then so is the set $B^A$ of all functions from $A$ to $B$—is a questionable intuition. Let $A$ be the set of some $5000$ spaces for symbols on a blank sheet of typewriter paper, and let $B$ be the set of some $80$ symbols of a typewriter; then perhaps $B^A$ is infinite. Perhaps it is even incorrect to think of $B^A$ as being a set. To do so is to postulate an entity, the set of all possible typewritten pages, and then to ascribe some kind of reality to this entity—for example, by asserting that one can in principle survey each possible typewritten page. But perhaps it is simply not so. Perhaps there is no such number as $80^{5000}$; perhaps it is always possible to write a new and different page.” -He believes that finite numbers are closed under addition and multiplication, but not exponentiation: he thinks you can have two numbers, like $80$ and $5000$, which are both finite, but where $80^{5000}$ in infinitely large! -My question is, can we illustrate this view using a nonstandard model of arithmetic? Specifically, how would we construct a nonstandard model of $Q$, containing an initial segment closed under successor, addition, and multiplication, but not exponentiation? Preferably, I'd like a computable nonstandard model. -Any help would be greatly appreciated. -Thank You in Advance. -EDIT: As @JoelDavidHamkins pointed out, exponentiation need not be total in a model of $Q$, so such a model wouldn't illustrate the phenomenon Edward Nelson believes is possible, of the exponentiation of finite numbers being infinite. So let me clarify: I'm looking for a computable nonstandard model of $Q$ + "exponentiation is total", such that the model has an initial segment closed under successor, addition, and multiplication, but not exponentiation. -EDIT 2: @EmilJeřábek has pointed out that different formulations of "exponentiation is total" in the language of arithmetic aren't provably equivalent in $Q$. How you formalize this assertion doesn't really concern me, so rather than talk about models of $Q$ + "exponentiation is total", let me talk about models $M$ of $Q$ equipped with a binary operation on $M$ that satisfies the basic properties of exponentiation: $a^{b+c} = a^b a^c$, $(a^b)^c = a^{bc}$, $a^1 = a$, $a^{0} = 1$, and $0^b = 0$. I also want addition and multiplication to be commutative and associative, multiplication to distribute over addition, and addition to obey the cancellation property that $a + c = b + c$ implies $a = b$. An example of a computable nonstandard model of $Q$ that satisfies at least these properties of addition and multiplication is the set of polynomials $\{P = a_nx^n + \ldots + a_0 \in \Bbb Z[x] , a_n > 0 \}$, together with $0$, with polynomial addition and multiplication, and the successor just involving adding $1$. - -REPLY [35 votes]: Thanks to Timothy Chow for informing me of this discussion. -To avoid vagueness, let Q* be Q with the usual relativization schemata adjoined. Construct a formal system F by adjoining an unary predicate symbol $\psi$, the axiom $\psi(0)$, and the rule of inference: from $\rm\psi(a)$ infer $\rm\psi(Sa)$ (for any term a). I think this is an adequate formalization of the concept of an "actual number". Is $\psi(80^{5000})$ a theorem of F? I see no reason to believe so. Of course, one can arithmetize F in various theories, even Q*, and prove a formula $\exists p[p \hbox{ is an arithmetized proof in F of } -`\psi(80^{5000})\hbox{'}]$, but to conclude from this that there is a proof in F itself of $\psi(80^{5000})$ appears to me to be unjustified. -Contrast F with the theory T obtained by adjoining to Q* a unary predicate symbol $\phi$ and the two axioms $\phi(0)$ and $\phi(0)\;\&\;\forall x'[\phi(x')\to\phi({\rm S}x')]\to\phi(x)$. Then one can easily prove in T $\phi(80^{5000})$ or even -$\phi(80^{5000...^{5000}})$. -The ellipsis means that the iterated exponential term is actually written down.<|endoftext|> -TITLE: Smooth sums of coprime smooth integers -QUESTION [7 upvotes]: Observe that for any $\epsilon > 0$ there are infinitely many triples of -$c^\epsilon$-smooth coprime positive integers $a$, $b$ and $c$ such -that $a + b = c$. -- Considering triples of the form $(2^n-1,1,2^n)$ -and the factorizations of the polynomials $x^n-1 \in \mathbb{Z}[x]$ into -cyclotomic polynomials, this holds since the set of quotients $n/\varphi(n)$ -for positive integers $n$ is unbounded. -How much can this obvious observation be improved, i.e. how much can the -smoothness bound $c^\epsilon$ be lowered such that there are still infinitely -many such triples which satisfy that bound? -Or to be more concrete: is there an $\epsilon > 0$ such that there are -infinitely many triples of $e^{(\ln c)^{1-\epsilon}}$-smooth coprime -positive integers $a$, $b$ and $c$ satisfying $a + b = c$? --- And if yes, which is the supremum of the set of values of $\epsilon$ -for which this holds? - -REPLY [4 votes]: This is related to the $xyz$ conjecture - and the $abc$ conjecture. The $xyz$ is much stronger than your question. -Let $S(X,Y,Z)$ be largest prime factor of $ X Y Z$ and -$H(X,Y,Z)=\max(|X|,|Y|,|Z|)$. -From page (2): -$xyz$ conjecture (weak form-1). There exists a positive constant $\kappa_0$ -such that the following hold. -(a) For each $\epsilon > 0$ there are only finitely many primitive solutions $(X, Y, Z)$ -to the equation $X + Y = Z$ with -$$S(X, Y, Z) < (\log H(X, Y, Z))^{\kappa_0 - \epsilon}$$ -(b) For each $\epsilon > 0$ there are infinitely many primitive solutions $(X, Y, Z)$ -to to the equation $X + Y = Z$ with -$$S(X, Y, Z) < (\log H(X, Y, Z))^{\kappa_0 + \epsilon}$$ -From page (7) -Assume GRH. -There are infinitely many primitive solutions $X,Y,Z$ to $X+Y+Z=0$ s.t. -$$S(X, Y, Z) < (\log H(X, Y, Z))^{8 + \epsilon}$$ -and in particular $\kappa_0 \le 8$.<|endoftext|> -TITLE: The number of conjugacy classes of the simple group PSL(2,q) -QUESTION [7 upvotes]: If $q=p^a$ , where $p$ is a prime number, then I would like to know the number of conjugacy classes related to elements of order $p$ and $2$ in the simple group $PSL(2,q)$ . - -REPLY [2 votes]: To amplify my comments (in community-wiki style: -1) Note that it's OK when discussing just unipotent elements (here those of order $p$) to work instead with the matrix group $G:=\mathrm{SL}(2,q)$, since $\mathrm{PSL}(2,q)$ is just the quotient by the center $\{\pm I\}$ which consists of semisimple elements (and is trivial for $p=2$). Also, the methods are basically the same for all $q$ with $p$ fixed. -2) Usually your question is part of a more general computation of conjugacy classes : size of each and total number of classes. This may be easier to organize, since the total number of elements i $|G|=q(q-1)(q+1)$. In any case, the tally of classes is most often part of the search for ordinary characters. (For a short exposition, see my old paper in Amer. Math. Monthly 82 (1975), 21--39.) Though I don't have most textbooks at hand, I'd also suggest looking at the arguments on page 230 of L. Dornhoff Group Representation Theory, Part A, Dekker, 1971. This is now out-of-print but has useful short chapters on many topics including this standard example. -3) Maybe it's worth emphasizing that the difference between $p=2$ and other primes for this purpose is that all nonzero elements of $\mathbb{F}_q$ are squares in the first case but only half the elements in the second case. Thus a unipotent matrix in $G$ is conjugate for odd $p$ to just one of the matrices $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & \nu \\ 0 & 1 \end{pmatrix}$$ with $\nu$ a fixed nonsquare. There are $q^2$ unipotent matrices in all (special case of a general result of Steinberg), half in each class for $p$ odd. The usual canonical form theory just has to be adapted a bit over a finite field. -4) If you also want to look at elements of order 2 when $p$ is odd, you have to work a little more. Again it's probably easiest to determine (as in Dornhoff) all classes of semisimple, unipotent, or mixed elements, along with the sizes of their classes. It's not clear to me what motivates the format of your question, however. -ADDED: To be more explicit about elements of order 2 when $p$ is odd, just look at the standard list of classes in $G$ and note there is a unique one containing elements of order 4 (hence order 2 in the projective group). Such elements are semisimple, but may be diagonalizable over $\mathbb{F}_q$ or not depending on which of $q-1$ or $q+1$ is divisible by 4.<|endoftext|> -TITLE: Restriction to Levi Subgroups and the Affine Grassmannian -QUESTION [8 upvotes]: Let $G$ be a complex reductive group, $L\subset G$ a Levi subgroup and $Rep(G)$ the category of rational representations of $G$. - -My Question: - What is the geometric analogue of the restriction functor $Res^G_L:Rep(G)\to Rep(L)$? - -To be a little bit more precise: -Let $\check{G}$ be the dual group of $G$. As usual define $\mathcal{K}:=\mathbb{C}$ and $\mathcal{O}:=\mathbb{C}[[t]]$. Let further -$Gr_{\check{G}}:=\check{G}(\mathcal{K})/\check{G}(\mathcal{O})$ denote the affine Grassmannian and $P_{\check{G}(\mathcal{O})}(Gr_{\check{G}})$ the category of $\check{G}(\mathcal{O})$-equivariant perverse sheaves on $Gr$. -The geometric Satake Isomorphism gives an equivalence of tensor categories categories -$$P_{\check{G}(\mathcal{O})}(Gr_{\check{G}}) \cong Rep(G)$$ - -Is there a nice (from the geometric viewpoint) functor $\check{Res}:P_{\check{G}(\mathcal{O})}(Gr_{\check{G}})\to P_{\check{L}(\mathcal{O})}(Gr_{\check{L}})$ which corresponds under the geometric Satake isomorphism to the restriction functor $Res^G_L$? - -REPLY [5 votes]: On the level of sheaves, the construction is indeed a pull-push formula. I believe it was first worked out by Beilinson and Drinfeld in section 5.3 of their preprint "Quantization of Hitchin's Hamiltonians and Hecke eigen-sheaves". For published references, look at section 2.4 in Braverman and Gaitsgory's paper "Crystals via the affine Grassmannian" (Duke Math. J. 107 (2001), 561–575) or at section 1.3 in Vasserot's paper "On the action of the dual group on the cohomology of perverse sheaves on the affine Grassmannian" (Compositio Math. 131 (2002), 51–60).<|endoftext|> -TITLE: Analogues of the curve complex for Out(F) -QUESTION [11 upvotes]: Let $F$ be a finitely generated free group. - -Question: Is there an authoritative survey of analogues of the curve complex for $\mathrm{Out}(F)$? If not, as seems likely, would a passing expert be willing to write a brief summary answering the following questions. - -Which analogues are thought most important? -Why? -How do they relate to each other? -What's known about them? (eg are they of infinite-diameter, contractible, Gromov-hyperbolic etc?) - - -Background -A curve complex $\mathcal{C}(\Sigma)$ is a certain simplicial complex associated to a compact surface $\Sigma$, first defined by Bill Harvey. The definition is simple and well known: the vertices are simple curves on $\Sigma$, and curves $(\gamma_i)$ span a simplex if and only if they're mutually disjoint. They turn out to be very useful when studying mapping class groups $\mathrm{Mod}(\Sigma)$, and are an area of intensive research. By the way, all the information is actually carried by the 1-skeleton, the curve graph. -(I'm a little shocked to see that there's no wikipedia article on curve complexes.) -In analogy with $\mathrm{Mod}(\Sigma)$, we are also interested in the group of outer automorphisms $\mathrm{Out}(F)$, where $F$ is a free group, and a nice strategy is to define analogous objects in this context. For instance, Culler–Vogtmann Outer Space is the analogue of Teichmüller space etc. -Of course, analogies aren't always perfect, and sometimes on object in one context may have several analogues in another context. In this case, this problem seems to have occurred many times over! There is a bewildering variety of analogues of the curve complex and the curve graph for $\mathrm{Out}(F)$. A brief literature search found the following: - -The free splitting complex (Handel--Mosher) -The free factor complex (Hatcher--Vogtmann) -The dual free splitting graph (Kapovich--Lustig) -The sphere complex (Hatcher) -The cyclic splitting complex (Mann) -The edge splitting graph (Sabalka--Savchuk) -The primitivity graph (Kapovich--Lustig) -The cut graph (Kapovich--Lustig) -The dual cut graph (Kapovich--Lustig) -The ellipticity graph (Kapovich--Lustig) -The dominance graph (Kapovich--Lustig) - -The names just indicate people who have worked on these - they're not meant to be authoritative. -Although this question may admit multiple answers, I'd like to resist making it community wiki. The accepted answer should either be a reference to a survey, or a fairly detailed summary of the leading analogues, their relationship to each other, and what's known about them. An answer of the latter sort would require a fair bit of work, and so would deserve credit. - -REPLY [7 votes]: I'll take a crack at this (this will be preliminary; I will correct if, as I expect, I forget things). -There is no survey that I know of. Knowledge is moving fast on this topic. It is perhaps too early to stretch too far at guessing which complexes will be important and which will not be, so I will stick to a more descriptive overview of what has been proved, focussing on hyperbolicity and other properties in the question. -The complexes known to be hyperbolic are: - -The free factor complex ($\approx$ several others, including some on the Kapovich-Lustig list). Proof by Bestvina and Feighn. -The free splitting complex ($\approx$ sphere complex). Proof by Handel and me. -The cyclic splitting complex. Proof by Brian Mann. - -The original hyperbolicity proof for the free splitting complex was expressed in the language of $F_n$ actions on trees. Hilion and Horbez reworked that proof in the language of Hatcher's sphere systems, introducing some simplifications. Bestvina and Feighn reworked the proof again, back in the language of $F_n$ actions on trees, introducing other simplifications. A major effect of these different proofs is to exhibit different classes of reparameterized quasigeodesics in the free splitting complex. In the case of the free splitting complex, the class of quasigeodesics called "fold paths" is given an explicit quasigeodesic parameterization. -Important relations amongst these complexes are natural equivariant Lipschitz maps from the free splitting complex to both the free factor complex and the cyclic splitting complex. Kapovich and Rafi exploited the first of these maps to give a new proof of hyperbolicity of the free factor complex, deriving it from hyperblicity of the free splitting complex. Mann subsequently used the same method in proving hyperbolicity of the cyclic splitting complex. -The $Out(F_n)$ action on each of the above hyperbolic complexes is known to contain loxodromic elements (i.e. having a quasi-axis), and so in particular these complexes are all of infinite diameter. In general the subset of $Out(F_n)$ acting loxodromically on the free splitting complex contains the sets acting loxodromically on the free factor and cyclic splitting complexes, because of the existence of an $Out(F_n)$-equivariant Lipschitz map from the free splitting complex to the other two. Also, the loxodromic sets for these three complexes are pairwise distinct, hence none of these complexes is $Out(F_n)$-equivariantly isomorphic to the other, nor even equivariantly quasi-isometric. Here are some details. - -In the free factor complex, the loxodromic elements are the fully -irreducible outer automorphisms (see Bestvina-Handel). -In free splitting complex, the loxodromic elements form a strictly larger class, namely those outer automorphisms having an attracting lamination that fills $F_n$ (Handel lectured on this in summer 2013 in Oberwohlfach; we hope to post this on the arXiv "soon"). -In the cyclic splitting complex, Mann describes outer automorphisms that are loxodromic here but not in the free factor complex. - -The edge-splitting complex of $F_n$ was proved by Sabalka and Savchuk to be non-hyperbolic, by showing that it contains quasiflats of arbitrarily high dimension. There is an analogue to this in Schleimer's proof that the separating curve complex of a surface is not hyperbolic, again with flats but not of arbitrarily high dimension. -As for topology/homotopy theory, here's what I know about: - -The free factor complex of $F_n$ is homotopy equivalent to a wedge of spheres of dimension $n-2$. Proof by Hatcher and Vogtmann (MR1660045 (99i:20038)). -The sphere complex ($\approx$ free splitting complex) is contractible. Proof by Hatcher (MR1314940 (95k:20030)).<|endoftext|> -TITLE: is this von Neumann algebra a tensor product? -QUESTION [9 upvotes]: Let $H_1$ and $H_2$ be Hilbert spaces. Let $A\subset B(H_1)$ be a factor and $A'$ its commutant. -If a von Neumann algebra $M\subset B(H_1\otimes H_2)$ contains $A\otimes 1$ and commutes with $A'\otimes 1$, is it then necessarily of the form $M=A\otimes B$ for some von Neumann algebra $B\subset B(H_2)$? -I think that I know how to prove this if $A$ is hyperfinite, and I wonder if it's true in general. - -REPLY [14 votes]: What about $A = A_1 \oplus A_2$ and $M = (A_1\otimes 1) \oplus (A_2 \otimes B(H_2))$? -It seems like your condition just says that $A\otimes 1 \subseteq M \subseteq A \otimes B(H_2)$, did you leave something out? -Edit: in case $A$ is a factor, the answer is yes. Ge and Kadison, On tensor products of von Neumann algebras, Invent. Math. 123 (1996), 453-466.<|endoftext|> -TITLE: Are finite (levelwise) homotopy limits of spectra homotopy invariant? -QUESTION [10 upvotes]: I found an easy proof that the (levelwise) homotopy limit of a pointwise equivalence of finite diagrams of orthogonal spectra is an equivalence, without assuming that the spectra in the diagrams are fibrant. This makes me a bit nervous. Does anyone know if this is in fact true? - -REPLY [7 votes]: It's perhaps a little strange to answer this question after three and a half years, but I've thought about this before too and couldn't resist posting. If the category $C$ indexing your diagram is finite in the sense that the classifying space $BC$ is a finite CW complex (or equivalently $C$ has finitely many composable strings of morphisms) then the Bousfield-Kan homotopy limit preserves stable equivalences, even if you don't make the spectra in the diagram fibrant first. -The first way to see this is to argue that the levels of your spectra can be made cofibrant first without changing the homotopy type of the levels or their homotopy limits. Then you observe that fibrant replacement can be achieved by a sequential colimit of loopspaces of the levels. But the kind of finite homotopy limit discussed above commutes with filtered homotopy colimits, so you're done. -A second argument is by induction up the coskeletal filtration of the cosimplicial object that defines the homotopy limit, see for instance section 4 of these notes for a space-level version. There are only finitely many such levels by our assumption on $C$. You end up only needing that homotopy pullback constructions and finite product constructions preserve stable equivalences of spectra, even if you don't make the spectra fibrant before plugging them in. The first is true because homotopy pullback squares are always equivalent to homotopy pushout squares without any point-set assumptions, and the second is true because a finite product commutes with the sequential colimit that defines the homotopy groups of a spectrum. -Having finitely many morphisms in $C$ isn't good enough. A counterexample is when $C$ has one object and morphisms $\mathbb Z/2$, and your spectrum is the sphere spectrum in the category of prespectra (or symmetric or orthogonal spectra), so the $n$th level is the space $S^n$. The homotopy limit over $C$ doesn't commute with fibrant replacement. If I remember correctly, the verification of this uses both the Segal conjecture and the Sullivan conjecture.<|endoftext|> -TITLE: Insightful books about elementary mathematics -QUESTION [36 upvotes]: What are some books that discuss elementary mathematical topics ('school mathematics'), like arithmetic, basic non-abstract algebra, plane & solid geometry, trigonometry, etc, in an insightful way? I'm thinking of books like Klein's Elementary Mathematics from an Advanced Standpoint and the books of the Gelfand Correspondence School - school-level books with a university ethos. - -REPLY [3 votes]: Two great ones are: - -Fuchs, Tabachnikov: Mathematical Omnibus and -Arnold: Lectures and Problems: A Gift to Young Mathematicians.<|endoftext|> -TITLE: The importance of generating series in Algebraic Geometry -QUESTION [7 upvotes]: I asked this question on MSE but got no answer. I was advised to put it here. I am sorry if it is not suitable for MO. -What follows is a very nebulous question. I just seek for some help in understanding a "technique" which has proven itself very powerful. -I noticed that very often generating series appear in Algebraic Geometry. I am referring to enumerative geometry, for instance Gromov-Witten theory or Donaldson-Thomas theory. I am wondering: why can they be a so powerful tool? What is their secret, which makes them so helpful even if they look, at a glance, so intractable? -What is actually happening when we have computed some numbers, enumerative invariants, classes in a Grothendieck ring... whatever data we are interested in, and we put them all together in a generating series? -Example. Think about Witten's conjecture. One can compute Gromov-Witten invariants -$$\int_{\overline M_{g,n}}\psi_1^{a_1}\cup\dots\cup\psi_n^{a_n}$$ and put them all together in the generating series -$$F_g=\sum_{n\geq 0}\frac{1}{n!}\sum_{a_1,\dots,a_n} \Big(\int_{\overline M_{g,n}}\psi_1^{a_1}\cup\dots\cup\psi_n^{a_n}\Big)t_{a_1}\dots t_{a_n}.$$ -Not yet satisfied, one takes the generating series over all genera $F=\sum_g F_gh^{2g-2}$ and then the exponential $e^F$ of $F$. Witten's conjecture is equivalent to $L_ne^F=0$ for all $n$, where $L_n$ are certain differential operators. -I am sorry for the vagueness of the above. Any insight and concrete example is very welcome. - -REPLY [2 votes]: Your example from Witten makes one point: generating functions can make differential operators summarize combinatoric/algebraic information -- basically by making differential operators express recurrence relations on the series of coefficients. And generating functions often have nice closed forms, as for example the long known expression of the power generating function for Fibonacci numbers as $x/(1-x-x^2)$. A closed form captures the series as a whole. Closed forms are at least concise information. They can reveal the effect of differential operators, or algebraic relations. Wilf's free book Generatingfunctionology http://www.math.upenn.edu/~wilf/DownldGF.html gives many examples of all this. Because you can use power generating functions, exponential generating functions, Dirichlet series, and more, you have some choice in tailoring the series to meet the application. -Treating generating functions as functions in the usual sense can be useful but is not always possible since the series are not always convergent. As noted Jeff Harvey's reply to Does Physics need non-analytic smooth functions? an important kind of power generating function in physics can have radius of convergence 0. -I would be surprised (and delighted) to see a concise yet comprehensive explanation of all the reasons generating functions work so well. For a brilliant, concise, avowedly not comprehensive effort at this see the preface of Wilf's book. He gives exceptionally clear exposition and motivation throughout that book.<|endoftext|> -TITLE: Generators for exact representations of 3-manifold groups -QUESTION [7 upvotes]: Does anyone have a list of matrix generators for a bunch of hyperbolic 3-manifold groups? I am testing an algorithm and am looking for a collection of test cases. I am looking for exact values of matrix elements as algebraic numbers, and not approximate values. - -REPLY [8 votes]: Grant Lakeland was kind enough to compute exact $PSL(2,\mathbb{C})$ representations for tetrahedral groups. The methods and formulas are written up on his webpage here. -Maclachlan and Reid give complete list of the arithmetic and non-arithmetic tetrahedral groups in $\S$13.1 and $\S$13.2 of their book, "The arithmetic of hyperbolic 3-manifolds." -As a warning if one should be careful about observing general phenomena from these groups. They are special a lot of ways. -However, if you want more that manifolds that not just covers of tetrahedral orbifolds, there are a few options. -In the rest of section 13 of Maclachlan and Reid, (especially 'the arithmetic zoo'), they give information on the invariant trace fields of a number of manifolds. From this exact value, one can often guess and verify the algebraic numbers (most of the time these will be algebraic integers) for the entries of the generators. -Also, for general 3-manifold groups you probably want to use snap or you can use SAGE's snappy package (documentation is here) to get exact representation. Again, one has to extrapolate information from the trace field to get the entries of the generators express as algebraic numbers.<|endoftext|> -TITLE: A complete surface with $K\to -\infty$ -QUESTION [7 upvotes]: Does anyone know an example of a complete surface (certainly not necessarily embedded in $\Bbb R^3$) with negative curvature unbounded below? Any example I have computed so far is either not complete or has curvature bounded below. Any light that might be shed would be welcome. (Sadly, this corner of geometry is far from my expertise.) - -REPLY [13 votes]: Kazdan and Warner (MR0343206 (49 #7950) Reviewed -Kazdan, Jerry L.; Warner, F. W. -Curvature functions for open 2-manifolds. -Ann. of Math. (2) 99 (1974), 203–219. -53C20 (35J05 58G15) ) -show that a smooth function on a $\mathbb{R}^2$ is the curvature of a complete metric if and only if $\lim_{r\rightarrow \infty} \inf_{|p| > r} K(p) \leq 0.$ So, presumably this should give plenty of examples of the sort you seek, and maybe if you read the paper it discusses how you might construct such.<|endoftext|> -TITLE: Le Haut Commissariat qui surveille rigoureusement l'alignement de ses Grandes Pyramides -QUESTION [11 upvotes]: Yesterday I came across the following one-paragraph summary of the history of the Law of Quadratic Reciprocity in Roger Godement's Analyse mathématique, IV, p.313 (perhaps the only treatise on Analysis which contains a statement of the Law in question). - -Legendre a deviné la formule et Gauss est devenu instatanément - célèbre en la prouvant. En trouver des - généralisations, par exemple aux anneaux d'entiers - algébriques, ou d'autres démonstrations a constitué un - sport national pour la dynastie allemande suscité par Gauss - jusqu'à ce que le reste du monde, à commencer par le Japonais - Takagi en 1920 et à continuer par Chevalley une dizaine - d'années plus tard, découvre le sujet et, après 1945, le - fasse exploser. Gouverné par un Haut Commissariat qui surveille - rigoureusement l'alignement de ses Grandes Pyramides, c'est - aujourd'hui l'un des domaines les plus respectés des - Mathématiques. -(English translation: Legendre guessed the formula [the quadratic reciprocity law] and Gauss instantly became famous by proving it. Finding generalizations, for example to rings of algebraic integers, or other proofs was a "national sport" for the German dynasty sparked by Gauss until the rest of the world, starting with the Japanese Takagi in 1920 and then Chevalley about ten years later, discovered the subject and, after 1945, made it "explode". Governed by a High Commission that rigorously monitors the alignment of its Great Pyramids, today it is one of the most respected areas of Mathematics.) - -Which Haut Commissariat is he referring to ? Or is it just a joke ? - -REPLY [10 votes]: I disagree with Michael Grünewald's interpretation, which by the way doesn't answer the initial question: who Godement is he referring too? I think this is a joke made without acrimony. "Thought police", "innovation preventing", are much too strong phrases to translate Godement's light ironical quotation. -To a french-spaking ear, "Haut Commissariat" in this context evokes the "Commissariat Général au Plan", created by the administration led by de Gaulle in 1946 (and including a large political spectrum, from right wing to communists). It was an institution without real power but which was supposed to prepare non-compulsory "plans" to develop the economy for the next five years, the idea being -to take advantage of whatever was thought efficient in soviet-like planning while staying essentially a free-market economy. (Of course there are other -institutions with that name, like UN's "haut-commissariat aux réfugiés", but really that the plan one that comes to mind). -So back to quadratic reciprocity, I may be completely wrong but I imagine that the Haut-Commissaire in question might be R.P. Langlands and his huge program that has provided a non-compulsory, but hugely influential, planning for the research in "higher class field theory" since more than 40 years.<|endoftext|> -TITLE: What is the categorical significance of the trivial $\mathfrak{g}$-module in the category of $\mathfrak{g}$-mod? -QUESTION [8 upvotes]: This question may be trivial for experts. -Let $\mathfrak{g}$ be a Lie algebra over a field $k$ and consider the category $\mathfrak{g}$-mod of $\mathfrak{g}$ modules. We can add suitable conditions, like finite dimension, semisimple, if necessary. -We call the 1-dimensional $k$-space with $\mathfrak{g}$ acts by $0$ the trivial $\mathfrak{g}$-module. This $\mathfrak{g}$ module is kind of special in the category $\mathfrak{g}$-mod. But it is not the initial or terminal object (the $0$-module is both initial and terminal). -$\textbf{My question}$ is: is there any categorical significance of this trivial $\mathfrak{g}$ module? Do we need to take the tensor structure into account? - -REPLY [4 votes]: To address the category O issue: no. Every finite dimensional simple in category O is "the same," in the sense there's a functor sending one to the other which isn't an equivalence on all of category O (you could cook such a thing up, but it wouldn't be very natural), but is an equivalence between the blocks of the two corresponding simples: the translation functors do this.<|endoftext|> -TITLE: Is there an inner model between two distinct inner models of ZFC? -QUESTION [7 upvotes]: Definition (1): A‎n‎‎ ‎inner ‎model ‎of ‎‎$‎ZFC‎$ ‎is a‎ ‎tarnsitive proper class ‎model ‎‎of $‎‎ZFC$ ‎which ‎contains ‎all ‎ordinal numbers. ‎Informally ‎we ‎denote ‎the ‎collection ‎of ‎all ‎inner ‎models ‎of ‎‎$‎ZFC‎$ ‎by ‎‎$‎‎Inn(ZFC)$.‎ -‎ -Now ‎if we ‎‎consider the partial order (reflexive and transitive) $‎\langle ‎Inn(ZFC),\subseteq ‎\rangle‎‎‎‎$ ‎with ‎end ‎points ‎‎$‎L‎$ ‎and ‎‎$‎‎V$ ‎then ‎there ‎are ‎some ‎natural ‎questions ‎about ‎other properties of this partial order:‎ -‎ -Question (1): ‎Is‎ $‎\langle ‎Inn(ZFC),\subseteq ‎\rangle‎$ ‎linear? ‎In the other words‎, ‎is ‎the ‎following ‎statement ‎true?‎ -‎ -‎‎$‎‎‎\forall ‎M,N\in Inn(ZFC)‎~~~M\subseteq N~~\vee~~N\subseteq M$‎ -‎ -Question (2): ‎Is‎ $‎\langle ‎Inn(ZFC),\subseteq ‎\rangle‎$ ‎dense?‎ ‎In the other words‎, ‎is ‎the ‎following ‎statement ‎true?‎ -‎‎ -$‎‎‎\forall ‎M,N\in Inn(ZFC)‎~~~(M\subsetneq N \Longrightarrow ‎\exists ‎P\in Inn(ZFC)~~~M\subsetneq P \subsetneq N‎)$‎ - -REPLY [8 votes]: Your question is related to the concept of degrees of constructibility, where we say that $c\equiv_L d$ if and only if $L[c]=L[d]$. When $c$ is well-ordered in $L[c]$, then this will be an inner model of ZFC, and so the structure theory of the degrees of construtibility are a part of your informal treatment of inner models. -The inner models can be linear, for example, if you add a Sacks real generically over $L$, then there are precisely two inner models, $L$ and $L[s]$, because of the minimality property of the Sacks real. This also shows that the inner models may not be dense. Indeed, if one adds a Sacks real over any $V$, then there are no models between $V$ and $V[s]$, which is exactly the Sacks minimality property. Meanwhile, as Miha pointed out, it is easy to make it non-linear, and in fact many other patterns are possible. -Theorem. If $c$ is an $L$-generic Cohen real, then the inner models of $L[c]$ are densely ordered by inclusion, but not linearly ordered. -Proof. Every inner model $M$ with $L\subset M\subset L[c]$ is a forcing extension by a subalgebra of Cohen forcing, and all such subalgebras are themselves isomorphic to Cohen forcing. So the inner models are just of the form $L[d]$ for a Cohen real $d$, and the quotient forcing $L[d]\subset L[c]$ is still countably-dense and hence also isomorphic to Cohen forcing. But if we add a Cohen real $r$ over any model $V$, then we can let $r_0$ be the even digits of $r$, and find $V\subsetneq V[r_0]\subsetneq V[r]$. So we get density precisely because all the forcing extensions arise by Cohen forcing. Namely, if $L[d_0]\subsetneq L[d_1]$, then $L[d_1]=L[d_0][e]$ for some Cohen real $e$, and we may take every-other digit of $e$ to find an intermediate model between $L[d_0]$ and $L[d_1]$. The order is not linear, since we may split $c$ into even and odd parts, and find mutually generic reals $c_0$ and $c_1$, whose extensions $L[c_0]$ nad $L[c_1]$ are incomparable, yet have meet $L$ and join $L[c]$. QED -In answer to the related question What can the degrees of constructibility be?, I posted the following: -The article Initial segments of the degrees of constructibility by Marcia Groszek and Richard Shore (Israel Journal of Mathematics -June 1988, Volume 63, Issue 2, pp 149-177) shows that - - -Any constructible, constructibly countable, (dual) algebraic lattice is isomorphic to the degrees of constructibility of reals in some generic extension of L. - - -And there is a lot of further work on this topic by Groszek, Slaman and others. - -REPLY [5 votes]: The answer to the second question is negative. Take, for example, Sacks forcing over $L$. The result is $L[r]$ where $r$ is a real number and the following property is true: $$\forall x(x\in L\lor r\in L[x])$$ -This shows that there is no intermediate model between $L$ and an extension by one Sacks real. This property is called minimality, and it is not unique to the Sacks real. -The first answer is also negative, as others have said. One can force over $L$ both a Cohen real $c$ and a random real $r$ (as a product of the forcing, that is), and then $L[r],L[c]$ are two different inner models of $L[r\times c]$ which are incomparable.<|endoftext|> -TITLE: The relationship between P vs NP problem and "Kolmogorov complexity with time" -QUESTION [6 upvotes]: Let $P$ - polynomial($P(x) \ge x$), $n \in \mathbb{N}$, $l < log(n)$. -Problem1: "Is there program with length $\le l$ that print $n$ by using $\le P(log(n))$ time?" -Is it Problem1 $\in NP$-complete? -Now I'll explain why I think that this problem hasn't a polynomial solution. -Define "Kolmogorov complexity with time" $K(P, n)$ as minimal program that work $\le P(log(n))$ time. -Kolmogorov complexity is not a computable function. A natural analogue of this statement: -"Kolmogorov complexity with time" $\notin P$ (1) -It is easy to see, that if Problem1 $\in P$ than "Kolmogorov complexity with time" $\in P$. So, if Problem1 $\in NP$-complete than (1) $\iff P \not=NP$ - -REPLY [5 votes]: Let's assume that Mark's description of the problem in the comments is correct, so that we have a fixed polynomial $P$, and the decision problem is: given $l$ and $n$, such that $l\lt \log(n)$, decide if there is some program $e$ with length $\leq l$ such that $e$ prints $n$ using time $\leq P(\log(n))$. -This problem is in NP, because we can simply guess the program that works and check that it works in polynomial time. Note that since $n$ is an input, for NP here we are guided by polynomial time in the length of $n$. For the NP algorithm, on input $(l,n,e)$, we verify whether $e$ is a successful candidate or not, simply checking the length requirement and then running $e$ for $P(\log(n))$ steps to see whether it writes $n$ or not. If it does, we accept, and otherwise reject. The point is that this verification process is polynomial time in our input $(l,n)$. The input $(l,n)$ is acceptable in your problem if and only if there is a successful candidate $e$ making it acceptable in my algorithm, and this shows that your problem is in NP.<|endoftext|> -TITLE: Strange (or stupid) arithmetic derivation -QUESTION [49 upvotes]: Let us consider the following operation on positive integers: $$n=\prod_{i=1}^{k}p_i^{\alpha_i} \qquad f(n):= \prod_{i=1}^{k}\alpha_ip_i^{\alpha_i-1}$$ (Is it true that if we apply this operation to any integer multilpe times, it will eventually get into a finite cycle?) Is there a constant $K$ such that any integer will fall into a cycle after $K$ steps? -Edit4: We managed to settle affirmatively the question of Mark Sapir, whether a cycle of arbitrary length exists: http://www.math.bme.hu/~kovacsi/Pub/arithmetic_derivation_v04.pdf -Edit3: I proposed two questions (in retrospect, it was a minor mistake), one of them was answered. To appreciate this, i accept Mark Sapir's answer, and alter the original text by putting the unanswered stuff into parentheses. Making the answered one the main question. -Edit2: István Kovács pointed out that there is a nice formula for $f(n)$ using the 'number of divisors' function: $$ f(n):= d \left( \frac{n}{ \prod_{i=1}^{n}p_i } \right) \frac{n}{ \prod_{i=1}^{n}p_i } $$ from which it fillows that for any $\varepsilon >0 , \quad f(n)=o(n^{1+\varepsilon})$. -I think that the answer to the first question is yes, but to the second no. We tested the first $10000$ integers and every integer fell into a cycle after at most $6$ steps. -Edit: @MarkSapir proved that the answer to the second question is no. His proof raises the (third) question: How long can such a cycle be? - -REPLY [5 votes]: I did some computations with the integers up to 400000 and I got the following conclusions: -1) The following are cycles with more than one element (i.e. non-fixed points) - -[32,80]=[2⁵, 2⁴·5] -[864,2160]=[2⁵·3³, 2⁴·3³·5] -[708588,2598156,787320]=[2²·3¹¹, 2²·3¹⁰·11, 2³·3⁹·5] -[5832,17496,61236,20412]=[2³·3⁶, 2³·3⁷, 2²·3⁷·7, 2²·3⁶·7] - -2) All the integers up to 400000 end up falling in a fix point or one of the previous cycles (or maybe another 2-cycle that I have missed) except for $2^{16}$, $2^{17}$, $2^{16}\cdot 3$, $2^{16}\cdot 5$, $2^{18}$, $2^{17}\cdot 3$, $2^{12}\cdot 3^4$, $2^4\cdot 3^9$, $2^9\cdot 3^6$. -These exceptional values end up reaching some number that my computer cannot handle. As an example the first six exceptional values (all of those with the exponent of 2 being bigger or equal to 16) lead to $2^{18}\cdot 3^5$, which after 10 more steps or so computed by hand keeps increasing and doesn't look like falling into a cycle soon. This agrees with Felipe's comment. -3) I found all the above using bad programming: I computed the actual numbers instead of keeping track of primes and exponents as one does by hand. Reprogramming everything using this idea should give much better results. -Addendum 1: Studying the case $2^{16}$ I could find some more 2-cycles, such as - -$[2^{63}\cdot 3^{13}\cdot 7\cdot 31,2^{62}\cdot 3^{14}\cdot 7\cdot 13]$ ($2^{16}$ reaches this 2-cycle after 33 steps). -$[2^{279}\cdot 3^{61}\cdot 31\cdot 139,2^{278}\cdot 3^{62}\cdot 31\cdot 61]$ -$[2^{15}\cdot 3^{33}\cdot 7\cdot 17,2^{14}\cdot 3^{34}\cdot 5\cdot 11]$ - -The left term of all of them is of the form $2^{2p+1}\cdot 3^{2q-1}\cdot p\cdot q$ with $(2p+1)(2q-1)=9\cdot c$ where $p,q$ are primes different from $2,3$ and $c$ is a square-free integer coprime with $2,3$. More generally we have the following. - -Let $p,q,r,s$ be pairwise distinct primes such that $(pr+1)(ps-1)=q^2\cdot c$ where $c$ is a square-free integer coprime with $p,q$. Then $[p^{pr}\cdot q^{ps}\cdot c,p^{pr+1}\cdot q^{ps-1}\cdot r\cdot s]$ is a $2$-cycle of $f$. - -N.B.: If $r,s\neq 2$, the hypothesis force either $p=2$ or $q=2$. So a much easier result is: - -Let $p>2,3$ be a prime such that $2p=3c+1$, where $c$ is a square-free integer coprime with $3$. Then $[2^3\cdot 3^{2p-1}\cdot p,2^2\cdot 3^{2p}\cdot c]$ is a $2$-cycle of $f$. - -E.g. $p=11,17,29,47,53$ do the trick. -Addendum 2: I have already coded everything properly (please tell me if you are interested in the code). At the moment I'm looking for cycles as long as possible. If found out some 5-cycles and 6-cycles: for example 2²·3⁴⁷·5⁶ is the beginning of a 5-cycle and 2⁶⁷·3¹⁰⁵·5⁵·53·67 is the beginning of a 6-cycle. Of course one wonders if there are cycles of arbitrary length and how to find them. It would be nice to find a proposition extending the one I gave in Addendum 1 to $f^k$ for arbitrary $k$. -Addendum 3: I got similar results for $k=4$: - -Let $p>2,3$ be a prime such that $6p+1$ is a square-free integer. Then $[2^3\cdot 3^{6p}\cdot p, 2^3\cdot 3^{6p+1}\cdot p, 2^2\cdot 3^{6p+1}\cdot (6p+1),2^2\cdot 3^{6p}\cdot(6p+1)]$ is a $4$-cycle of $f$. - -All the primes between 5 and 97 satisfy the conditions of this proposition except for 29 and 79. - -Let $p>2,3$ be a prime such that $p-1=6c$ for a square-free integer $c$. Then $[2^2\cdot 3^p\cdot p, 2^2\cdot 3^{p-1}\cdot p, 2^3\cdot 3^{p-1}\cdot c, 2^3\cdot 3^p\cdot c]$ is a $4$-cycle of $f$. - -Some primes which satisfy the conditions are 31, 43, 67 and 79.<|endoftext|> -TITLE: Finite groups with few double cosets with respect to abelian subgroup -QUESTION [11 upvotes]: The following question is motivated by the study of certain tensor categories, namely integral near-group categories. -Let $G$ be a finite group and $H\subset G$ be a subgroup. Is it possible to give a complete classification (or at least give some examples) satisfying the following conditions: -1) $H$ is abelian; -2) $G$ is the union of the normalizer $N$ of $H$ and exactly one more double coset with respect to $H$. -Pairs $(G,H)$ satisfying the additional condition below is of special interest: -3) the double coset not contained in $N$ consists of $|H|^2$ elements (which is maximal -possible size) -Known examples satisfying 1), 2) and 3): -a) $G=F_q\rtimes F_q^*$ and $H=F_q^*$ (here $F_q$ is a finite field) -b) $G=S_4$ and $H$ is a non-normal subgroup of order 4. -An easy observation is that condition 2) implies that $G-$action on $G/N$ is 2-transitive but I don't know how to use it. - -REPLY [10 votes]: We can construct such a group in the following way: -First take $\mathbb F_q \rtimes \mathbb F_q^{\times}$. -Then take a product of two isomorphic abelian groups $X \times Y$, where $X\cong Y$. (This notation appears strange but is more convenient.) -Choose an action of $\mathbb F_q \rtimes \mathbb F_q^{\times}$ on $X\times Y$ by automorphisms where $\mathbb F_q^{\times}$ fixes $X$ but some element sends $X$ to $Y$. -Then choose an extension $1 \to X \times Y \to G \to \mathbb F_q \rtimes \mathbb F_q^{\times}$ where previous action is the action by conjugation and the inverse image of $\mathbb F_q^{\times}$ has a normal subgroup $H$ containing $X$ with complement $Y$. -$G,H$ is a pair of the type you desire. These are the only type of pair satisfying conditions 1 to 3. -The choice of extension is the only fiddly bit in terms of such a classification. I don't know how to classify these types of extensions - maybe by some type of group cohomology? However, extensions of $1 \to X \times Y \to N \to \mathbb F_q^{\times}$ can be easily classified - they consist of pairs of an extension $1 \to X \to H \to \mathbb F_q^\times$ (itself classified by maps $\mathbb F_q^{\times} \to X$) and an action of $Y$ on $H$ which fixes the components, which is just a map $Y \otimes \mathbb F_q^\times \to X$. But I don't know how many lifts each of these will have to the whole group. -But note that lots of easy extensions, like semidirect products, do satisfy these conditions! - -To prove that this is actually a pair of the type you desire, first note that $H$ is an extension of a cyclic group by an abelian group where the abelian group is central, hence it is abelian. Then note that the order of $G$ is $q(q-1)|X|^2=|H|^2 +(q-1)|X|^2$, but the normalizer of $H$ contains the inverse image of $\mathbb F_q^{\times}$ so it has size at least $(q-1)|X|^2$, so we just need to find a double coset of size $|H|^2$. Let $g\in g$ be such that $gXg^{-1}=Y$. Then I claim $gHg^{-1} \cap H=1$, producing the coset. Since $g$ cannot be in the inverse image of $\mathbb F_q^{\times}$, the image of $H \cap gHg^{-1}$ in $\mathbb F_q \rtimes \mathbb F_q^{\times}$ is trivial, so any intersection must lie in $X \times Y$. But $H \cap X \times Y = X$, and $gXg^{-1}=Y$, and $X \cap Y=1$, so the intersection is trivial. - -To prove that every pair satisfying conditions 1,2, and 3 can be described in this way, take the coset $HgH$ of size $|H|^2$. Since it can also be described as the complement of $N$, it is closed under the action of $N$, so $NgH$ has order $|H|^2$ as well. Let $X$ be the subgroup of $H$ which acts trivially on $Ng$ on the right, and let $Y$ be the subgroup of $N$ that acts trivially on $gH$ on the left. Then $X$ is conjugate to $Y$ by $g$, $|X|=|Y|=|N|/|H|$, and $Y \cap H=1$. -Since $Y\subset N$ and $X \subset H$, $Y$ normalizes $X$. Since $Y\cap X=1$, $XY$ is a subgroup of order $|N/H|^2$. Next we will check that $XY$ acts trivially on $G/N$. $X \subset H$, and $H$ acts on $G/N$ with a single nontrivial orbit, and is abelian, so any subgroup of it that fixes one point fixes every point, so $X$ acts trivially on $G/N$. $Y$ is conjugate to $X$, so $XY$ acts trivially on $G/N$. -Next we see that $XY$ is everything that acts trivially on $G/N$. Indeed, every element that acts trivially must be in $N$, but $|NgN|=|H|^2$, so only $|N|^2/|H|^2=|XY|$ elements can act trivially. -Now we check that $G/(XY)$ is a group of this type. Indeed, the image of $H$ in $G/(XY)$ is abelian, and the action of $H$ on $G/(XY)H= G/N$ is a single nontrivial orbit plus a single trivial orbit, so $G/XY$ is a group of this type and, moreover, a Frobenius group, so as I showed in my comment, it is $\mathbb F_q \rtimes \mathbb F_q^{\times}$. -Now we are done. $XY$, the product of two isomorphic abelian groups, is a normal subgroup with quotient $\mathbb F_q \rtimes \mathbb F_q^{\times}$, which acts by conjugation, with one element sending $X$ to $Y$, but $\mathbb F_q^{\times}$ fixing $X$, since they are both part of the abelian subgroup $H$. $H$ is a normal subgroup of $N$, containing $X$, and $Y$ is a complement, since $Y \cap H=1 -$ and $|Y|=|N|/|H|$.<|endoftext|> -TITLE: The amplituhedron minus the physics -QUESTION [72 upvotes]: Is it possible to appreciate the geometric/polytopal properties of the amplituhedron without delving into the physics that gave rise to it? -All the descriptions I've so far encountered assume familiarity with quantum field theory, -but perhaps there exist more purely geometric explications...? -If so, I would appreciate a reference—Thanks! - -(Image from Quanta Magazine) -(Added). Here is a snapshot from p.64 of the paper which jc cited, -"Scattering Amplitudes and the Positive Grassmannian." -(I am guessing that what they called the "positroid" in this paper -is either closely related or perhaps identical to what -was later named the "amplitudedron"?) - -REPLY [4 votes]: Alexander Postnikov gave a detailed series of lectures on the positive Grassmanian at the Hebrew University of Jerusalem (see links below). He also briefly referred to the amplituhedron. -Folowwing Postinkov, I can briefly explain the situation as follows: -A) The stratification of the positive Grassmanian: -Regard the Grassmanian as represented by equivalence classes of -$m$ by $n$ matrices ($m \le n$) under row operators. -The positive grassmanian is the set of totally-non negative matrices namely those where all $m$ by $m$ minors are nonnegative. It has an important cell-like structure (the open cells are known to be homeomorphic to open balls, their closures are conjectured to be homehomorphic to closed balls) that can be defined in two equivalent ways as follows: -Given an ordering of the $n$ columns the cell of the Shubert stratification indexed by a subset $S$ (of size $m$) are matrices whose lexicographically first non-zero minor (w.r.t. the ordering) is $S$.The matroidal stratification is the common refinement of these Shubert decomposition with respect to all permutations. -The cell structure studied by Postnikov can be seen -a) as the matroidal stratification reduced to this part of the grassmanian; -b) Start with the common refinement of Shubert cells with respect to a cyclic family of permutations (for the whole Grassmanian this lies between the Shubert stratification and the matroid one). Then restrict it to the positive Grassmanian. -This cell structure has beautiful combinatorial description in terms of certain planar graphs and statistics of permutations. -B) The amplituhedron: -Recall that every polytope is a projection of a simplex and projections with respect to totally positive matrices give precisely the cyclic polytope. -Now, replace the simplex by the positive Grassmanian: The amplitutahedron is a projection of the positive Grassmanian based on a totally positive matrix. So the amplituhedron is a common generalization of the positive grassmanian and the cyclic polytope. - -Video1, Video2, Video3, Video4 -Postnikov's ten minutes explanation of the amplituhedron starts here. -Update: Some more details and links can be found in the blog post The simplex, the cyclic polytope, the positroidron, the amplituhedron and beyond.<|endoftext|> -TITLE: Puzzle on deleting k bits from binary vectors of length 3k -QUESTION [50 upvotes]: Consider all $2^n$ different binary vectors of length $n$ and assume $n$ is an integer multiple of $3$. You are allowed to delete exactly $n/3$ bits from each of the binary vectors, leaving vectors of length $2n/3$ remaining. The number of distinct vectors remaining depends on which bits you delete. Assuming your aim is to leave as few remaining different vectors as possible, how few can you leave as a function of $n$? -Example, $n=3$. You can leave only the two vectors $11$ and $00$. -Following comments at the math.se site (in particular by Jack D'Aurizio), in general for larger values of $n$ you can replace any block of three consecutive bits by either $00$ or $11$. This gives an upper bound of $2^{n/3}$. Is this in fact the correct answer? -Now I have some code to solve small instances, we can start to fill in a table of optimal results. We use the notation $H(n,b)$ to indicate the smallest number of distinct vectors that results from starting with all vectors of length $n$ and removing $b$ from each. -$$12 \leq H(15,5) \leq17$$ -$$H(12,4) = 10$$ -For $n=10$ and $b = 1,2,3,4$ we have $\leq140,\leq 31,10, 4$ -For $n=9$ and $b = 1,2,3,4$ we have $70,18,6,2$ -For $n=8$ and $b = 1,2,3$ we have $40,10,4$ -For $n=7$ and $b = 1,2,3$ we have $20,6,2$ -For $n=6$ and $b = 1,2$ we have $12,4$ -For $n=5$ and $b = 1,2$ we have $6,2$ -$$H(4,1) = 4, H(3,1) = 2, H(2,1) = 2$$ -If we only allow symmetric solutions then $H(10,2)=32$. This implies (assumnig no error in the calculations) that for some instances there may be no symmetric optimal solutions as we already have $H(10,2) \leq 31$. - -REPLY [4 votes]: I think it may be helpful and/or intersting to consider the problem for fractions other than $1/3$. -Specifically, if $n$ is a multiple of $q$, let $f(p/q,n)$ be the minimal size of a set of $n- (p/q)n$ bit strings such that deleting $(p/q)n$ bits from each $n$-bit string produces an element of the set. Then set -$$g(p/q) = \lim _{n\to \infty} \frac{\log f\left(\frac{p}{q},n\right)}{n}$$ -The limit exists because, by the divide-into-independent-blocks argument, any particular value is a bound for the $\lim\sup$. -It is clear that $g$ is monotonic. It's not immediately obvious to me if we can prove that $g$ is continuous. -Clearly $g(x)=0$ for $x \geq 1/2$. -For a simple upper bound on $g$ for $x<1/2$, we can assume with the loss only of a constant that there are fewer $0$s than $1$s. Then divide the $n$ bits into $q$ equally sized parts, and delete all the $0$s in the $2p$ parts with the fewest ones. This takes $2^{ (1-2p/q)n }$, giving an upper bound $g(x) \leq (1-2x) \log 2$. -Thanks to Brendan McKay and Yury, we have a lower bound on $g$. If I understand this bound correctly, it is that $g(x) \geq \log 2 + x \log x + (1-x) \log(1-x)$. (We can easily check that the maximal term in the $x=1/3$ case remains maximal for all $0 -TITLE: Independent families versus generators -QUESTION [9 upvotes]: I asked this question on M.SE a while ago and got no answers, so I'm asking it here. -Let $\kappa$ be an infinite cardinal. A family $\mathcal{A}\subseteq\mathcal{P}(\kappa)$ is independent if for any $A_1,\ldots,A_n\in\mathcal{A}$ and $i_1,\ldots,i_n\in\{0,1\}$, we have -$$ \left|\bigcap_{k=1}^n A_k^{i_k}\right| = \kappa $$ -where $A^0 = A$ and $A^1 = \kappa\setminus A$. -Question: Is there an independent family $\mathcal{A}$ such that the Boolean algebra generated by $\mathcal{A}$, along with the subsets of $\kappa$ of size $< \kappa$, is all of $\mathcal{P}(\kappa)$? -I am particularly interested in the case $\kappa = \omega_1$, though an answer for any $\kappa$ would be interesting. - -REPLY [12 votes]: I think the answer is always no. Consider the algebra $\mathcal{B}$ generated by the independent family and the ideal $J$ of subsets of $\kappa$ of size $<\kappa$. Then look at $\mathcal{C} = \{ X \subseteq \kappa : \exists Y \in \mathcal{B}, X \triangle Y \in J \}$. It is easy to see that $\mathcal{C}$ is closed under set operations, and every member of $\mathcal{C}$ is obtained by a finite boolean combination of members of $\mathcal{A}$ and $J$, so it is the algebra generated by $\mathcal{A}$ and $J$. If $\mathcal{C} = \mathcal{P}(\kappa)$, then consider the homomorphism $h : \mathcal{C} \to \mathcal{P}(\kappa)/J$ given by $X \mapsto [X]_J$. Now we know $\mathcal{P}(\kappa)/J$ has antichains of size $\kappa^+$, but $\mathcal{B}$ is a free algebra and thus has the c.c.c. By the independence of $\mathcal{A}$, $\mathcal{B} \cong \mathcal{B}/J$, so $h[\mathcal{C}] \cong \mathcal{B}$, and thus $\mathcal{P}(\kappa)/J$ has the c.c.c., contradiction.<|endoftext|> -TITLE: Kolmogorov complexity is the strongest noncomputable function -QUESTION [19 upvotes]: Yury I. Manin says that Kolmogorov complexity (in some nontrivial sense) is the strongest noncomputable function ("Колмогоровская сложность... невычислима... она во многих интересных смыслах заслуживает титул универсальной невычислимой функции... в некотором смысле слова(нетривиальном) это такая самая сильная невычислимость которая может существовать":http://youtu.be/nnZPqnwoD64?t=15m39s). -What is the exact wording of this statement? -upd: my translate of dialog on this video: -designations and definitions: -Let $u$ - is partial recursive function between $\mathbb{Z}_+$ and $X$, where $X$ is countable set. -$K_u(x) = min \{ m \in \mathbb{Z}_+ | u(m) = x\}$ or infinity -Statement: $\exists u ($optimal Kolmogorov numeration$): \forall v($function as$ u) \exists c(u, v) > 0 \forall x \in X: K_u(x) \le K_v(x)$ -Kolmogorov's order$(u) \mathbb{Z}_+ \to X$ - in order of order Lower_to_higher -Dialog: -Misha Verbitsky: Is $u$ bijection? -Yury Manin: No, and it is focus and big trap. $u$ is not define on many $n$, many -Misha Verbitsky: And no for any $m$ too? -Yury Manin: It get all $x$. -Michael Tsfasman: No, because sometimes -Yury Manin: Optimal, optimal get all $x$. Not all $m$ are programs only some. And its non-computable... Furthermore it (in many interesting ways) deserves the title of the Universal noncomputable function. -Somebody: Complexity? -Yury Manin: Complexity and Kolmogorov's order too. -Somebody: Order? It is order of increasing complexity? -Yury Manin: Order of increasing complexity. They are non-computable. In a manner (nontrivial) it is the strongest noncomputable. If you have oracle that give you things in Kolmogorov's order then very mach things became computable (I'll show it for codes). I wrote somewhere that civilization is such oracle, that we do produce scientific knowledge in order of increasing its Kolmogorov complexity... - -REPLY [4 votes]: . - - -'Now, a Turing-degree is the set of all problems that are Turing-equivalent to a given problem. What are some examples of Turing-degrees? Well, we've already seen two examples: (1) the set of computable problems, and (2) the set of problems that are Turing-equivalent to the halting problem[1]. Saying that these Turing-degrees aren't equal is just another way of saying that the halting problem isn't solvable. -Are there any Turing-degrees above these two? In other words, is there any problem even harder than the halting problem? Well, consider the following "super halting problem": given a Turing machine with an oracle for the halting problem, decide if it halts! Can we prove that this super halting problem is unsolvable, even given an oracle for the ordinary halting problem? Yes, we can! We simply take Turing's original proof that the halting problem is unsolvable, and "shift everything up a level" by giving all the machines an oracle for the halting problem. Everything in the proof goes through as before, a fact we express by saying that the proof "relativizes."'----Lecture by Scott Aaronson - - -[1]Finding the Kolmgorov complexity is Turing equivalent to deciding the halting problem.(For how,look at the comments by @usul below this post here.) -So we see there is an arithmetical hierarchy of oracles ( and even intermediate ones) and as such there is no strongest Oracle or its equivalent non computable function (as an example this non computable function can be taken as Chaitin's $\omega$ for its class.) -In my opinion ; all problems which are "natural" and have actually arisen in practice -; the hardest akin to NP hard; can be solved by the halting oracle.. hence perhaps Kolmogorov complexity in the strongest noncomputable function which occurs in * practice*. - We will have to qualify the above as it is a subjective statement eg the function AreThereInfinitelyManyTwinPrimes() can not be calculated by the Halting oracle and hence more in-computable than Kolmogorov complexity K() here on MO<|endoftext|> -TITLE: Units in a group algebra -QUESTION [6 upvotes]: Let k be a field and let G be a finite group. I would like to know if there is any nice description of the group of units in the group algebra kG. (If there is no nice answer in this generality, assume that the characteristic of k is p and that G is a p-group.) - -REPLY [12 votes]: If $k$ has characteristic $p$ and $G$ is a finite $p$-group, then the augmentation ideal $J$ of $kG$ is nilpotent and of codimension one, hence coincides with the Jacobson radical. This means that the units of $kG$ are precisely the complement $kG \setminus J = \{ \sum_g c_g g \mid \sum_g c_g \neq 0 \}$ of the augmentation ideal. -Is this the description you asked for, or do you rather want a procedure for determining the isomorphy type of the group of units (which is the product of $k^{\times}$ and a $p$-group of order $|k|^{|G|-1}$) from that of $G$? When $G$ is abelian, this problem is completely solved in R. Sandling's paper Units in the modular group algebra of a finite abelian $p$-group (J. Pure Appl. Algebra 1984). In the non-abelian case there are many papers in the literature devoted to particular $p$-groups $G$, but as noted in Stefan Kohl's answer, the problem is not understood in full generality. - -REPLY [11 votes]: If $k$ is a field of characteristic $p>0$ and $G$ is a finite $p$-group then it is well-known that the Jacobson radical of $kG$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. The group of units of $kG$ is given by $kG\backslash \omega(kG)$ and is isomorphic to the direct product $k^\times \times (1+\omega(kG))$, where $k^\times=k\backslash\{0\}$. Some other results on the group of units of group algebras can be found, e.g., in the books "S.K. Sehgal: Topics in group rings", "D. Passman: The algebraic structure of group rings", "I. Passi: group rings and their augmentation ideal", and in many papers of the existing literature on the subject.<|endoftext|> -TITLE: Formal adjoint of the covariant derivative -QUESTION [14 upvotes]: Let $E \to M$ be a vector bundle over some Riemannian metric $(M, g)$ and endow it with some fibre metric. Assume that covariant derivative $\nabla$ is compatible with the metric. -It is essentially an application of Stokes theorem to derive the following identity -$$\nabla^*_X = - \nabla_X - div(X)$$ -for the formal adjoint of $\nabla_X$ with respect to the usual $L^2$-scalarproduct. -Question 1: Is there a closed expression for the formal adjoint of $\nabla$ regarded as a map $\Gamma^\infty(E) \to \Gamma^\infty(T^*M \otimes E)$, i.e. $\nabla^*: \Gamma^\infty(T^*M \otimes E) \to \Gamma^\infty(E)$. -I'm especially interested in the case where one drops the compatibility condition on $\nabla$ and the metric. The background for this question lies in the equation $\nabla^* \nabla \phi = - \text{Tr}_g (\nabla^2_{\cdot, \cdot} \phi)$ for the Bochner Laplacian. I can see where it steams from for metric connections, but not for arbitary $\nabla$. Is this forumla just taken as a generalization of the compatible case? -Question 2: The word 'formal' refers to the fact that the adjoint is only a 'true' adjoint operator after extension to appropriate Hilbert spaces. Exists a definition of adjointness on Fréchet spaces such that one can view $\nabla^*$ as a true adjoint operator? (Assume that $M$ is compact so that the section spaces are Fréchet spaces). I had the idea to represent $\Gamma^\infty(E)$ as the projective limit of Hilbert spaces and then piece the the adjoints on the building blocks together, but failed working out the details. - -REPLY [4 votes]: Being a bit late for the party, here is nevertheless a small answer. In fact, there is a rather explicit way to compute adjoints of every differential operator (any order) between (smooth, compactly supported) sections of vector bundles. Of course, this is (in general) not the Hilbert space adjoint, here you need a bit more analysis ;) -The main idea is to use a symbol calculus based on a covariant derivative. I explain here the scalar version (trivial vector bundles) but the whole thing can be done in full generality as well. First, you choose a covariant derivative, say torsion-free and you choose a density on your manifold in order to have an integration measure. As you probably know, the symmetrized covariant derivative allows you to establish a $C^\infty(M)$-linear bijection between symbols, i.e. smooth functions on the cotangent bundle being polynomial in the fiber directions, and differential operators. Note that this is a real bijection, not just taking into account the leading symbol. Of course, the symbol depends on the chosen covariant derivative. -In a second step you compute once and for all the adjoint of a differential operator $D$ with symbol $f$ by zillions of integrations by parts. The funny thing is that there is a fairly simple way how the symbol of the adjoint looks like. You need two ingredients for that: -First, the covariant derivative allows you to define a horizontal lift which in turn determines a maximally indefinite pseudo Riemannian metric on the cotangent bundle (horizontal spaces are in bijection to tangent spaces at the base point, vertical spaces are in bijection to the cotangent space, thus there is a natural pairing). This metric has a Laplace operator (better: d'Alembert operator) $\Delta$ with which you can act on the symbol $f$. In the flat situation this is just -\begin{equation} -\Delta_{\mathrm{flat}} = \frac{\partial^2}{\partial q^i \partial p_i} -\end{equation} -for a Darboux chart on $T^*M$ induced by a chart on $M$. In general, there are a couple of Christoffel symbols needed to make this globally defined ;) -Second, the density $\mu$ of your integration might not be covariantly constant. In any case, it defines a one-form by -\begin{equation} -\alpha(X) = \frac{\nabla_X \mu}{\mu}, -\end{equation} -which is now be used to cook up a new differential operator on $T^*M$. You can lift $\alpha$ vertically to a vector field $F(\alpha)$ on $T^*M$, completely canonical. -Having these two ingredients, the adjoint of $D^*$ has the following symbol -\begin{equation} -f^* = \exp\left(\frac{1}{2i}(\Delta + F(\alpha))\right) \overline{f}. -\end{equation} -The prefactor in the exponential depends a bit on your conventions concerning the assignment of symbol to operator. With this formula it is typically really just a computation to get adjoints of all kind of operators. In many cases, you can chose your density to be covariantly constant, so $\alpha = 0$. -You can find all this in much detail in my book on Poisson geometry, based on some old papers on quantization of cotangent bundles in the late 90s (together with Bordemann and Neumaier). -In the case of interesting bundles, the formula is essentially the same: you only have to choose covariant derivatives for the two vector bundles in question and modify the Laplace operator accordingly. Then you can proceed in the same way. This generalization is in a paper of mine with Bordemann, Neumaier and Pflaum.<|endoftext|> -TITLE: Gauss linking integral and quadratic reciprocity -QUESTION [20 upvotes]: In the setting of Mazur's "primes and knots" analogy, prime ideals $\mathfrak p\subset\mathcal O_K$ correspond to "knots" $\operatorname{Spec}\mathcal O_K/\mathfrak p$ inside a "3-manifold" $\operatorname{Spec}\mathcal O_K$. The law of quadratic reciprocity $(\frac pq)=(-1)^{(p-1)(q-1)/4}\cdot(\frac qp)$ is thus analogous to the symmetry of the linking number $\operatorname{lk}(\gamma_1,\gamma_2)=\operatorname{lk}(\gamma_2,\gamma_1)$. -THANKS to S. Carnahan for linking some other questions (see below) which give lots of interesting background on Mazur's analogy. -I have read in a few places (e.g. the book Knots and Primes by Masanori Morishita p59 Remark 4.6) that in this analogy the Gauss linking integral: -$$\operatorname{lk}(\gamma_1,\gamma_2)=\int_{K_1}\int_{K_2}\omega(x-y)\,dxdy$$ -for a certain 2-form $\omega$ on $\mathbb R^3\setminus 0$ (in particular, its interpretation as the $U(1)$ Chern--Simons path integral as in Witten) is analogous to the Gauss sum expression for the Legendre symbol $(\frac pq)$. -Can someone give more details on how exactly the Gauss sum is analogous to the abelian path integral (or a reference)? - -REPLY [8 votes]: This is a beautiful aspect of the analogy. In fact, I think it's important and cool to view it as a consequence of previously-established aspects of the analogy, rather than a new analogy-by-fiat. Under this light, the relationship between quadratic residues and abelian integrals is rather simple. As with many ideas in arithmetic topology, the ideas is to replace something analytic (in this case, Gauss's linking integral) with something sufficiently algebraic that we can port it over to the number field situation. The short answer is that both linking numbers and quadratic residue symbols can be computed as cup products in appropriate cohomology groups (here, "appropriate" might be taken to mean that they have already been established as analogous by previous aspects of the analogy). The symmetry of the linking number thus becomes completely analogous to the statement of quadratic reciprocity for primes congruent to 1 mod 4 (for which the analogy is most precise/applicable for ramification reasons). -Here are some rough details: -We begin by observing that in the framework of algebraic topology, the mod-2 linking number of two knots $K$ and $L$ can be computed as a cup product in the relevant cohomology groups: We have -$$ -\text{lk}(K,L)=[K]\cup [\Sigma_L]\in H_c^3(S^3-L,\mathbb{Z}/2)\cong \{\pm 1\}, -$$ where $[K]\in H_c^2(S^3-L,\mathbb{Z}/2)$ and $\Sigma_L$ is the Siefert surface of $L$, so $[\Sigma_L]\in H^1(S^3-L,\mathbb{Z}/2)$. -With an algebraically-analogizable object in hand, we turn to the number field situation. Here, for primes $p$ and $q$, playing the role of the knots, we have a well-established body of arithmetic-topology analogies. Here, instead of $S^3-K$ we have $\text{Spec}(\mathbb{Z}-\{p\}$, whose (etale$^*$) cohomology with coefficients in $\mathbb{Z}/2$, $H^1(\text{Spec}(\mathbb{Z}-\{p\},\mathbb{Z}/2)\cong \text{Hom}(\text{Gal}(\mathbb{Q}(\sqrt{p})/\mathbb{Q}),\mathbb{Z}/2)$ contains the class $[\Sigma]$ of the "Seifert surface" $\Sigma$ corresponding under that isomorphism to the traditional Kummer character $\chi_p$ of $\text{Gal}(\mathbb{Q}(\sqrt{p})/\mathbb{Q}).$ Now, in what amounts to not much more than the standard analysis of quadratic reciprocity using the language of cohomology, we identify an element of $H^2(\text{Spec}(\mathbb{Z}-\{p\}),\mathbb{Z}/2)$ by taking the class of $[q]\in \mathbb{Q}_p^{\times}/\mathbb{Q}_p^{\times 2}\cong H^1(\mathbb{Q}_p,\mathbb{Z}/2)$, and then using the map $H^1(\mathbb{Q}_p,\mathbb{Z}/2)\to H^2(\text{Spec}(\mathbb{Z}-\{p\}),\mathbb{Z}/2).$ Then, finishing the analogy, we have -$$ -\left(\frac{q}{p}\right)=[p]\cup [\Sigma_q]\in H^3(\text{Spec}(\mathbb{Z}-\{p\}),\mathbb{Z}/2)\cong \{\pm 1\}. -$$ -Comparing the two identities for $\text{lk}(K,L)$ and $\left(\frac{q}{p}\right)$ should make the whole thing apparent. -(*): Actually, you need some slight variant of etale cohomlogy which properly compactifies the infinite primes, the details of which I am fuzzy on. All of the ideas here are, to my knowledge, due to Morishita, and best addressed in his article "Analogies between knots and primes, 3-manifolds and number rings," though I'd suggest some other of his papers as prerequisites.<|endoftext|> -TITLE: Group scheme counterexample -QUESTION [6 upvotes]: Could someone give me an example of a finite group scheme $G$ (over some base $S$) so that $G$ minus a point is still a group scheme over $S$, but not affine over $S$? -Oort mentions that there are examples of this kind in his book "Commutative Group Schemes", but doesn't give an example. - -REPLY [2 votes]: Here's a minor variant on Margaux's comment from a year ago. Let $S = \mathbb{A}^2$ (say, over your favorite field), and let $\pi: G \to S$ be the constant $S$-group scheme $(\mathbb{Z}/2\mathbb{Z})_S$. As a scheme, it is a disjoint union of two copies of $S$. -If we remove a point $g$ from the non-identity section, we get an $S$-group scheme $G - g$, whose underlying scheme is a disjoint union of $S$ and $S - \pi(g)$ (note that $(G-g) \times_S (G-g)$ is a disjoint union of an identity section $S$ and 3 copies of $S - \pi(g)$). This is not affine over $S$, since $S$ is affine and $G - g$ is not.<|endoftext|> -TITLE: motivation for multiplier ideal sheaves -QUESTION [18 upvotes]: What is the origin of multiplier ideal sheaves?It was introduced ny Nadel.Yum Tong Siu,his advisor in his plenary lecture in 2002 icm mentions some thing that it arose in pde.Can anyone kindly elaborate on the motivation behind defining multiplier ideal sheaves.I think there are lots of experts here in mathoverflow who are experts in these things like diverio and many others.http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/trieste.pdf this is I think one of the most standard places to learn about it. - -REPLY [15 votes]: There's a parallel history of multiplier ideals (especially of the non-dynamic multiplier ideal sheaves on algebraic varieties, say as described in Lazarsfeld's book). -From this perspective, for $\mathfrak{a}$ an ideal sheaf on $X$, the multiplier ideal of $(X, \mathfrak{a}^t)$ is defined as follows (assuming $K_X$ is $\mathbb{Q}$-Cartier, which always holds if $X$ is smooth). Choose $\pi : Y \to X$ a log resolution of $(X, \mathfrak{a})$ with $\mathfrak{a} \cdot O_Y = O_Y(-G)$. Then -$$ -\mathcal{J}(X, \mathfrak{a}^t) = \pi_* O_Y( \lceil K_Y - \pi^* K_X - t G\rceil) -$$ -These ideal sheaves are older than Nadel's work. For instance, they were extremely common in the work of Esnault and Viehweg in the early 1980s (see for instance their notes which survey some of this work Lectures on vanishing theorems), also see the works of Kawamata and Kollar. Indeed, these sheaves and slight variants appeared frequently whenever Kawamata-Viehweg vanishing theorems were applied throughout the 1980s. Essentially, the reason why they show up in this context is as follows. You want to prove some Kodaira-type vanishing theorem on a variety that is either non-smooth or with respect to a not-necessarily-ample line bundle. The multiplier ideal lets you correct for this. -If you assume that $\mathfrak{a} = O_X$ and if you remove the $\pi^* K_X$ from the definition, then you get a subsheaf of $\omega_X$. This subsheaf appeared in the work Grauert and Riemenschneider (1970) and was used frequently by Lipman in his work in the 1970s especially in his work on resolution of singularities of excellent two-dimensional rings (the fact that the multiplier submodule of $\omega_X$ is not equal to $\omega_X$ is a measure of singularities). -In the case that $t = 1$ and $X$ is regular, this appeared in the work of Lipman in the 1980s and 1990s (especially in relation to questions of integral closure of powers of ideals). - -REPLY [14 votes]: Here is a sketch of Nadel's original motivation. Classical results of Aubin and Yau imply the existence of Kahler-Einstein metrics on manifolds with ample canonical bundle and and for all polarisations of Calabi-Yau manifolds. The method involved is a continuity method for the complex Monge-Ampère equation (see for example Tian's Canonical metrics in Kaehler geometry for an introduction to this stuff), together with certain a priori $C^0$ estimates. -When one searches for Kaehler-Einstein metrics on Fano manifolds ($-K_X$ ample), things are harder. In Nadel's time, certain obstructions were known (for example Matsushima showed the lie algebra of the automorphism group must be reductive), but few sufficient conditions were known. However, on Fano manifolds without a Kaehler-Einstein metric, the continuity method must fail. Nadel's idea was to study consequences of the failure of the continuity method. Specifically, Nadel showed that if the continuity method fails, then there must exist a singular hermitian metric written locally $h=h_0e^{-\phi}$ on $-K_X$, where $h_0$ is a genuine smooth hermitian metric, and $\phi$ satisfies some mild regularity assumptions, such that $h$ has semipositive curvature current and $\phi$ has non-trivial multiplier ideal sheaf $\mathcal{I}(\gamma \phi)$ for all $\gamma \in (\frac{n}{n+1},1)$. Here, one views the multiple ideal sheaf as the functions where certain integrals don't converge (equivalently, if certain integrals converge, the continuity method doesn't fail and there is a Kaehler-Einstein metric). Moreover, one can assume that for any compact $G\subset Aut(X)$, $h_0$ and $\phi$ are $G$-invariant. -Nadel combined this with his vanishing result: $H^q(X,\mathcal{I}(\gamma \phi))=0$ for all $q>0$. Here we're using that $h$ is a singular hermitian metric on $-K_X$. This form of Nadel vanishing has strong geometric consequences: associating a $G$-invariant subscheme $Z_{\gamma}$ to $\mathcal{I}(\gamma \phi)$, this implies that $H^q(Z_{\gamma}, \mathcal(O_{Z_{\gamma}}))=0$ for all $q>0$ and equals $\mathbb{C}$ for $q=0$. A simple corollary is that $Z_{\gamma}$ is connected, so if $G$ acts without fixed points, cannot be of dimension $0$. Then, if $X$ is of dimension $3$, $Z_{\gamma}$ must be $1$ dimensional, and Nadel showed $Z_{\gamma}$ must be a tree of rational curves, the existence of which can sometimes be ruled out. Nadel's construction therefore gave new examples of Fano manifolds with Kaehler-Einstein metrics. -One can also think about multiplier ideal sheaves as follows. This probably isn't how Nadel thought about them at the time, however it is slightly more appealing algebro-geometrically. Given an anti-canonical divisor $D$, one can naturally associate a singular hermitian metric on $-K_X$. One property of the pair $(X,D)$ is whether or not it is log canonical - algebraically this means it is not too singular, analytically this tells you a certain integral converges. The multiplier ideal sheaf associated to $D$ refines this, essentially giving a scheme structure to the set at which the pair $(X,\gamma D)$ is not log canonical, for all $\gamma$. Nadel vanishing then tells you, for example, that the set at which $\gamma D$ is not log canonical (i.e. is highly singular) is connected. In this case then one can view Nadel's result on Kaehler-Einstein metrics as saying that the non-existence of such a metric implies the existence of a highly singular anti-canonical divisor, and moreover the "highly-singular" locus of this divisor satisfies certain geometric conditions which can be ruled out in certain cases (at least in the case that the singular hermitian metric in Nadel's theorem arises from an anti-canonical divisor - I suspect this is the case due to certain approximation results). -I think a good reference for this is section $6$ of the Demailly-Kollár paper "Semi-continuity of complex singularity exponents and Kaehler-Einstein metrics on Fano orbifolds". It explains what I have described above fully and precisely (gives definitions etc.), and proves Nadel's result on Kaehler-Einstein metrics in a simpler way than Nadel originally did.<|endoftext|> -TITLE: Is there an algorithm to solve quadratic Diophantine equations? -QUESTION [19 upvotes]: I was asked two questions related to Diophantine equations. - -Can one find all integer triplets $(x,y,z)$ satisfying $x^2 + x = y^2 + y + z^2 + z$? I mean some kind of parametrization which gives all solutions but no points which do not satisfy the equation. -Is there an algorithm that will determine, given any quadratic $Q(x_1,\ldots,x_n)$ as input, all integer points of this quadratic? In the case of existence of solution, there is an algorithm, https://math.stackexchange.com/questions/181380/second-degree-diophantine-equations/181384#comment418090_181384 - -REPLY [3 votes]: S. Carnahan (Sep 24 '13 at 18:33) showed parametrization by cosets of congruence subgroups -of SL(2,Z). What is missing is parametrization of SL(2,Z) and its congruence subgroups which can be found in -L.N. Vaserstei. Polynomial parametrization for the solutions of Diophantine equations and -arithmetic groups , Annals of Math. 171:2 (March of 2010), 979–1009. -MR2630059. Zbl 05712747<|endoftext|> -TITLE: Centralizers in amalgamated free products -QUESTION [11 upvotes]: An old result of Karrass and Solitar from 1970 says that if $g$ is a nontrivial element in an amalgamated free product $G=A*_HB$, with $H$ malnormal in $G$, then the centralizer of $g$ in $G$ is either infinite cyclic or it is in a conjugate of a factor. -Is it sufficient to assume that $H$ is malnormal in $A$ or in $B$ ? Does the same result hold if we assume that $H$ is almost or weakly malnormal in $G$ ? -$H -TITLE: What is the probability of an arbitrary nonlinear dynamical system to be chaotic? -QUESTION [5 upvotes]: Particularly, how to characterize a set of chaotic nonlinear dynamical systems as a subset of nonlinear dynamical systems with respect to the set cardinality? -To explain the question more, a simple similar question would be "if I point to an arbitrary number on a real number line, what is the probability of this number to be irrational". Here, this would be 1 which is explained by the cardinalities of rational and of the irrational subsets of a real set. -Though it's interesting in general, the fact that linear dynamical systems can be chaotic as well is dismissed in the question since that, if I'm right, requires infinite dimensionality. I'm doing theoretical neuroscience research and here nonlinear dynamics are largely used to model motor control and perceptual phenomena. -I can't find the answer within research nor textbook so far, and I'm also new to dynamical systems. - -REPLY [8 votes]: This is an important and interesting question; unfortunately it's very hard to say anything in much generality. -First one needs to establish within which class of dynamical systems the question is being asked, and then to establish a notion of "largeness" within that class. Cardinality is not the most useful because whatever precise definition of "system" and "chaotic" we settle on, both classes are likely to be uncountable. So we may try to impose a measure on the space of systems, and ask if a positive measure set (or full measure set) of systems are chaotic. (Again, deferring for the moment the question of just what chaotic means.) Or we may ask whether the set of chaotic systems is residual (countable intersection of open dense sets), so that a topologically generic system is chaotic. -One immediate difficulty is that the space of systems is infinite-dimensional. More precisely, if "dynamical system" means "diffeomorphism of a given manifold", then it's not clear what measure to use on the space of systems. There are notions of prevalence that address this issue, but I won't go into them here. -We still haven't said what chaotic means. Let's agree that a dynamical system is a smooth map $f$ from some manifold $M$ to itself, and that chaotic means "$f$ has a physical measure with non-zero Lyapunov exponents". This means that there is an $f$-invariant probability measure $\mu$ on $M$ such that - -all the Lyapunov exponents of $\mu$ are non-zero ($\mu$ is a hyperbolic measure), and at least one is positive; moreover, -there is a set of points $G_\mu\subset M$ such that $G_\mu$ has positive volume (with respect to the Lebesgue measure induced by a Riemannian structure compatible with the smooth structure of $M$) and every $x\in G_\mu$ is generic for $\mu$, meaning that it satisfies the conclusion of the Birkhoff ergodic theorem for every continuous observable $\phi$. - -Now one can ask: within the space of smooth maps, how large is the set of maps for which the above conditions hold? This at least gives a more precise formulation of the question. Here's what I know. - -The answer is very different depending on whether one considers the space of $C^1$ maps or the space of $C^2$ maps. I won't go into the reasons for this. Let's consider $C^2$ maps, which is where I know more - other people are more familiar with the $C^1$ case. -There is a series of conjectures due to Jacob Palis which address the question in this general formulation (see "A global view of dynamics and a conjecture on the denseness of finitude of attractors", in Géométrie complexe et systèmes dynamiques (Orsay, 1995), Astérisque, 2000, pp. xiii-xiv, 335-347). These conjectures are very hard and not too much is known in full generality. -Rather more can be said in specific classes of examples. In particular, one-dimensional maps are reasonably well-understood. If one considers the case where $M=[0,1]$ is an interval and $\{f_\lambda \mid \lambda\in [a,b]\}$ is a one-parameter family of smooth maps satisfying certain technical conditions -- in particular, the family of logistic maps $f_\lambda(x) = \lambda x(1-x)$ satisfies these with $\lambda\in [0,4]$ -- then the following is known. - -Write $R\subset [a,b]$ for the set of parameters $\lambda$ where $f_\lambda$ has regular behaviour, meaning that every orbit eventually is attracted to a stable periodic orbit (and the system is non-chaotic). Then $R$ is open and dense. -Write $S\subset [a,b]$ for the set of parameters $\lambda$ where $f_\lambda$ has stochastic behaviour, meaning that there is a hyperbolic physical measure as discussed above. Then $S$ has positive Lebesgue measure. -$R\cup S$ has full Lebesgue measure. -A very nice account of all this is given by Lyubich in the October 2000 Notices of the AMS, entitled "The Quadratic Family as a Qualitatively Solvable Model of Chaos". - - -Numerical experiments suggest that the behaviour observed for one-dimensional maps -- namely, existence of chaotic behaviour for a positive measure set of parameters together with open sets of parameters with regular behaviour -- occurs more broadly than just in this setting. But rigorous results are quite hard to come by. -One final remark: You don't need nonlinearity to get chaotic behaviour, but for a linear system you do need non-trivial topology. The doubling map $x\mapsto 2x \pmod 1$ has all the chaotic behaviour you could want, but relies on the non-trivial topology of the circle; similarly, the Arnold cat map $x\mapsto \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}x \pmod {\mathbb{Z}^2}$ as a map on the two-torus is both linear and chaotic.<|endoftext|> -TITLE: Homotopy groups of an infinite wedge of 2-spheres -QUESTION [8 upvotes]: I know Hilton's result about a finite wedge of spheres, and I know that certain homotopy groups (such as the third homotopy group) can be directly calculated for an infinite wedge too. -My question is -- Is there some general result that gives the homotopy groups of an (uncountable) infinite wedge of 2-spheres in terms of the homotopy groups of 2-spheres (or other simpler spaces)? -Thanks. - -REPLY [14 votes]: For any infinite wedge $X=\bigvee_{i\in I} X_i$, the homotopy groups are just the colimit $$\pi_n(X)=\operatorname{colim}_F \pi_n\left(\bigvee_{i\in F} X_i\right),$$ -where $F$ ranges over all finite subsets of $I$. The reason for this is simple: $S^n$ is compact, so the image of any map $S^n\to X$ must be contained in $\bigvee_F X_i$ for some finite $F$; this means the natural map from the colimit to $\pi_n(X)$ is surjective. Similarly, since $S^n\times I$ is compact, any homotopy also lands in a finite wedge, so the natural map is injective. The same reasoning would apply if you replaced $S^n$ with any compact space, and the infinite wedge with any colimit for which any compact subset factors through some stage (eg, any filtered system of sub-CW complexes).<|endoftext|> -TITLE: Probability that a stick randomly broken in five places can form a tetrahedron -QUESTION [59 upvotes]: Edit (June 2015): Addressing this problem is a brief project report from the Illinois Geometry Lab (University of Illinois at Urbana-Champaign), dated May 2015, that appears here along with a foot-note saying: An expanded version of this report is being prepared for possible publication. -Below is an excerpt (though, to be clear, this question on MO is about finding a closed-form solution; moreover, (a) the main finding in part i agrees with an earlier MSE response, and (b) I have relayed the typographical error of "math.overflow.net" to the write-up's corresponding faculty mentor). - - -The following problem was brought to my attention by a doctoral dissertation on Mathematics Education, but - as far as I know - the solution remains unknown. -I have already asked this question on MSE, where the post has garnered over 90 votes, but still no canonical answer in the more-than-a-year since it has been there. -Please add or suggest different tags if it seems warranted. - -Randomly break a stick in five places. -Question: What is the probability that the resulting six pieces can form a tetrahedron? -Clearly satisfying the triangle inequality on each face is a necessary but not sufficient condition. -Furthermore, the question of when six numbers can be edges of a tetrahedron is related to a certain $5 \times 5$ determinant, namely, the Cayley-Menger determinant. (See, e.g., Wirth, K., & Dreiding, A. S. (2009). Edge lengths determining tetrahedrons. Elemente der Mathematik, 64(4), 160-170. A more recent article by these authors is cited in the comments below: Wirth, K., & Dreiding, A. S. (2013). Tetrahedron classes based on edge lengths. Elemente der Mathematik, 68(2), 56-64.) -Obviously, this problem is far harder than the classic $2D$ "form a triangle" one. I would welcome any progress on finding a solution or a reference to one if it already exists in the literature. - -REPLY [14 votes]: This is far from a complete answer, but it may be helpful progress. -The marked problem, where the numbers are labeled, looks much simpler than the unmarked version. The region of lengths which can be assembled into tetrahedron in some order can be viewed as a union of $6!$ copies of the region for marked lengths, although this can be reduced by the symmetries of the tetrahedron. So, let's look at the simpler case of marked lengths. -There are linear conditions from the triangle inequality on each face of the tetrahedron, plus one quadratic condition from the positivity of the square of the volume. Ignoring the quadratic condition gives us an upper bound on the probability the edges form a tetrahedron. -The collection of $12$ triangle inequalities and one equation, that the sum of the lengths is $1$, produces a $5$-dimensional polytope $P$ which I analyzed with the help of qhull. Given the simplicity of the result, perhaps there is a way to read off the structure more directly. There are $7$ vertices. Four of these vertices give $1/3$ length to the $3$ edges meeting at a vertex, and $0$ length to the other three edges forming a triangle. Three of the vertices give $1/4$ length to a cycle of length $4$, and $0$ length to two opposite edges. If the tetrahedron includes faces with lengths $\lbrace a, b, c \rbrace$ and $\lbrace a, d, e \rbrace$ then the vertices include $(0, 0, 0, 1/3, 1/3, 1/3)$ and $(0, 1/4, 1/4, 1/4, 1/4, 0).$ With only $2$ more vertices than the dimension, the combinatorial structure of $P$ is relatively simple, and is determined by the fact that the convex combination (in fact average) of $4$ vertices equals a convex combination of the other $3$ vertices. Much as the bipyramid in $3$ dimensions can be triangulated with either $2$ or $3$ tetrahedra, $P$ can be triangulated with either $3$ or $4$ symmetric simplices. $P$ is related to the normal disks in a tetrahedron. -The volume of $P$ is $1/54$ of the volume of the simplex of edge lengths summing to $1$. This gives an upper bound on the probability that breaking a stick into marked lengths produces the edge lengths of a tetrahedron, although this is far from the $1/79$ observed numerically by Kirill. It is interesting that the quadratic condition rules out a large fraction of the lengths which satisfy the triangle inequalities. -I think the quadratic condition can be added by considering how the surface intersects the tetrahedra of one of the triangulations of $P$. This is much simpler than taking the intersection of a quadratic inequality with an arbitrary polytope.<|endoftext|> -TITLE: What exactly does this diagram of Omar Khayyam represent? -QUESTION [24 upvotes]: Evidently Omar Khayyam (1048-1131) was quite the mathematician. He did groundbreaking work on finding geometric solutions to the cubic equation, which is all the more notable since he did not have a good system of notation to work with. -As an example, suppose you want to solve $x^3 + 5x + 1 = 0$. Substituting $y = x^2$, one obtains $x(y + 5) + 1 = 0$, and so the solution lies at the intersection of a parabola and a hyperbola, which can be easily graphed. -More about his work can be found, e.g., here and here, and the ideas in his work also appear in modern papers such as this one of Wright and Yukie, and follow-up papers by Bhargava. -I was able to find this picture on Khayyam's Wikipedia page, which is possibly in Khayyam's own handwriting. - -I would love to use it in a talk. But what precisely does this picture represent? Judging from the sources I quoted (among others), it seems possibly related to the solution to $x^3 + 200x = 20x^2 + 2000$, but it is not clear exactly how. -Do any MO users read Arabic or otherwise know what this picture is? - -REPLY [6 votes]: These figures (second exact, first with an extra line) are discussed in two places (PP. 98-99 & 165-166) in "Omar Khayyam the Mathematician" ISBN 0-933273-46-0, by Rashed & Vahabzadeh, Bibliotheca Persica Press, NY. I have not tried to digest the discussion.<|endoftext|> -TITLE: What is the "Tangle" at the Heart of Quantum Simulation? -QUESTION [8 upvotes]: The following questions generalize and naturalize the question that was originally asked. Provisional answers largely due to Will Sawin are now included. -As was discussed in the question originally asked, let $$\mathcal{L}(r,\{a_i\})\ {\colon}{=}\ \sigma_r\big(\text{Seg}(\mathbb{P}^{a_1-1}\times\cdots\times\mathbb{P}^{a_n-1})\big) \subset \ \mathbb{P}^{(\Pi_{i=1}^n a_i) -1}$$ be the usual Segre immersion of the rank-$r$ secant join. Regarding $\mathbb{P}^{(\Pi_{i=1}^n a_i) -1}$ as an $n$-qudit Hilbert space denoted $\mathcal{H}$ for concision, and furthermore regarding $\mathcal{H}$ as a Kähler manifold, we specify $\omega$ on $\mathcal{L}$ as the symplectic form associated to the Kähler potential $\kappa = \langle\psi|\psi\rangle$, given first as a real-valued (bilinear/biholomorphic) function $\kappa\colon \mathcal{H}\to\mathbb{R}$, and thus specified as $\kappa\colon \mathcal{L}\to\mathbb{R}$ by pullback onto the Segre immersion $\mathcal{L}\subset\mathcal{H}$; similarly let $g=\langle\psi|G|\psi\rangle$ be the real-valued (bilinear/biholomorphic) symbol function that is specified by a general hermitian operator $G$, with $g$ similarly pulled-back to $\mathcal{L}$ from $\mathcal{H}$. -Answers to these questions (or better-constructed questions) were sought: - -Q1  Is the Zariski closure of $\mathcal{L}(r,\{a_i\})$ endowed with a differential structure that is globally smooth? -A1  The answer is no (not in general). For example (as explained by Will Sawin) the manifolds $\mathcal{L}(a-1,\{a,a\})$ are determinantal, and thus are known to be generously endowed with singular points. -=== -Q2  In the event that $\mathcal{L}$ is equipped with (one or more) smooth differential structure(s), is it the case that $dg \in \text{span}\ \hat\omega$?  Physically, are hermitian operators on $\mathcal{H}$ generically associated to singularity-free symplectomorphic flows on differentially smooth immersed $\mathcal{L}$-manifolds? -A2  Formally the answer is no, in that $\mathcal{L}$ generically has no smooth differential structure, per answer A1. -=== -Q3  In the event that $\mathcal{L}$ has no defect-free differential structure(s), what smoothness-defects appear in the symplectomorphic flows on $\mathcal{L}$ that are induced by pullback of (bilinear/biholomorphic) Hamiltonian potential functions from $\mathcal{H}$? -A3  Despite the algebraic singularities that are mentioned in A1–2, the answer is conjectured to be "symplectic defects are dynamically occult", in a concrete sense that will be explained in an auxiliary answer (to be posted in the next day or two). In brief, the Hamiltonian structure that is pulled back from $\mathcal{H}$ is conjectured to induce dynamical maps $M(t)\colon \mathcal{L}\to\mathcal{L}$ that are exact symplectic isomormorphisms, despite the singular points of $\mathcal{L}$, and despite the rank-deficit of the pulled-back symplectic form $\omega$ at those singular points. Physically speaking this, means that the varietal singularities of $\mathcal{L}$ do not obstruct computational simulations of thermodynamical physics associated to symplectomorphic dynamical flow. - -Engineering motivation  Engineers "know" — from concrete computations — that $\mathcal{L}$ is richly endowed with singularities in Riemannian/Ricci/scalar curvature. And yet experience teaches that these curvature singularities are seldom (never?) associated to computational pathologies in simulating quantum trajectories as integral curves of Hamiltonian flows. Thus the questions asked are motivated by a postulate that the structure of $\mathcal{L}$ is differentially smooth with respect to symplectic forms and Hamiltonian potentials that are pulled-back from $\mathcal{H}$, even though $\mathcal{H}$ induces metric curvature on $\mathcal{L}$ that is singular. -The broad question  In what mathematical sense(s) — if any — does algebraic geometry teach that $\mathcal{L}$'s symplectic geometry is differentially smooth, even though $\mathcal{L}$'s metric geometry is singular? In purely practical terms, why don't our quantum trajectory simulation codes break more often? Advice in framing these questions more clearly and naturally is very welcome. - -The questions originally asked  The questions asked concern one-dimensional closed-loop submanifolds of Hilbert space called LangreTangles. The specific questions asked are: - -What is the large-$n$ probability $P(n)$ that an $n$-dimensional Hilbert space supports LangreTangle trajectories? -What is the large-$n$ length $l(n)$ of a generic LangreTangle trajectory? - -Numerical evidence suggests $P(n) \gtrsim 5/(\ln n)$  and  $\ln l(n)\sim \mathcal{O}(n)$. -In physical terms LangreTangles are emergently-quantum trajectories on "foamy" dynamical manifolds; thus natural physics-oriented questions include "What experimental techniques are suited to the observation of LangreTangle trajectories?" - -Caption  A numerically-integrated portion of a typical LangreTangle trajectory $L(\psi_0,4,\{3,3,3\})$, drawn as a projective map $S^3\,{\to}\,R^3$. It is conjectured that the full LangreTangle would fill $R^3$ with a closed-loop "tangle". Physically this particular LangreTangle is associated to the rank-$4$ secant join of the product state of three spin-$1$ particles. Further LangreTangles have been computed as follows: - -a full (closed-loop) LangreTangle for the simplest-of-all determinantal varieties $\mathcal{L}(1,\{2,2\})$ ( image here ); -a partial LangreTangle for the determinantal variety $\mathcal{L}(4,\{5,5\})$ ( image here ) -a partial LangreTangle for the unit-zabacity (is it nondeterminantal?) variety $\mathcal{L}(9,\{2,2,2,2,2,2\})$ ( image here ); this is an example of a nontrivial unit-zabacity variety that is nondefective in the sense of Landsberg. - - -Remark  The $\mathcal{L}(a-1,\{a,a\})$ representation of determinantal varieties is defective for all $a\gt2$, and the $\mathcal{L}(4,\{3,3,3\})$ variety of the original question is defective also; the latter by a result of Strassen per Landsberg p. 130. -Background  With reference to a recent much-discussed preprint *Scattering Amplitudes and the Positive Grassmannian( by Nima Arkani-Hamed, Jacob L. Bourjaily, Freddy Cachazo, Alexander B. Goncharov, Alexander Postnikov, and Jaroslav Trnka (arXiv:1212.5605 [hep-th]), coauthor Arkani-Hamed asserted the following research objective in an associated lecture The Amplituhedron: - -We can't just keep making equivalences between ideas that were essentially handed to us from the early part of the 20th century. We have to find really new things! - -As a conjectural candidate for "a really new thing," Arkani-Hamed et al. describe a class of non-Hilbert dynamical systems in which unitarity and locality are encoded emergently, in geometric objects called amplituhedrons — for details see Gil Kalai's MathOverflow question "What is the amplituhedron?" — that have been depicted in the non-specialist literature as "a jewel at the heart of quantum physics" -Engineering considerations  Quantum systems engineers are naturally sympathetic to the notion of dynamical systems in which unitarity and spatial locality are emergent rather than fundamental, in that large-scale quantum systems commonly are simulated by algorithms in which unitarity is only approximate (in that trajectories are integrated on non-Hilbert tensor network manifolds) and spatial locality is emergent (via Lindblad/Carmichael trajectory unravelings that dynamically quench macroscopic Schröedinger cats). -Nomenclature  To concretely visualize these non-Hilbert quantum trajectories, we adopt the notation of Robert Landsberg's text Tensors: Geometry and Applications, in which the geometric objects of primary interest are secant varieties of Segre varieties -$$ -\mathcal{L}(r,\{a_i\}) \equiv -\sigma_r\big(\text{Seg}(\mathbb{P}^{a_1-1}\times\cdots\times\mathbb{P}^{a_n-1})\big) \subset \mathbb{P}^{(\Pi_{i=1}^n a_i) -1} -$$ -Here the manifold $\mathcal{L}(r,\{a_i\})$ is regarded as Segre-embedded in a Hilbert space of (normalized) states $\mathbb{P}^{(\Pi_{i=1}^n a_i)-1}$. Then the zabacity function $\text{zab}(r,\{a_i\})$ is defined to be the dimensional deficit -$$ -\text{zab}(r,\{a_i\}) = \big((\Pi_{i=1}^n a_i)-1\big) - \dim \mathcal{L}(r,\{a_i\}) -$$ -Unit-zabacity manifolds  Henceforth we focus our attention upon unit-zabacity manifolds (that is, $\text{zab}(r,\{a_i\})=1$). Physically these are tensor network manifolds that almost fill the embedding Hilbert space, being dimensionally deficient by precisely one complex dimension, which is to say, deficient by precisely two real dimensions. Among the most celebrated unit-zabacity manifolds is $\mathcal{L}(4,\{3,3,3\})$ (see for example Theorem 5.5.2.1 of Landsberg's text) which physically is the rank-$4$ secant join of the product state of three spin-$1$ particles. -Visualizing unit-zabacity geometry  The geometry of $\mathcal{L}(4,\{3,3,3\})$, regarded as a submanifold of complex dimension 25 immersed in a normalized Hilbert space $\mathbb{P}^{3^3-1}$ of dimension $3^3-1=26$, can be concretely visualized by intersecting $\mathcal{L}(4,\{3,3,3\})$ with any convenient three-dimensional submanifold of the immersing $\mathbb{P}^{3^3-1}$ Hilbert space; an obvious and geometrically natural candidate for such a three-dimensional submanifold is the rotation group $\text{SO}(3)\sim S^3$. Then we have: - -Definition  The LangreTangle trajectory $L(\psi_0,r,\{a_i\})$ that is associated to a unit-zabacity manifold $\mathcal{L}(r,\{a_i\})$, and to a fiducial point $p_0\in \mathcal{L}(r,\{a_i\})$, with $\psi(p_0)=\psi_0$, is by definition the one-dimensional submanifold of points $q\in S^3$ such that the rotated state $\psi(q)= R(q)\circ\psi_0$ satisfies $\psi(q) = \psi(p)$ for some $p\in \mathcal{L}(r,\{a_i\})$. - -Numerical and graphical considerations  Closed-form arbitrary-dimension representations of the quantum rotation matrix $R(q)$ (which are due to Wigner) are given in standard textbooks such as Gottfried's Quantum Mechanics (1966); quaternionic coordinates on $S^3$ are well-suited to efficient trajectory integration; graphical representations of the Hopf fibration commonly employ the simple-to-compute projective map $S^3\to R^3$; thus the exhibition of concrete LangreTangle trajectories is computationally straightforward. -LangreTangled trajectories  For generic starting-states $\psi_0$, numerical integration of the LangreTangle trajectories for $L(\psi_0,4,\{3,3,3\})$ shows a striking "tangle", and higher-dimension unit-zabicity manifolds (not shown) are found to support even more tightly tangled trajectories. This tangled geometry motivates both the name of the zabacity dimension-measure (from the Persian zabd meaning "foam-like," and zubdat meaning "cream-like," as skimmed from a body of liquid) and the portmanteau word LangreTangle, which derives from "Robert LANdsberg" and "Corrado SeGRE*" and "quantum enTANGLEment". -Open questions  The questions asked in the beginning then follow naturally from the preceding considerations (along with many more mathematical questions, needless to say). In particular, the strikingly high prevalence of unit-zabicity large-dimension Hilbert spaces apparently is associated to the numeric properties of Cunningham numbers, which cause unit-zabacity Hilbert spaces of large dimension to be exponentially more common than heuristic factor-counting arguments would suggest (in that, for various number-theoretic reasons, the Cunningham integers $a^b\pm 1$ commonly have small factors); the question-asker would be grateful for mathematical illumination in regard to this connexion. -Physics perspectives  From a physics perspective, a natural question is simply: "By what observations, either in principle or in practice, can (flat) Hilbert-space dynamical manifolds be experimentally distinguished from ("foamy") finite-zabacity dynamical manifolds, even for systems as simple (seemingly) as three interacting spin-$1$ particles?" Again, references and/or experimental suggestions are welcome. - -Acknowledgements  The continuing stimulus of comments and questions by Gil Kalai]in regard to issues of quantum unitarity, locality, and algebraic geometry is gratefully acknowledged. - -REPLY [8 votes]: $\mathcal L(a-1,\{a,a\})$ is the hypersurface of the determinant $0$ matrices in the rank $a \times a$. I'm afraid that this hypersurface is not a topological manifold, and hence also not a smooth one. -To see this, it is sufficient to look in a small neighborhood of a rank $a-2$ matrix. Were the hypersurface a manifold, that space would have to be topologically a ball, as we see in the neighborhood of a cuspidal curve singularity like the singularity $(x,y,z) = (0,0,1)$ of $y^2z-x^3=0$. However, our neighborhood will not be a ball. -The topology of an algebraic variety in the neighborhood of a point is determined only by the leading terms of the polynomial equation that cuts it out. In the neighborhood of the point: -$\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$ -(here depicted with $a=6$) the leading terms come are the terms involving all $a-2$ nonzero entries. Since there is one term for each permutation, there are only two leading terms : the identity permutation and the transposition of the last two entries. -This means that locally, the hypersurface looks like the equation $x_1x_2-x_3x_4$ in the variables $x_1,x_2,\dots, x_{a^2-1}$, where $x_1$, $x_2$, $x_3$, $x_4$ are the entries of the bottom-right $2 \times 2$ submatrix. The vanishing set of the equation $x_1 x_2 -x_3x_4$ in just the variables $x_1,x_2,x_3,x_4$ is topologically isomorphic to the cone on an $S^1$ bundle on $S^2 \times S^2$, which is the cone on a non-sphere, so is not a topological manifold. Since we can detect this failure homologically, adding more variables, which just corresponds to taking the product with $\mathbb R^{2(a^2-5)}$, does not make it a manifold.<|endoftext|> -TITLE: Does a bifunctor that's monoidal in each argument take pairs of monoids to a commutative monoid? -QUESTION [7 upvotes]: Let $\mathcal{C}, \mathcal{D}, \mathcal{E}$ be (symmetric?) monoidal categories, and $H : \mathcal{C} \times \mathcal{D} \to \mathcal{E}$ be a functor that is monoidal in both arguments, ie. $H(C,-)$ and $H(-,D)$ are (strong) monoidal for all objects $C$ and $D$. -And take two monoids $M : \Delta \to \mathcal{C}$ and $N : \Delta \to \mathcal{D}$ (where $\Delta$ is the free monoidal category on a monoid, 1). -$H(M 1, N 1)$ is a monoid in two different ways, as $H(M1, -)$ and $H(-,N1)$ both preserve monoids. -I'd like to use an Eckmann-Hilton style argument to prove that the two monoids coincide and are commutative. -Do I need additional assumptions to prove this is true? The concrete example I had in mind is: $\mathcal{C} = h\mathrm{Top}_*^{op}$, $\mathcal{D} = h\mathrm{Top}_*$, $\mathcal{E} = \mathrm{Set}$ and $H = \mathrm{Hom}$, $M$ is the "cogroup" on $S^1$ used to define the fundamental group and $N$ is some topological group and I'd like to conclude that $\pi_1(N,id_N) = \mathrm{Hom}_{h\mathrm{Top}_*}((S^1,1),(N,id_N))$ is abelian. -Also, is there a category $\mathcal{K}$ representing the functors like $H$, ie. -$$\mathrm{Hom}_\mathrm{MonCat}(\mathcal{K},\mathcal{E}) \cong \{ H : \mathcal{C} \times \mathcal{D} \to \mathcal{E} \text{ st. } H(C,-) \text{ and } H(-,D) \text{ are monoidal for all } C, D \}$$ sort of like a tensor product? Is it of any use? - -REPLY [2 votes]: First of all, we have to require that the monoidal structure on $H$ in each variable is natural with respect to the other variable: $1 \cong H(A,1)$ and $H(A,B \otimes C) \cong H(A,B) \otimes H(A,C)$ should be natural in $A \in \mathcal{C}$, similarly for the other variable. -In the context of symmetric monoidal categories, I think you need the following compatibility conditions: - -The two isomorphisms $1 \cong H(1,1)$ agree. -For all $A,B \in \mathcal{C}$ and all $C,D \in \mathcal{D}$ the diagram - -$${\small \begin{array}{c} H(A \otimes B,C \otimes D) & \rightarrow & H(A,C \otimes D) \otimes H(B,C \otimes D) \\ \downarrow && \\ H(A \otimes B, C) \otimes H(A \otimes B,D)&& \downarrow \\ \downarrow & & \\ H(A,C) \otimes H(B,C) \otimes H(A,D) \otimes H(B,D) & \rightarrow & H(A,C) \otimes H(A,D) \otimes H(B,C) \otimes H(B,D)\end{array}}$$ -commutes. Now if $A$ and $B$ are monoids, then 1. implies that the two units of $H(A,B)$ agree, and 2. implies that the two multiplications of $H(A,B)$ agree and are commutative. I haven't checked the details. EDIT: As suggested by Chris Heunen, 1. already follows from 2., using ${\small A=B=C=D=1}$. -If $\mathcal{C}$ and $\mathcal{D}$ are presentable, then I'm pretty sure that there is some classifying symmetric monoidal category $\mathcal{K}$. But I doubt that it will be easy to describe.<|endoftext|> -TITLE: Does every countably infinite interval-finite partial order embed into the integers? -QUESTION [8 upvotes]: A partially ordered set $(S,\le)$ is called interval finite if the open intervals $(x,z):=\{y|x\le y\le z\}$ are finite for all choices of $x,z$ in $S$. An embedding $(S,\le)\rightarrow(S',\le')$ of partially ordered sets is an injective order-preserving map. Does every countably infinite interval finite partially ordered set admit an embedding into the integers? This is equivalent to extending the partial order to a linear suborder of the integers. If so, where can I find the proof? If not, can you give a counterexample? - -REPLY [8 votes]: $\newcommand{\P}{\mathbb{P}} - \newcommand{\Z}{\mathbb{Z}}$ -The answer is yes. First, let's prove a lemma. By order preserving, I assume that you mean forward-preservation of the order: $p\leq q\implies f(p)\leq' f(q)$. -Lemma. Every countable interval-finite partial order $\P$ has -a convex enumeration, an enumeration $\langle p_0,p_1,p_2,\ldots\rangle$ of -$\P$, all of whose initial segments are convex sets in $\P$. -Proof. If we have a finite convex subset of $\P$, and new point -$p$ to be added, then by convexity $p$ does not appear in any -interval of points we already have. If $p$ is above some points we -have already, then it is not below any point that we have already, -and so we can look at the intervals $(q,p)$ determined by a point -$q$ we have already and the new point $p$. By convexity, none of -these new points can be below any point we already have, and so we -can simply add them from the bottom while maintaining convexity. A -similar argment works if the new point is only below points we -already have. And if $p$ is incomparable to the points we already -have, then we can simply add it to the list. QED -Now, we can prove the theorem. -Theorem. Every countable interval-finite partial order embeds -into $\Z$. -Proof. Suppose that $\P$ is a countable -interval-finite partial order. By the lemma, it has a convex -enumeration $p_0,p_1,p_2,\ldots$. Suppose by induction that we have -mapped $p_k\mapsto m_k$ in an injective order-preserving manner, for $k\lt -n$. Consider the next point $p_n$. Since the order so far is -convex and adding $p_n$ maintains convexity, it follows that -either $p_n$ is above some points $p_k$ for $k\lt n$ and not below any, or -below some such $p_k$ and not above any, or incomparable to them all. In any -case, we can easily extend the map to define $p_n\mapsto m_n$ in -such a way to still be order preserving and injective. QED<|endoftext|> -TITLE: The Teichmüller's algebraic interpretation of $H^3$ in group cohomology -QUESTION [17 upvotes]: In the book "Cohomology of Groups" of Kenneth S. Brown, it is told in the introduction that Teichmüller arrived to $H^3$ in an algebraic context, i.e. that Teichmüller worked with an algebraic interpretation of $H^3$ in group cohomology. I have tried to find the original article Über die sogenannte nichtcommutative Galoische Theorie und die Relation $\xi_{\lambda,\mu,\nu}\xi_{\lambda,\mu\nu,\pi}\xi^\lambda_{\mu,\nu,\pi}=\xi_{\lambda,\mu,\nu\pi}\xi_{\lambda\mu,\nu,\pi}$ of Teichmüller, but I haven't found it. -So here is my question, which is this algebraic interpretation of $H^3$ in group cohomology? Or, in another way, which was the algebraic context in which Teichmüller was working to arriving to $H^3$? - -REPLY [3 votes]: In "Normality of algebras over commutative rings and the Teichmüller class. I" (II, III, arXiv version 1512.07264), Hübschmann constructs an explicit crossed module representing the Teichmüller cocycle and, among other things, reinterprets in terms of it the Deuring extension obstruction which seems to have been the original motivation of Teichmüller: very roughly, given an algebra $A$ with a group action, one seeks an embedding $A\subset C$ into a larger algebra which realizes the action via conjugation by (invertible) elements of $C$. You can find lots of references to previous work there, although, it seems, nobody before considered the corresponding crossed modules. -What I find especially interesting, in the third of these papers Hübschmann explains how this construction relates to Dixmier-Douady classes and work of Jones, Rieffel and others concerning group actions on C*-algebras, crossed products of von Neumann algebras and such.<|endoftext|> -TITLE: What one really can do with fractals built from L-systems? -QUESTION [8 upvotes]: For any L-system one can naturally associate a fractal. Why these fractals are (mathematically) useful apart that they are a source of nice pictures? - -REPLY [5 votes]: The Wikipedia page for L-systems links to the Thue-Morse sequence for which there are a large number of applications. In particular it can be interpreted as the sequence of distances between atoms in a 1D quasicrystal. Hence I suppose you might think about the work on substitution tilings and higher dimensional quasicrystals as an application of similar ideas. -Thinking even more broadly, the recent work by Borrelli, Jabrane, Lazarus and Thibert on implementing convex integration for a $C^1$ embedding of a flat torus in $R^3$ uses explicitly an L-system-like rewriting process to add corrugations.<|endoftext|> -TITLE: What is the probability two random maps on n symbols commute? -QUESTION [28 upvotes]: It is well known that two randomly chosen permutations of $n$ symbols commute with probability $p_n/n!$ where $p_n$ is the number of partitions of $n$. This is a special case of the fact that in a group, the probability that two elements chosen uniformly at random (with repetition allowed) is the number of conjugacy classes divided by the size of the group. - - -Question. What is the probability that two mappings of $n$ symbols chosen uniformly at random commute? - - -I suspect an exact answer would be difficult and would be happy to learn of reasonably tight asymptotic results. -Added. This probability should go to zero quickly because Misha Berlinkov recently showed that with probability going to 1 as $n$ goes to infinity, two random elements generate a subsemigroup containing a constant map and so if they commute they generate a unique constant map. This should happen almost never (and most likely has been proven). -Added based on Brendan McKay's answer. Computing the probability that an element of a monoid $M$ commutes with an element of its groups of units $G$ is no harder than the commuting probability in a group. Namely, $G$ acts on $M$ by conjugation; let's call the orbits conjugacy classes. Then the probability that an element of $G$ commutes with an element of $M$ is the number of conjugacy classes of $M$ divided by the number of elements of $M$. The proof is the same as for groups. If $Fix(g)=\{m\in M\mid gmg^{-1}=m\}$, then -$$\frac{|\{(g,m)\in G\times M\mid gm=mg\}|}{|G||M|} = \frac{1}{|M|}\frac{1}{|G|}\sum_{g\in G}|Fix(g)| = \frac{\text{number of conjugacy classes}}{|M|}$$ by the Cauchy-Burnside-Frobenius orbit counting formula. -For $M=T_n$ the monoid of all mappings on $n$ symbols and $G=S_n$ the symmetric group, conjugacy classes correspond to isomorphism classes of functional digraphs on $n$ vertices. A functional digraph is a digraph (loops allowed) in which each vertex has outdegree $1$. Each mapping $f$ gives a functional digraph by drawing an edge from $i$ to $f(i)$. It is obvious that $f,g$ are conjugate iff their corresponding digraphs are isomorphic (it is the same proof that permutations are determined up to conjugacy by cycle type). -According to the book of Flajolet and Sedgewick, the number of unlabelled functional digraphs grows likes $O(\rho^{−n}n^{−1/2})$ where $\rho\approx .29224$. So the probability of a random mapping commuting with a random permutation is pretty small. Brendan raises the nice question of how different the probability of a random permutation commuting with a random mapping is from the probability of a random mapping commuting with a random mapping. My guess is the latter goes to $0$ qualitatively faster. - -REPLY [3 votes]: The expected value in Brendan McKay's answer on the probability of $f(g(1))=g(f(1))$ is correct. Just count the quintuples $(a,b,c,f,g)$ where $g(1)=a$, $f(1)=b$, $f(a)=c$, $g(b)=c$. For instance, there are $n(n-1)^2$ triples $(a,b,c)$ with $a\ne1$, $b\ne1$, and for each such tripel there are $n^{n-2}$ possibilities for $f$ and $g$ each, contributing $n(n-1)^2n^{2n-4}$ to the possibilities. In the cases $a\ne1$, $b=1$ we must have $c=a$, so the contribution is $(n-1)n^{2n-3}$. The same for $a=1$, $b\ne1$. Finally, if $a=b=1$, then $c=1$, and that case contributes $n^{2n-2}$.<|endoftext|> -TITLE: Short lattice vectors orthogonal to a random vector -QUESTION [5 upvotes]: Let $N$ be some prime number. -Suppose I draw $s$ elements $g_1,..., g_s$, -where each $g_i\in [N]$ is taken uniformly from some interval $I_i$ of size, say -$\sqrt{N}$. -Is it possible to provide a lower-bound (which works on average, or w.h.p.) on the minimal length of a vector $h\in \mathbf{Z}^s$, for which $h \cdot g = 0 (mod N)$.? -Here I refer to the length of a vector as the magnitude of its largest coordinate. -At least intuitively, I would say the minimal length $h$ for a "typical" $g$ is $\Omega(\sqrt{N})$. -One can assume that $s$ is much smaller than $log^a(N)$, where $a<1$, so that there is negligible chance of two disjoint subsets of $g_i$'s having the exact same sum. - -REPLY [4 votes]: I left my old answer below. For very small $s$ I think that the right answer is about $M=N^{1/s}$ (for non-negative entries.) My reasoning is that one can pick all but the last entry of $h$ freely and then the last entry is forced and uniformly distributed in $[0,N-1].$ So if we run over the $k=M^{s-1}$ ways to make those choices keeping the entries under $M$, we expect the smallest possibility for that last entry to be of order $\frac{N}{M^{s-1}}$. -Here is a small experiment with $N=1009$ and $s=3$. Ten times I picked 3 random elements and then looked for the minimal vector $h$ (using non-negative entries) I expected about $p^{1/3} \approx 10$ to be enough most of the time. It got a bit higher than that but nowhere near $\sqrt{N} \gt 31.$ -[855, 752, 433], [8, 7, 7] -[872, 804, 715], [4, 0, 5] -[862, 647, 603], [7, 1, 9] -[764, 731, 897], [7, 14, 13] -[352, 811, 776], [12, 8, 7] -[285, 653, 876], [13, 11, 6] -[334, 502, 752], [7, 10, 9] -[840, 788, 333], [15, 2, 15] -[48, 476, 627], [6, 1, 2] -[526, 55, 580], [7, 6, 7] -Allowing $h$ to have entries in the range $(1-p)/2,(p-1)/2]$ should (and does in similar experiments) make the max about half as big. -OLDER If $h\in \mathbf{Z}^s$ then $h$ might be said to have length $s$. Obviously that is not what you mean, but what do you mean? The separation between the first and last non-zero entries? The sum of the squares of the entries? -For either of those there is a "short" solution when $s \gt 2\sqrt[4]{N}$. Then there are $\binom{s}{2} \ge 2\sqrt{N}+\sqrt[4]{N}$ sums $g_i+g_j$ all falling in an interval of length $2\sqrt{N}.$ Hence some two are equal and there is sure to be an appropriate vector $h$ with all entries $0$ except two $+1$ and two $-1$. -If I compute correctly, then, for $s = 2\sqrt[4]{N}$, there is about an $85\%$ chance that there is $i \ne j$ with $g_i=g_j$ which allows only two non-zero entries, each $\pm 1$. For larger $s$ this becomes highly likely. -The cases above have $h\cdot g=0$ in $\mathbf{Z}$ -If $s \gt \log_2{N}$ then there will have to be (disjoint) subsets with equal sum $\mod N$ and hence some appropriate vector $h$ with all entries $-1,0,1$. Probably we can keep $s$ much smaller and have such a solution with high probability. With $s \gt (1+\epsilon)\log_2{N}$ and $g_i$ chosen from $[0,N-1]$ one could even be sure to have an equal sum in $\mathbf{Z}.$ -Later Thanks for clarifying. My argument for magnitude $1$ when $s \gt \log_2{N}$ still applies. -I am note sure what happens if the vector $h$ needs entries from $\mathbf{N}.$ That seems more natural (npi) to me. -Perhaps you do want to just choose the $g_i$ from $[0,N-1]$, otherwise the choice of $I_i$ matters. -Perhaps you meant fixed $s$ although it seems likely to depend on $s$. for $s=1$ one has the uniform distribution in $[0,N-1]$ or $[0,\frac{N}{2}]$ (non-negative vs integer case).<|endoftext|> -TITLE: Combinatorial spin structures -QUESTION [24 upvotes]: I would like to know how to define spin structures combinatorially, for an oriented smooth manifold equipped with a triangulation. In the case of a 2d manifold, spin structures correspond to equivalence classes of Kasteleyn orientations (i.e. orientations of edges so that every face has an odd number of clockwise oriented edges). This fact is important in the theory of dimer models. -In general (for arbitrary dimensions) some sort of construction is proposed in http://arxiv.org/pdf/1306.4841.pdf but it is horribly complicated, I do not even see how it is related to Kasteleyn orientations in the 2d case. I would be quite happy to see a combinatorial description of spin structures for 3d and 4d triangulations. -Upd: there is an obvious generalization of the 2d definition to n dimensions. One could assign orientations to (n-1)-dimensional simplices so that for a given n-simplex an odd number of its faces have the orientation opposite to that of the n-simplex. Does this do the job, or is it wrong for some reason? - -REPLY [9 votes]: Here's the standard answer, but perhaps you are looking for something different? -A spin structure on a manifold $M$ is determined by a framing of the tangent bundle $TM$, restricted to the 1-skeleton. (i.e. the framing is only defined on the 1-skeleton.) This framing is required to be the bounding framing on the boundary of each 2-cell (i.e. the framing can be extended to the 2-skeleton). Two such 1-skeleton framings determine the same spin structure if and only if they are homotopic. Such homotopies are generated by local moves which change the framing at a 0-cell and at all of the 1-cells adjacent to that 0-cell. -Kirby and Taylor have a nice paper on low-dimensional Spin and Pin structures -- http://www3.nd.edu/~taylor/papers/PSKT.pdf - -[Added later] -(1) As Ryan points out in a comment, making the above idea explicit enough to be implemented on a computer is the point of his paper (which is linked to in the original question). -(2) If you are willing switch to the dual space -- functions on spin structures rather than spin structures themselves -- then there is the following relatively simple description which works in any dimension. Define a ribbon in $M$ be be a 1-dimensional submanifold $S \subset M$ equipped with a framing of $TM|_S$. Consider (finite linear combinations of) isotopy classes of ribbons in $M$ modulo the following three moves. - -If $S$ and $S'$ differ by a framed saddle move, then $S \sim -S'$. -If $S$ and $S'$ differ by adding a single "kink" or "twist" to the framing, then $S \sim -S'$. -If $S$ and $S'$ differ by adding/removing a small unknotted circle with standard (non-bounding) framing, then $S \sim -S'$. - -One can show that the above vector space is canonically isomorphic to the vector space of functions on (equivalence classes of) spin structures on $M$. -Note that if the three occurrences of $S \sim -S'$ above are changed to $S \sim +S'$, then the vector space is canonically isomorphic to finite linear combinations of elements of $H_1(M; \mathbb Z/2)$, which in turn is isomorphic to functions on $H^1(M; \mathbb Z/2)$. This is the counterpart to the fact that the set of spin structures is a torser for $H^1(M; \mathbb Z/2)$. -The above vector space naturally extends to a (fully extended) TQFT.<|endoftext|> -TITLE: Is the filtered colimit of sheaves of abelian groups a sheaf? -QUESTION [7 upvotes]: This might be embarrassingly simple, but I want to be 100% sure I am not missing any subtleties. Let $F_i$, $i\in I$ be a filtered inductive system of sheaves of abelian groups on some site. Take the presheaf colimit $F$. It seems to me that $F$ is a sheaf. Is this true? -It seems to be a consequence of the following: let $\{U_j\longrightarrow U\}_{j -\in J}$ be a covering. Then, for each $i\in I$, $F_i$ being a sheaf, we know that: -$$F_i(U)=lim(\prod_{j\in J}F_i(U_j)\overset{\longrightarrow}{\underset{\longrightarrow}{ }} \prod_{j,j'\in J}F_i(U_j\times_UU_{j'})$$ -Now, because filtered colimits commute with finite limits for abelian groups, -it follows that $F$ is a sheaf. Did I miss something? - -REPLY [6 votes]: Your proof works if the covering is finite. Hence, the claim is true for noetherian sites. The corresponding statement for sheaf cohomology is also true, see Stacks Project, Tag 0739. But in general filtered colimits (of abelian groups, doesn't really matter) don't commute with infinite products. For example, the canonical map $\mathrm{colim}_n ~\prod_{i \in I} \frac{1}{n} \mathbb{Z} \to \prod_{i \in I} \mathrm{colim}_n \frac{1}{n} \mathbb{Z} = \prod_{i \in I} \mathbb{Q}$ is not surjective when $I$ is infinite, since the denominators of a rational sequence don't have to be bounded. As soon as you have an example for $\mathrm{colim}_n \prod_{i \in I} A_{n,i} \not\cong \prod_{i \in I} \mathrm{colim}_n A_{n,i}$, this is is witnessed by sheaves: Take $X = \sqcup_{i \in I} X_i$ with connected $X_i$ (or just points) and consider the sheaves $\underline{A_{n,i}}$ on $X_i$ which glue to a sheaf $A_n$ on $X$. Then the presheaf colimit $\mathrm{colim}_n A_n$ is not a sheaf.<|endoftext|> -TITLE: Two pullback diagram -QUESTION [11 upvotes]: Suppose whole square and the left square in the diagram below are pullbacks, then we may wonder whether the right square is a pullback. It is usually not the case. - -Now we seek some addition condition on $X\to Y$ that forces the right square is a pullback too. -My question: is epic a sufficient condition? (If the category is Sets, then yes.) -Added: Let $P$ be the pullback of the right square, then there exists $B\to P$, and the square $A\to P \to Y$ // $A\to X \to Y$ is a pullback, so we have the following diagram in which the bottom and the whole squares are pullback, so is the upper square. If the category is Sets, $X\to Y$ is surjective then $A\to P $ is also surjective. Since the pullback of $B\to P$ along a surjective map is an bijection, $B\to P$ must be a bijection. This shows the right square of the original diagram is a pullback. We can also see why we consider some nice condition on $X\to Y$. - -REPLY [11 votes]: I tried to write the explanation of the comment about a dozen of times, but was never satisfied with the result. Finally, I decided to write a full note (I will try to put it on arXiv in a few minutes; here it is) describing the natural setting for such questions: -http://www.mimuw.edu.pl/~mrp/the_other_pullback_lemma.pdf -(the note still needs some improvements, but I am running out of time now...) -I found it easier to characterise your condition by "extremal epimorphisms" rather than "strong epimorphisms" (notice however, that in case of finite connected limits, these concepts coincide). Here is the formal statement: -Let us assume that finite connected limits exist. The following are equivalent: - -your condition along $e \colon X \rightarrow Y$ holds -$e \colon X \rightarrow Y$ is an extremal morphism stable under pullbacks.<|endoftext|> -TITLE: Do taking a general hyperplane section and taking a colon ideal commute? -QUESTION [7 upvotes]: Let $I$ be an ideal and $f$ be an element of $R = \mathbb{C}[x_1,\ldots,x_n]$, where $\mathbb{C}$ is an algebraically closed field of characteristic $0$. Does -$$ -(I+H):f = (I:f)+H -$$ -hold for a general hyperplane in $\operatorname{Spec}(R)$, i.e., for $H = \langle \text{general linear function} \rangle$? - -REPLY [5 votes]: This is true for a general affine-linear form. First we assume $(I,f)\neq R$. The point is that $(I,f)$ has finitely many associated primes, so a general $H$ would be a non-zerodivisor on $R/(I,f)$. -For any ideal $I$ and any element $f$, there is an exact sequence: -$$0\to R/(I:f) \to R/I \to R/(I,f) \to 0$$ -(The first map is simply multiplication by $f$. Now tensor the exact sequence with $R/(H)$, keep in mind that $H$ is nzd on $R/(I,f)$, so $Tor_1^R(R/(I,f), R/H)=0$, we get an exact sequence: -$$0\to R/(I:f, H) \to R/(I,H) \to R/(I,f,H) \to 0$$ -On the other hand, the first exact sequence with $I$ replaced by $(I,H)$ says that the leftmost term should be isomorphic to $R/(I,H):f$, so we are done. -Note that the exact sequence also takes care of the case $(I,f)=R$, since it implies in that case that $(I:f)=I$. Since $(I+H):f=(I+H)$ then, the equality is clear.<|endoftext|> -TITLE: Perfect matchings between levels in products of posets -QUESTION [8 upvotes]: Let $P$ be a finite poset. Assign the following diagram to it - put the maximal element of $P$ on the first level, the maximal of the rest to the second level, etc. Assume that the diagram is connected, that is, that between each pair of elements in the diagram we have a chain connecting them. -Let $L_n=P\times P\times\cdots\times P$ ($n$-times) to be the product poset, that is, one element in $L_n$ is larger than another if we have all inequalities coordinate-wise. Is it true that there exists and $N=N(P)$ such that for $n>N$ the following is true: -In the diagram of L_n there is a perfect matching between all consecutive levels (from the one with smaller size to the one with larger size). -Note: due to the useful feedback of Andres Caicedo and David Speyer the question was altered. - -REPLY [6 votes]: The statement is still false when $P$ is connected and graded. It is even false if we add in addition that $P$ has a unique minimal and maximal element. -Theorem For any graded finite poset $P$, let $\rho(P)$ be the average of the ranks of the elements of $P$. Suppose that $Q$ is an upper ideal of $P$ and $\rho(Q) < \rho(P)$. (And, equivalently, $\rho(P) < \rho(P \setminus Q)$.) For all sufficiently large $n$, there is no perfect matching between levels of $P^n$ and $Q^n$. -Proof Let $P$ be graded with $p_i$ elements in grade $i$. Let $p(t) = \sum p_i t^i$. So $P^n$ is graded with corresponding polynomial $P(t)^n$. I'll write $P(t)^n = \sum p^n_i t^i$. Define $Q(t)$ similarly and $q^n_i$. Suppose that $p^n_i \leq p^n_{i+1}$ but $q^n_i > q^n_{i+1}$. Then there could be no matching between levels $i$ and $i+1$ of $P^n$. The first inequality would force the matching to be upwards (level $i$ injecting into level $i+1$), but the second inequality would imply that there were not enough elements to match level $i$ of $Q^n$ to. -Now, by a result of Odlyzko and Richmond, for sufficiently large $n$, the coefficients of $P(t)^n$ will be unimodal. Also, by the central limit theorem, the coefficient of $t^k$ in $P(t)^n$ will be much larger when $k$ is near $\rho(P) n$ then anywhere else. Combining these results, the coefficients of $P(t)$ increase up to $ (\rho(P) + o(1))n$ and then decrease down the other side. -Similarly, the coefficients of $Q(t)^n$ increase up to $(\rho(Q)+o(1)) n$ and -then decrease. Since $\rho(Q) < \rho(P)$, this shows that the largest coefficient of $Q(t)^n$ occurs before the largest coefficient of $P(t)^n$ for $n$ large. Thus, for $n$ large, there is no matching at the levels between $(\rho(Q)+o(1)) n$ and $(\rho(P)+o(1)) n$. $\square$ -Here is are explicit counterexamples. The poset $P$ has $6$ elements, denoted by the symbols $Q$ and $p$; the poset $Q$ is the $3$ elements labeled $Q$: - -Then $P(t) = 3+3t$ and $P(t)^n$ has its largest coefficient at $t^{n/2}$, while $Q(t) = 2+t$ and has its largest coefficient at $t^{n/3}$. So, choosing $i$ between $n/3$ and $n/2$, there is no matching between level $i$ and level $i+1$. -Here is a similar example where $P$ has a maximal and minimal element. - -We have $\rho(P) = 3/2=1.5$ and $\rho(Q) = (5/7) 1 + (1/7) 2 + (1/7) 3 = 10/7 \approx 1.4$ as desired.<|endoftext|> -TITLE: f-vectors of Pure Complexes and Eulerian Complexes -QUESTION [5 upvotes]: Let $\Delta$ be a simplical complex. -Call $\Delta$ pure if all the maximal faces have the same dimension. -Call $\Delta$ Eulerian if it is pure and $\chi (lk (F))= \chi (S^{dim (lk(F))})$ for any $F \in \Delta$. -Question 1: From what I understand characterizing the $f$-vectors of pure simplicial complexes is hopeless at the moment. -What if however I give you an $f$ (or $h$)-vector? What are some necessary conditions that I can check for it to be the $f$-vector of a pure simplicial complex. -Say I give you something like this: $(1,19,99,276, 504, 630, 546, 324, 126, 28)$. One can check the Kruskal-Katona bounds (which hold in this case) and even produce the shifted complex, however this is not generally pure even if the $f$-vector is pure. Are there any similar constructions of these type? -Questions 2: What can one say about the $h$-vectors of Eulerian complexes other that they are symmetric? -Thank you! - -REPLY [3 votes]: As Gil remarks in his comment, Corollary 1 of the paper which I mentioned does not in fact imply the upper bound conjecture except when one additionally assumes isolated singularities. Still, I hope that this paper and its references may be of some use to the OP. - -At least partial answers to both your questions may be found in the following paper: - -Patricia Hersh and Isabella Novik, A short simplicial $h$-vector and the upper bound theorem, Discrete & Computational Geometry, 28 (3): 283-289 (2002). - -In particular, the main theorem implies (via Corollary 1) that if the dimension $d$ is odd, then the first $d-1$ entries of your $f$-vector will be bounded above by those of the cyclic polytope $C(n,d)$. For additional details, look for the "upper bound conjecture" as well as the references in this paper. -I don't think much is known when $d$ is even.<|endoftext|> -TITLE: Where does the notation $\pi_1(X,x)$ for the fundamental group first appear? -QUESTION [7 upvotes]: I've spent the last half hour browsing Stillwell's translation of Poincaré's Analysis Situs and Dieudonné's History of Algebraic and Differential Topology, and I haven't found the source of this notation. The "1" in the subscript would lead one to believe this notation was introduced at the same time as $\pi_n(X,x)$ for the higher homotopy groups as well. -In my search, I have learned that the name "Fundamental group" comes straight from Poincaré, and from the "Fundamental region" of a group action. (And, incidentally, that "torsion" in algebra comes from the homology of non-orientable surfaces, which are twisted in on themselves. A bit of a tangent, but I'm still marveling.) - -REPLY [11 votes]: Notice that in Poincare's setting there is no base point $x$, since he regards the fundamental group as a group of permutations of fundamental regions -- though he also knew the path interpretation, of course, as it arose in integration on Riemann surfaces. -The first occurrence of the notation $\pi_n(X)$ for the $n$-th homotopy group of space $X$ is Hurewicz's paper -Beiträge zur Topologie der Deformationen: Höherdimensionale Homotopiegruppen, Proceedings of Koninklijke Akademie van Wetenschappen te Amsterdam vol. 38 (1935) pp. 112-119. -which is translated in the book Collected Works of Witold Hurewicz. Selected pages of that, including this notation, are on Goggle books. Search "homotopy group." -If you want the base point $x$ as well, then I have failed. It seems not to appear anywhere in Math Reviews before 1945 but I did not find when it did occur.<|endoftext|> -TITLE: The fundamental group of a complex, quasi-affine variety -QUESTION [6 upvotes]: Can the fundamental group of a quasi-affine variety over $\mathbb{C}$ be a torsion group? - -REPLY [6 votes]: A less interesting but personally checkable answer: consider $f : ({\mathbb C}^2 \setminus 0)/\{\pm 1\} \hookrightarrow {\mathbb C}^3$, $(x,y) \mapsto (x^2,xy,y^2)$. The image is the punctured quadric cone and obviously has fundamental group $\{\pm 1\}$.<|endoftext|> -TITLE: Intuition behind the ricci flow -QUESTION [37 upvotes]: I hope you don't shoot me for this question. -I try to understand among other things the Ricci flow. However I have no idea of the intuition behind the definition. So my questions is: - -What is the intuition behind the Ricci flow? - -I would appreciate any illustrative example. - -REPLY [25 votes]: I've combined two answers into one (out of order). -Second answer: some historical intuition. This is only a partial answer. Assume that you have stumbled upon the equation -$\frac{\partial}{\partial t}g=-2\operatorname{Ric}$ and that you are -interested in whether you can use it to deform metrics to better metrics on -closed manifolds. -PDE intuition. The first question is that of short time existence given a -$C^{\infty}$ initial metric $g_{0}$. So one linearizes the operator -$g\mapsto-2\operatorname{Ric}_{g}$ and computes its symbol and finds that it -is weakly elliptic. In fact $\operatorname{Ric}_{\varphi^{\ast}g} -=\varphi^{\ast}\operatorname{Ric}_{g}$, accounts for the kernel of the symbol. -By breaking the diffeomorphism invariance of the Ricci flow in the right way, -DeTurck simplified Hamilton's proof of short time existence by obtaining an -equivalent equation for the metric which linearizes to a heat-type equation. -To see if the metric gets better, one computes the evolutions of geometric -quantities associated to $g(t)$. If $Q=Q[g]$ is such a quantity, assuming the -variation $\frac{\partial}{\partial t}g=-2\operatorname{Ric}$ of the metric, -one computes the corresponding variation $\frac{\partial Q}{\partial t}$. One -immediately sees heat-type equations everywhere. For example, the scalar -curvature evolves by $\frac{\partial R}{\partial t}=\Delta -R+2|\operatorname{Ric}|^{2}$. Since the global method of the maximum principle -relies on local calculations, it applies to closed manifolds. So $R_{\min -}(t)=\min_{x\in M}R(x,t)$ is nondecreasing. One finds other examples of Ricci -flow preferring positive curvature over negative curvature. Basically, any -polynomial of the curvature and its covariant derivatives, whether it be a -function or more generally a tensor, satisfies a heat-type equation. E.g., -derivative of curvature estimates follow from the maximum principle. -Having obtained some control of the metric as it evolves, one then aims to -prove convergence. In dimension two, this is always possible after rescaling -to normalize the volume to be constant. Generally, an Einstein metric shrinks, -is stationary, or expands according to whether $R$ is positive, zero, or -negative, respectively. -Quantities satisfying heat-type equations. The full curvature tensor -$\operatorname{Rm}$ satisfies an equation of the form $\frac{\partial -}{\partial t}\operatorname{Rm}=\Delta\operatorname{Rm}+q(\operatorname{Rm})$, -where $q$ is a quadratic polynomial. Since $\operatorname{Rm}$ is a symmetric -bilinear form on the vector space $\wedge^{2}T_{x}^{\ast}M$ at each point $x$, -we have the notion of nonnegativity of $\operatorname{Rm}$. Since -$q(\operatorname{Rm})$ satisfies a property sufficient for the maximum -principle for systems to be applied, $\operatorname{Rm}\geq0$ is preserved -under the Ricci flow. Generally, we can analyze the behavior of -$\operatorname{Rm}$ by the maximum principle under various hypotheses. -Geometric application. In particular, when $n=3$ and $\operatorname{Ric} -_{g_{0}}>0$, we have $\pi_{1}(M)=0$ and hence the universal cover $\tilde{M}$ -is a homotopy $3$-sphere. Encouraged by this, Hamilton proved that the -solution to the normalized Ricci flow exists for all time and converges to a -constant positive sectional curvature metric; thus $M$ is diffeomorphic to a -spherical space form. The main gonzo estimate is $\frac{|\operatorname{Ric}% --\frac{R}{3}g|^{2}}{R^{2}}\leq CR^{-\delta}$ for some $C$ and $\delta>0$. -Intuitively, we expect $R\rightarrow\infty$ and hence $\operatorname{Ric} --\frac{R}{3}g\rightarrow0$. -Singularities. Schoen and Yau proved that if an orientable $M^{3}$ admits a -metric with $R>0$, then it is a connected sum of quotients of homotopy -$3$-spheres and $S^{2}\times S^{1}$'s.\ Yau proposed to Hamilton that in this -case Ricci flow should be able to produce surgeries to obtain a connected sum -of spherical space forms and $S^{2}\times S^{1}$'s. One first sees, that by -the strong maximum principle, the universal cover of singularities of the -Ricci flow often split as products of $\mathbb{R}$ with a solution on a -surface. This is a motivation to study the Ricci flow on surfaces, to rule out -the formation of the cigar soliton. -Inspired by his corresponding results for the curve shortening flow and the -Ricci flow on surfaces, Hamilton proved that the Li-Yau differential Harnack -method extends to the Ricci flow assuming $\operatorname{Rm}\geq0$. Since -$3$-dimensional singularity models have $\operatorname{Rm}\geq0$, Hamilton was -able to classify certain singularities as steady gradient Ricci solitons and -cylinders. Provided the Little Loop Lemma is true, or more aptly, no local -collapsing is true, for finite time singular solutions one obtains round -cylinder $S^{2}\times\mathbb{R}$ limits unless one is one a spherical space -form. At this point, one can begin to believe that Ricci flow does indeed -perform the surgeries that Yau proposed. - -First answer: Ricci flow as a heat-type equation. Remark about Hamilton's statement: "The Ricci flow is the heat equation for metrics". -(The following is a well known calculation.) The Ricci tensor is given in local coordinates by -\begin{align*} --2R_{jk} & =-2\left( \partial_{q}\Gamma_{jk}^{q}-\partial_{j}\Gamma_{qk}% -^{q}+\Gamma_{jk}^{p}\Gamma_{qp}^{q}-\Gamma_{qk}^{p}\Gamma_{jp}^{q}\right) \\ -& =-g^{qr}\partial_{q}\left( \partial_{j}g_{kr}+\partial_{k}g_{jr}% --\partial_{r}g_{jk}\right) +g^{qr}\partial_{j}\left( \partial_{q}% -g_{kr}+\partial_{k}g_{qr}-\partial_{r}g_{qk}\right) \\ -& \quad\;+\left( g^{-1}\right) ^{\ast2}\ast\left( \partial g\right) -^{\ast2}\\ -& =\Delta\left( g_{jk}\right) -g^{qr}\left( \partial_{q}\partial_{j}% -g_{kr}+\partial_{q}\partial_{k}g_{jr}-\partial_{j}\partial_{k}g_{qr}\right) -+\left( g^{-1}\right) ^{\ast2}\ast\left( \partial g\right) ^{\ast2}\\ -& =\Delta\left( g_{jk}\right) -g_{k\ell}\partial_{j}\left( g^{qr}% -\Gamma_{qr}^{\ell}\right) -g_{j\ell}\partial_{k}\left( g^{qr}\Gamma -_{qr}^{\ell}\right) +\left( g^{-1}\right) ^{\ast2}\ast\left( \partial -g\right) ^{\ast2}. -\end{align*} -In harmonic coordinates $\{x^{i}\},$ $0=g^{ij}\Gamma_{ij}^{k},$ so then -$-2R_{jk}=\Delta\left( g_{jk}\right) +Q\left( g^{-1},\partial g\right) ,$ -where $Q$ is quadratic in both arguments. (From line to line, various terms are absorbed in the lower order quadratic term.) -In normal coordinates $\{x^{i}\}$ centered at $p$ we have $g_{ij}\left( -x\right) =\delta_{ij}-\frac{1}{3}R_{i\ell mj}\left( p\right) x^{\ell}% -x^{m}+O\left( r^{3}\right) $, where $r=d\left( x,p\right) =(\sum_{i}% -(x^{i})^{2})^{1/2}$. Then $\Delta\left( g_{ij}\right) \left( p\right) -=-\frac{2}{3}R_{ij}\left( p\right) $. Note that $\partial_{i}g_{jk}\left( -p\right) =0$. -Hamilton likes to joke that when he first wrote down the Ricci flow equation, he wrote: $\frac{\partial}{\partial t}g_{ij} = 2 R_{ij}$, having a preference for positivity over negativity. -December 13, 2013. Answer to Qfwfq's question. $\partial g$ denotes a -nonspecific factor of the form $\partial_{i}g_{jk}$. $\ast$ denotes a product, -possibly together with contractions (summing over a pair of repeated indices, -one upper and one lower). For example, -\begin{align*} -2\partial_{i}\Gamma_{jk}^{\ell} & =\partial_{i}(g^{\ell m}(\partial_{j} -g_{km}+\partial_{k}g_{jm}-\partial_{m}g_{jk}))\\ -& =g^{\ell m}(\partial_{i}\partial_{j}g_{km}+\partial_{i}\partial_{k} -g_{jm}-\partial_{i}\partial_{m}g_{jk})\\ -& \quad-g^{\ell p}g^{qm}\partial_{i}g_{pq}(\partial_{j}g_{km}+\partial -_{k}g_{jm}-\partial_{m}g_{jk})\\ -& =g^{\ell m}(\partial_{i}\partial_{j}g_{km}+\partial_{i}\partial_{k} -g_{jm}-\partial_{i}\partial_{m}g_{jk})+(g^{-1})^{\ast2}\ast(\partial -g)^{\ast2}. -\end{align*}<|endoftext|> -TITLE: A question on residues mod an even integer -QUESTION [5 upvotes]: I posted the question here, but it seems to be more difficult than I expected. So I think it may be suited for MO. (Another reason is that the answer may hopefully give solution to the question on this site.) -Let $k$ and $l$ be integers greater than $4$. My question is to determine the set $S$ of $k$ elements in $\mathbb{Z}/2l\mathbb{Z}$ satisfying the following three conditions: -(1) if $a\in S$, then $a+l\not\in S$; -(2) for any $a\in S$, there exist $b\neq c\in S$ such that $2a=b+c$; -(3) for any $b\neq c\in S$ such that $b+c$ is even (this makes sense since $2l$ is even), there exists $a\in S$ such that $2a=b+c$. -An observation is that whenever $S$ is a set satisfying the above conditions, $S+m$ is also such a set for all $m\in\mathbb{Z}/2l\mathbb{Z}$. -It is easy to verify that when $k$ is odd and $l=kn$, the set $\{t,t+2n,t+4n,\dots,t+2(k-1)n\}$ satisfies the three conditions for all $t\in\mathbb{Z}/2l\mathbb{Z}$. But I'm not aware of any other examples yet. (If these are the only satisfactory sets, then I am able to give an affirmative answer to this question.) - -REPLY [3 votes]: I think there is some structure to your problem, but I haven't fully thought this through. For example, if $\ell$ is prime then I think $k$ must be $1$ or $\ell$. Here are some thoughts on this case. Consider the odd elements $S_{o}$ in $S$ and the even elements $S_{e}$ in $S$. Your conditions impose the restrictions that the sumsets $S_o+S_o$ and $S_e+S_e$ are both at most as big as $S$ in size. If one views these sets just in ${\Bbb Z}/\ell {\Bbb Z}$ (relations are preserved in just viewing the congruences $\pmod \ell$) then this is very close to the extremal case of the Cauchy-Davenport theorem in set addition (since one of $S_o$ or $S_e$ must contain at least half the elements in $S$). In this situation the extremal cases are well understood -- this is known as Vosper's theorem and generalizations, and you would then be able to get the structure of $S_o$ and $S_e$. Essentially they look like progressions with maybe one term omitted. (Note that Zieve's examples look like this.) So you should be able to push this through to obtain a theorem at least for $\ell$ prime. For composite $\ell$ things are more complicated since there are more subgroups, and I don't think precise extensions of Vosper's theorem are known. There is an extensive literature on such inverse problems in additive number theory, and recent work by Serra, Zemor, Hamidoune, Rodseth and many others on this topic. Let me give a pointer to http://arxiv.org/pdf/math/0507561.pdf where you will find other references; also Serra's homepage will contain other papers. -Update: As the OP pointed out my guess that when $\ell$ is prime $k$ must be $1$ or $\ell$ is not correct. What does follow from the inverse theorems referred above I think is that when $\ell$ is prime and $k<\ell$ then $S$ decomposes as the union of $S_o$ and $S_e$ and that each of these sets forms an arithmetic progression missing possibly one term. -More precise version: Here's a more precise version of my comments above. I will assume below that $\ell$ is an odd prime. First we reformulate the problem. Let $S$ be a subset of cardinality $k$ in ${\Bbb Z}/2\ell {\Bbb Z}$ such that no two elements in $S$ are congruent $\pmod \ell$. Let $S_o$ and $S_e$ denote the odd and even elements in $S$. Denote by $2\times S$ the set $\{ 2s: s\in S\}$, so that $2\times S$ also has cardinality $k$. The problem asks for the structure of $S$ given that $(S_o+S_o)\cup (S_e+S_e) = 2\times S$. -If $k =\ell$ there are many such sets and there is no structure. To see this choose a random set of $(\ell+1)/2$ residue classes $\pmod \ell$ and lift these to a set $S_o$ of odd residue classes $\pmod {2\ell}$. Take the complementary set of $(\ell-1)/2$ residue classes $\pmod \ell$ and lift these to the set $S_e$ of even residue classes $\pmod {2\ell}$. With high probability $S_o+S_o =S_e+S_e = \{0, 2, 4, \ldots, 2\ell -2\}$. -If $k=\ell -1$ there is a similar construction. From every pair $i$, $\ell-i$ -($1\le i\le (\ell-1)/2$) choose a random number, and lift these $(\ell-1)/2$ numbers to a set $S_o$. Take the complementary choice and lift that to a set $S_e$. Then with high probability $S_o+S_o = S_e+S_e =\{2,4,6,\ldots, 2\ell -2\}$. -If $k$ is not too close to $\ell$ then one gets structure. This follows from the Hamidoune-Rodseth version of Vosper's theorem that I mentioned above. Precisely suppose that $8\le k\le \ell -4$. Suppose that $S_o$ contains at least $k/2$ elements (similar argument if $S_e$ is the bigger set). From the Cauchy-Davenport theorem one gets that in fact $S_o$ must be of size $\lceil k/2\rceil$ (and so $S_e$ contains $\lfloor k/2\rfloor$ elements). Then from the Hamidoune-Rodseth theorem we find that $S_o$ must be an arithmetic progression missing possibly one term. If $k$ is even, the same conclusion applies to $S_e$. If $k$ is odd (so that $S_o$ has $(k+1)/2$ elements and $S_e$ has $(k-1)/2$ elements) then one needs the stronger result of Serra, Zemor and Hamidoune, and here $S_e$ can omit two elements from an arithmetic progression (if $\ell$ is at least $53$).<|endoftext|> -TITLE: Does the Hasse principle hold for the square root problem on symmetric matrices? -QUESTION [12 upvotes]: Let $S$ be an $n \times n$ symmetric matrix with rational entries. -It is known that the equation $XX^T=S$ has a solution in $\text{M}_n(\mathbb{Q})$ -if and only if it has a solution in $\text{M}_n(\mathbb{Q}_p)$ for all $p \in \mathcal{P} \cup \{\infty\}$ (indeed, one easily reduces the problem to the case when $S$ is non-singular, and then the problem amounts to characterize the situation when $S$ is rationally congruent to $I_n$). -Does a similar Hasse principle hold for the equation $X^2=S$ with unknown $X \in \text{S}_n(\mathbb{Q})$? - -REPLY [5 votes]: The answer is yes, by a series of reductions. -Given a polynomial $f(x)$, there is a unique polynomial $g(x)$ of the same degree whose roots are the squares of the roots of $f$. We can compute its coefficients with the equation $g(x^2)=f(x)f(-x)$. Suppose $g(x)$ is the characteristic polynomial of $X$. we can easily construct a diagonalizable matrix $S'$ with characteristic polynomial $f(x)$. Then $S'^2$ will have characteristic polynomial $g(x)$. Since $S'^2$ and $X$ will both be diagonalizable, they will be conjugate over $\mathbb C$, hence, by rational canonical form, they will be conjugate over $\mathbb Q$. (A symmetric matrix over $\mathbb Q$ is diagonalizable over $\mathbb R$.) Say $A S'^2 A^{-1}=X$, then $(AS'A^{-1})^2=X$, so set $S = A S'A^{-1}$, then $S^2 = (AS'A^{-1})^2=X$. -Conversely, if such an $S$ exists, then its characteristic polynomial is such an $f(x)$. So having a solution to the equation is equivalent to having a polynomial $f(x)$ such that $g(x)$ is the characteristic polynomial. So your question is equivalent to: Does the Hasse principle hold for the problem of, given a polynomial $g(x)$, finding a polynomial $f(x)$ such that $f(x)f(-x)=g(x^2)$? -Given a polynomial $g(x)$, we can factor it into irreducibles. $f(x)$ must handle each irreducible separately, unless some irreducible appears with even multiplicity, since $g(x^2) g( (-x)^2) = \left(g(x^2)\right)^2$. So it is sufficient to check the Hasse principle for an irreducible $g(x)$. -For an irreducible $g(x)$, this equation has a solution over $\mathbb Q$ if and only if the element $x$ in the number field $\mathbb Q(x)/g(x)$ is a perfect square. Otherwise, the extension that comes from adjoining its square root will have a degree too large, and there will be no irreducible polynomial defining its square root. This equation has a solution over $\mathbb Q_p$ if and only if the element $x$ in the product of local fields $\mathbb Q_p(x)/g(x)$ is a perfect square. This is because, in $\mathbb Q_p$, $g(x)$ factors as a product of distinct terms, hence each term must individually have a square root. -So the equation has a solution over each place of $\mathbb Q$ if and only if the equation $y^2=x$ has a solution over each place of $\mathbb Q(x)/g(x)$. By the Chebotarev Density Theorem, this final equation satisfies the Hasse principle.<|endoftext|> -TITLE: Does Dehn filling always decrease Gromov norm? -QUESTION [6 upvotes]: As an immediate consequence of Proposition 6.5.2 of Thurston's notes, we have that, if $M$ is a compact 3-manifold with toric boundary and $\tilde M$ is obtained from $M$ via Dehn filling, then for the 'variant' $\|\cdot\|_0$ of the Gromov norm it holds that $\|[\tilde M,\partial\tilde M]\|_0\leq \|[M,\partial M]\|_0$. Under which assumptions is it also true that $\|[\tilde M,\partial\tilde M]\|\leq \|[M,\partial M]\|$, $\|\cdot\|$, being the 'proper' Gromov norm? This holds, for instance, if $M$ is hyperbolic (Lemma 6.5.4), but what if we drop this hypothesis? - -REPLY [4 votes]: As mentioned in another answer, one may deduce the desired result from the following facts: - -(Gromov's Equivalence Theorem): If every component of the boundary of $M$ has amenable fundamental group, then for every $\varepsilon>0$ there exists a fundamental cycle $z$ for $M$ such that $\|z\|\leq \|M,\partial \|+\varepsilon$ and $\|\partial z\|\leq \varepsilon$. -(Matsumoto-Morita): If $T$ is a boundary component of $M$ and the fundamental group of $T$ is amenable, then any two fundamental cycles $z,z'$ for $T$ differ by a boundary $c$ such that $\|c\|\leq K\cdot (\|z\|+\|z'\|)$. - -Putting together (1) and (2) one easily gets that the simplicial volume is subadditive with respect to gluings along boundaty components having amenable fundamental groups. -Gromov's proof of the Equivalence Theorem is based on the quite technical theory of multicomplexes, which is studied and developed in the preprint by Kuessner cited above. A different proof (using Ivanov's and Monod's approach to bounded cohomology) is given in our recent preprint -http://arxiv.org/abs/1305.2612 -(joint work with Bucher, Burger, Iozzi, Pagliantini, Pozzetti).<|endoftext|> -TITLE: Almost linearly dependent functions -QUESTION [18 upvotes]: Everyone knows that analytic functions $f_1,\ldots,f_n$ are linearly dependent if and only if their Wronski determinant is identically equal to zero. There are several proofs of this, -one in Polya-Szego, vol. 2 Ch. 7, probl. 60. -Is this result stable? More precisely, is the following true: - -For every $\epsilon>0$ there exists $\delta>0$ with this property: if analytic - functions $f_1,\ldots,f_n$ in the unit disc satisfy - $|W(f_1,\ldots,f_n)(z)|<\delta \| f\|^n$ for $|z|<1$, then there exists a unit vector - $a=(a_1,\ldots,a_n)$ such that $|a_1f_1+\ldots+a_nf_n|<\epsilon\| f\|$ for $|z|<1/2$. - Here $\| f\|=\sqrt{|f_1|^2+\ldots+|f_n|^2}$. - -If this is true, I would like to know something about $\delta$ as a function of -$\epsilon$. This is easy to prove when $n=2$ with $\delta\approx\epsilon$. It is also -easy to prove for formal power series that if the Wronskian is small in "p-adic norm", -then there is a linear combination that is small in "p-adic norm". I can also prove the converse statement for every $n$: if the functions are almost linearly dependent, then their Wronskian is small. -EDIT. I would be glad to see any other statement meaning that "if the Wronskian is small, -then there is a linear combination which is small". - -REPLY [2 votes]: A simple remark. Let $\mathbb H$ be an Euclidean space with dimension $N$ and let $u_1,\dots, u_{N-1}$ be an orthonormal set of vectors. Let $x$ be a vector in $\mathbb H$: it belongs to the hyperplane $V$ generated by $u_1,\dots, u_{N-1}$ if and only if -$$ -u_1\wedge\dots\wedge u_{N-1}\wedge x=0,\quad\text{i.e. }\det(u_1,\dots,u_{N-1}, x)=0 -$$ -Now, for $x\in \mathbb H$,$\vert\det(u_1,\dots,u_{N-1}, x)\vert$ is the Euclidean distance from $x$ to $V$: the smallness of that determinant is simply the smallness of the distance to the hyperplane.<|endoftext|> -TITLE: Does the strong law of Large Number hold for an infinite dimensional Brownian motion? -QUESTION [10 upvotes]: For finite-dimensional Brownian motion $W_t$, it is well known that -\begin{equation} -\lim_{t\to \infty}\frac{W_t}{t}=0,\text{ a.s. }\ \ \ \ \hspace{1cm} \langle 1\rangle -\end{equation} -Now suppose we are given an $L^2(\mathcal{D})$-valued Brownian motion $W_t$ defined by -$$W_t:=\sum_{k=1}^{\infty}\sqrt{\sigma_k}W_t^k\phi_k(x),$$ -where $\mathcal{D}$ is bounded domain in $\mathbb{R}^d$, $\{\phi_k(x)\}$ forms the complete orthogonal basis of $L^2(\mathcal{D})$, $\{W_t^k\}_{k\in \mathbb{N}^{+}}$ are mutually independent one-dimensional standard BM, and $\sigma_k$ satisfies -$$\sum_{k=1}^{\infty}\sigma_k<\infty.$$ -I wonder does $\langle1\rangle$ still holds for the infinite-dimensional BM introduced above? - -REPLY [8 votes]: One can see it much easier by using the subadditive ergodic theorem, which is a standard tool for dealing with the law of large numbers. -Since the second moment of $W_t$ is obviously finite (this is precisely summability of $\sigma_k$), its first moment is also finite, so that the subadditive ergodic theorem implies that there exists a constant $C$ such that $\|W_t\|/t \to C$ a.e. and in $L^1$. In particular, $\|W_t\|^2/t^2\to C^2$ in probability. On the other hand, $\mathbb E\|W_t\|^2=t\,\mathbb E\|W_1\|^2$ grows linearly, so that $C=0$.<|endoftext|> -TITLE: Does every linear group admit a subgroup of dimension 1? -QUESTION [6 upvotes]: Suppose that $G$ is a linear group of positive dimension, defined over some field $k$. Is that true, that $G$ admits a (closed) one-dimensional subgroup? -I'm pretty much sure this is true in characteristic 0, or at least for $k=\mathbb{C}$. It seems that the main obstacle in positive characteristic is that there may not exist elements of infinite order. -EDIT1: As @Daniel Loughran pointed out in the answer below, one need at least to assume that $k$ is algebraically closed -(or it may not even be necessary, see the comments of @Marguax abaut $k$ being separably closed, or real closed in characteristic 0). -EDIT2: Since my first statement of the problem seems to be genereting a lot of confusion (after getting a comment from @Jim Humphreys I'm even a little ashamed), here is the final version of the question: -Suppose that $G$ is a linear algebraic group, of positive dimension, defined over an algebraically closed field $k$ (but of arbitrary characteristic). Is it true, that $G$ posses a one dimensional (closed) subgroup? (so in fact, either $\mathbb{G}_a$ or $\mathbb{G}_m$) - -REPLY [10 votes]: Every positive dimensional linear algebraic group $G$ over an algebraically closed field has a one dimensional subgroup. -Case 1 $G$ is reductive. In that case, $G$ contains a torus $T$. Since we are over an algebraically closed field, $T \cong \mathbb{G}_m^r$ for some $r>0$. In particular, $\mathbb{G}_m \subseteq T \subseteq G$. -Case 2 $G$ has a nontrivial unipotent radical $U$. In turn, $U$ has a nontrivial center $Z$. We have $Z \cong \mathbb{G}_a^s$ for some $s>0$ (not quite right, see Peter McNamara's correction below). In particular, $\mathbb{G}_a \subseteq Z \subseteq U \subseteq G$. - -REPLY [7 votes]: Over an algebraically closed field, the answer is yes. I assume that by "linear group" you mean a smooth affine algebraic group scheme. Such a group has a filtration whose quotients (modulo finite group schemes) are successively a unipotent group, a torus, and a semisimple group, so it is only a question of checking each of these cases.<|endoftext|> -TITLE: Does the Feferman-Schutte analysis give a precise characterization of Predicative Second-Order Arithmetic? -QUESTION [8 upvotes]: A definition is called impredicative if it involves quantification over a domain that contains the thing being defined. For instance, if you define hereditary property to be a property which applies to $n + 1$ whenever it applies to $n$, and you define an inductive number to be any number that has all the hereditary properties of $0$, then the definition of "inductive" involves quantification over all hereditary properties, including the property "inductive". If you're sufficiently Platonistic concerning sets of natural numbers, that won't concern you, because you can say for instance that $0$ obviously has ALL the properties of $0$, so in particular it has all the hereditary properties of $0$, so it's inductive. But predicativists like Weyl and Poincare would say that in order to verify that $0$ is inductive, we need to know what hereditary properties it has, and in order to know what hereditary properties it has, we need to verify that it has the hereditary property known as inductiveness! -In the context of second-order logic, impredicative comprehension refers to any instance of the second-order comprehension schema that involves bound set variables. If we take second-order arithmetic and allow only predicative comprehension, i.e. we only apply the comprehension schema to formulas with no bound set variables, then we get a subsystem of second-order $PA$ known as $ACA_0$, which is conservative over first-order $PA$. But disallowing bound set variables is not the only way to ensure predicativity. We can instead adopt the ramified theory of types, which breaks the comprehension schema into levels. The comprehension schema for level $0$ sets allows no bound set variables. The schema for level $1$ sets allows quantification over level $0$ sets. In general, the schema for level $n + 1$ sets allow for quantification over sets of level $n$ and below. This way, you can still quantify over sets, but you won't risk quantifying over the set you happen to be defining. -And there's no particular reason to stop at finite levels. We can define a comprehension schema for level $\omega$ sets, for instance, allowing quantifies to range over sets of finite level. And so on, going to bigger and bigger transfinite ordinals. Godel apparently showed that if you carried this process all the way up to $\omega_1$, the first uncountable ordinal, you wouldn't need to go any further. But this is somewhat problematic, since there are presumably countable ordinals whose existence aren't even provable by predicative means. In other words, for sufficiently great ordinals $\alpha$, it may not be predicatively provable that there exists a well-ordering of the natural numbers with order-type $\alpha$. Feferman and Schutte remedied this by only allowing a comprehension-schema for level $\alpha$ sets if the ordinal $\alpha$ is in fact predicatively acceptable (using the comprehension-schemata for lower-level sets). See pg 15-16 of this paper for a brief explanation. In any case, they reached the following conclusion: if $RA_\alpha$ denotes second-order arithmetic with a comprehension schemata for sets of level $\alpha$ and below, then $RA_\alpha$ is only predicatively acceptable if $\alpha$ is less than a certain ordinal known as $\Gamma_0$, the Feferman-Schutte ordinal. -My question is, assuming the Feferman-Schutte analysis of predicativity is accurate, can we give a precise criterion for what is and isn't provable in predicative second-order arithmetic. That is, given a sentence $S$ in the language of second-order arithmetic, under what conditions does there exist an $\alpha < \Gamma_0$ such that $S$ is provable in $RA_\alpha$? Of course, such a characterization of what truths are predicatively acceptable would not itself be predicatively acceptable. It's similar to how the truths provable in Edward Nelson's system of predicative first-order provable are precisely the bounded consequences of $Q$ + bounded induction + "exponentiation is total", even though Nelson himself would reject that characterization because he doesn't think exponentiation IS total. -It seems to me that just like $Q$ + bounded induction + "exponentiation is total" is an "upper bound" for Nelson's predicative first-order arithmetic, the system $ATR_0$, Arithmetical Transfinite Recursion, is an "upper bound" for the Feferman-Schutte conception of predicative-second order arithmetic. After all, the proof-theoretic ordinal of $ATR_0$ is $\Gamma_0$, the least impredicative ordinal, and it proves the consistency of $ACA_0$, a predicative subsystem of arithmetic. So can we characterize the set of predicatively acceptable theorems as some appropriate subset of the set of theorems of $ATR_0$, distinguished by some syntactic property, just like we could take the bounded consequences of $Q$ + bounded induction + "exponentiation is total"? -Any help would be greatly appreciated. -Thank You in Advance. -EDIT: As @AndreasBlass pointed out, on page 18 this paper, Feferman says that $ATR_0$ is "locally predicatively justifiable", meaning that a predicativist would reject the system as a whole as impredicative, but would accept each individual theorem of the system as predicatively acceptable, i.e. each theorem of $ATR_0$ is provable in $RA_\alpha$ for some $\alpha < \Gamma_0$ (where $\alpha$ can vary). So now the question becomes, is $ATR_0$ "maximally locally predicatively justifiable"? In other words, are there predicatively acceptable statements of second-order arithmetic not provable in $ATR_0$? And if so, is there some larger system that IS maximal? I assume any such system would still have proof-theoretic ordinal $\Gamma_0$. (Am I right about that?) - -REPLY [2 votes]: All of the traditional "locally predicatively justifiable" theories are supposed to have equivalent $\Pi^1_1$ theorems; but they differ for statements of higher complexity. So they would not, even under the traditional reading, characterize all of second-order predicative arithmetic. Presumably, one would need a predicative version of the Axiom of Projective Determinacy to characterize the provable higher-complexity statements (and of course the analysis would depend on the classical version of this axiom). -But, in fact, something seems to have gone wrong somewhere, because $\mathcal{ATR}_0$ is finitely axiomizable (Feferman points this out on page 21); so it cannot be locally predicatively justifiable. For an explicit example, "For any $\alpha<\Gamma_0$, the Axiom of Reducibility is relatively consistent over $RA_\alpha$" is a $\Pi_2^0$ sentence provable in $\mathcal{ATR}_0$ but not in $RA_\alpha$ for any $\alpha < \Gamma_0$.<|endoftext|> -TITLE: Cohomology of the classifying space of $Ss(4m)$ -QUESTION [6 upvotes]: Let $Ss(4m)$ be the $Z/2Z$ quotient of $Spin(4m)$ which is not $SO(4m)$. (This group is somtimes called the semi-spin group.) Its $Z/2Z$ cohomology was determined e.g. by Baum and Browder MR article. Is the $Z/2Z$ cohomology of its classifying space determined somewhere? What is -\begin{equation} -H^*(BSs(4m),Z/2Z) ? -\end{equation} -Update: In the string theory application I have in mind, it would be enough to know it up to degree 11. Does this make the determination any easier? - -REPLY [11 votes]: Tetsu Nishimoto kindly performed the computation, and allowed me to reproduce it here. --Yuji - -Proposition: -The mod-2 cohomology $H^*(BSs(16m);\mathbb Z/2)$ of the classifying space of the Lie group $Ss(16m)$, is isomorphic to the following algebra up degree $ \leq 11$: -$$ - \mathbb Z/2[x_2, x_3, x_5, x_9, y_4, y_6, y_7, y_{10}, y_{11}] - /(x_2y_7+x_3y_6+x_5y_4+x_2x_3y_4). -$$ -Within $\ast \leq 11$, the action of $Sq^k$ is given by -$$ - \begin{array}{|c|c|c|c|c|c|} - \hline - & Sq^1 & Sq^2 & Sq^3 & Sq^4 & Sq^5 \\ - \hline - x_2 & x_3 & x_2^2 & & & \\ - x_3 & 0 & x_5 & x_3^2 & & \\ - x_5 & x_3^2 & 0 & 0 & x_9 & x_5^2 \\ - x_9 & x_5^2 & 0 & & & \\ - y_4 & 0 & y_6 & y_7 & y_4^2 & \\ - y_6 & y_7 & 0 & 0 & y_{10} & y_{11} \\ - y_7 & 0 & 0 & 0 & y_{11} & \\ - y_{10} & y_{11} & & & & \\ - \hline - \end{array}. -$$ -Let us describe the outline of the computation. -First we need to quote the structure of $H^*(Ss(16m);\mathbb Z/2)$ as a Hopf algebra from -Proposition 4.1 of - -Hopf Algebra Structure of mod 2 Cohomology of Simple Lie groups - K. Ishitoya, A. Kono and H. Toda - Publ. RIMS, Kyoto Univ. 12 (1976) 141-167 electric version - -The proposition states that, in the range $\ast \leq 10$, - $H^*(Ss(16m);\mathbb Z/2)$ is isomorphic as an algebra to -$$ - \Delta (w_3, w_5, w_6, w_7, w_9, w_{10}) \otimes \mathbb Z/2[\bar{v}] -$$ -where $\deg w_i = i$, $\deg \bar{v} = 1$. -The generators other than $w_7$ are primitive, while the coproduct of $w_7$ is given by -$$ - \bar\psi (w_7) = \bar{v} \otimes w_6 + \bar{v}^2 \otimes w_5 - + \bar{v}^4 \otimes w_3. -$$ -The action of $Sq^k$ within the range $\ast \leq 10$ is given by -$$ - \begin{array}{|c|c|c|c|c|c|c|c|} - \hline - & Sq^1 & Sq^2 & Sq^3 & Sq^4 & Sq^5 \\ - \hline - w_3 & 0 & w_5 & w_6 = w_3^2 & & \\ - w_5 & w_6 & 0 & 0 & w_9 & w_{10} = w_5^2 \\ - w_6 & 0 & 0 & 0 & w_{10} & \\ - w_7 & m\bar{v}^8 & w_9 & w_{10} & & \\ - w_9 & w_{10} & & & & \\ - \hline - \end{array}. -$$ -Next, we consider the Rothenberg-Steenrod spectral sequence -$$ - E_2 = \mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)} - (\mathbb Z/2, \mathbb Z/2) - \Longrightarrow H^*(BSs(16m);\mathbb Z/2). -$$ -We first need to compute the $E_2$ term. -Here we use May's spectral sequence -$$ -E'_1=\mathrm{Cotor}_{A'}(k,k) \Longrightarrow \mathrm{Cotor}_{A}(k,k). -$$ Here, $A'$ is a Hopf algebra such that it is isomorphic as an algebra with $A'$ such that every generator is primitive. -When the characteristic of $k$ is $2$, $\mathrm{Cotor}_{A'}(k,k)$ is a polynomial ring whose generators are in one-to-one correspondence with the primitive elements of $A'$. -Here we take $k=\mathbb{Z}/2$ and $A=H^*(Ss(16m);\mathbb Z/2)$. -Then, up to degree 11, we have -$$ -\mathrm{Cotor}_{A'}(k,k) = \mathbb{Z}/2[[v],[v^2],[v^4],[v^8],[w_3],[w_5],[w_6],[w_7],[w_9],[w_{10}]] -$$ -The differential at $E_2$ is given by -$$ -d_2([w_7]) = [v][w_6]+[v^2][w_5]. -$$ -Note that from the construction of May's spectral sequence the term $[v^4][w_3]$ vanish. -As all the other differentials are zero, $E'_\infty$ up to degree 11 is isomorphic to -$$ -\mathbb Z/2[[v],[v^2],[v^4],[v^8],[w_3],[w_5],[w_6],[w_9],[w_{10}]]/([v][w_6]+[v^2][w_5]). -$$ -It is easily seen that in $\mathrm{Cotor}_A(k,k)$ the relation corresponding to $[v][w_6]+[v^2][w_5]$ is $[v][w_6]+[v^2][w_5]+[v^4][w_3]$. Therefore $\mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)} -(\mathbb Z/2, \mathbb Z/2) $ is given by -$$ - \mathbb Z/2 [[\bar{v}], [\bar{v}^2], [\bar{v}^4], [\bar{v}^8], - [w_3], [w_5], [w_6], [w_9], [w_{10}]]/ - ([\bar{v}][w_6]+[\bar{v}^2][w_5]+[\bar{v}^4][w_3]) -$$ up to degree 11 as algebras. -We note here that $[\bar{v}^{2^j}] \in E_2^{1,2^j}$, $[w_i] \in E_2^{1,i}$, -and that these generators all correspond to primitive elements. -When the degrees are higher this is not necessarily the case. The relation came from the coproduct of $w_7$, as we saw above. -The differentials are given in the range $r \geq 2$ as -\begin{align*} - & d_r : E_r^{1,1} \longrightarrow E_r^{1+r,1-(r-1)} = - E_2^{1+r,1-(r-1)} = 0, \\ - & d_r : E_r^{1,3} \longrightarrow E_r^{1+r,3-(r-1)} = - E_2^{1+r,3-(r-1)} = 0. -\end{align*} -Therefore, $[\bar{v}]$ and $[w_3]$ are permanent cycles. -In general, when $x$ is a permanent cycle, for any cohomology operation $\theta$ - $\theta x$ is also a permanent cycle. - Other generators can be written as -\begin{align*} - & [\bar{v}^2] = Sq^1[\bar{v}], - \quad - [\bar{v}^4] = Sq^2Sq^1[\bar{v}], - \quad - [\bar{v}^8] = Sq^4Sq^2Sq^1[\bar{v}], \\ - & [w_5] = Sq^2[w_3], - \quad - [w_6] = Sq^3[w_3], - \quad - [w_9] = Sq^4Sq^2[w_3], - \quad - [w_{10}] = Sq^5Sq^2[w_3] -\end{align*} -and therefore these are also permanent cycles. -Therefore, up to degree 11, we have $E_{\infty} = E_2$. -Let us now define elements of $H^*(BSs(16m);\mathbb Z/2)$. -Let $x_2$ be a representative of $[\bar{v}]$, -and $y_4$ be a representative of $[w_3]$. -$x_2$ is uniquely determined but there are two elements $y_4$ and $y_4+x_2^2$ representing $[w_3]$. This freedom is used below when we fix the relations. -Let us further set -\begin{align*} - & x_3 = Sq^1 x_2, - \quad - x_5 = Sq^2 x_3, - \quad - x_9 = Sq^4 x_5, \\ - & y_6 = Sq^2 y_4, - \quad - y_7 = Sq^1 y_6, - \quad - y_{10} = Sq^4 y_6, - \quad - y_{11} = Sq^1 y_{10} -\end{align*} -then they are representatives of $[\bar{v}^2]$, $[\bar{v}^4]$, $[\bar{v}^8]$, -$[w_5]$, $[w_6]$, $[w_9]$, $[w_{10}]$, respectively. -Using the Adem relation, we can determine how $Sq^k$ acts on these elements, giving the table shown above. -Finally let us determine the relation in $H^*(BSs(16m),\mathbb Z/2)$ corresponding to -the relation $[\bar{v}][w_6]+[\bar{v}^2][w_5]+[\bar{v}^4][w_3]$ of - $\mathrm{Cotor}_{H^*(Ss(16m);\mathbb Z/2)} -(\mathbb Z/2, \mathbb Z/2)$. -When $k \geq 3$, the basis of $E_2^{k,9-k}$ can be given by -$$ - x_2^2x_5, \quad x_3^3, \quad x_2^3x_3, \quad x_2x_3y_4 -$$ -and therefore the degree-9 relation $r$ in $H^*(BSs(16m);\mathbb Z/2)$ can be given by -$$ - r = x_2y_7 + x_3y_6 + x_5y_4 + a_1x_2^2x_5 + a_2x_3^3 - + a_3x_2^3x_3 + a_4x_2x_3y_4 - \quad (a_i \in \mathbb Z/2). -$$ -From -$$ - Sq^1 r = (1+a_4)x_3^2y_4 + (a_1+a_3)x_2^2x_3^2 -$$ -we have $a_4 = 1$ and $a_1 = a_3$. From -\begin{align*} - Sq^2 r & = x_2^2y_7 + (a_1+a_2)x_3^2x_5 + a_1x_2^4x_3 - + a_1x_2x_3^3 + a_1x_2^3x_5 + x_2^2x_3y_4 + x_2x_5y_4 - + x_2x_3y_6 \\ - & = (a_1+a_2)x_3^2x_5 - + (a_1+a_2)x_2x_3^3 -\end{align*} -we have $a_1 = a_2$. Then the relation is given by -$$ - r = x_2y_7 + x_3(y_6+a_1x_3^2) + (x_5+x_2x_3)(y_4+a_1x_2^2). -$$ -When $a_1 = 1$, we exchange $y_4$ by $y_4+x_2^2$. -Then $y_6$ is exchanged with $y_6+x_3^2$ and $y_{10}$ is exchanged with $y_{10}+x_5^2$, while all the other generators are fixed. -The relation then becomes -$$ - r = x_2y_7 + x_3y_6 + (x_5+x_2x_3)y_4 -$$ -and has the same form as the relation for the case $a_1 = 0$. -This completes the determination of the relation.<|endoftext|> -TITLE: Equivariant resolution of singularities -QUESTION [8 upvotes]: I am looking for some references on equivariant resolution of singularities. In most references quoted on mathoverflow (for instance : Reference on an equivariant resolution of singularities), they only talk about finite group action (if I am not mistaken). -I was wondering if it is known that equivariant resolutions do not exist in general for larger group (are there counter-examples?). I am especially interested when the group is $\mathbb{C}^*$. -Thanks in advance. - -REPLY [5 votes]: To any variety $X$ over a field of characteristic zero, say $k$, one can attach a resolution of singularities, say $X'\to X$, with the following properties: - -$X'\to X$ is an isomorphism over $X_{\mathrm{reg}}$. -if $\Theta:X\rightarrow Y$ is an isomorphism of schemes (not necessarily defined over $k$), then $\Theta$ can be extended to an isomorphism $\Theta':X'\rightarrow Y'$ compatible with $\Theta$ and the resolutions. - -It follows from 1. and 2. that if $\Psi:Y\rightarrow Z$ is a second isomorphism of schemes, then $(\Psi\circ\Phi)'$ and $\Psi'\circ\Phi'$ coincide on the inverse image of $X_{\mathrm{reg}}$ in $X'$, and hence they coincide on $X'$. With a similar argument, we have $(\mathrm{id}_X)'=\mathrm{id}_{X'}$. In particular, if $G$ is any group of automorphisms of $X$ (not necessarily defined over $k$), the action of $G$ on $X$ can be lifted to an action on $X'$. -For a reference, see: -O. Villamayor: Equimultiplicity, algebraic elimination, and blowing-up. Adv. in Math. 262 (2014) 313-369. (ArXiv link, DOI link)<|endoftext|> -TITLE: What is the relation between the Lie bracket on $TX$ as commutator and that coming from the Atiyah class? -QUESTION [14 upvotes]: Let X be a complex manifold and $TX$ its tangent bundle. The Atiyah class $\alpha(E)\in \text{Ext}^1(E\otimes TX, E)$ for a vector bundle $E$ is defined to be the obstruction to the global existence of holomorphic connections on $E$. In particular if we take $E=TX$ to be the tangent bundle of $X$ itself, we can prove that the Atiyah class $\alpha(TX)\in \text{Ext}^1(TX\otimes TX, TX)$ and this gives a Lie algebra structure -$$ -TX[-1]\otimes TX[-1]\rightarrow TX[-1] -$$ - in the derived category $D(X)$. -For the details see M. Kapranov's paper "Rozansky–Witten invariants via Atiyah classes" (arXiv) and N. Markarian's paper "The Atiyah class, Hochschild cohomology and the Riemann-Roch theorem" (arXiv). -And it is well-known that we have the Lie bracket as commutator on the tangent vector fields -$$ -\Gamma(X,TX)\otimes \Gamma(X,TX)\rightarrow\Gamma(X,TX). -$$ -Of course these two kinds of Lie bracket are very different. $\textbf{My question}$ is: is there any relation between them? More precisely, could we regard the Lie bracket of Atiyah class as a kind of "higher lift" of the naive Lie bracket as commutators? - -REPLY [9 votes]: Starting from $X$ one can construct a sequence $(X_0,X_1,\dots)$ of groupoid objects in spaces, all having $X$ as space of objects (I am writing "space" in order not to specify the geometric context I am working with, but I actually mean "derived scheme"): - -$X_0$ is the pair groupoid $X\times X$ of $X$. The Lie algebroid of this groupoid is $TX$. -$X_1=X\times^h_{X_0}X$ is the derived self-intersection of the diagonal in $X_0$ (i.e. the derived loop space of $X$). The Lie algebroid of this groupoid is $TX[-1]$ with the Atiyah Lie bracket. It happens to be a Lie algebra (because the derived loop space is actually a group). -In general $X_{n+1}=X\times^h_{X_n}X$ and its Lie algebroid $\mathfrak g_{n+1}$ is $TX[-n-1]$, and is an abelian Lie algebra as soon as $n\geq1$. - -In other words, we have that $\mathfrak g_n=\Omega^n_0(TX)=TX[-n]$. Hence, in particular $TX[-1]$ is a group in Lie algebroids, and thus must be a Lie algebra. You get that the Atiyah-Kapranov Lie bracket on $TX[-1]$ is completely dertmined by the Lie algebroid structure on $TX$. -One also gets that the Lie algebroid $\mathfrak g_2$ of $X_2$ is going to be $\Omega_0(TX[-1])$, and thus is a group in Lie algebras... hence it must be an abelian Lie algebra. -The hierarchy Lie algebroid -> Lie algebra -> abelian Lie algebra -> ... one gets is analogous to the following one we get in algebraic topology: set -> group -> abelian group -> ...<|endoftext|> -TITLE: Is "small" dependence enough for central limit theorem? -QUESTION [5 upvotes]: Writing down a paper about some estimation of some combinatorial quantities, i realized that i would have much more precise results if these two questions have positive answer: -1) Suppose you have a sequence of random variables $X_n$(boolean random variable with $Pr(X_i=1)=1/2$), such that $Cov(X_n,X_{n+1})=c$(in my case $c=-1/12$) and $X_i,X_j$ are independent whenever $|i-j|>1$. What can be said of $X_1+...+X_n$? Does the central limit theorem hold, altough there is no complete indipendence but "almost"? -2) Consider the well known relation ${2n} \choose {n} $$\sim\frac{4^n}{\sqrt \pi n}$. Is there a purely probabilistic way(using the central limit theorem or something near there) to prove this? The setting i've in mind is of course Bernoulli of parameter 1/2 $X_1,...,X_{2n}$ independent each other. And i can see that it would suffice to prove the convergence in 0 of the discrete densities to the gaussian in 0. -Moreover if the answer in 1) is positive, does it allows, also there, the same asymptotics for the central term of $X_1+...+X_n$? -Thanks for any explanation! - -REPLY [6 votes]: For point 1, search for -``CLT for mixing sequences'' -you will be drowned by the number of hits. -Also, there are CLT's for stationary sequences in many textbooks - Hall and Heyde's book has a section on that. -For point 2, search for -``local CLT for lattice variables''<|endoftext|> -TITLE: Rigidity of secondary characteristic classes -QUESTION [6 upvotes]: For a representation $\rho:\pi_1M\rightarrow GL(n,C)$ and the associated flat $GL(n,C)$-bundle $E_\rho\rightarrow M$ one has the Cheeger-Chern-Simons classes -$$\hat{c}_k(E_\rho)\in H^{2k-1}(M,R/Z)$$ -as defined in http://www.jstor.org/stable/1971013 -It is claimed at several places in the literature that $\hat{c}_k(E_\rho)$ is constant on components of the representation variety. (For example in Reznikov's proof of the Bloch conjecture http://arxiv.org/pdf/dg-ga/9407007.pdf where this fact is called rigidity of secondary characteristic classes and is attributed to the above-linked paper of Chern and Simons. However I was not able to find such a statement in that paper.) -Question: is there are a reference for this fact, or is there some well-known fact from which this local constantness follows? - -REPLY [3 votes]: To answer my own question: the wanted rigidity follows from Theorem 3.4 in http://arxiv.org/pdf/math/9904131.pdf together with the main result from P. Ntolo, Homologie de Leibniz d'algébres de Lie semi-simples, Comptes Rendus Acad. Sci. 318 (1994) -UPDATE: the details of Reznikov‘s original argument have been worked out in a recent paper of Pitsch and Porti: https://arxiv.org/pdf/1704.01321.pdf<|endoftext|> -TITLE: Best and worst centrally symmetric convex covering shapes -QUESTION [6 upvotes]: Suppose you have a centrally symmetric convex 2D shape $C$ of area $A$, and you randomly throw -down copies of $C$ on the plane so that each $C$-center lies within a given unit square $S$, -until $S$ is entirely covered, i.e., every point of $S$ lies in some copy of $C$. -For example, below (left) it took $9$ disks to cover the blue unit square, and (right) -$10$ squares of the same area to cover the blue unit square (scaled differently): - -Let $\rho(C)$ be the expected number of such randomly placed copies of $C$ needed to -cover $S$. My questions are: - -Q1. Is the disk the most efficient such covering shape, in that it minimizes $\rho(C)$ over -all centrally symmetric convex bodies $C$ of the same area? If not, which shape is the best? -Q2. What shape is the worst covering shape, achieving the maximum of $\rho(C)$ over such shapes? -Q3. Do the best and worst shapes depend upon the choice to cover a square rather than to cover some other convex shape? - -The same questions can be asked in any dimension. - -REPLY [3 votes]: Q2: there is no worst shape, but arbitrarily bad shapes. Take a very thin and long rectangle: then any intersection of a displacement of it with the blue square has area bounded by some $a$, which can be made arbitrarily small. In consequence, at least $1/a$ copies are needed in any case to cover the blue square. -Variant of the question: search $C$ maximizing $\inf_A \rho(A\cdot C)$ where $A$ runs over linear maps.<|endoftext|> -TITLE: Is Nijenhuis–Richardson bracket a BV bracket? -QUESTION [7 upvotes]: Let $g$ be a finite dimensional Lie algebra, and let me denote $A=(\bigwedge g^* \otimes g, d)$ the Chevalley-Eilenberg complex that calculates cohomology of the Lie algebra with coefficients in the adjoint representation. On the complex $A$ there is Nijenhuis–Richardson bracket and the resulting DGLA could be used to describe deformations of the given algebra. -Suppose that $g$ is unimodular i.e. determinant of the adjoint representation is trivial representation. In such situation there are isomorphisms -$$ -\bigwedge^i g^* \otimes g \cong \bigwedge^{n-i} g \otimes g, -$$ -thus differential in the complex for homology of the lie algebra induces one more map on the complex (this map goes in the opposite direction on $A$). I want to mimic Ginzburg's construction of the Batalin-Vilkovisky algebra structure on the Hochshild cohomology complex of a CY algebra and such induced map should play a role of BV differential, and unimodularity condition plays role of the CY condition. But one important ingredient is missing in this situation, namely there is no associative algebra structure on $A$. -I wounder is it possible to introduce some multiplication on $A$ such that Nijenhuis–Richardson bracket become a BV bracket? - -REPLY [2 votes]: There is one silly answer to your question, and you are probably aware of it: if you use as coefficients $S(g)$, the symmetric algebra of $g$, and not just $g$, everything will work wonderfully. Alas I have never seen anything smaller in this context.<|endoftext|> -TITLE: A map from the coinvariants of the dual to the dual of the invariants for a G-module -QUESTION [13 upvotes]: Suppose $G$ is a group and $X$ is a $\mathbb{Z}[G]$-module. Recall that the augmentation ideal $I \subset \mathbb{Z}[G]$ is generated by elements of the form $g - 1$ for $g \in G$, the coinvariants are defined as $X_G := X / IX$, and the invariants are given by $X^G := \{x \in X : gx = x, \forall g \in G\}$. -If $G$ is finite then the trace map $X_G \to X^G$ given by $x \mapsto \sum_{g\in G} gx$ is used in Tate cohomology, but this map doesn't make sense in the case that $G$ is infinite. -Suppose instead that we take a $\mathbb{Z}$-module $A$ and set $X^* := \operatorname{Hom}_{\mathbb{Z}}(X, A)$. Then there is a well-defined map -\begin{align*} -\rho:(X^*)_G &\to (X^G)^* \\ -[f] &\mapsto f|_{X^G} -\end{align*} -even when $G$ is infinite. Is there a nice description of the kernel and cokernel of $\rho$? - -REPLY [5 votes]: I will answer the question in the case that $G$ is cyclic or pro-cyclic. Write $t$ for a (topological) generator of $G$. Then we can describe $X^G$ and $X_G$ as the kernel and cokernel of $t-1$: -$$ -0 \to X^G \to X \xrightarrow{t-1} X \to X_G \to 0. -$$ -Write $Y = (t-1)X$ for the augmentation ideal. We can split into two short exact sequences -\begin{align*} -&0 \to X^G \to X \to Y \to 0 \\ -&0 \to Y \to X \to X_G \to 0. -\end{align*} -Note that the map $X \to Y$ is given by the action of $t-1$, while the map $Y \to X$ is just the inclusion. Dualizing, we get sequences -\begin{align*} -&0 \to Y^* \xrightarrow{t-1} X^* \to (X^G)^* \to \operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A)\\ -&0 \to (X_G)^* \to X^* \longrightarrow Y^* \to \operatorname{Ext}^1_{\mathbb{Z}}(X_G, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A). -\end{align*} -Since the cokernel of $Y^* \xrightarrow{t-1} X^*$ is $(X^*)_G$ by definition, we get a sequence -$$0 \to (X^*)_G \to (X^G)^* \to \operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A).$$ -The first map in this sequence is $\rho$, so $\rho$ is injective in this case, with cokernel equal to the kernel of $\operatorname{Ext}^1_{\mathbb{Z}}(Y, A) \to \operatorname{Ext}^1_{\mathbb{Z}}(X, A)$. -If $A$ is divisible as an abelian group ($A = \mathbb{C}^\times$ for example), then $\operatorname{Ext}^1_{\mathbb{Z}}(-, A)$ vanishes and $\rho$ is an isomorphism. In this case we also get an isomorphism $(X_G)^* \to (X^*)^G$.<|endoftext|> -TITLE: An extension of Morera's Theorem -QUESTION [7 upvotes]: Morera's Theorem states that - -If $f$ is continuous in a region $D$ and satisfies $\oint_{\gamma} f = 0$ for - any closed curve $\gamma$ in $D$, then $f$ is analytic in $D$. - -I have two questions: - -If $f$ is continuous in $D$ and $\oint_C f = 0$ for any circle $C$ in $D$, -can we deduce that $\oint_{\gamma} f = 0$ for any closed curve $\gamma$ in $D$? -(more ambitiously) If $f$ is continuous and $\oint_C f = 0$ for any circle $C$ in $D$, is $f$ analytic in $D$ ? - -Partial ansers for question 2 seem to be here, but I doubt their argument, specificly, the construction of the original function. - -REPLY [11 votes]: First of all, your questions 1 and 2 are equivalent (by the usual Morera theorem). -Second, even stronger generalizations of Morera are available (one does not need all circles). -There is an old nice survey of Zalcman, Offbeat Integral Geometry, in the Monthly. -In particular it contains the following result for the case $D=C$: if the integrals over all -circles of two fixed radii $r_1$ and $r_2$ are zero, then the function is analytic, -unless the ratio of these radii is a zero of Bessel's function $J_1$. -On some more modern research on the topic, I recommend the papers of Hansen, Nadirashvili -and Tumanov MR2046196. - -REPLY [7 votes]: The answer is yes, and a proof can be found for example on this webpage: http://anhngq.wordpress.com/2009/07/20/a-generalization-of-the-morera%E2%80%99s-theorem/ -A brief summary: Suppose $f$ is continuous and $\int_C f = 0$ for every circle $C$, but $\int_\gamma f \neq 0$ for some closed curve $\gamma$. By convolving $f$ with a smooth approximation to the identity, we may assume $f$ is smooth. But then by applying Green's formula to $\int_C f = 0$ for small circles $C$, we see that $f$ must satisfy the Cauchy-Riemann equations, so $\int_\gamma f = 0$, a contradiction.<|endoftext|> -TITLE: distortion of cyclic subgroups of linear groups -QUESTION [6 upvotes]: In an informal talk I heard a statement: -"Any cyclic subgroup in a linear group is at most exponentially distorted" -with a vague reference to a work of Lubotzky with coauthors. -The works of Lubotzky that may be relevant to this quiestion and that I was able to find reference for: -Lubotzky, Alexander; Mozes, Shahar; Raghunathan, M. S. -Cyclic subgroups of exponential growth and metrics on discrete groups. -C. R. Acad. Sci. Paris Sér. I Math. 317 (1993), no. 8, 735–740. -http://www.ams.org/mathscinet-getitem?mr=1244421 -Lubotzky, Alexander; Mozes, Shahar; Raghunathan, M. S. -The word and Riemannian metrics on lattices of semisimple groups. -Inst. Hautes Études Sci. Publ. Math. No. 91 (2000), 5–53 (2001). -http://www.ams.org/mathscinet-getitem?mr=1828742 -seem to deal with lattices in Lie groups, not with arbitrary linear groups. So the question is: -Is the statement above true in such generality? If not, then for which classes of linear groups cyclic subgroups are at most exponentially distorted? -EDIT: As far as I understood, the distortion of a cyclic subgroup $H$ inside $G\subset GL(n,F)$ is meant here to be with respect to the respective word metrics on $H$ and $G$, not the underlying metric on $GL(n,F)$. -EDIT2: Sorry for the possible confusion. I guess, the distortion here is meant in the following sense: -Let $H$ be a subgroup of $G$ and fix some generating sets for $H$ and $G$. Consider a ball $B(n,1)$ of radius $n$ in the Cayley graph of $G$ centered at $1$, i.e. the set of all elements of $G$ having word length $\le n$ with respect to the generating set of $G$. Define a function -$$ -f(n):=\max\{|h|_H \mid h\in H\cap B(n,1)\}, -$$ -where $|h|_H$ is the length of the element $h$ with respect to the generating set of $H$. -Then $f(n)$ measures the distortion of subgroup $H$ inside $G$, and its order of growth is independent on the choice of the generating sets. -For example, in the group $G=\langle a,b\mid aba^{-1}=b^2\rangle$, subgroup $H=\langle b\rangle$ has exponential distortion function, as the element $h=a^mba^{-m}=b^{2^m}$ lies in the ball $B(2m+1,1)$ in $G$, but the word length of $g$ in $H$ is $2^m$. -But if we consider a group $G=\langle a,b,c\mid aba^{-1}=b^2, bcb^{-1}=c^2\rangle$ then a similar reasoning shows that $H=\langle c \rangle$ has double exponential distortion in $G$: $f(n)=2^{2^n}$. -The statement I am inquiring about stipulates that cyclic subgroups $H$ of linear groups $G\subset GL(n,F)$ cannot have distortion functions that grow faster than exponential functions. - -REPLY [12 votes]: Yes it's true. First recall the definitions: -1) if $G$ is a group (discrete), a finitely generated subgroup $H$ is at most exponentially distorted (resp. undistorted) if for some/every finite generating subset $S$ of $H$ and any finite subset $T$ of $G$ such that $H \subseteq \langle T \rangle$, there exists a function with exponential growth (resp. with linear growth) $f$ such that $|g|_S\le f(|g|_T)$ for all $g\in H$. [If $G$ is finitely generated it's enough to check with a given $T$.] Note it is trivially true if $H$ is finite. -2) A linear group is a subgroup of $\mathrm{GL}_d(K)$ for some $d$ and field $K$. -Now to answer your question: -Proposition: in a linear group, every cyclic subgroup is at most exponentially distorted. -Lemma: let $\mathbf{K}$ be a nondiscrete locally compact field, and $g\in\mathrm{GL}_d(\mathbf{K})$. Endow the latter with the word length $\ell$ with respect to some/any compact generating subset. Then exactly one of the following holds: -a) $(\ell(g^n))_{n\in\mathbf{Z}}$ is bounded. This holds iff all eigenvalues of $g$ (over a finite extension) have modulus one, and in case $\mathbf{K}$ is Archimedean you require in addition that $g$ is semisimple. -b) $(\ell(g^n))$ grows logarithmically. This holds iff $\mathbf{K}$ is Archimedean, all eigenvalues of $g$ have modulus 1, and $g$ is not semisimple. -c) $(\ell(g^n))$ grows linearly. This holds iff $g$ has an eigenvalue (over some finite extension) of norm $\neq 1$. -I leave the proof of the lemma as an instructive exercise. -Now let $G\subset\mathrm{GL}_d(K)$ be a linear group and $\langle g\rangle$ be a cyclic subgroup and let us show the result. Clearly, we can assume that $\langle g\rangle$ is infinite and $G$ is finitely generated; actually we can then assume that $K$ is a finitely generated field. -If some eigenvalue of $g$ is not a root of unity, we can embed $K$ in a nondiscrete locally compact field so that this eigenvalue has norm $\neq 1$ (this is a theorem of Tits, used in the proof of Tits' alternative). It follows that $\langle g\rangle$ is undistorted. -If otherwise every eigenvalue of $g$ is a root of unity, after passing to a power we can suppose that $g$ is unipotent. Then since $g$ has infinite order, we deduce that the characteristic of $K$ is 0 and $g$ is not semisimple. There exists an field embedding of $K$ into $\mathbf{C}$, so by the lemma the growth of $g^n$ in $\mathrm{GL}_d(\mathbf{C})$ is logarithmic. So $g$ is at most exponentially distorted. -Note that the proof also shows that in a linear group in positive characteristic, every cyclic subgroup is undistorted.<|endoftext|> -TITLE: How do you not forget old math? -QUESTION [104 upvotes]: I am trying to not forget my old math. I finished my PhD in real algebraic geometry a few years ago and then switched to the industry for financial reasons. Now I get the feeling that I want to do a postdoc and I face the dilemma that I actually have forgotten a lot of my algebraic geometry classics. Sometimes they are minor things and if I browse a book I recall everything, and sometimes they are major (I don't really think I have a mental lapse though!). So I see myself reading a lot of books in order to recall some of my old math. -It gets frustrating that I have to repeat reading 80% of the article that I once used to read and understand. Maybe the new math that I have been feeding myself should be blamed too (I tried learning more differential geometry and fractal theory after doing algebraic geometry and hardly looked back at algebraic geometry after that). I have never tried avoiding to forget old math, especially parts that I do not use in a daily basis (esp. now that I work in the industry). But this can and will be fatal if I do apply for a postdoc. So now I want to read again, yes, but I don't want to forget again. -Is there a magic recipe for this? Usually I do find it helpful to always connect even the most abstract of mathematics with something that is tangible as an example, either in real life or in easier math (e.g. connect invertible sheaves and Picard group to line bundles, vector bundles to tangent bundles and tangent spaces .. etc.). This usually helps me not to forget things, but some of the math that I used to learn is too abstract to make such a connection, or maybe I just didn't learn correctly to apply such a connection. So my approach now, when I start reading something new or old, is to find a practical example ASAP, or ask myself why the originator of the theory first thought of developing this in the first place, before even getting any deeper into the subject. I must be honest though, sometimes this is very difficult to do (esp. if you read references for which such connection is not made). - -REPLY [6 votes]: This is not an answer, rather information I think might be useful to you for better clarifying this question and for possibly changing your personal views on what you believe you have forgotten. To preface my response, your brain automatically stores most memories by cues for later retrieval, encountering these cues invokes said memory, for example the smell of chocolate cake could be a retrieval cue for a memory of your sixth birthday. -With that said, if a person describes something and you are unable to recall it, that doesn’t necessarily mean you forgot it, rather that their description was not one of the retrieval cues you used to store it. For example, people often wake up thinking they “forgot” a dream they had that night, but when they encounter some random object during the day it cues back the memory and the contents of their dream come flooding back to them. We all have stored memories that are unconscious to us - that is data our brain has recorded (sort of) but which can only be accessed by certain cues (cues we often don't know exist). -In your case you might believe you have forgotten say Bob's theorem in algebraic geometry simply because the enunciation of "Bob's theorem" does not trigger a recall of its contents. Thus it may be that despite having read a proof of Bob's theorem three years ago, your brain is no longer using the name of that theorem as a memory cue. However this does not rule out the possibility that other cues still exist for content related to Bob's theorem, for example it may be that some obsecure lemma used in the proof is actually a memory cue for the theorem and thus though you can't recall Bob's theorem now, if someone were to show you the lemma used in the proof, you would then be able to recall part of Bob's theorem. -For an overview, see: https://en.wikipedia.org/wiki/Cue-dependent_forgetting -Now lastly I want to comment that as we can see here, even common terms like "forget" are often not rigorous enough to describe what is happening cognitively to people in everyday situations, to make matters worse there are currently many psychological theories about memory that roughly say the same thing but which use different terminology or worse theories that in my opinion say and do nothing at all apart from create complex terminology to state elementary observations. Though like a lot of cognitive science I personally think this is because its almost always impossible to get useful biological measurements from within the brain (with our current knowledge/technology) as so far as they relate to what we see, thus instead we have to reverse engineer what exact processes are occurring within the brain based on the results of how test subjects perform on various tasks, like say someone trying to figure out server side code (this would be some cognitive process) by sending search queries (these would be the psychological tests) to an interface on a website (this would be the test subject) - its an incredibly difficult task as one can create many different scripts (biological processes in the brain) that all respond the same way to search queries (produce the same results in psychological tests) i.e. we have no way of knowing what script(s) (biological processes) are responsible for what. Like a doctor trying to diagnose a patient with only knowledge of their symptoms and not being allowed to administer any tests that involve physically interacting with them. However this is clearly an oversimplification (also I'm getting off-topic) and I should preface this last paragraph by saying that I am by all means a layperson at the will of only my own judgement and online research. The little I know is from reading articles/texts online as well as half a lay persons' book (which I undoubtedly am) by Matthew Walker during an unfortunate stint in a substance abuse program where apart from studying math without the internet that was all I found to do. I would not dare call myself a mathematician as I've not even completed my formal undergraduate education yet, though I can say for myself that worrying about my poor memory/mental-health with respect to my peers has brought me a fair amount of misery so (for whatever my advice is worth) I don't recommend dwelling on it, there are things I and I imagine everyone else has done/experienced be it simply aging or whatever that provide no net benefit no matter how you look at the situation, but I guess that's life.<|endoftext|> -TITLE: Isometrically-invariant measures and dilation of the Cantor set -QUESTION [8 upvotes]: Let $C$ be the Cantor middle-thirds set. Let $\mu$ be a finitely-additive isometrically-invariant measure on all subsets of $\mathbb R$. Then $\mu(3C)=2\mu(C)$, where $aB = \{ ax : x \in B \}$. Thus if $a$ is a power of $3$, $\mu(aC) = a^{\log_3 2} \mu(C)$. -Question 1: Is it the case for all $a\in (0,\infty)$ that $\mu(aC)=a^{\log_3 2} \mu(C)$, if $\mu$ is a finitely-additive isometrically-invariant measure on $\mathcal P\mathbb R$? -The remaining questions are predicated on an affirmative answer, though currently I'm suspecting a negative answer to 1. -Question 2: Is this still true if isometric-invariance is replaced by translation-invariance? -Question 3: Has there been any work on using the the relationship between $\mu(aC)$ and $\mu(C)$ for invariant measures $\mu$ on $\mathbb R^n$ (perhaps not defined on all of the powerset, but just on the Borel sets) to define a dimension and comparing it to Hausdorff and box dimensions? - -REPLY [2 votes]: Negative to the first question, so the others are moot. -Let $G_1$ be the isometries of $\mathbb R$. -First note that no finite number of translates (or reflections, but that doesn't add anything) of $(1/2)C$ covers $C$. This can be seen by playing around with base three expansions. -Let $I$ be the ideal in $\mathcal P(\mathbb R)$ generated by $(1/2)C$ and its translates. Let $\mathcal B$ be the quotient boolean algebra $\mathcal P(\mathbb R)/I$. Because $C$ isn't covered by translates of $(1/2)C$, $[C]\ne 0$. The action of $G_1$ on $\mathbb R$ induces an action of $G_1$ on $\mathcal B$. Since $G_1$ is supramenable (i.e., doesn't have nonempty paradoxical subsets), it follows from a theorem of Mycielski ("Finitely additive invariant measures. I", Colloq. Math. 42 (1979), 309–318) that there is a finitely additive measure $\mu_1$ on $\mathcal B$ invariant under $G_1$ with $\mu_1([C])=1$. Let $\mu(A)=\mu_1([A])$ for $A\subseteq\mathbb R$. Then $\mu(C)=1$ and $\mu((1/2)C)=0$ and the answer to Question 1 is negative for $a=1/2$. -This uses the Axiom of Choice.<|endoftext|> -TITLE: Polynomiality of functions over residue rings -QUESTION [6 upvotes]: Suppose $\mathbb{Z}/m \mathbb{Z}$ is a residue ring for some $m \in \mathbb{N}$. If $m=p$ is a prime number then every function $f:\mathbb{Z}/p \mathbb{Z} \rightarrow \mathbb{Z}/p \mathbb{Z}$ is a polynomial $f(x) \in (\mathbb{Z}/p \mathbb{Z})[x]$. -Question: Are there any simple criteria of polynomiality of $f$ if $m$ is composite (particularly, if $m$ is a power of a prime number or $m$ is a product of two distinct primes)? -Update: Recently I found the article On polynomial functions $\text{(mod $m$)}$ by D. Singmaster; there are also a series of papers by Z. Chen (1 and 2) about polynomial functions. One interesting result is the following: - -Let $f$ be a polynomial function $\text{(mod $m$)}$. Then $f$ has a unique polynomial representation $$f=\sum_{k=0}^{n-1} b_k x^k\quad \text{with}\quad 0 \leq b_k \leq m/(k!,m)$$ and $n$ is the least integer s.t. $m | n!$. - -REPLY [8 votes]: If $p$ and $q$ are distinct primes, then a function $f:\mathbb{Z}/pq\mathbb{Z}\to\mathbb{Z}/pq\mathbb{Z}$ is induced by a polynomial if and only if $f$ induces well-defined functions mod $p$ and mod $q$: in other words, if and only if $f(c+p)\equiv f(c)\pmod{p}$ and $f(c+q)\equiv f(c)\pmod{q}$ for all $c\in\mathbb{Z}/pq\mathbb{Z}$. That polynomials have this property follows by reducing mod $p$; conversely, if a function has this property then there are polynomials $f_1,f_2\in\mathbb{Z}[x]$ such that $f_1(c)\equiv f(c)\pmod{p}$ and -$f_2(c)\equiv f(c)\pmod{q}$ for all $c$. Then the function $f$ is represented by the polynomial $qu f_1(x) + pv f_2(x)$ where $u,v\in\mathbb{Z}$ satisfy $qu\equiv 1\pmod{p}$ and $pv\equiv 1\pmod{q}$. -The problem is much more difficult for functions $f:\mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}$ when $m$ is a prime power, say $m=p^n$. One constraint is that if $f$ comes from a polynomial then $f$ must induce well-defined functions mod $p^i$ for all $i -TITLE: Is there an asymptotic formula for an inverse function of the binomial coefficient? -QUESTION [5 upvotes]: Fix a positive integer $k$. Let $$ f(n):= \frac{k!\binom{n}{k}}{n^k} $$ Then $\lim_{n\to \infty} f(n) = 1$. Hence $f(n) \ge 1-\epsilon$ for large $n$. -Define $n_0(\epsilon)$ as the least positive integer such that $n\ge n_0$ implies $f(n)\ge 1 -\epsilon$. Is there an asymptotic for $n_0(\epsilon)$ as $\epsilon\to 0$? A lower bound for $n_0(\epsilon)$ for fixed small $\epsilon$ would be useful too. - -REPLY [3 votes]: The approximation coming from the normal approximation of the binomial distribution is -$$ k = \sqrt{-2n\ln(1-\epsilon)}.$$ -To get more terms write $\Delta=-\ln(1-\epsilon)$, then by expanding $-\ln(1-i/n)$ as a Taylor series in $i$ and summing over $i=0\ldots k-1$, you get a series -$$ \Delta = -1/2\,{\frac {k\, ( k-1 ) }{n}}+1/12\,{\frac {k\, ( 2\, -k-1) ( k-1 ) }{n^2}}+1/12\,{\frac {{k}^{2} - ( k-1) ^{2}}{{n}^{3}}} + \cdots -$$ -which converges if $k=O(n^c)$ for $c<1$ (but is useless if $c$ is too close to 1). The dominant term for moderately small $k$ is $k^2/2n$. You can use a series reversion method to get more terms of the inverse. I think (without much checking) that -$$ k = \sqrt{2n\Delta} + \frac{3-2\Delta}{6} + \text{smaller terms},$$ -and note that we are already running into the issue of $k$ being an integer.<|endoftext|> -TITLE: Proving $\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}$ -QUESTION [12 upvotes]: I found the following formula in a book without any proof: -$$\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.$$ -This does not seem to follow immediately from the basic binomial identities. I would like to know how to prove this, and any relevant references. -Remark : This question has been asked previously on math.SE without receiving any complete answers. - -REPLY [28 votes]: Here is a short proof of the more general identity -$$ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k} \binom{x}{k}\binom{x}{2m-k} = (-1)^m \binom{2m}{m} \binom{x+m}{2m}. $$ -Considered as polynomials in $x$, both sides have degree $2m$. If $x = m$ then $\binom{x}{k}\binom{x}{2m-k}$ is non-zero only when $k=m$, and so both sides equal $(-1)^m \binom{2m}{m}$. If $x \in \{0,1,\ldots,m-1\}$ then both sides are zero. If $x = -r$ where $r \in \{1,\ldots, m\}$ then, using that $\binom{-r}{k} = \binom{r+k-1}{r-1}$, -the left-hand side becomes -$$ \sum_{k=0}^{2m} (-1)^k \binom{2m}{k} f(k) $$ -where $f(y) = \binom{r+y-1}{r-1} \binom{r+2m-y-1}{r-1}$. Since $f$ is a polynomial of degree $2(r-1) < 2m$ in $y$, its $2m$-th iterated difference is zero. So both sides are again zero. This shows that the two sides agree at $2m+1$ values of $x$, and so they must be equal as polynomials in $x$. -This identity is a specialization of (2) in the Gessel-Stanton paper linked to above. -It is (6.56) in Volume 4 of Gould's tables. To get Dixon's identity, take $x=2m$ and use that $\binom{2m}{2m-k} = \binom{2m}{k}$.<|endoftext|> -TITLE: What is the amplituhedron? -QUESTION [74 upvotes]: The paper ”Scattering Amplitudes and the Positive Grassmannian” by Nima Arkani-Hamed, Jacob L. Bourjaily, Freddy Cachazo, Alexander B. Goncharov, Alexander Postnikov, and Jaroslav Trnka, introduces a remarkable new way for computations in quantum field theory based on volume computations of certain polyhedra. This and related works may also have profound implications for the foundation of particle physics. It is related to various beautiful mathematics, and, in particular, to the combinatorics of certain stratifications of the Grassmanians.(See also the Quanta Magazine article, A Jewel at the Heart of Quantum Physics, by Natalie Wolchover, and Nima Arkani-Hamed’s on-line SUSY 2013 video lecture The Amplituhedron.) -The amplituhedron is remarkable new subsequent geometric object which comes in this study extending the notion of "positive grassmanian." (I did not see it explicitly defined in the above paper, at least not by this name.) It is very briefly described in Arkani-Hamed's lecture. - -My question is quite simple: -what is the mathematical definition of the amplituhedron? -An older-sister MO question: The amplituhedron minus the physics -Update: The paper The amplituhedron by Nima Arkani-Hamed and Jaroslav Trnka is now on the arxive. -Update: See also this blog post by Trnka http://www.preposterousuniverse.com/blog/2014/03/31/guest-post-jaroslav-trnka-on-the-amplituhedron/ - -REPLY [12 votes]: There is now an AMS Notices article whose title is the same as the title of this question (and which therefore may be enlightening to anyone on this page): http://www.ams.org/journals/notices/201802/rnoti-p167.pdf<|endoftext|> -TITLE: $L^p$-norm of Fourier series in terms of coefficients, $p \neq 2$ -QUESTION [6 upvotes]: It is known that the $L^2$-norm of a Fourier series equals the $l^2$-norm of the coefficients. Are there similar results in the case of $L^p$-norm for $p\neq 2$? Can it be expressed explicitly in terms of the coefficients? - -REPLY [20 votes]: Consider a function $f \in L^2$ and $f \notin L^{p}$ (for $p>2$). Now multiple the Fourier coefficients by random signs. Almost surely, the new function, $g$, will be in $L^{p}$ (by Khinchin's inequality and Fubini's theorem). Thus we have two functions, $f$ and $g$, both of whose Fourier coefficients have the same absolute values, one of which has finite $L^{p}$ norm and the other of which has infinite $L^{p}$ norm. Thus, no expression involving only the absolute values of the Fourier coefficients can compute (or even bound!) the $L^p$ norm of a function.<|endoftext|> -TITLE: Logarithm of the hypergeometric function -QUESTION [7 upvotes]: For $F(x)={}_2F_1 (a,b;c;x)$, with $c=a+b$, $a>0$, $b>0$, it has been proved in [1] that $\log F(x)$ is convex on $(0,1)$. -I numerically checked that with a variety of $a,\ b$ values, $\log F(x)$ is not only convex, but also has a Taylor series in x consisting of strictly positive coefficients. Can this be proved? -[1] Generalized convexity and inequalities, Anderson, Vamanamurthy, Vuorinen, Journal of Mathematical Analysis and Applications, Volume 335, Issue 2, -http://www.sciencedirect.com/science/article/pii/S0022247X07001825# - -REPLY [4 votes]: Here's a sketch of a proof of a stronger statement: the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients. -To see this we first note that -$$\begin{aligned} -\frac{d\ }{dx} \log {}_2F_1(a,b;a+b+c;x) &= -\frac{\displaystyle -\frac{d\ }{dx}\,{}_2F_1(a,b;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}\\[3pt] - &=\frac{ab}{a+b+c}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)}. -\end{aligned} -$$ -Then -$$ -\begin{gathered} -\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b;a+b+c;x)} - = \frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)} \\ - \hfill\times - \frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)}.\quad -\end{gathered} -$$ -We have continued fractions for the two quotients on the right. -Let $S(x; a_1, a_2, a_3, \dots)$ denote the continued fraction -$$\cfrac{1}{1-\cfrac{a_1x} -{1-\cfrac{a_2x} -{1-\cfrac{a_3x} -{1-\ddots} -}}} -$$ -Then -$$\begin{gathered}\frac{{}_2F_1(a+1,b+1;a+b+c+1;x)}{{}_2F_1(a,b+1;a+b+c;x)} - = S \left( x;{\frac { \left( b+1 \right) \left( b+c \right) }{ \left( a -+b+c+1 \right) \left( a+b+c \right) }}, -\right.\hfill\\ -\left. -{\frac { \left( a+1 \right) - \left( a+c \right) }{ \left( a+b+c+2 \right) \left( a+b+c+1 \right) -}}, -{\frac { \left( b+2 \right) \left( b+c+1 \right) }{ \left( a+b+c+3 - \right) \left( a+b+c+2 \right) }}, - \right.\\ -\hfill -\left. - {\frac { \left( a+2 \right) - \left( a+c+1 \right) }{ \left( a+b+c+4 \right) \left( a+b+c+3 - \right) }},\dots \right) - \end{gathered} -$$ -and -$$\begin{gathered} -\frac{{}_2F_1(a,b+1;a+b+c;x)}{{}_2F_1(a,b;a+b+c;x)} -=S \left( x,{\frac {a}{a+b+c}}, -{\frac { \left( b+1 \right) \left( b+c - \right) }{ \left( a+b+c+1 \right) \left( a+b+c \right) }}, - \right.\hfill\\ - \left. - {\frac { - \left( a+1 \right) \left( a+c \right) }{ \left( a+b+c+2 \right) - \left( a+b+c+1 \right) }}, - {\frac { \left( b+2 \right) \left( b+c+1 - \right) }{ \left( a+b+c+3 \right) \left( a+b+c+2 \right) }}, - \right.\\ - \hfill - \left. - {\frac { - \left( a+2 \right) \left( a+c+1 \right) }{ \left( a+b+c+4 \right) - \left( a+b+c+3 \right) }}, \dots\right) -\end{gathered} -$$ -The first of these continued fractions is Gauss's well-known continued fraction, and the second can easily be derived from the first. It follows from these formulas that the coefficients of the Taylor series for $\log{}_2F_1(a,b;a+b+c;x)$ are rational functions of $a$, $b$, and $c$ with positive coefficients.<|endoftext|> -TITLE: Finite homotopy limits commute with sequential homotopy colimits -QUESTION [6 upvotes]: I would like to know for what kind of model category finite homotopy limits commute with sequential homotopy colimits. Would cofibrantly generated and finitely locally presentable be enough? It seems to be true in simplicial sets. -I found a proof of something close in Stephan Schwede's paper "Spectra in model categories and applications to the algebraic cotangent complex", but I wasn't able to deduce the statement above from his. -Jacob Lurie has a proof of the analogous statement for colimits and limits of quasi-categories, in "Higher topos theory". -Anything related would be much appreaciated! - -REPLY [7 votes]: In combinatorial model categories finite limits commute with (sufficiently large) filtered homotopy colimits. Suppose, for simplicity, that the combinatorial model category is simplicial and generating cofibrations have $\lambda$-presentable domain and codomain. In this case $\lambda$-filtered colimits are homotopy colimits. Suppose, in addition, that the underlying locally presentable model category is $\lambda$-locally presentable. Then $\lambda$-filtered colimits commute with $\lambda$-small limits. Finite limits are $\lambda$-small for all $\lambda$. -Lets say that the category $J$ indexing the homotopy limit is finite if it has finitely many objects and morphisms, and the diagram $EJ$ of simplicial sets serving as a cofibrant replacement of the constant diagrams of points indexed by $J$ in the projective model structure on ${\cal S}^J$ has a finite simplicial set in each entry. For example, a finite group is not a finite category by this definition. A finite homotopy limit is a homotopy limit over a finite diagram. -Suppose that $\cal M$ is a simplicial $\lambda$-combinatorial model category and $F\colon J\to \cal M$ a finite diagram. Then $\mathrm{holim}_J F$ may be computed as a weighted limit with the weight $EJ$, in other words this is an end construction: -$$ -\mathrm{holim}_J F = \mathrm{hom}(EJ, F), -$$ -which is a finite weighted limit commuting with $\lambda$-filtered colimits, hence, commuting with $\lambda$-filtered homotopy colimits. In particular, if $\lambda=\aleph_0$, then filtered homotopy colimits commute with finite homotopy limits.<|endoftext|> -TITLE: Deformation of foliation -QUESTION [6 upvotes]: Suppose $\kappa$ is a no-where vanishing 1-form, then its kernel is integrable is equivalent to condition $d\kappa \wedge \kappa = 0$. -My question is, can such foliation smoothly deformed such that actually $d\kappa = 0$? -EDIT: Smooth means real $C^\infty$, and I want the $\kappa$ to be nowhere-vanishing along the deformation. Thanks Loïc Teyssier pointing out the lack of context. -Thank you. - -REPLY [15 votes]: Although jvp's argument is correct, there's a more elementary argument that a foliation on $S^3$, or any closed integral homology sphere $M$ ($b_1(M)=0$), cannot admit a foliation defined by a closed (nowhere zero) 1-form. The point is that if $d\kappa=0$, then since $H_1(M;\mathbb{R})=0$, there exists a smooth function $f:M\to \mathbb{R}$ so that $\kappa=df$. Let $p\in M$ be a point at which the maximum of $f$ is achieved, then $\kappa_p=df_p = 0$, contradicting that $\kappa$ is nowhere zero. -In fact, there is a complete characterization of foliations defined by closed nowhere-zero 1-forms. Suppose $\kappa$ is a closed, nowhere zero 1-form. Then $ker(\kappa)$ defines an integrable plane field, which integrates to a foliation $\mathcal{F}$ with a transverse measure given by integration agains $\kappa$. A closed manifold with a measured foliation must fiber over $S^1$, and $\kappa$ may be deformed to a closed nowhere zero 1-form $\alpha$ with integral periods, which defines a submersion to $S^1$, and therefore a fibration $\Sigma \to M \to S^1$, where $\Sigma$ is a closed surface tangent to $ker(\alpha)$. -The deformation argument is simple. Take $k=b_1(M)$ loops $c_1,\ldots,c_k$ which generate $H_1(M;\mathbb{Z})/Torsion$. For any closed 1-form $\beta$, we obtain a $k$-tuple of periods $(\beta(c_1),\ldots,\beta(c_k))\in \mathbb{R}^k$. Choose $k$ closed 1-forms $\alpha_1,\ldots,\alpha_k$ generating $H^1(M;\mathbb{R})$. Then for small $\epsilon$, we have $\kappa+\sum t_i \alpha_i$ is a nowhere zero closed 1-form for $|t_i|< \epsilon$. Moreover, the period map takes these forms to a small open set in $\mathbb{R}^k$. This set must contain a point $\kappa+\sum t_i\alpha_i$ with rational coordinates, so we choose $\alpha=m(\kappa +\sum t_i \alpha_i )$, where $m$ is the $lcm$ of the denominators of the coordinates. Clearly $\alpha$ smoothly deforms through closed nonzero 1-forms to $\kappa$, and $\alpha$ has integral periods. -The foliations described by closed 1-forms may be characterized up to isotopy: there is a unique such foliation for each element in the projective cone over the fibered faces of the Thurston norm unit ball. If two nowhere zero closed 1-forms are cohomologous, one can show that the corresponding foliations are isotopic, and the same if one is a multiple of the other (by rescaling). -Associated to each face of the Thurston norm unit ball, Thurston showed that there is a canonical element of $H^2(M)$ which realizes the Thurston norm when evaluated on norm-minimizing surfaces in that face. For a nowhere zero integrable 1-form $\alpha$ defining a foliation (so $\alpha \wedge d\alpha=0$), the Euler class of $ker(\alpha)$ has the property that it gives the Euler characteristic of closed leaves. Deformation of integrable 1-forms preserves the Euler class. Thus, if an integrable 1-form is deformable to a closed 1-form, then it must induce the same Euler class as the class associated to a fibered face of the Thurston norm unit ball. For example, associated to each homology class on the boundary of a fibered face, there is a depth one foliation realizing the Euler class of the fibered face, but which is not itself cohomologous to a fibration. To completely answer your question, one would therefore like to know if integrable 1-forms which define the same Euler class are deformations of each other. This question is partially resolved. Suppose that two foliations of a 3-manifold are deformable, then their tangent bundles are homotopic. Eynard-Bontemps proved a converse, that if two oriented foliations have homotopic tangent bundles, then they are deformable, but only through $C^1$ foliations. If the two foliations are taut, then the deformation may be made smoothly. In your case, at one end the foliation is taut (when defined by a closed 1-form), but the initial foliation defined by $\kappa$ might not be taut.<|endoftext|> -TITLE: Can we always permute Cohen reals? -QUESTION [6 upvotes]: Consider the Cohen forcing, and suppose that $\dot x,\dot y$ are names for reals, which are not in the ground model (i.e. $1$ forces that neither is in the ground model). -Can we always find an automorphism mapping $\dot x$ to $\dot y$? -The answer is negative, as Andreas Blass points out. But let me refine the question a lot more. -Suppose $p$ is a condition which forces that $\dot x$ and $\dot y$ are both generic with respect to the restriction of the forcing to $A$ (i.e. take only conditions whose domain is a subset of $A$ and complete that to the subalgebra of the Cohen forcing). Is there an automorphism $\pi$ such that $p\Vdash\pi\dot x=^*\dot y$? -Of course, this depends on how you consider the Cohen forcing, so let's take the "richest" way, and consider it as the completion of functions from finite sets of integers to $\{0,1\}$. - -REPLY [10 votes]: No, because reals not in the ground model can be different in definable (with parameters from the ground model) ways. For example, $\dot x$ might be (forced by all conditions to be) a Cohen-generic subset of $\omega$, while $\dot y$ is the intersection of $\dot x$ with the set of even numbers. Then $\dot y$ is disjoint from an infinite ground-model set of natural numbers but $\dot x$ is not. -EDIT to answer the edited version of the question: The answer is still negative, if $A$ is an infinite, coinfinite, ground-model subset of $\omega$. Let $\dot c$ be (the canonical name of) the Cohen subset of $\omega$ directly added by the forcing. Let $\dot x$ be the intersection of $\dot c$ with $A$. Let $\dot y$ be a copy on $A$ of all of $\dot c$, i.e., the $n$-th element of $A$ is in $\dot y$ iff $n\in\dot c$. Then both $\dot x$ and $\dot y$ are generic with respect to the Cohen forcing over $A$. They are not related by an automorphism of the forcing, because $\dot y$ generates (over the original ground model) the whole forcing extension (the same as $\dot c$) while $\dot x$ generates only an intermediate model.<|endoftext|> -TITLE: Is $SL_1(D)$ toplogically finitely generated, for $D$ a division algebra over a local field? -QUESTION [5 upvotes]: I've been struggling with this one all day, and I was wondering if someone can give me a hand with the proof. I'm not even sure if the group in question is finitely generated, so I would appreciate if anyone will tell me otherwise as well. -remark 1 I've posted this question on Math.stackexchange as well, and am posting here since I don't seem to get any replies. I don't know if this question is "MO standard", but I guess there's no harm in trying :-P.. -remark 2 For infinite topological groups, I write $G$ is finitely generated to mean that $G$ has a dense subgroup which is finitely generated (i.e. $G$ is topologically finitely generated). -Let $F\supseteq \mathbb{Q}_p$ be a local field, and let $D$ be a finite dimensional division algebra over $F$, of degree $n$ (i.e $\dim_F D=n^2$). -It is known that $D$ contains an unramified extension $E$ of $F$, such that $E/F$ is cyclic Galois. It follows that $D$ contains an element $u\in D^\times$, and that there is some $\pi\in F$ such that $D=\bigoplus_{j=0}^{n-1}u^j E$, $u^n=\pi$ and such that the restriction of the map $x\mapsto u^{-1}xu$ to $E$ is a generator of the Galois group $\mathbf{G}(E/F)$. -Thus, $D$ embeds into the $F$ algebra $M_n(E)$ of $n\times n$ matrices over $E$, via the left regular action $x\mapsto \lambda_x$ where $\lambda_x:E^n\to E^n$ is defined by $$\lambda_x(y_0,\ldots,y_{n-1})=x\cdot(\sum_{j=0}^{n-1}u^jy_j).$$ -Once this is done, one can define the reduced norm on $D$ by $$Nrd_{D/F}(x)=\det\lambda_x.$$ -My interest specificaly is in the groups $G:=SL_1(D):=\lbrace x\in D\mid Nrd_{D/F}(x)=1\rbrace$. I'm trying to understand whether this group is finitely generated or not. -It is clear that the embedding defined above maps $G$ into the group $SL_n(E)$. Once we use the fact that the group $SL_n(E)$ is generated by transvections (i.e matrices with $1$s along the diagonal and at most one additional non-zero entry), it is not hard to use the local structure of $E$ to prove that $SL_n(E)$ is finitely generated. -Unfortunately, a subgroup of a finitely generated group need not be finitely generated in general. If anyone knows of a criterion that might prove why this case is special that would be a great help. -On additional possibility, is to use some heavy tools from asymptotic group theory (even though I must admit this turns out a bit cumbersome..): -We call a group $G$ positively finitely generate (abbreviated PFG), if there exists a number $k\in\mathbb N$ such that the probability that an arbitrary $k$-tuple $(x_1,\ldots, x_k)\in G^k$ generates $G$, is positive. Clearly, any PFG group is finitely generated (although the converse is not true, e.g the free group in two generators is not PFG). -For any $n\in\mathbb N$ we define $m_n(G)$ to be the number of maximal subgroups of $G$ with index $n$. It is a theorem of Avinoam Mann that the group $G$ is PFG if and only if $m_n(G)$ is bounded by some polynomial in $n$. So it would suffice to prove that the (maximal) subgroups of $G$ have polynomial growth. -As I said, this sort of proof seems a bit of an overkill, but if someone knows of a reference (or a direct proof) to why the group $G$ has polynomial (maximal) subgroup growth that will be swell :-). -Anyway, this question turns out to be a bit lengthy, so I'll stop here. I would very much appreciate If someone can offer any hints as to whether or not $SL_1(D)$ is finitely generated. -Thank you very much. - -REPLY [6 votes]: Yes, this follows from finiteness properties of arithmetic groups and the strong approximation theorem for simply connected groups. -Choose a global field $K$ with a non-archimedean place $v$ such that $K_v \simeq k$; this can be found via elementary approximation arguments. By our knowledge of the Brauer groups of $K$ and $k$ via class field theory, there is a central division algebra $A$ over $K$ such that $A_v \simeq D$ over $k$. -Let $G$ be the absolutely simple semisimple $K$-group informally denoted as ${\rm{SL}}_1(A)$ (to be precise, $G$ represents the functor $R \rightsquigarrow -{\rm{SL}}_1(A \otimes_K R)$ on the category of $K$-algebras), so $G(k) = {\rm{SL}}_1(D)$. Hence, it suffices to show that $G(k)$ contains a dense finitely generated subgroup. -By general "spreading out" arguments, for any connected semisimple $K$-group $H$ there is a finite set of places $S$ of $K$ containing the archimedean places and such that $H$ extends to a semisimple group scheme $\mathcal{H}$ over $O_{K,S}$, so $H$ is quasi-split at all places outside $S$ and hence (by the Borel--Tits structure theory) has positive rank at all places outside $S$. Thus, by enlarging $S$ to contain at least two such places, we can arrange that $$\sum_{w \in S} {\rm{rank}}(H_{k_w}) \ge 2.$$ -We apply this with $H = G$ and therefore denote $\mathcal{H}$ as $\mathcal{G}$, and we enlarge $S$ if necessary so that $v \in S$. -Consider the arithmetic $S$-group $\Gamma = \mathcal{G}(O_{K,S})$ inside $G(K)$. In the number field case this is finitely generated, by general finiteness results for arithmetic $S$-subgroups of the group of rational points of absolutely simple semisimple group over number fields. In the function field case the same holds because of the rank-sum condition arranged above (by a theorem of Behr). So it suffices to prove that $\Gamma$ has dense image in $G(k) = G(K_v)$. -Let $S' = S - \{v\}$ (non-empty). Since $G$ is absolutely simple and simply connected, by the strong approximation theorem we know that $G(K)$ is dense in $G(\mathbf{A}_K^{S'}) = G(K_v) \times G(\mathbf{A}_K^S)$, where $\mathbf{A}_K^{S'}$ is the adele ring of $K$ with the $S'$-factors removed. Let $\widehat{O}_{K,S} \subset \mathbf{A}_K^S$ be the open subring $\prod_{w \not\in S} O_{K_w}$, so $\mathcal{G}(\widehat{O}_{K,S})$ is an open subset of $G(\mathbf{A}_K^S)$ that meets $G(K)$ in exactly $\mathcal{G}(O_{K,S}) = \Gamma$. -Hence, if $U \subset G(K_v)$ is a non-empty open subset then $U \times \mathcal{G}(\widehat{O}_{K,S})$ is an open subset of $G(\mathbf{A}_K^{S'})$ whose intersection with $G(K)$ is identified (via ${\rm{pr}}_v$) with $U \cap \Gamma$ inside $G(K_v)$. But recall that $G(K)$ is dense in $G(\mathbf{A}_K^{S'})$, so its intersection with $U \times \mathcal{G}(\widehat{O}_{K,S})$ is non-empty. In other words, $U \cap \Gamma$ is non-empty inside $G(K_v)$. Since $U$ was arbitrary, this says that $\Gamma$ is dense in $G(K_v) = G(k)$ as desired. -This argument is quite robust with respect to the permitted generality of the initial connected semisimple $k$-group, the main point being the existence of the "globalization" of some $K$. One is then led to ask: if $G$ is a connected semisimple group over $k$ that is absolutely simple then does $G$ admit a "globalization" relative to some pair $(K,v)$ consisting of a global field $K$ and place $v$ such that $K_v \simeq k$ (and if so then can this be proved without case-checking). But this refinement was not part of the question, so I'll stop here.<|endoftext|> -TITLE: Efficient rank-two updates of an eigenvalue decomposition (or more generally SVD) -QUESTION [14 upvotes]: Let $A$ be a symmetric matrix with eigenvalue decomposition $UDU^T$. Golub, et al.1 and Bunch, et al.2 have shown that given such an $A$, the eigenvalue decomposition of $A+\rho xx^t$ may be computed efficiently. -Has anything similar been done for the case where the update is of the form $A+B$, where $B=uv^t+vu^t$ is a rank-two symmetric matrix (note we can't just do two rank-one symmetric updates)? -What if we relaxed the insistence that $B$ be symmetric and asked instead for an efficient computation of the SVD of the update $A+B$? -1Golub, Gene H., Some modified matrix eigenvalue problems, SIAM Rev. DOI: 10.1137/1015032, JSTOR, ZBL0227.65025. -2Bunch, J.R., Nielsen, C.P. & Sorensen, D.C. Rank-one modification of the symmetric eigenproblem. Numer. Math. 31, 31–48 (1978). DOI:10.1007/BF01396012, eudml, ZBL0369.65007 - -REPLY [9 votes]: Just wanted to note that if one knows that the rank two update is of the form $uv^t + vu^t$ and one knows $v$ and $u$ then it is easy to find $x,y$ such that $uv^t + vu^t = xx^t - yy^t$. -$x,y = \sqrt{\frac{|v|}{2|u|}} (u \pm \frac{|u|}{|v|}v)$. Once we know $x,y$ then the problem reduces to two symmetric rank one updates and we can use the symmetric rank one update methods cited in the question. -proof : $ xx^t - yy^t = \frac{|v|}{2|u|}( u + \frac{|u|}{|v|}v)(u + \frac{|u|}{|v|}v)^t - \frac{|v|}{2|u|}( u - \frac{|u|}{|v|}v)(u - \frac{|u|}{|v|}v)^t = \frac{|v|}{2|u|}(\frac{|u|}{|v|} (2uv^t + 2vu^t)) = uv^t + vu^t$ -I haven't checked if doing so would be more efficient than the more general method of Brand but I suspect that it would be.<|endoftext|> -TITLE: normalized laplacian spectrum of trees -QUESTION [8 upvotes]: Is it known for which class of graphs the normalized laplacian has only simple eigenvalues (i.e., with multiplicity one)? In particular, are there trees (or perhaps a specific class of trees) whose normalized laplacian has only simple eigenvalues? -Thanks. - -REPLY [4 votes]: It is a partial answer for your question: -For $P_n$, the form of normalized laplacian matrix is three diagonal and with some calculations, we can show that all its normalized laplacian eigenvalues are simple. For example, the normalized laplacian spectrum of $P_n$ for $n=2,3,4,5$ are: -$P_2:$ $[0,2]$, -$P_3: [0,1,2]$, -$P_4: [0,2,\frac{1}{2},\frac{3}{2}]$, -$P_5: [0,1,2,1-\frac{\sqrt2}{2},1+\frac{\sqrt2}{2}]$. -In the paper with name: -"Eigenvalues of normalized Laplacian matrices of fractal trees and dendrimers: Analytical results and applications", -There are some useful results about fractal and Cayley tree and their normalized laplacian spectrum and their degeneracy. Specially, in section $III$ part $B$, you can find some useful calculation. For example, the fractal $F_1$ for $m=0$ has simple normalized spectrum and is an other example for your question. I wrote a Maple program that can calculate the requested spectrum. If it is useful, I can send it for you. -I tested many trees up to 10 vertices with package RandomTree in Maple, and it seems $star like$ trees have the property that you want. These graphs are interesting, since they studied a lot for determining are they $DS$ or not. -Also, since almost all trees have diameter 2 and these trees are $star$, so almost all trees do not have the property that you want. In contrast, a lot of trees up to 10 vertices have the property that you want.<|endoftext|> -TITLE: Are all complete finitely axiomatizable first order theories $\aleph_0$-categorical? -QUESTION [10 upvotes]: Suppose $T$ is a complete first order theory with a finite axiomatization. Must $T$ be $\aleph_0$-categorical? If not are there any simple examples of finitely axiomatized complete first order theories which are not $\aleph_0$-categorical? - -REPLY [16 votes]: The answer is no: for a simple example, take $Th(\mathbb Z,<)$. The axioms are: - -$<$ defines a linear order; -every element has a immediate successor and an immediate predecessor. - -Getting examples which are lower in the classification (e.g. stable) is much harder, and this is the point of the exercise in Marker's book which Garrett mentioned in the comments above.<|endoftext|> -TITLE: “is topologically mixing” vs. “is topologically transitive” in the defition of chaos -QUESTION [9 upvotes]: This question is cross-posted from MSE, since it hasn't gotten an answer there for over 72 hours. - -Wikipedia gives essentially "is topologically mixing and has dense periodic periodic orbits" - -as the definition of chaos, and this paper shows that its (the paper's) definition of chaos -is equivalent to "is topologically transitive and has dense periodic orbits". - -Clearly, every topologically mixing map is topologically transitive, and there are topologically transitive maps that are not topologically mixing (such as an irrational rotation of the circle). -Is there a map that is topologically transitive and has dense periodic orbits - -but is not topologically mixing? $\:$ If yes, can its space be metrizble? - -REPLY [5 votes]: Another important example of a topologically transitive and not topologically mixing system (although it does not answer exactly your question) is the suspension flow over a topologically mixing system. -Imagine that $f:X\to X $ is mixing. Then construct the suspension space -$X\times\mathbb{R}$, and the space $Y$ obtained after quotient by the equivalence relation $(x,1)\sim (f(x),0)$. And consider the flow $\phi_t:(x,s)\in Y\mapsto (x,t+s)$. -This dynamical system is topologically transitive, has dense orbits if it is the case for $f:X\to X$, but is not topologically mixing. -But of course, it is a flow, not a map.<|endoftext|> -TITLE: Menon’s identity -QUESTION [6 upvotes]: I also put this question in stackexchange, but remained unanswered. https://math.stackexchange.com/questions/506996/menons-identity -Let $G$ be a group of order $n$. Consider an action of $U_n$, the group of invertible residues modulo $n$, on $G$. With each $s\in U_n$ we associate the permutation of $G$, $\psi_s$, defined by -$$\psi_s(g)=g^s,$$ -for all $g\in G$. -Under this action two elements belong to the same orbit if and only if they generate the same cyclic subgoup. Thus the number of orbits, which we shall denote by $c(G)$, is equal to the number of cyclic subgroups of $G$. By Burnside's Lemma we have the relation: - $$c(G)=\frac{1}{\varphi(n)}\sum_{s\in U_n}|F(s)|.$$ -Here $\varphi(n)$ is the Euler totient function, while $F(s)$ is the fixed set of $\psi_s$, that is, the set of elements in $G$ which satisfy the equation $x^{s-1}=1$. -If $G$ is a cyclic group or order $n$, then $c(G)$ is equal to $d(n)$, the number of divisors of $n$, and $|F(s)|$ is equal to $(s-1, n)$. We conclude that -$$d(n)=\frac{1}{\varphi(n)}\sum_{s\in U_n}(s-1, n).$$ -This identity is known as Menon’s identity and this dedution appears in the paper -"A remark on the number of cyclic subgroups of a finite group" by I. M. Richards. -In this paper the the author states without proof that for a polynomial $f(x)$ in $\mathbb Z[x]$ we have -$$\sum_{s\in U_n}(f(s), n)=\varphi(n)\sum_{d|n}|\{r\in U_d; f(r)\equiv 0 \;\;mod\;d\}|.$$ -How can we prove this last identity? - -REPLY [4 votes]: [This is a corrected version of my previous answer, in which I incorrectly said that the identity was false. I apologize for getting it wrong the first time.] -Here is a proof of the identity. Since both sides are multiplicative functions of $n$, it suffices to prove the identity when $n$ is a prime power, say $n=p^k$. Write -$$ -C_i = \lvert\{r\in U_{p^i}: f(r)\equiv 0 \bmod p^i\}\rvert. -$$ -Then the number of $s\in U_n$ for which $(f(s),n)=p^k$ is $C_k$, and for $0 -TITLE: Is there a forcing closure? -QUESTION [7 upvotes]: The main theorem of forcing says that for any c.t.m of $ZFC$ like $M$ and for all partial order $\mathbb{P}$ and $\mathbb{P}$-generic $G$ over $M$, there is a c.t.m of $ZFC$, like $N$ such that $N$ is the least (by inclusion order) c.t.m of $ZFC$ which $M\subseteq N$ and $G\in N$. Now the question is: -Question (1): Let $I$ be an index set and $\lbrace M_{i}\rbrace_{i\in I}$ a family of c.t.m s of $ZFC$, $\lbrace \mathbb{P}_{i}\rbrace_{i\in I}$ a family of partial orders and $\lbrace G_{i}\rbrace_{i\in I}$ a family of sets such that for all $i\in I$, $G_{i}$ is a $\mathbb{P}_{i}$ - generic filter over $M_{i}$. -Is there a c.t.m $N$ of $ZFC$ such that $N$ be the least c.t.m of $ZFC$ (by inclusion order) with the property: $\forall i\in I~~~~~M_{i}\subseteq N~~\wedge~~G_{i}\in N$? -We call $N$ the forcing closure of $\lbrace (M_{i},\mathbb{P}_{i},G_{i})~|~i\in I\rbrace$. -Question (2): Is any c.t.m of $ZFC$, a forcing closure of some family $\lbrace (M_{i},\mathbb{P}_{i},G_{i})~|~i\in I\rbrace$? - -REPLY [10 votes]: $\newcommand{\of}{\subset}\newcommand{\Q}{\mathbb{Q}}$ -In general, there may be no such $N$, even when $I$ has only two elements. To see this, suppose that we have a countable transitive model $M$, and suppose there is no countable transitive model with more ordinals. Let $z$ be a real coding the ordinals of $M$ as a relation on $\omega$, so that $z$ is a forbidden real, which cannot be added to $M$ without collapsing all the ordinals. Now, build two $M$-generic Cohen reals $c$ and $d$, in stages. First, enumerate the dense sets in $M$ as $D_0,D_1,\ldots$, and let $c_0$ be any element of $D_0$. Let $d_0$ be all-zeros, of exactly the same length, followed by $1$, followed by the digit $z(0)$, followed by an extension putting it into $D_0$. Now extend $c_0$ by adding $0$s until it has the same length as $d_0$, followed by $1$, and then make an extension putting it into $D_1$. Now extend $d_0$ to $d_1$ by adding $0$s, then a $1$, then $z(1)$, and then extend into $D_1$. Continue in this way. Each of $c=\bigcup c_n$, $d=\bigcup d_n$ is an $M$-generic Cohen real, so we have the forcing extensions $M[c]$ and $M[d]$, but if $c$ and $d$ appear together in a model $N$, then by inspecting the blocks of zeros, we can define the number $z$, which collapses all the ordinals of $M$. In this case, $N$ would have to be taller than $M$, and this would violate our assumption. -Basically, the example shows that without a mutual-genericity hypothesis, the forcing extensions of a given model $M$ are not upward directed. I find it more natural to look for an extension $N$ with the same ordinals as $M$, and furthermore, to look for an $N$ that is itself a forcing extension of $M$. Thus, one is looking at the generic multiverse of $N$ (the collection of all models that are obtainable from $M$ by going to a forcing extension or a ground). And here there are some interesting things to say. -In my paper Set-theoretic geology, joint with Gunter Fuchs and Jonas Reitz, we prove the following: -Theorem. Suppose that $W$ is a countable model of ZFC, -and $$W[G_0]\of W[G_1]\of\cdots\of W[G_n]\of\cdots$$ is an increasing -sequence of forcing extensions of $W$, with $G_n\of\Q_n\in -W$ being $W$-generic. If the cardinalities of the $\Q_n$'s -in $W$ are bounded in $W$, then there is a set-forcing -extension $W[G]$ with $W[G_n]\of W[G]$ for all -$n<\omega$. -The bounded cardinality assumption is very natural, since otherwise one could collapse more and more cardinals of $M$, or add more and more Cohen reals over $M$, and these extensions could clearly not be amalgamated.<|endoftext|> -TITLE: What is the set of possible densities of pointless members in a family of rational curves over $\mathbb{Q}$? -QUESTION [9 upvotes]: This is a sequel to the following previous MathOverflow posts, but the question itself appears to be of a different flavor, being limited to rational curves: -Are most curves over Q pointless? -How many curves in a family possess a rational point? -Are most cubic plane curves over the rationals elliptic? -Let $B$ be a geometrically irreducible quasi-projective variety over $\mathbb{Q}$ such that $B(\mathbb{Q})$ is infinite, and consider a flat family $X \to B$ of geometrically irreducible (projective) rational curves. Let $h : B(\bar{\mathbb{Q}}) \to \mathbb{R}$ be a Weil height associated to a projective embedding of $B$, and consider -$$ -c := \limsup_{T \to +\infty} \frac{\big\{ b \in B(\mathbb{Q}) \mid h(b) \leq T, \, X_b(\mathbb{Q}) = \emptyset \big\}}{\big\{ b \in B(\mathbb{Q}) \mid h(b) \leq T \big\}} \in [0,1]. -$$ -[Most likely, the limit actually exists. ] -The values of $c$ which arise in this way (with some $X/B$ and $h$) form a certain countable subset of $[0,1]$ which includes the two endpoints. -Question. What is this set $\{c\} \subset [0,1]$? -[Remark. Obviously this question has many variants, although it is conceivable that the answer could be the same in all of them. For example, we may restrict to families over $B = \mathbb{A}_{\mathbb{Q}}^1$. ] - -REPLY [8 votes]: Edit: I have clarified a bit more the relationship between conic bundles and quaternion algebras and the relationship to weak approximation, and tidied up some typos. -I have myself recently started studying problems of this type, and it turns out that there is much to be said and such questions lead to very rich research problems. As far as I am aware, the first person to study such problems in any kind of generality was Serre [1]. -I work in the following setting: Let $B$ be a projective variety over $\mathbb{Q}$ with a Zariski dense set of rational points and let $X$ be a non-singular variety together with a flat morphism $\pi:X \to B$ such that every fibre is a non-singular projective curve of genus zero. Then it is well-known that such a curve is isomorphic to a plane conic, and flatness implies that all the fibres are plane conics. Hence we obtain a conic bundle. To such a conic bundle one may asssociate an element $Q_\pi$ of $Br_2(\mathbb{Q}(B))$. If for example the generic fibre of $\pi$ is given by -$$ax^2 + by^2 = z^2, (*)$$ -where $a,b \in \mathbb{Q}(B)$, then $Q_\pi$ is the class given by the quaternion algebra $(a,b)$. This construction is essentially a bijection, namely given a quaternion algebra $(a,b)$ over $\mathbb{Q}(B)$ one may associate to it in an obvious way the conic bundle $(*)$. This element controls most of the arithemtic of the family. One can show that $Q_\pi$ is trivial if and only if $\pi$ admits a rational section (we will use this fact later to address Felipe's question). Following -Serre, I instead consider the limit -$$C(B,\pi,H) = \limsup_{T \to +\infty} \frac{\big\{ b \in B(\mathbb{Q}) \mid H(b) \leq T, \, X_b(\mathbb{Q}) \neq \emptyset \big\}}{\big\{ b \in B(\mathbb{Q}) \mid H(b) \leq T \big\}},$$ -where $H$ is a choice of height function on $B$. -Serre [1] basically only considers the case where $B = \mathbb{P}^n$. Let $H$ be the usual naive height on $\mathbb{P}^n$. Then Serre shows that -$$C(\mathbb{P}^n, \pi, H)=0, $$ -as soon as there exists a divisor $D$ in $\mathbb{P}^n$ such that the residue of $Q_\pi$ at $D$ is non-trivial (he in fact shows something more precise, see [1]). -In particular in such cases, "most" conics in the family do not contain a rational point. Moreover, one knows that an element $A \in Br(\mathbb{Q}(T_1,..,T_n))$ admits a non-zero residue if and only if $A \not \in Br(\mathbb{P}^n_{\mathbb{Q}}) = Br(\mathbb{Q})$. In particular, we see that $C(\mathbb{P}^n,\pi,H) >0$ if and only if $Q_\pi=0$, which by the above remark happens if and only if $\pi$ obtains a rational section (which in particular implies that we must have $C(\mathbb{P}^n,\pi,H)=1$). I believe this answers your question in this case. Your specific conic bundle -$$qx^2 -y^2 = z^2, (**)$$ -corresponds to the class of the quaternion algebra $(t,-1)$ over $Br(\mathbb{Q}(t))$. In this case, Serre's more general result in fact shows that -$$\#\{q \in \mathbb{Q}: H(q) \leq T, (**) \text{ has a solution}\} -\ll \frac{T^2}{\log T}.$$ -Here is how one may obtain conic bundles which do not admit rational sections, but for which $C(B,\pi,H) =1$. Suppose that $B$ admits a Brauer group element $A \in Br(B)$ which is the class of a non-trivial quaternion algebra, for which $A \otimes_{\mathbb{Q}_v} \mathbb{Q}_v = 0$ for every place $v$ of $\mathbb{Q}$ (such elements may e.g. be given by elements of the Tate-Shaferavich groups of algebraic tori, see Sansuc's classical paper [2]). As by assumption $A \otimes_{\mathbb{Q}_v} \mathbb{Q}_v = 0$ for all $v$, we see that every conic in the family is everywhere locally soluble. The Hasse-Minkowski theorem therefore implies that every conic in the family contains a rational point, hence $C(B,\pi,H) =1$. However, since $A$ was non-trivial we see that $\pi$ does not admit a rational section. -Your question in the general case is highly non-trivial. Assuming that the results proved by Serre hold in any kind of generality, one would expect that $C(B,\pi,H)=0$ if $Q_\pi \not \in Br(B)$. To obtain positive values of $C(B,\pi,H)$ which are not $1$, one would need a quaternion algebra $A$ over $\mathbb{Q}(B)$ for which there exists some place $v$ of $\mathbb{Q}$ and some $b_1,b_2 \in B(\mathbb{Q})$ such that $A_v(b_1) \neq 0 \in Br(\mathbb{Q}_v)$ but $A_v(b_2) = 0 \in Br(\mathbb{Q}_v)$. This would force some conics in the family to fail to have $\mathbb{Q}_v$-points (hence not have $\mathbb{Q}$-points), but at least make sure that there are also some conics in the family which still have $\mathbb{Q}_v$-points (hence could still have $\mathbb{Q}$-points). Such elements are exactly the ones which give rise to a failure of weak approximation for $B$ via the Brauer-Manin obstruction. So the values of $C(B,\pi,H)$ which may be obtained should intimately depend on the extent to which $B$ fails weak approximation, at least if the Brauer-Manin obstruction is the only one to weak approximation (for more details on the Brauer-Manin obstruction, see [2]). -Problems of this type naturally admit numerous generalistions e.g. to Brauer-Severi varieties and to number fields, where many of the above remarks still hold true. -[1] Serre - Spécialisation des éléments de $Br_2(\mathbb{Q}(T_1,..,T_n))$. -[2] Sansuc - Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.<|endoftext|> -TITLE: Reverse Ricci Flow and Longtime Existence -QUESTION [7 upvotes]: The usual Ricci flow and normalized Ricci flow for surfaces are -$$ \partial_t g = -2Kg $$ -and -$$ \partial_t g = -2Kg + 2sg,$$ -where $K$ is the Gaussian curvature and $s$ is its average. -The latter equation can be solved on the interval $[0,\infty)$ and converges to a metric of constant Gaussian curvature. -Now consider the reverse flow -$$\partial_t = 2Kg - 2sg,$$ -which should function as some sort of reverse flow. What can you say about its long-time existence? -I had a discussion about this with someone, who argued that there should be no time-reversal of that flow as the flow would not know "where to go" if you start in, say, a sphere of constant curvature (because of the above convergence result; this flow should move the metric away from a constant curvature metric). -But clearly, at least for round spheres the above equation has a unique solution, as $\partial_t g = 0$. So why can't a round sphere, for instance, be a repelling fix point of the flow... Writing this, I realize that this flow, if it existed, would have one crucial disadvantage: $g(t)$ would not depend continuously on $g$ in the $C^k$-topology. - -REPLY [7 votes]: I've combined two answers into one. -Bando and Kotschwar's real analyticity in space and time. As Ian Agol wrote, Shigetoshi Bando, by way of summable derivative of -curvature estimates of Bernstein type, proved that for a solution $(M,g(t)),$ -$t\in(0,T)$, to the Ricci flow on a closed manifold, at each $t$ the metric -$g(t)$ is real analytic (in space). These summable estimates and the -consequent real analyticity result can be localized; see arXiv.1111.0355 by -Brett Kotschwar. Also, in arXiv.1210.3083, Kotschwar proves that $g(t)$ is -real analytic in both space and time. This involves careful estimates for -$t^{k+2\ell}|\nabla^{k}\partial_{t}^{\ell}\operatorname{Rm}|^{2}$. As Thomas -Richard wrote, backwards uniqueness was obtained by Kotschwar in arXiv.0906.4920. -The King-Rosenau solution is the only nonround ancient solution $S^2$. The following somewhat unrelated facts come to mind regarding long time existence backwards. By this I mean an ancient solution to the Ricci flow on a closed surface. Either it is flat or $R>0$. In the latter case, if it is also Type I, Hamilton proved that the solution must be a round shrinking sphere or its $\mathbb{Z}_2$ quotient. If it is Type II, then Daskalopoulos, Hamilton and Sesum proved that it must be the King-Rosenau solution or its $\mathbb{Z}_2$ quotient. Note that the King-Rosenau solution is rotationally symmetric and invariant under a reflection. Since (loosely speaking) its backwards limits are two opposing cigar steady solitons, one may think of the King-Rosenau solution as a heteroclinic orbit joining two cigars (fixed points modulo conformal diffeomorphisms) to the round sphere (a fixed point modulo scaling). - -The cigar is the only nonflat noncompact 2-dimensional ancient solution. This additional comment is on $2$-dimensional noncompact long time solutions -to the backward Ricci flow, i.e., ancient solutions to the forward Ricci flow. -We indicate a key idea in the proof of Daskalopoulos and Sesum (Intern. Math. Res. Notices 2006) of their result -that any complete noncompact nonflat ancient solution to the Ricci flow -$(M^{2},g(t))$ with bounded curvature and finite width must be a cigar soliton. -Since $R>0$, there exists $f$ such that $\Delta f=-R$ (e.g., see A. Huber, -Comment. Math. Helv. 1957). Because $n=2$ implies $\operatorname{Ric}=\frac -{R}{2}g$, we can write a Bochner formula as -$$ -\Delta\left( R+|\nabla f|^{2}\right) =2|\operatorname{Ric}+\nabla^{2} -f|^{2}+4\frac{|\operatorname{div}(\operatorname{Ric}+\nabla^{2}f)|^{2}}{R}+H, -$$ -where the trace Harnack $H\doteqdot\Delta R+2|\operatorname{Ric}|^{2}-\frac -{1}{2}\operatorname{Ric}^{-1}(\nabla R,\nabla R)\geq0$ is nonnegative since -the solution is ancient. Remarkably, the right side of the display is the sum -of three nonnegative terms, whereas the left side is a divergence. Now take a -suitable exhaustion $\Omega_{i}$ of $M$. Integrating the display yields that -$\int_{\partial\Omega_{i}}\langle\nabla(R+\left\vert \nabla f\right\vert -^{2}),\nu\rangle ds$ is nonnegative, where $\nu$ is the unit outward normal to -$\partial\Omega_{i}$. It can be shown that this boundary integral tends to -zero as $i\rightarrow\infty$. This implies $\operatorname{Ric}+\nabla^{2}f=0$ -and, by the classification of $2$-dimensional steady Ricci solitons, we must -be on a cigar. -See S.-C. Chu (Comm. Anal. Geom. 2007) for the case where the width is not finite. -Added December 13, 2013. Consider $2$-dimensional complete noncompact nonflat -ancient solutions with bounded curvature. -(1) By Richard Hamilton, there are no such Type I solutions. -(2) By Sun-Chin Chu, there are no such solutions with infinite width. His work -is based on the works of Wan-Xiong Shi and of Lei Ni and Luen-Fai Tam. -It would be interesting to see if there are other proofs of (2). For example, -assume that $u$ on $M^{2}\times(-\infty,0)$ satisfies $\frac{\partial -u}{\partial t}=-\Delta u$, $u>0$, and $\lim_{t\rightarrow0}u(t)=\delta_{x_{0} -}$. -(i) Can one prove that $\frac{d}{dt}\int uRd\mu\leq0$? Note that $\int -\frac{\partial}{\partial t}(uRd\mu)=\int(-R\Delta u+u\Delta R)d\mu$ is likely -to be zero (can we prove spatial decay of the backward heat kernel?). -(ii) Does there exist $\alpha(t)$ with $\lim_{t\rightarrow-\infty}\alpha(t)=0$ -such that $u(x,t)\leq\alpha(t)$? Presumably one needs to use the infinite -width assumption here. Note that $\int(-\frac{\partial}{\partial t} -)(ud\mu)=\int(\Delta u+uR)d\mu$, while $\int uRd\mu\geq0$ is a bad sign for -showing $\int ud\mu$ decays backward in time. -If the answers to (i) and (ii) are yes, then $R(x_{0},0)\leq\alpha(t)\int -Rd\mu(t)\rightarrow0$ as $t\rightarrow-\infty$, which would imply that the -solution is flat, a contradiction.<|endoftext|> -TITLE: Origin of the Socle of a module -QUESTION [10 upvotes]: Where does the notation $\mbox{Soc}(M)$ (the sum of all simple submodules of a module $M$) first appear? - -REPLY [3 votes]: Just to establish some more reference points: -In the book -[Curtis, Charles W.; Reiner, Irving. Representation theory of finite groups and associative algebras. Pure and Applied Mathematics, Vol. XI Interscience Publishers, a division of John Wiley & Sons, New York-London 1962], -the socle of a module is defined, but no special notation is used. -In the paper -[Curtis, C. W.; Jans, J. P. On algebras with a finite number of indecomposable modules. Trans. Amer. Math. Soc. 114 1965 122--132.], -the notation $S(M)$ is used. -In the book -[Anderson, Frank W.; Fuller, Kent R. Rings and categories of modules. Graduate Texts in Mathematics, Vol. 13. Springer-Verlag, New York-Heidelberg, 1974], -the notation ${Soc} \; M$ is used.<|endoftext|> -TITLE: Intersection of compact sets in the unit interval -QUESTION [17 upvotes]: Let $\mathscr K$ be an uncountable set such that every $K\in\mathscr K$ is a compact subset of $[0,1]$ with positive Lebesgue measure. Does it then follow that there exists an uncountable $\mathscr A\subseteq\mathscr K$ with $\bigcap\,\mathscr A\not=\emptyset$ ? - -REPLY [13 votes]: Assuming $MA_{\aleph_1}$, the answer is positive. -Let $P$ be the collection of all positive measure finite intersections of elements of $\mathcal{K}$, ordered by inclusion. Then $P$ is a ccc uncountable partial order so (by $MA_{\aleph_1}$) it contains an uncountable centered subset $Q \subseteq P$. If we let $\mathcal{A}$ be the collection of all elements of $\mathcal{K}$ that contain some element of $Q$, it follows that $\mathcal{A}$ has the finite intersection property and therefore $\bigcap \mathcal{A} \neq \emptyset$. -Edit: A subset $Q$ of a poset $P$ is called centered if any finite $F \subseteq Q$ has a lower bound in $P$. It was proved by Velickovic and Todorcevic in "Martin's axiom an partitions" (1987) that $MA_{\aleph_1}$ is equivalent to the statement that every ccc uncountable partial order contains an uncountable centered subset.<|endoftext|> -TITLE: Integral and conformal mappings -QUESTION [5 upvotes]: Let $f$ be a conformal mapping of the unit disk $U$ into $C$. Is the following integral convergent $$\int_U \frac{dx dy}{|f'(z)|}?$$ - -REPLY [4 votes]: Problems of this type are part of the so-called Brennan's conjecture. More precisely, suppose that $f:\mathbb{D} \to \mathbb{C}$ is univalent. Brennan's conjecture states that -$$\int_{\mathbb{D}}|f'|^p dA < \infty$$ -for $-2 -TITLE: "Concretely" writing down elements in a free profinite group -QUESTION [12 upvotes]: Let $r$ be a natural number. The elements of the free group $F_r$ on $r$ generators have a nice concrete description as "words" in the $r$ generators (and their inverses). I'd like to know if there is anything analogous for its profinite completion. -Namely, let $\hat{F}_r$ be the free profinite group on $r$ generators. -Then $\hat{F}_r$ contains the free group $F_r$ as a dense subgroup, and every element of $\hat{F}_r$ can be written as an infinite product $\prod_{i=1}^\infty x_i$ where each $x_i \in F_r$ and the $x_i$ tend to $1$ in the profinite topology. -In other words, we can arrange that $x_i \to 1$ in the following sense: for every $N$, the $x_i$ have the property that almost all of them are mapped to $1$ by every homomorphism $F_r \to G$ where $G$ is any group of cardinality $\leq N$ (for $i \gg 0$, depending on $N$). This suggests a natural descending filtration on $F_r$ via the intersection of the kernels of all maps $F_r \to G$ where $|G|\leq N$. Given a nice description (e.g., generators) of the elements in this filtration, one would have a somewhat explicit procedure of describing elements in the profinite completion via "infinite" words. -There are some natural elements in filtration $N$ in $F_r$, namely the $c(N)$th powers where $c(N)$ is the least common multiple of $\{1, 2, \dots, N\}$. -Question: Is the $N$th filtration in $F_r$ generated by the $c(N)$th powers? That is, if $a \in F_r$ is an element that goes to $1$ in every finite group of cardinality $\leq N$ a (finite) product of $c(N)$th powers? -Question${}^\prime$: If the above is too much to expect, are the two possible filtrations on $F_r$ above at least "commensurable"? - -REPLY [10 votes]: Let me answer the question in the title rather than in the body. First of all for the free profinite group on one generator $\widehat Z$ we have a direct product of p-adic integers over all primes and so we know how to write down elements. Now if $w$ is any element of a profinite group and $\lambda\in \widehat Z$ then define $w^\lambda$ to be the image of $\lambda$ under the map sending the generator of $\widehat Z$ to $w$. So we can build up some elements by starting with words and closing under products and infinite powers. -Here is an example. Let $p$ be a prime. The sequence $p^{n!}$ converges in $\widehat Z$ to a generator, denoted $p^{\omega}$, of the $p'$-component (i.e., the product of all q-adic groups with $q\neq p$). Hence a profinite group is pro-$p$ if and only if it satisfies the profinite identity $x^{p^{\omega}}=1$ because $x^{p^{\omega}}$ generates the p'-prime component of the pro-cyclic subgroup generated by $x$. -Jorge Almeida, principally in the context of free profinite monoids, but also for free profinite groups, came up with the following method to generate further elements. Let $M=End(\widehat F_r)$ be the endomorphism monoid. It is a profinite monoid. Given any endomorphism $\phi$ one has that $\phi^{\omega}=\lim_{n\to \infty} \phi^{n!}$ is an idempotent endomorphism. One can create interesting and useful elements which are recursive (in several senses, eg,there is an re sequence converging to it and one can compute its image in any finite quotient group) as follows. -Take an endomorphism $\phi$ of $F_r$. The take a generator $x$ of $F_r$ and consider the element $\phi^{\omega}(x)$. This element is easily computed in the above senses and can be interesting. For example $\phi(x)=x^p$ is an endomorphism of the free group on one generator. The map $\phi^{\omega}$ takes $x$ to $x^{p^{\omega}}$. -Define $\psi\colon F_2\to F_2$ by $\psi(x)=[x,y]$ and $\psi(y)=y$ where $x,y$ are free generators. Then $\psi^{\omega}(x)$ is like an infinite iterated commutator $[x,y,y,...]$. Almeida denotes this element $[x,_{\omega}y]$ and observes a profinite group is pro-nilpotent iff it satisfies the profinite identity $[x,_{\omega} y]=1$. -There is also a 2-variable profinite identity defining pro-solvable groups, but this is more complicated and if memory serves relies on Thompson's classification of minimal non-sovable groups and maybe even the classification of finite simple groups. -Here are some Almeida surveys. - -http://cmup.fc.up.pt/cmup/jalmeida/preprints/finite-semigroups-and-dynamics-pdf.pdf -http://cmup.fc.up.pt/cmup/jalmeida/preprints/Engel-CMUP.pdf<|endoftext|> -TITLE: Absoluteness of completeness -QUESTION [6 upvotes]: Suppose $V_0, V_1$ are (not necessarily well-founded) models of ZFC and suppose $\varphi$ is a first order sentence in a finite language $L$ (in our background model of set theory). Because every true finite set is an element of the well-founded part of both $V_0$ and $V_1$ we can consider $\varphi$ as an element of both $V_0$ and $V_1$. -Is it the case that: -$V_0 \models$ The deductive closure of $\varphi$ is a complete $L$-theory -if and only if -$V_1 \models$ The deductive closure of $\varphi$ is a complete $L$-theory? -Thanks. - -REPLY [5 votes]: The deductive closure of $\phi$ could be complete in one model and incomplete in the other. -Here's one way to see that (perhaps not the most elegant). Take any reasonable finitely axiomatized theory $T$ in a finite language $L$ capable of sufficient coding to carry out the proof of Gödel's Theorem and not implying the consistency of $ZFC$. (My favorite choice is $I\Sigma_1$, but others might prefer something in the language of set theory.) Fix your favorite encoding of $Con(ZFC)$ in this theory. -Let $T^-$ be some other finitely axiomatized theory on the same language which is complete and so that $T^-\vdash\neg Con(ZFC)$. It need not have any relation to $T$, so it could just interpret all functions as projections onto the first coordinate and all relations as empty, assuming that deduces $\neg Con(ZFC)$. (Note that $\neg Con(ZFC)$ has no meaning in this theory---it's just some arbitrary existential sentence.) -Now consider the sentence $\phi$ which is -$$(Con(ZFC)\rightarrow T)\wedge(\neg Con(ZFC)\rightarrow T^-)$$ -(Interpreting $T$ and $T^-$ as the sentences which are the conjunctions of all their axioms.) -In any model of $ZFC$ where $Con(ZFC)$ holds, this is incomplete because $T$ is incomplete. (And also because it's basically a disjunction of two theories and can't tell which case holds.) -In a model $V_1$ of $ZFC$ where $Con(ZFC)$ fails, $T$ can actually prove $\neg Con(ZFC)$, because this is a $\Sigma_1$ formula. Therefore the $T$ branch of the disjunction is self-invalidating: $V_1\vDash T\vdash\neg Con(ZFC)$, so $V_1\vDash \phi\vdash Con(ZFC)\rightarrow\neg Con(ZFC)$, so $V_1\vDash\phi\vdash T^-$. And (the deductive closure of) $T^-$ is complete.<|endoftext|> -TITLE: Does "Higher Infinite" have a volume II? -QUESTION [16 upvotes]: Kanamori in the introduction of his famous book "The Higher Infinite" says that his book is the first volume of a complete book and the second volume is about large cardinals and forcing. I saw several papers which refer to its content. -Question (1): When the volume II of "The Higher Infinite" will be published? -Question (2): Does anybody have a preprint of this volume? - -REPLY [14 votes]: There are some papers in which "The higher infinite II" is given as references: -1) Forcing Axioms and the Continuum Problem -Sakae Fuchino -2) The mathematical development of set theory from Cantor to Cohen-Kanamori, -3) Distributivity properties on $P_\omega(\lambda)$-Matet. -But I have no idea about if the book is going to be published or not. -Even the following content is presented for the book (note that the book is a continuation of volume I, so it starts with chapter VI)!!!! -Chapter VI. Higher Combinatorics -Kurepa's Hypothesis and Chang's Conjecture -Combinatorial Principles -Subtle Properties - The Tree Property -Chapter VII. Forcing with Strong Hypotheses I -Master Conditions (Silver's upward Easton forcing, -Kunen's saturated ideal) -Ultrafilters (structure theory, combinatorics) -Singular Cardinals Problem (intro, Solovay's -result, ultrafilters, Silver's result) -Strong vs. Supercompactness -Singular Cardinals Problem Forcing (Magidor's earlier -results) -Precipitous Ideals (combinatorics, equiconsistency -with measurable) -Chapter VIII. Covering and the Core Model -The Covering Theorem for $L$ -The Core Model -Models and Mice -The Covering Theorems for $K$ and for $L[U]$ -Applications of $K$ -Chapter IX. Higher Combinatorics II -Reflecting stationary sets, Shelah's LM result, etc. -Chapter X. Forcing with Strong Hypotheses II -Radin forcing, Proper forcing, forcing axioms -Chapter XI. Consistency Results about AD<|endoftext|> -TITLE: When is $f(x_1, \dots, x_n)+c$ an irreducible polynomial for almost all constants $c$? -QUESTION [16 upvotes]: Let $f\in \mathbb C[x_1, \dots, x_n]$, $n\ge 1$, be a non-constant polynomial. Consider the polynomial $f+t\in \mathbb C[t, x_1,\dots, x_n]$. This is an irreducible polynomial in $\mathbb C(t)[x_1, \dots, x_n]$. -Let $\mathbb C(t)^{alg}$ be an algebraic closure of $\mathbb C(t)$. -My question is: - -Under which condition $f+t$ remains irreducible in $\mathbb C(t)^{alg}[x_1, \dots, x_n]$ ? - -More precisely: we can construct a non-example as follows. If $f=P(g)$ for some $g\in \mathbb C[x_1, \dots, x_n]$ and $P(T)\in \mathbb C[T]$ of degree $\deg P(T)\ge 2$. Then $f+t$ is reducible in $\mathbb C(t)^{alg}[x_1, \dots, x_n]$. My precise question: - -Is the above non-example the only one ? - -So far I only found a necessary condition for $f+t$ to be reducible over $\mathbb C(t)^{alg}$: if we decompose $f$ into homogeneous components $f=f_d+\cdots +f_1 + f_0$, then there must be a non-constant $h\in \mathbb C[x_1, \dots, x_n]$ such that $h\mid f_{d-1}$ and $h^2\mid f_d$. -Motivation: Standard arguments of algebraic geometry show that $f+t$ is irreducible in $\mathbb C(t)^{alg}[x_1, \dots, x_n]$ if and only if for all but finitely many $c\in \mathbb C$, $f+c$ is irreducible in $\mathbb C[x_1,\dots,x_n]$. - -REPLY [17 votes]: [Answer rewritten] -Yes, that non-example is the only one. As Terry Tao points out, this follows from Bertini's second theorem. In fact there are quantitative versions of this result, following work of Yosef Stein, Dino Lorenzini, Angelo Vistoli, Ewa Cygan, and Salah Najib. Here is a consequence of the formulation from Najib's paper "Une generalisation de l'inegalite de Stein-Lorenzini" (J.Algebra 292 (2005), 566-573): let $K$ be an arbitrary algebraically closed field, and let $f\in K[x_1,\dots,x_n]$ be a polynomial which cannot be written as $P(g)$ with $P(T)\in K[T]$ of degree at least $2$ and $g\in K[x_1,\dots,x_n]$. Then there are fewer than $\deg(f)$ values $c\in K$ for which $f-c$ is reducible. This can be refined to take into account the extent of reducibility of the various $f-c$, namely the number of distinct irreducible factors. Namely, for $c\in K$, let $s(c)$ be the number of distinct irreducible factors of $f-c$. Then -$$ -\sum_{c\in K} (s(c)-1) < \deg(f). -$$<|endoftext|> -TITLE: The universal property of the Liouville $1$-form -QUESTION [10 upvotes]: I am not totally sure if this question is appropriate for MathOverflow, or if it more adeguate to MathStackexchange. -As usual any feedback is welcome. -Introduction -Given an arbitrary smooth manifold $Q$, on the cotangent bundle $T^\ast Q$ there exists a $1$-form $\lambda_Q$, which is variously known as the Liouville $1$-form, or the tautological $1$-form. -Local expression in fibered coordinate -For any local coordinate system $q_i$ on $Q$, let $(q_i,p_i)$ be the associate coordinate on $T^\ast Q$. -Then, locally, $\lambda_Q$ can be given by $$\lambda_Q=\sum_i p_i \cdot dq_i.\tag{$\star$}$$ -These local descriptions can be correctly patched together to give a global $1$-form on $T^\ast Q$. -Intrinsic expression -For any $1$-form $\phi$ on $Q$, we have also $$\phi^\ast\lambda_Q=\phi,\tag{$\star \star$}$$ where in the left-hand side we are looking at $\phi$ as a section $\phi:Q\to T^\ast Q$ of the cotangent bundle $\tau_Q^\ast:T^\ast Q\to Q$. -Indeed this condition is enough to completely determine $\lambda_Q\in\Omega^1(T^\ast Q)$ as its unique solution. -Question -In some references (cfr. these lecture notes on page 8), I have found condition $(\star\star)$ referred to as the universal property of the Liouville $1$-form. -All the examples I know of mathematical objects characterized (up to isomorphisms) by a certain universal property, can be recast in the language of category theory, as universal objects of some category (cfr. for example here). -Now my question is: - -The universal property $(\star ~ \star)$ of the Liouville $1$-form can be recast in the language of category theory? or otherwise, in what sense can it be called a universal property? - -REPLY [6 votes]: I'm not an expert in differential geometry, but it seems to me that the property of the Liouville $1$-form only talks about the given manifold and its tangent bundle, no other manifolds are involved, hence this isn't a universal property. But I think that we can generalize the property as follows: -Let $\mathsf{Mfd}/X$ denote the category of smooth manifolds $Y$ equipped with a smooth map $Y \to X$. Consider the functor $(\mathsf{Mfd}/X)^{\mathrm{op}} \to \mathsf{Vect}$ which maps $Y \to X$ to $\Omega^1(Y)$. Then I claim that this functor is represented by the Liouville $1$-form $(T^* X \to X,\lambda_X)$. This means: Given $Y \to X$ and $\omega \in \Omega^1(Y)$, there is a unique smooth map $f : Y \to T^* X$ over $X$ such that $f^* \lambda_X = \omega$. -In fact, one defines $f$ to be the composition $Y \xrightarrow{\omega} T^* Y \to T^* X$. Then $f^* \lambda_X = \omega^* \lambda_Y = \omega$. -EDIT: As mentioned in the comments, one has to take relative differential forms $\Omega^1(Y/X)$.<|endoftext|> -TITLE: Is it known whether every $\omega$-tree with an infinite antichain has an infinite chain in $\mathsf{ZF}$? -QUESTION [12 upvotes]: In this paper by Good and Tree, the following result is mentioned without proof as part of Proposition 6.5: - -Each of the following statements imply those beneath it. - -The countable union of finite sets is countable. - -Every $\omega$-tree has either (sic) an infinite chain or an infinite antichain. - -Every countable collection of [non-empty] finite sets has a choice function. - - - -By "tree," it is meant "single-rooted tree" in this sense. An $\omega$-tree is a tree of height $\omega$ with all of its levels finite. An antichain is a set of mutually incomparable elements of the tree. -I know that the first and (edited) last statements are equivalent, so that these statements should all be equivalent, according to the paper. -Proving that the first implies the second (with "either" removed) is not difficult. Given an $\omega$-tree $T$, we know $T$ is countably-infinite since each of its countably-infinitely-many non-empty levels is finite. Let $f:\omega\to T$ a bijection. Supposing that $T$ has no infinite antichain, let $A$ be the set of all nodes of $T$ without successor. Since this is readily an antichain, then it is finite, so, put $$m=\max\bigl(\{0\}\cup\{k<\omega:A\cap T_k\ne\emptyset\}\bigr).$$ Let $c_0\in T_{m+1}.$ Given $c_n$ with height greater than $m$, we have by definition of $A$ and $m$ that $c_n$ has a successor, and letting $$c_{n+1}=f\bigl(\min\{k<\omega:c_n0}.$ For any $n\in\Bbb Z_{>0}$ and any $\overline i\in\Bbb Z/p_n\Bbb Z,$ we let $a(n,\overline i)$ indicate a unique atom. For each $n\in\Bbb Z_{>0},$ we let $A_n:=\left\{a(n,\overline i):\overline i\in\Bbb Z/p_n\Bbb Z\right\},$ and let $A:=\bigcup_{n\in\Bbb Z_{>0}}A_n.$ -Consider the group $G:=\prod_{n\in\Bbb Z_{>0}}(\Bbb Z/p_n\Bbb Z),$ and define the action of $G$ on $A$ by $\pi\bigl(a(n,\overline i)\bigr):=a(n,\overline i+\pi_n).$ -Now, we define a set $V_\alpha$ for each ordinal $\alpha$ by transfinite recursion as follows: - -$V_0:=A,$ - -$V_{\alpha+1}:=V_\alpha\cup\mathcal P(V_\alpha),$ and - -for limit ordinals $\lambda,$ $V_\lambda:=\bigcup_{\alpha<\lambda}V_\alpha.$ - - -The class $\mathscr{V}$ given by the union of all $V_\alpha$ is a model of $\mathsf{ZFA}.$ The rank of an element $x\in\mathscr V$ is the least ordinal $\alpha$ such that $x\in V_\alpha.$ Thus, the atoms are precisely the elements of rank $0.$ We extend the action of $G$ on $A$ to all of $\mathscr{V}$ by $\pi(x):=\{\pi(y):y\in x\},$ using recursion on the rank of $x.$ -A few quick facts are readily seen: - -$\pi(\emptyset)=\emptyset$ for any $\pi\in G.$ By recursion on rank, we see that for any ordinal $\alpha,$ $\pi(\alpha)=\alpha.$ - -More generally, if $x\in\mathscr{V}\setminus A$ and the transitive closure of $x$ is disjoint from $A,$ then $\pi(x)=x$ for all $\pi\in G.$ We call such $x$ "pure sets," and the class of all pure sets "the kernel of $G.$" This class is readily identical to the class of all hereditary sets in the standard cumulative heirarchy of $\mathsf{ZF}.$ - -For any $\pi\in G,$ the map $x\mapsto\pi(x)$ is a permutation $\mathscr{V}\to\mathscr{V}.$ - - -Next, consider the set $\mathcal{S}$ of subsets of $\Bbb Z_{>0}$ of zero upper density. The following facts are readily shown: - -$\emptyset\in\mathcal S.$ - -If $S\in\mathcal S$ and $S'\subseteq S,$ then $S'\in\mathcal S.$ - -If $S,S'\in\mathcal S,$ then $S\cup S'\in\mathcal S.$ - -If $A\subset\Bbb Z_{>0}$ and $|A|<\aleph_0,$ then $A\in\mathcal S.$ - - -Given any $S\in\mathcal S,$ define $$G(S):=\{\pi\in G:\forall n\in S,\pi_n=\overline 0\},$$ which is readily a subgroup of $G,$ and let $$\mathcal F:=\{H0}$ such that $a\in A_n,$ and noting that $G(\{n\})=\{\pi\in G:\pi(a)=a\},$ then by the fourth fact about $\mathcal S$ above, $\{\pi\in G:\pi(a)=a\}\in\mathcal F.$ Thus, $\mathcal F$ is a normal filter on $G$ (see p. 46 of Thomas Jech's The Axiom of Choice). -Now, given any $x\in\mathscr{V},$ we say that $x$ is "symmetric" if $\{\pi\in G:\pi(x)=x\}\in\mathcal F,$ and "hereditarily symmetric" if $x$ is symmetric and all elements of $x$ are hereditarily symmetric (by recursion on rank, hereditary symmetry is well-defined). Let $\mathscr{N}$ be the class of all hereditarily symmetric elements of $\mathscr{V}.$ By Theorem 4.1 from Jech, $\mathscr{N}$ is a transitive model of $\mathsf{ZFA}$ containing all elements of the kernel of $G$ as well as all elements of $A.$ - -Claim A: Form 10 fails to hold in $\mathscr N.$ -Proof: To show that Form 10 fails, we show that $A$ is not countable in $\mathscr{N},$ but that $\{A_n:n\in\Bbb Z_{>0}\}$ is countable and each $A_n$ is finite in $\mathscr{N}.$ -Note that for any $n\in\Bbb Z_{>0},$ we have by definition of the action of $G$ on $A$ that $\pi(A_n)=A_n$ for all $\pi\in G.$ Since $\omega$ and each of its elements lie in the kernel of $G,$ then the function $f:\omega\to\{A_n:n\in\Bbb Z_{>0}\}$ given by $k\mapsto A_{k+1}$ is thus a bijection and $f\in\mathscr{N},$ so that $\{A_n:n\in\Bbb Z_{>0}\}$ is countable in $\mathscr{N}.$ -Fixing any $n\in\Bbb Z_{>0}$, we have for any $\pi\in G(\{n\})$ and any $i\in p_n$ that $\pi(a(n,\overline i))=a(n,\overline i),$ so the function $p_n\to A_n$ given by $i\mapsto a(n,\overline i)$ is both a bijection and an element of $\mathscr{N},$ whence each $A_n$ is finite in $\mathscr{N}.$ -Now consider any function $g:\omega\to A$ such that $g\in\mathscr{N}.$ We show that $g$ is not a surjection $\omega\to A,$ and so $A$ is uncountable in $\mathscr{N}.$ Since $g\in\mathscr{N},$ then by definition, there is some $S\in\mathcal S$ such that for all $\pi\in G(S),$ we have $\pi(g)=g.$ Since $S\in\mathcal S,$ then $S\subsetneq\Bbb Z_{>0},$ so there is some $n\in\Bbb Z_{>0}$ such that $n\notin S,$ and so there is some $\pi\in G(S)$ such that $\pi_n\neq \overline 0.$ Take any $\langle k,a\rangle\in g,$ so that $\pi(k)=k$ (since $k$ is in the kernel of $G$), and so $$\langle k,\pi(a)\rangle=\langle \pi(k),\pi(a)\rangle=\pi\bigl(\langle k,a\rangle\bigr)\in\pi(g)=g.$$ Since $g$ is a function and $\langle k,a\rangle,\langle k,\pi(a)\rangle\in g,$ it follows that $\pi(a)=a.$ Since $\pi_n\neq \overline 0,$ it follows that $a\notin A_n.$ Thus, the range of $g$ is disjoint from $A_n,$ so $g:\omega\to A$ is not a surjection, as was to be shown. $\Box$ -Lemma 2: Suppose that $S\in\mathcal S$ and $x\in\mathscr{N}$ such that the set $\{\pi(x):\pi\in G(S)\}$ has finite cardinality $m.$ Then there is a $\subseteq$-least finite set $X\subsetneq\Bbb Z_{>0}$ such that for all $\pi\in G(S\cup X),$ $\pi(x)=x.$ In particular, $X=\{k\in\Bbb Z_{\geq 0}:p_k\mid m\},$ and $$m=\prod_{k\in X}p_k.$$ -Proof: If $m=1,$ then immediately, $X=\emptyset$ satisfies all of the desired properties, so suppose that $m>1,$ whence $X:=\{k\in\Bbb Z_{>0}:p_k\mid m\}$ is immediately finite and non-empty since $n\mapsto p_n$ is an enumeration of the primes. -By orbit-stabilizer theorem, $\operatorname{Stab}(x):=\{\phi\in G(S):\phi(x)=x\}$ is a (normal) subgroup of $G(S)$ of index $m,$ so the order of each element of $G(S)/\operatorname{Stab}(x)$ divides $m.$ Thus, for any $\pi\in G(S),$ $m\pi+\operatorname{Stab}(x)=\operatorname{Stab}(x),$ meaning that $m\pi\in\operatorname{Stab}(x),$ and so $mG(S)\subseteq\operatorname{Stab}(x).$ Readily, $mG(S)$ is a subgroup of $G,$ so a subgroup of $\operatorname{Stab}(x),$ and since $G$ is abelian, then $mG(S)$ is a normal subgroup of $\operatorname{Stab}(x).$ Thus, $$[G(S):mG(S)]=[G(S):\operatorname{Stab}(x)][\operatorname{Stab}(x):mG(S)]=m\cdot[\operatorname{Stab}(x):mG(S)]\ge m.$$ -Take any $\pi\in G(S),$ meaning that $\pi\in G$ and that $\pi_n=\overline 0$ for all $n\in S,$ and consider $m\pi.$ Given any $n\in X,$ we have by definition that $p_n\mid m,$ so $p_n\mid m\pi_n\equiv_{p_n}(m\pi)_n,$ and so $(m\pi)_n=\overline 0.$ Also, for $n\in S$ we immediately have $(m\pi)_n\equiv_{p_n}m\pi_n=m\overline 0=\overline 0.$ Hence, $(m\pi)_n=\overline 0$ for all $n\in S\cup X,$ meaning that $m\pi\in G(S\cup X),$ whence $mG(S)\subseteq G(S\cup X).$ -On the other hand, suppose $\phi\in G(S\cup X)$ with $\phi$ not the identity element, and take any $n\in\Bbb Z_{>0}\setminus (S\cup X)$ such that $\phi_n\neq \overline 0.$ Since $n\notin X,$ then $p_n\not\mid m,$ and so there exists $k\in\Bbb Z_{>0}$ such that $p_n\mid(mk-1).$ Let $k_n$ be the least such $k,$ so since $\phi_n\neq \overline 0$ and $k_n$ is coprime with $p_n,$ then $k_n\phi_n\neq\overline 0.$ -We show that $\phi\in mG(S),$ so that $G(S\cup X)=mG(S)$ by double-inclusion. Define $\pi\in G$ as follows. For each $n\in(\Bbb Z_{>0}\setminus S)\cap(\Bbb Z_{>0}\setminus X)$ such that $\phi_n\neq \overline 0,$ let $\pi_n=k_n\phi_n\neq\overline 0.$ For each $n\in(\Bbb Z_{>0}\setminus S)\cap(\Bbb Z_{>0}\setminus X)$ such that $\phi_n=\overline 0,$ let $\pi_n=\overline 0.$ For each $n\in S,$ let $\pi_n=\overline 0,$ so that $\pi\in G(S).$ For each $n\in(\Bbb Z_{>0}\setminus S)\cap X,$ let $\pi_n=\overline 1.$ -Note that $\bigl\{S,(\Bbb Z_{>0}\setminus S)\cap X,(\Bbb Z_{>0}\setminus S)\cap(\Bbb Z_{>0}\setminus X)\bigr\}$ is a partition of $\Bbb Z_{>0},$ whence $\pi\in G$ is well-defined. As noted above, $\pi\in G(S);$ furthermore, $(m\pi)_n=\overline 0=\phi_n$ for all $n\in S,$ since $\phi\in G(S\cup X)\subseteq G(S).$ Now, for $n\in (\Bbb Z_{>0}\setminus S)\cap X,$ we have that $p_n\mid m,$ so $(m\pi)_n=m\pi_n=\overline 0,$ and since $\phi\in G(S\cup X)\subseteq G(X)$ and $n\in X,$ then $(m\pi)_n=\overline 0=\phi_n.$ For $n\in\Bbb Z_{>0}\setminus(S\cup X),$ we have that $\pi_n=\overline 0$ iff $\phi_n=\overline 0,$ so for $n\in\Bbb Z_{>0}\setminus(S\cup X)$ such that $\pi_n=\overline 0,$ we have that $m\pi_n=\overline 0,$ whence $(m\pi)_n=\phi_n.$ Finally, if $n\in\Bbb Z_{>0}\setminus(S\cup X)$ such that $\pi_n\neq \overline 0,$ then $\pi_n=k_n\phi_n.$ Since $p_n\not\mid m$ by definition of $X,$ this is equivalent to $m\pi_n=mk_n\phi_n\,$ so since $p_n\mid(mk_n-1),$ this is equivalent to $m\pi_n=\phi_n,$ whence $(m\pi)_n=\phi_n.$ Thus, $m\pi=\phi,$ so $\phi\in mG(S),$ so $G(S\cup X)\subseteq mG(S),$ and so $mG(S)=G(S\cup X)$ by double inclusion. -By definition, $G(S)\cong\prod_{n\in \Bbb Z_{>0}\setminus S}(\Bbb Z/p_n\Bbb Z),$ so $$\begin{eqnarray*}mG(S) & \cong & \left(\prod_{n\in (\Bbb Z_{>0}\setminus S)\cap X}m(\Bbb Z/p_n\Bbb Z)\right)\times\left(\prod_{n\in \Bbb Z_{>0}\setminus(S\cup X)}m(\Bbb Z/p_n\Bbb Z)\right)\\ & = & \left(\prod_{n\in (\Bbb Z_{>0}\setminus S)\cap X}0(\Bbb Z/p_n\Bbb Z)\right)\times\left(\prod_{n\in \Bbb Z_{>0}\setminus(S\cup X)}(\Bbb Z/p_n\Bbb Z)\right).\end{eqnarray*}$$ Hence, $$G(S)/mG(S)\cong\prod_{n\in (\Bbb Z_{>0}\setminus S)\cap X}(\Bbb Z/p_n\Bbb Z),$$ so $$[G(S):mG(S)]:=\bigl|G(S)/mG(S)\bigr|=\prod_{n\in (\Bbb Z_{>0}\setminus S)\cap X}\bigl|\Bbb Z/p_n\Bbb Z\bigr|=\prod_{n\in (\Bbb Z_{>0}\setminus S)\cap X}p_n,$$ so $[G(S):mG(S)]\mid m$ by definition of $X.$ Hence, $m\leq[G(S):mG(S)]\leq m,$ so $[G(S):mG(S)]=m.$ Moreover, since $\prod_{n\in X}p_n$ divides $m$ by definition of $X,$ and since $m=\prod_{n\in (\Bbb Z_{>0}\setminus S)\cap X}p_n,$ then $X\subseteq(\Bbb Z_{>0}\setminus S)\cap X,$ whence $X\subseteq \Bbb Z_{>0}\setminus S,$ so that $S\cap X=\emptyset,$ and $m=\prod_{n\in X}p_n.$ -Since $m=[G(S):mG(S)]=m\cdot[\operatorname{Stab}(x):mG(S)],$ then $[\operatorname{Stab}(x):mG(S)]=1,$ so $G(S\cup X)=mG(S)=\operatorname{Stab}(x).$ -Finally, suppose that $Y\subseteq\Bbb N$ such that $G(S\cup Y)$ fixes $x,$ whence $G(S\cup Y)$ is a subgroup of $\operatorname{Stab}(x)=G(S\cup X),$ and so $G(S\cup Y)\subseteq G(S\cup X).$ Take any $n\in X$ and any $\pi\in G(S\cup Y),$ so since $\pi\in G(S\cup X)=G(S)\cap G(X),$ then $\pi_n=0.$ Since $\pi\in G(S\cup Y)$ was arbitrary, then $n\in S\cup Y,$ but $n\in X$ and $S\cap X=\emptyset,$ so $n\in Y.$ Thus, $X\subseteq Y,$ as desired. $\Box$ -Lemma 3: Suppose that $\langle T,\sqsubset\rangle\in\mathscr V$ and that there is some $S\in\mathcal S$ such that $G(S)$ fixes $\langle T,\sqsubset\rangle.$ Then for each $\pi\in G(S),$ $\pi\restriction_T$ is a tree automorphism of $\langle T,\sqsubset\rangle.$ -Proof: Since each such $\pi$ fixes $\langle T,\sqsubset\rangle,$ then each such $\pi$ fixes $T$ and $\sqsubset.$ -Fix any $\pi\in G(S).$ Since $\pi$ is a permutation of $\mathscr{V}$ and $\pi(T)=T,$ then $\pi\restriction_T:T\to T$ is a bijection. For any $\bigl\langle a,b\bigr\rangle\in\;\sqsubset$, we have $\bigl\langle\pi(a),\pi(b)\bigr\rangle=\pi\bigl(\langle a,b\rangle\bigr)\in\pi(\sqsubset)=\;\sqsubset.$ Analogously, for any $\bigl\langle c,d\bigr\rangle\in\;\sqsubset$, we have $\langle\pi^{-1}(c),\pi^{-1}(d)\rangle\in\;\sqsubset,$ so if $\bigl\langle\pi(a),\pi(b)\bigr\rangle\in\;\sqsubset,$ then letting $c=\pi(a)$ and $d=\pi(b),$ we have $\langle a,b\rangle\in\;\sqsubset.$ Therefore, $\pi\restriction_T$ is an order automorphism of $\langle T,\sqsubset\rangle,$ as desired. $\Box$ -Lemma 4: Suppose $X$ is an infinite subset of $\Bbb Z_{\geq 0}.$ Then $X$ has an infinite subset $S$ of zero upper density. -Proof: For each $n\in\Bbb Z_{\geq 0},$ we have $n^2\in\Bbb Z_{\geq 0}.$ -Clearly, $0^2=0$ isn't an upper bound of $X,$ so let $s_0:=\min\{k\in X:k>0\}.$ Given $s_n,$ we have that $\max\bigl\{(n+1)^2, s_n\bigr\}$ is not an upper bound of $X,$ so let $s_{n+1}:=\min\left\{k\in X:k>(n+1)^2,k>s_n\right\}.$ By recursion, $n\mapsto s_n$ is an increasing map $\Bbb Z_{\geq 0}\to X,$ so $S:=\{s_n:n\in\Bbb Z_{\geq 0}\}$ is an infinite subset of $X.$ Moreover, $s_n>n^2$ for all $n\in\Bbb Z_{\geq 0},$ so the upper density of $S$ is $$\limsup_{n\to\infty}\frac{n}{s_n}\leq\limsup_{n\to\infty}\frac{n}{n^2+1}=0.$$ Since upper density is always a nonnegative number, then $S$ has zero upper density, as desired. $\Box$ -Lemma 5: Suppose that $\langle T,\sqsubset\rangle\in\mathscr N$ is an $\omega$-tree, that $T'$ is a subtree of $\langle T,\sqsubset\rangle,$ and that $A$ is the set of $\sqsubset$-minimal elements of $T\setminus T'.$ Then $A\in\mathscr N$ if and only if $T'\in\mathscr N.$ -Proof: Since $T\in\mathscr N,$ then $T\subset\mathscr N,$ so $A\subseteq\mathscr N$ and $T'\subseteq\mathscr N.$ It remains to show that $A$ is symmetric iff $T'$ is symmetric. -Since $\langle T,\sqsubset\rangle\in\mathscr N,$ then there is some $S_T\in\mathcal S$ such that each $\pi\in G(S_T)$ fixes $\langle T,\sqsubset\rangle.$ Given any $S\in\mathcal S,$ we have that $S_T\cup S\in\mathcal S,$ whence $G(S_T\cup S)=G(S_T)\cap G(S)$ is a subgroup of $G(S_T),$ and so each $\pi\in G(S_T\cup S)$ fixes $\langle T,\sqsubset\rangle.$ Any such $\pi$ fixes $T'$ iff it fixes $T\setminus T',$ since $\pi\restriction_T:T\to T$ is a bijection. Further, by Lemma 3, any such $\pi$ maps $\sqsubset$-minimal elements of $T'$--that is, elements of $A$--to $\sqsubset$-minimal elements of $\pi(T').$ -If such a $\pi$ fixes $T',$ then $\pi(T')=T',$ so by the above, $\pi(A)=A.$ On the other hand, suppose that $\pi(A)=A$ for some such $\pi.$ Since $T'$ is a subtree of $\langle T,\sqsubset\rangle,$ then $T'$ is non-empty and has the root of $T$ as an element. In addition, every element of $A$ is of finite height, so since the root of $T$ is not in $A,$ then each element of $A$ is a successor of some (unique) element of $T'.$ Put another way, each element of $T'$ is a predecessor of an element of $A$, so by Lemma 3, is necessarily fixed by $\pi.$ The Lemma then follows. $\Box$ -Claim B: Suppose that Form 10 holds in $\mathscr V.$ Then Form 216 holds in $\mathscr{N}.$ -Proof: Take any $\omega$-tree $\langle T,\sqsubset\rangle\in\mathscr V,$ and suppose that there is some $S_T\in\mathcal S$ such that $G(S_T)$ fixes $\langle T,\sqsubset\rangle,$ so that $\langle T,\sqsubset\rangle\in\mathscr N.$ Further suppose that if $f\in\mathscr V$ such that $f:\langle\omega,\in\rangle\to\langle T,\sqsubset\rangle$ is an embedding, then $f\notin\mathscr N.$ To show that Form 216 holds in $\mathscr N,$ it suffices (by Lemmas 1 and 5) to show that $\langle T,\sqsubset\rangle$ has an infinite subtree $T'$, every node of which is bounded above by some element of $T\setminus T'.$ -Since Form 10 holds in $\mathscr V,$ then there exists $f:\omega\to T$ such that for all $n\in\omega,$ $f(n)\sqsubset f(n+1).$ Note that we may assume without loss of generality that for each $n\in\omega,$ $f(n)$ is on the $n$th level of $\langle T,\sqsubset\rangle$--that is, that the image of $f$ is an infinite branch of $\langle T,\sqsubset\rangle$--since if not, then the downward closure of the image of $f$ is such an infinite branch, and we may enumerate the elements of that branch in increasing order as desired. Let $x_n:=f(n)$ for each $n\in\omega.$ By Lemma 3, we have that $G(S_T)$ fixes each level of $\langle T,\sqsubset\rangle,$ and since each such level has finite cardinality, then for each $n\in\omega,$ $\{\pi(x_n):\pi\in G(S_T)\}$ has finite cardinality, so by Lemma 2, we can let $X_n$ be the unique, minimal, finite subset $X\subseteq \Bbb Z_{>0}$ such that for all $\pi\in G(S_T\cup X),$ $\pi(x_n)=x_n.$ -Given $m,n\in\omega$ with $m -TITLE: Intuitive explanation of Dvoretzky's theorem -QUESTION [16 upvotes]: I am wondering if anyone has an enlightening explanation of why Dvoretzky's theorem (which says that a high-dimensional convex body has an almost round central section) is true -- there are a number of proofs but all of them seem a bit technical... - -REPLY [10 votes]: There is a more difficult proof than the quantitative finite dimensional proofs that gives only the qualitative version of Dvoretzky's theorem but is arguably more intuitive. You use Ramsey's theorem to prove that if $X$ is an infinite dimensional Banach space, then there is a Banach space $Y$ that has a monotonely unconditional basis $(e_n)$ s.t. $(e_n)$ is isometrically equivalent to every subsequence of itself and such that $Y$ is finitely representable in $X$ (meaning that for every $\epsilon > 0$, every finite dimensional subspace of $Y$ is $1+\epsilon$-isomorphic to a subspace of $X$). This is at the beginning of the Brunel-Sucheston spreading model theory and is elementary. So $Y$ looks a bit like the spaces $\ell_p$, $1\le p < \infty$, and $c_0$. Now $c_0$ is universal for finite dimensional spaces (up to $1+\epsilon$), and $L_p$ for all $p$ contains $\ell_2$ isometrically (span of IID $N(0,1)$ random variables when $p<\infty$), and $L_p$ is finitely representable in $\ell_p$, so this is a pretty good hint that Dvoretzky's theorem is true. To finish the proof, just apply Krivine's theorem, which says that for some $1\le p \le \infty$, the space $\ell_p$ is finitely represented in $Y$ (in fact, for each $n$ there are disjointly supported elements in $Y$ that are $1+\epsilon$-equivalent to the unit vector basis of $\ell_p^n$). Krivine's theorem is proved in the Springer Lecture Notes volume written by V. Milman and S. Schechtman. -If you are willing to settle for subspaces of $Y$ that are just uniformly isomorphic to $\ell_2^n$, you can replace Krivine's theorem with more elementary arguments. Tzfriri did that in -Tzafriri, L. On Banach spaces with unconditional bases. Israel J. Math. 17 (1974), 84–93. - -REPLY [4 votes]: To develop some intuition, the following argument might help, suggested (and dismissed) by K. Villaverde, O. Kosheleva, and M. Ceberio, Why Ellipsoid Constraints, Ellipsoid Clusters, and Riemannian Space-Time: Dvoretzky's Theorem Revisited. -A stronger version of Dvoretzky’s theorem (due to Milman) asserts that almost all low-dimensional sections of a convex set have an almost ellipsoidal shape. An $n$-dimensional section consists of points $(x_1,x_2,\ldots x_n)$ such that $g(x_1,x_2,\ldots x_n)\leq 0$. Generically, this function $g$ will be smooth and a Taylor expansion to second order would be a good approximation, -$$\sum_{i,j=1}^n a_{ij}x_i x_j+\sum_{i=1}^n b_i x_i \leq a_0,$$ -producing an ellipsoid.<|endoftext|> -TITLE: Brauer groups of punctured affine lines over a base -QUESTION [7 upvotes]: Let $R$ be a torsion-free regular noetherian ring. The Brauer group $Br(R)$ of $R$, defined equivalently (by a theorem of Gabber) as the group of Morita equivalence classes of Azumaya $R$-algebras or as the étale cohomology $H^2(X_{et}, \mathbb{G}_m)$, has the property that it is homotopy invariant: that is, $Br(R) \simeq Br( R[t])$. (I learned this from Auslander and Goldman's article "The Brauer group of a commutative ring." ) -Auslander and Goldman's argument is that $Br(R)$ is always a summand of $Br(R[t])$ (because of the natural retraction), so it suffices to show this homotopy invariance property when $R$ is a field $k$ of characteristic zero: this is because Brauer groups of regular domains inject into those of their quotient fields. In this case, -one can use Galois descent along $k \to \bar{k}$ together with "Tsen's theorem." (The reason for characteristic zero is simply that Tsen's theorem requires an algebraically closed, rather than separably closed, base field.) -Now this isn't true when the affine line is replaced by a punctured affine line, for instance, $\mathbb{G}_m$ or $\mathbb{A}^1 \setminus \{0, 1\}$: even the Galois descent argument (when $R$ is a field) already breaks down, because the units in the ring of functions on $R[t, t^{-1}]$ is not simply $R^{\times}$, but -rather $R^{\times} \oplus \mathbb{Z}$. In particular, if $R = k$ is a field, this will lead to a $H^2( G_k, \mathbb{Z})$ (which is the Pontryagin dual of the Galois group $G_k$) sitting inside the Brauer group of $k[t, t^{-1}]$. -I'm curious if there is a general procedure for calculating the Brauer group of a punctured affine line over a regular ring in terms of the Brauer group of the -base ring. (I have in mind something like a localization of $\mathbb{Z}$ as the base.) - -REPLY [8 votes]: This is a calculation using the local cohomology sequence for etale cohomology together with -Gabber's relatively recent proof of the absolute purity conjecture. Let $% -Y=\operatorname{Spec}(R)$, $X=\operatorname{Spec}( R[t]) $, and let $\sigma \subseteq X$ be the -image of a section so that $U=\operatorname{Spec}\left( R[t,t^{-1}]\right) $ is its -complement. Gabber's recent proof of absolute purity establishes -Grothendieck's purity conjecture for the Brauer group which states that $% -H_{\sigma }^{3}\left( X,\mathbb{G}_{m}\right) =H^{1}\left( \sigma ,\mathbb{Q}/\mathbb{% -Z}\right) $ if $\sigma \subseteq X$ is a regular subscheme of a regular -scheme. Moreover $R\rightarrow R[t,t^{-1}]$ has a section. Thus we have an -exact sequence coming from the local cohomology of $\mathbb{G}_{m},$ -\begin{equation*} -0\rightarrow Br\left( R\right) \rightarrow Br\left( R[t,t^{-1}]\right) -\rightarrow H^{1}\left( R,\mathbb{Q}/\mathbb{Z}\right) \rightarrow 0 -\end{equation*} -since $H^{3}\left( R,\mathbb{G}_{m}\right) =H^{3}\left( R[t],\mathbb{G}_{m}\right) -\hookrightarrow H^{3}(R[t,t^{-1}],\mathbb{G}_{m}).$ Note though that this uses Kummer -sequences and homotopy invariance for $\mu_n$ and so only applies to torsion in $Br(R[t,t^{-1}])$ of order -invertible in $R.$ In fact you can argue directly with $n$ torsion using Kummer sequences and purity for $\mu_n$ cohomology to get this result for $n$ torsion elements if $n$ is a unit in $R$. -Of course if you puncture the line in more than one point, you get multiple copies of $ H^{1}\left( R,\mathbb{Q}/\mathbb{Z}\right)$. -The easiest way to address $n$ torsion when $n$ is not a unit in $R$ is, I think, to look at the split exact sequence on $R_{et}$ -$$ -0\rightarrow k_*\mathbb{G}_{m,R[t]} \rightarrow p_*\mathbb{G}_{m,R[t,t^{-1}]} \rightarrow \mathbb{Z} \rightarrow 0 -$$ -where the last map is the degree map. The Leray spectral sequence for $k$ and $p$ yield a split exact sequence -$$ -0\rightarrow Br(R) \rightarrow Br_{sp}(R[t,t^{-1}]) \rightarrow H^1(R,\mathbb{Q}/\mathbb{Z})\rightarrow 0 -$$ -where $Br_{sp}(R[t,t^{-1}])$ consists of elenents in $Br(R[t,t^{-1}])$ that are split by passing to $R_{\mathfrak{p}}^{sh}[t,t^{-1}]$ as $R_{\mathfrak{p}}^{sh}$ runs through all strict henselizations of $R$ at prime ideals $\mathfrak{p}$.<|endoftext|> -TITLE: Efficient Hamiltonian cycle algorithms for graph classes -QUESTION [7 upvotes]: Generally speaking, finding a Hamiltonian cycle is NP-Hard and so tough. But if $G=L(H)$ is the line graph of $H$, then we can reduce the problem of finding a Hamiltonian cycle in $G$ to finding an Euler tour of $H$, which is easy. -Question is: are there similar shortcuts known for other graph classes? - -REPLY [4 votes]: In fact, it can be proved that Hamiltonian Cycle remains $\mathsf{NP}$-hard even for a very restricted subclass of line graphs: line graphs of $1$-subdivisions of planar cubic bipartite graphs. These graphs are cubic. Not surprisingly the same holds if we pass to $4$-regular line graphs: Hamiltonian Cycle remains $\mathsf{NP}$-hard for line graphs of planar cubic bipartite graphs. -Both results can be obtained by reductions from Hamiltonian Cycle restricted to planar cubic bipartite graphs: -T. Akiyama, T. Nishizeki, N. Saito, NP-Completeness of the Hamiltonian Cycle Problem for Bipartite Graphs, Journal of -Information Processing 3 (1980) 73-76.<|endoftext|> -TITLE: algorithm to compute the integral orthogonal group -QUESTION [10 upvotes]: Suppose I have an indefinite quadratic form over the integers, and I want to compute its orthogonal group. Is there an algorithm, or at least a heuristic? If yes, is there any implementation anywhere? - -REPLY [3 votes]: I am currently writing such an algorithm (in Magma) for Lorentzian lattices. It is almost finished and has already been used to answer this question on MO. -It furnishes a fundamental domain for the action of the group of automorphisms $G$ of the lattice on the hyperbolic space, and identifies the boudary components. Thus it may be used to obtain a finite presentation of $G$. -Perhaps this algorithm will be available directly on Magma someday ... Now if you have an example in head, just tell me, I can perform the computations for you (if your example in not too twisted ... computations become quickly heavy). -Edit : -For your example : the Gram matrix is -A= [0 1 1 1] - [1 0 1 1] - [1 1 0 1] - [1 1 1 0] - -A fundamental domain is the convex hull of the following 4 points : -a=(0,0,1,1), b=(0,1,1,1), d=(1,1,1,1), c=(0,0,0,1) - -(suitably rescaling a,b,c to make them fall in $\mathcal H_3$. The point c is a cusp.) -The reflections supported by the faces of this tetrahedron are -[-1 1 1 1] (a,b,c) -[ 0 1 0 0] -[ 0 0 1 0] -[ 0 0 0 1] - -[0 1 0 0] (a,c,d) -[1 0 0 0] -[0 0 1 0] -[0 0 0 1] - -[1 0 0 0] (b,c,d) -[0 0 1 0] -[0 1 0 0] -[0 0 0 1] - -[1 0 0 0] (a,b,d) -[0 1 0 0] -[0 0 0 1] -[0 0 1 0] - -The matrices I gave in the comments below are conjugated to these.<|endoftext|> -TITLE: A Ramsey avoidance game -QUESTION [21 upvotes]: Consider the following game: Given $K_n$ the complete graph on $n$ vertices, two players take turns coloring its edges. Initially no edges are colored. At his turn a player can color a prevoiusly not colored edge blue or red. The goal is to avoid monochromatic $K_k$. The first player who completes a monochromatic $K_k$ loses. If $n$ is large enough the Ramsey theorem ensures that there can be no draw. Therefore someone has a winning strategy. My question is that: -Is there any pair $(n,k)$ such that $n -TITLE: Uniform proof that a finite (irreducible real) reflection group is determined by its degrees? -QUESTION [24 upvotes]: Given a finite (irreducible real) group $G$ generated by reflections acting on euclidean $n$-space, it was shown by Chevalley in the 1950s that the algebra of invariants of $G$ in the associated polynomial algebra over $\mathbb{R}$ is itself a polynomial algebra with homogeneous generators of uniquely determined degrees $d_1, \dots, d_n$. (The converse is true for groups generated by "quasi-reflections", which applies in the complex setting.) Moreover, $|G|= \prod d_i$. Determination of the degrees is usually done via the classification of the possible groups. These are in fact the finite (irreducible) Coxeter groups, characterized by their Coxeter graphs. -The "crystallographic" ones are the Weyl groups familiar in Lie theory. -Textbook references include Bourbaki Groupes et algebres de Lie, Chap. V, $\S5$ and my book Reflection Groups and Coxeter Groups (Cambridge, 1990), Chapter 3. In my section 3.7 is a table giving the degrees for each irreducible type. Once one verifies this table, it's clear that the group $G$ is uniquely determined (up to isomorphism) by its degrees. When the degrees are listed in non-decreasing order, we have $d_1 =2$ and $d_n =h$, the Coxeter number of $G$. (Coxeter showed that the eigenvalues of a Coxeter element are the $m_i$th powers of a primitive $h$th root of 1, where it turns out that $m_i+1 = d_i$.) - -Is there a uniform way to prove (without using the classification) that $G$ is determined by its degrees? - -[EDIT: As the comment by Noam Elkies indicates, the following paragraph should be ignored.] -Though I'm less familiar with the behavior of complex (=unitary) groups generated by quasi-reflections, the same question seems to arise there. For a modern treatment, including the Shephard-Todd classification and a list of degrees in Appendix D, see the book by Lehrer and Taylor Unitary Refelction Groups (Cambridge, 2009). - -REPLY [3 votes]: As it is not clearly stated here so far: As far as I know there is still no conceptual explanation of this observation. -We were looking at this situation for unitary reflection groups in Proposition 2.1 in our last year's preprint http://arxiv.org/abs/1404.5522, tried to give an explanation and discussed possible approaches with David Bessis. However, this lack of understanding still seems to be the state of the art, even for finite Coxeter groups.<|endoftext|> -TITLE: Shelah's categoricity conjecture -QUESTION [7 upvotes]: Does Shelah's categoricity conjecture for abstract elementary classes have applications in other branches of mathematics? - -REPLY [3 votes]: I’m not sure if it counts as “applications in other branches of mathematics”. -The other branch could be category theory. The conjecture was translated in the category-theoretic framework: - -Beke, Tibor, and Jirí Rosický. "Abstract elementary classes and accessible categories." Annals of Pure and Applied Logic (2012) (arxiv link) - -The above work can be seen as a part of research in categorical model theory which was also mentioned on the MO before (see here and here).<|endoftext|> -TITLE: First occurrence of "by the usual compactness argument"? -QUESTION [19 upvotes]: In his blog, Jeff Shallit asks, what was the first occurrence of the exact phrase, "by the usual compactness arguments," in the mathematical literature? -He reports that the earliest appearance he has found was in a paper from 1953: it's on page 400 of John W. Green, Pacific Journal of Mathematics 3 (2) 393-402. -I found another example from that same year, on page 918 of F. A. Valentine, Minimal sets of visibility, Proc Amer Math Soc 4, 917-921. -So, is this year the 60th anniversary of the first appearance of that phrase? -If there was an earlier occurrence of the equivalent phrase in a language other than English, that, too, would be of interest. - -REPLY [26 votes]: There is a paper, written in German, called Ueber Räume mit verschwindender erster Brouwerscher Zahl. by Urysohn and Alexandroff from 1928. (Notice that Urysohn drowned while swimming with Alexandroff in 1924) -The following quote is from page 810 (emphasis added): - -Der Beweis dieser Tatsache ist wörtlich derselbe wie im Falle, wo $R$ der $R^n$ ist: es genügt zu zeigen, dass die erwähnte Trennungseigenschaft im Brouwerschen Sinne induktiv ist (was aus der Kompaktheit von $R$ in der üblichen Weise folgt), und dann den Phragmén-Brouwerschen Satz anzuwenden. - -And here is my translation of that quote: - -The proof of this fact is literally the same as in the case, where $R$ is $R^n$: it is sufficient to show that the mentioned separation property is inductive in the sense of Brouwer (which follows from the compactness of $R$ in the usual fashion), and then apply the Phragmén-Brouwer Theorem. - -I suspect, that as soon as there was a good notion of compactness, people used it in a routine way as an standard argument.<|endoftext|> -TITLE: Is there an extremal metric on toric Fano manifolds which have nonzero Futaki invariant? -QUESTION [5 upvotes]: According the work by Wang & Zhu, on toric Fano manifolds there exist Kaehler-Ricci solitons. If Futaki=0, there also exist CSCK metrics. But if the Futaki invariant does not vanish, what about extremal metrics? Is there some counterexample? - -REPLY [2 votes]: Some additional comment: -Let $(X,J,\omega)$ be Fano Kahler manifold with $\omega_0\in [\omega_0]=\kappa$, then by $\partial\bar\partial$-lemma any other Kahler 2-form can be written as $\omega_0+\sqrt{-1}\partial\bar\partial \varphi$. Now the set of Kahler metrics in $[\omega_0]$ can be identified with $$\mathcal H=\{\varphi \in C^\infty(X,\mathbb R)|\omega_0+\sqrt{-1}\partial\bar\partial \varphi>0\}/\mathbb R$$. Note that $T_\varphi\mathcal H=C^\infty(X,\mathbb R)/\mathbb R$ -Now, consider the Calabi- functional(which is sort of special case of Yang-Mills functional) -If we denote $S(\omega_\varphi)$ denotes to be the scalar curvature of the metric $\omega_\varphi$ and $\underline S$ be the average scalar -curvature -$$\mathcal C a:\mathcal H\to \mathbb R$$ -$$\varphi\to \mathcal Ca(\varphi)=\int_X (S(\omega_\varphi)-\underline S)^2\frac{\omega_\varphi^n}{n!}$$ -Now if you take $\psi\in T_\varphi\mathcal H$ , then we can write the variation of Calabi functional via using Lichnerowicz operator $\mathcal D_\varphi=\overline \partial\nabla$ as follows -$$(d\mathcal Ca(\varphi))\psi=2\int_X (\mathcal D_\varphi^*\mathcal D_\varphi S(\omega_\varphi)).\psi\frac{\omega_\varphi^n}{n!}$$ where $\mathcal D_\varphi^*$ is the $L^2(X,\omega_\varphi)$-adjoint of $\mathcal D_\varphi$. As an exercise integrating by part we have $\ker \mathcal D_\varphi^*\mathcal D_\varphi =\ker \mathcal D_\varphi$ -This means that $\omega_\varphi$ is extremal metric (which means the critical points of Calabi functional) if the gradient of scalar curvature $S(\omega_\varphi)$ define a vector field . -About your question this recent paper https://arxiv.org/pdf/1610.06865.pdf<|endoftext|> -TITLE: $ - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1}$ -QUESTION [7 upvotes]: Numerical evidence suggests: -$$ - \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} =? \sum_{n=1}^\infty \frac{1}{2^n+1} \approx 0.764499780348444 $$ -Couldn't find cancellation via rearrangement. -For the second series WA found closed form. - -Is the equality true? - -REPLY [14 votes]: Write each term of the right hand side as a geometric series, $1/3 = 1/2-1/4+1/8\dots$, $1/5 = 1/4-1/16+1/64-\dots$, $1/9=1/8-1/64+1/256-\dots$ etc. Now the sum of the first terms of each series is $1/2+1/4+1/8+\dots = 1$, the sum of all second terms is $-1/4-1/16-1/64-\dots = -1/3$ etc, giving the alternating sum of the left hand side. - -REPLY [2 votes]: This is just a formal proof using idea of Loïc Teyssier that appeared in the comments. We just rewrite both sides of the equation as double sums (using geometric series) and notice that they are equal after changing the order of summation. -Since $$\frac{q}{1-q} = \sum_{k=1}^\infty q^k,$$ we have for the left hand side -\begin{align} -- \sum_{n=1}^\infty \frac{(-1)^n}{2^n-1} & = - \sum_{n=1}^\infty \frac{(-1)^n2^{-n}}{1-2^{-n}} \\ -& = - \sum_{n=1}^\infty (-1)^n\sum_{k=1}^\infty (2^{-n})^k \\ -& = \sum_{k,n = 1}^\infty (-1)^{n+1} 2^{-nk}. -\end{align} -Similarly, for the right hand side we use $$\frac{q}{1+q} = \sum_{k=1}^\infty (-1)^{k+1} q^k $$ to get -\begin{align} -\sum_{n=1}^\infty \frac{1}{1+2^n} &= \sum_{n=1}^\infty \frac{2^{-n}}{1+2^{-n}}\\ -& = \sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{k+1}(2^{-n})^k \\ -& = \sum_{k,n=1}^\infty (-1)^{k+1} 2^{-nk}. -\end{align}<|endoftext|> -TITLE: Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$ -QUESTION [12 upvotes]: I'm working on solving the quartic Diophantine equation in the title. Calculations in maxima imply that the only integer solutions are -\begin{equation} - (r,s) \in \{(-3, -2), (-2, 3), (-1, 0), (0, -1), (0, 1), (1, 0), (2, -3), (3, 2)\}. -\end{equation} -Evidently, the set above are all solutions, and furthermore if $(r,s)$ is a solution then so is $(-r,-s)$; hence we only need to prove that there are no solutions with $r > 3$. I have factored the equation as both -\begin{align} - 4r^2s(3s-2r) &= (r-s)^4-1 = (r-s-1)(r-s+1)\bigl((r-s)^2+1\bigr) -\end{align} -and -\begin{align} - 4rs^2(3r+2s) &= (r+s)^4-1 = (r+s-1)(r+s+1)\bigl((r+s)^2+1\bigr), -\end{align} -but don't know where to go from that point. I am hoping there is an elementary solution, even if it's not particularly "simple". -Any help would be appreciated. -Thanks, -Kieren. -EDIT: Note that for all known solutions, $\lvert r + s\rvert = 1$ or $\lvert r - s\rvert = 1$. -EDIT: In a comment, Peter M. pointed out that this can be written as the Pell equation -$$ - (r^2+2rs-s^2)^2-2(2rs)^2=1. -$$ -Curiously — and perhaps not coincidentally — the fundamental solution to that Pell equation is $(3,2)$, which is also the largest [conjectured] positive integer solution. As the fundamental solution in this case is $(r^2+2rs-s^2,2rs)$, whereas the largest integer solution is $(r,s)$, perhaps there's a way of using that to force some sort of descent or contradiction? -EDIT: Adding $4r^4$ and $4s^4$ to both sides of the equation and factoring yields, respectively -\begin{align*} - (r-s)^2(r+s)(r+5s) &= (2s^2-2s+1)(2s^2+2s+1) -\end{align*} -and -\begin{align*} - (r-s)(r+s)^2(5r-s) &= (2r^2-2r+1)(2r^2+2r+1) -\end{align*} -Note that, in each case, the two factors on the right-hand side are relatively prime (because they're odd, and evidently $\gcd(r,s)=1$). So far, this is the most interesting factorization I've found. -EDIT: Considering the equation modulo $r-s$ and modulo $r+s$, one can (I believe) prove that if a prime $p \mid (r-s)(r+s)$, then $p \equiv 1\!\pmod{4}$. -EDIT: Still holding out for an elementary proof. In addition to the restriction -$$p \mid (r-s)(r+s) \implies p \equiv 1\!\pmod{4},$$ -I've found the following list of divisibility restrictions: -\begin{align} -r &\mid (s-1)(s+1)(s^2+1) \\ -s &\mid (r-1)(r+1)(r^2+1) \\ -(r-s) &\mid (4r^4+1) \\ -(r+s) &\mid (4s^4+1) \\ -(r+s)^2 &\mid (4r^4+1) \\ -(r-s)^2 &\mid (4s^4+1) \\ -(r-s-1) &\mid 4(s-2)s(s+1)^2 \\ -(r-s+1) &\mid 4(s-1)^2s(s+2) \\ -(r+s-1) &\mid 4(s-3)(s-1)s^2 \\ -(r+s+1) &\mid 4s^2(s+1)(s+3), -\end{align} -as well as a host of other [less immediately compelling] restrictions. Based on this, I'm hoping to prove that one of $r-s$ or $r+s$ must be $\pm 1$; bonus if I can show that the other divides $5$. -EDIT: I can show that $4s > 3r$. Calculations in maxima suggest that no numbers $r,s$ with $4 \le r \le 13000$ and $r > s \ge 1$ and $r$ odd and $s$ even and $r-s>1$ and $4s>3r$ also satisfy the six divisibility requirements -\begin{align} -r &\mid (s-1)(s+1)(s^2+1) \\ -s &\mid (r-1)(r+1)(r^2+1) \\ -(r-s-1) &\mid 4(s-2)s(s+1)^2 \\ -(r-s+1) &\mid 4(s-1)^2s(s+2) \\ -(r+s-1) &\mid 4(s-3)(s-1)s^2 \\ -(r+s+1) &\mid 4s^2(s+1)(s+3). -\end{align} -Note that I didn't even need all of the congruences in the previous list. Next I'll run $r$ up to $10^6$ or so. Hopefully, though, I can obtain an algebraic proof of all of this! -EDIT: So far, the best bounds I can prove are $4/3 < r/s < 3/2$. - -REPLY [6 votes]: Answered on stackexchange -where Kieren MacMillian linked to this MO question. Briefly: -This Thue equation -is equivalent to Ljunggren's equation $X^2+1 = 2Y^4$. -Ljunggren proved in 1942 that the only integer solutions are -$(X,Y) = (\pm 1, \pm 1)$ and $(\pm 239, \pm 13)$. -[Equivalently, $(a,b,c) = (119,120,169)$ is the unique Pythagorean -triple with $a-b = \pm 1$ and $c$ a perfect square.] -This implies the desired result that the only solutions of -$$ -Q(r,s) = r^4 + 4 r^3 s - 6 r^2 s^2 - 4 r s^3 + s^4 = 1 -$$ -are the known small ones with $r^2+s^2 = 1$ or $13$. The implication -we need is provided by the identity $Q(r,-s)^2 + Q(r,s)^2 = 2(r^2+s^2)^4$. -Ljunggren's proof is a difficult application of Skolem's $p$-adic method. -In the decades since then other techniques have been developed which -make the solution of such an equation routine; for example -this is how gp can calculate almost instantaneously -[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]] - -in response to the command -thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1) - -But none of these techniques is elementary either, -and as far as I know the problem of finding an elementary proof -of Ljunggren's theorem remains open.<|endoftext|> -TITLE: Does this sequence always give an integer? -QUESTION [23 upvotes]: It is known that the $k$-Somos sequences always give integers for $2\le k\le 7$. -For example, the $6$-Somos sequence is defined as the following : -$$a_{n+6}=\frac{a_{n+5}\cdot a_{n+1}+a_{n+4}\cdot a_{n+2}+{a_{n+3}}^2}{a_n}\ \ (n\ge0)$$ -where $a_0=a_1=a_2=a_3=a_4=a_5=1$. -Then, here is my question. -Question : If a sequence $\{b_n\}$ is difined as -$$b_{n+6}=\frac{b_{n+5}\cdot b_{n+1}+b_{n+4}\cdot b_{n+2}+{b_{n+3}}^2}{b_n}\ \ (n\ge0)$$$$b_0=b_1=b_2=b_3=1, b_4=b_5=2,$$ -then is $b_n$ an integer for any $n$? -We can see -$$b_6=5,b_7=11,b_8=25,b_9=97,b_{10}=220,b_{11}=1396,b_{12}=6053,b_{13}=30467$$ -$$b_{14}=249431,b_{15}=1381913,b_{16}=19850884,b_{17}=160799404$$ -$$b_{18}=1942868797,b_{19}=36133524445, \cdots.$$ -Remark : This question has been asked previously on math.SE without receiving any answers. -Motivation : I've been interested in seeing what happens when we change the first few terms. It seems true, but I can neither find any counterexample nor prove that the sequence always gives an integer. As far as I know, it seems that this question cannot be solved in the way which proved that the $6$-Somos sequence always gives an integer. - -REPLY [7 votes]: Andrew Hone in the articles Analytic solutions and integrability for bilinear recurrences of order six and Sigma-function solution to the general Somos-6 recurrence via hyperelliptic Prym varieties (with Yuri N. Fedorov) considered general Somos-6 recurrence -$$\tau_{n+6}\tau_{n}=\alpha\tau_{n+5}\tau_{n+1}+\beta\tau_{n+5}\tau_{n+2}+\gamma\tau_{n+3}^2$$ -with arbitrary coefficients $\alpha$, $\beta$, $\gamma$. He gave explicit analytic solution in the form -$$\tau_n=AB^n\dfrac{\sigma(\mathbf{v}_0+n\mathbf{v})}{\sigma(\mathbf{v})^{n^2}},$$ where $\mathbf{v}$, $\mathbf{v}_0\in\mathbb{C}^2$ and $\sigma$ is a Kleinian sigma function associated with some genus $2$ curve $\mu^2=4\nu^5+c_3\nu^3+c_2\nu^2+c_1\nu+c_0.$ From Baker's addition formula (see Baker's An introduction to the theory of multiply periodic functions (1907)) -$$\dfrac{\sigma(\mathbf{u}+\mathbf{v}) -\sigma(\mathbf{u}-\mathbf{v})}{\sigma(\mathbf{u})^2\sigma(\mathbf{v})^2}= -\wp_{22}(\mathbf{u})\wp_{12}(\mathbf{v})-\wp_{12}(\mathbf{u})\wp_{22}(\mathbf{v})+ -\wp_{11}(\mathbf{v})-\wp_{11}(\mathbf{u})$$ -follows that for some fumctions $f_k$, $g_k$ ($1\le k\le 4$) $$\tau_{m+n}\tau_{m-n}=\sum\limits_{k=1}^{4}f_k(m)g_k(n).$$ -It means that infinite matrix consisting from $A_{mn}=\tau_{m+n}\tau_{m-n}$ has rank at most $4$ and every minor of order $5$ vanishes. -Let's apply this theory to the given sequence. (From this point I'll follow David Speyer's solution.) -Taking two $5$-tuples of $m$'s and $n$'s $(m,22,21,20,19)$ and $(24,18,12,6,0)$ from -$$\left| -\begin{array}{ccccc} - \tau_{m-24} \tau_{m+24} & \tau_{m-18} \tau_{m+18} & \tau_{m-12} \tau_{m+12} & \tau_{m-6} \tau_{m+6} & \tau_m^2 \\ - \tau_{22-24} \tau_{22+24} & \tau_{22-18} \tau_{22+18} & \tau_{22-12} \tau_{22+12} & \tau_{22-6} \tau_{22+6} & \tau_{22}^2 \\ - \tau_{21-24} \tau_{21+24} & \tau_{21-18} \tau_{21+18} & \tau_{21-12} \tau_{21+12} & \tau_{21-6} \tau_{21+6} & \tau_{21}^2 \\ - \tau_{20-24} \tau_{20+24} & \tau_{20-18} \tau_{20+18} & \tau_{20-12} \tau_{20+12} & \tau_{20-6} \tau_{20+6} & \tau_{20}^2 \\ - \tau_{19-24} \tau_{19+24} & \tau_{19-18} \tau_{19+18} & \tau_{19-12} \tau_{19+12} & \tau_{19-6} \tau_{19+6} & \tau_{19}^2 \\ -\end{array} -\right|=0$$ -we get his first recurrence which proves that $\tau_{6n}$, $\tau_{6n+1}$, $\tau_{6n+2}$, $\tau_{6n+3}$ are always odd. -Taking two $5$-tuples of $m$'s and $n$'s as $(m,45,44,43,42)$ and $(48,36,24,12,0)$ we get his second formula which proves that $2$-adic valuation has period $24$. -(Thanks to David Speyer who found a gap in first version of this answer.)<|endoftext|> -TITLE: Calogero-Moser system: relationship between dual variables and the KKS construction -QUESTION [7 upvotes]: This is a question about the relationship between two ways of viewing the Calogero-Moser system. -$$\ddot x_i=2\sum_{j\neq i}\frac{1}{(x_i-x_j)^3}\qquad i=1,\ldots N$$ - -By introducing the $N$ variables $y_i$ ("dual variables" for want of a better term), one has the equivalent first order system - -$$i\dot x_j=\sum_{k\neq j}\frac{1}{x_k-x_j}-\sum_{k=1}^N\frac{1}{y_k-x_j}\\ -i\dot y_j=\sum_{k\neq j}\frac{1}{y_j-y_k}-\sum_{k=1}^N\frac{1}{y_j-x_k}\\$$ -I am not sure who to attribute these equations to, but for the case where the $x$'s and $y$'s are complex conjugate, they appear in KM Case, PNAS, 75, 3562 (1978) in the context of pole dynamics for the Benjamin-Ono equation. - -Alternatively, the construction of Kazhdan, Kostant, Sternberg (KKS) views the CM system as (to cut a long story short) the dynamics of the eigenvalues of the matrix $X+Pt$, where $P_{ij}=i/(x_i-x_j)$ for $i \neq j$ and the diagonal elements give the initial velocities. This is as an example of Hamiltonian reduction. - - -My question is: what is the relationship (if any) between these two viewpoints? Specifically, to what do the $y_i$ correspond in the matrix formulation? - -REPLY [3 votes]: I) Let us introduce a collective notation $z_i$, $i\in I$, for OP's $x_i$'s and $y_i$'s (which by the way do not have to be equal in numbers). Here $I$ is a finite index set. We assume that the map $z:I\to \mathbb{C}$ is injective. Also let us introduce a parity $\sigma: I\to \{\pm 1\}$, which is $+1$ for an $x_i$ and $-1$ for a $y_i$. Define a bi-linear skew-symmetric bracket$^1$ -$$\tag{1} \{z_i,z_j\}~:=~ \left\{ \begin{array}{ccl} \frac{1}{z_i-z_j} & \text{for} & i\neq j, \\ 0& \text{for} & i= j. \end{array} \right.$$ -Then OP's coupled first-order system can be written in Hamiltonian form -$$\tag{2} \mathrm{i}\dot{z}_j -~=~\sum_{i\in I\backslash\{j\}}\frac{\sigma_i}{z_i-z_j} -~\stackrel{(1)+(3)}{=}~\{z_j,H\},$$ -with Hamiltonian -$$\tag{3} H~:=~ -\sum_{i\in I} \sigma_i z_i. $$ -More generally, for a function $f=f(z)$, the time evolution is given as -$$\tag{4} \mathrm{i}\frac{d f}{dt} -~=~\{f,H\}+ \mathrm{i}\frac{\partial f}{\partial t}. $$ -II) The corresponding second-order system is the Calogero-Moser equations$^2$ -$$\tag{5} -\ddot{z}_j -~=~\sum_{i\in I\backslash\{j\}}\frac{1+\sigma_i\sigma_j}{(z_i-z_j)^3}. $$ -It is a major point that the sum on the rhs. of eq. (5) only runs over elements of the same kind, i.e. if ${z}_j$ on the lhs. is an $x_j$, then the non-zero terms in the sum on the rhs. is only over the $x_i$'s, i.e. independent of the $y_i$'s. Hence the evolution of $x_j$ only depend on the $y_i$'s via their initial conditions [1]. And vice-versa with the roles $x_i \leftrightarrow y_i$ exchanged. -III) The Calogero-Moser Hamiltonian in Darboux coordinates $(z_i,p_i)$ reads -$$ \tag{6}H_{CM}~=~\frac{1}{2}\sum_{i\in I}p_i^2 -+\frac{1}{4}\sum_{i,j\in I}^{i\neq j}\frac{1+\sigma_i\sigma_j}{(z_i-z_j)^2}. $$ -IV) In the spirit of the Kazhdan-Kostant-Sternberg (KKS) construction [2,3], let us define a position matrix -$$ \tag{7} Z~:=~\text{diag}(z_i),$$ -a parity matrix -$$ \tag{8} \Sigma~:=~\text{diag}(\sigma_i), \qquad \Sigma^2 = {\bf 1},$$ -and a momentum matrix -$$ \tag{9} P_{ij}~:=~\left\{ \begin{array}{ccl} -\frac{\mathrm{i}}{z_j-z_i} & \text{for} & i\neq j, \\ -p_i& \text{for} & i= j. \end{array} \right.$$ -The diagonal elements $p_i$ of the momentum matrix (9) have a physical interpretation as initial velocities. The momentum matrix (9) satisfies the following Canonical Commutation Relation (CCR) for finite-dimensional matrices: -$$ \tag{10} [Z,P]-\mathrm{i}{\bf 1}~=~\text{rank-one matrix}.$$ -The Calogero-Moser Hamiltonian (6) can then be written as -$$ \tag{11} H_{CM}~=~\frac{1}{4}{\rm Tr}(P^2+P\Sigma P\Sigma).$$ -The flow of $z_i(t)$ is given by the eigenvalues of the matrix -$$\tag{12} Z(t=0)+t\frac{P(t=0)+\Sigma P(t=0)\Sigma}{2}. $$ -The matrix (12) is block diagonal consisting of two blocks. Each block is just the standard KKS construction. In particular, the eigenvalues $x_i(t)$ and $y_i(t)$ do only talk to each other via the initial conditions. -References: - -M. Stone, I. Anduaga, and L. Xing, The classical hydrodynamics of the Calogero–Sutherland model, J. Phys. A: Math. Theor. 41 (2008) 275401, arXiv:0803.3735. -D. Kazhdan, B. Kostant, and S. Sternberg, Hamiltonian group actions and dynamical systems of Calogero type, Comm. Pure Appl. Math. 31 (1978) 481. -P. Etingof, Lectures on Calogero-Moser systems, arXiv:math/0606233. - --- -$^1$ Note that the bracket (1) does not satisfy the Jacobi identity. We suspect that the bracket (1) can be extended to a homotopy hierarchy of higher brackets, although we did not pursuit the matter, partly because the bracket (1) is not important for the rest of the answer. -$^2$ To prove Eq. (5) from eq. (2), the following identity is helpful: -$$\mathrm{i}(\dot{z}_i -\dot{z}_j) ~=~\{z_i-z_j,H\}$$ -$$\tag{13} ~\stackrel{(1)+(3)}{=}~-\frac{\sigma_i+\sigma_j}{z_i-z_j} -+(z_i-z_j)\sum_{k\in I\backslash\{i,j\}}\frac{\sigma_k}{(z_i-z_k)(z_j-z_k)}.$$<|endoftext|> -TITLE: Learning a little Motivic Cohomology -QUESTION [15 upvotes]: Simply because I find it interesting, I have spent some time studying motivic cohomology from the lectures by Mazza, Voevodsky and Weibel. However, I'm finding it hard to tell if the theory is something I could use for intuition or to prove interesting theorems. I was hoping somebody could give me a few examples if this is the case. -An example I would consider for K-theory could be that Bloch (Ch. 5 in Duke Lectures) proves Roitman's theorem using in part K-theoretic techniques. -$\textbf{Question:}$ For example, I've been told that one can, in a useful way, write down the Abel-Jacobi map using this theory. However I have no example to this effect. Does anyone know of an example where this is the case, or understand maybe why this is predicted to be the case? -$\textbf{Question:}$ There are several conjectures that one can state in complete generality if one uses this theory. However, what one can prove if they learn motivic cohomology? Could one prove (perhaps in an "easier" way) some theorems that an arithmetic algebraic geometer (interpreted however you like) could find interesting? Are there number theoretic things that people study using these tools? -$\textit{Edit}$: I should clarify: Beilinson's conjecture on special values of L-functions of course aims to explain special values of various L-functions -via motivic cohomology. Hence, it makes sense that in studying it people use the theory. I'm really interested if there are examples that are not of this form. (In the sense that the theory is used in work on conjectures not written in its language). Examples would be the question about the Abel-Jacobi map above, or in the case of K-theory Bloch's proof mentioned above. Milne's result on the the polynomials $P_{2r}$ appearing in the Weil conjectures (from the paper referenced by Andrew below) is also an example. Somehow I feel this can't be the only result of this kind ... - -There are some results that sound extremely interesting: For example, in http://arxiv.org/pdf/1309.4068.pdf Geisser and Schmidt construct a pairing between $\textrm{mod } m$ algebraic singular homology and $\textrm{mod } m$ tame etale cohomology group. This gives a kind of class field theory in a very general setting, and appears to formally resemble the topological situation a lot. -However, it is completely unclear to me if I could ever use such a result as someone who doesn't intend to specialize in this subject. - -REPLY [5 votes]: My understanding is that motivic cohomology has the ability to describe integral values of L-functions up to a constant (and specifically values of zeta functions up to a sign). An example would be Milne's paper "motivic cohomology and values of zeta functions" published in compositio in '88. So, the answer to your question in the bold is yes, but I am sorry I don't have a more detailed answer for you.<|endoftext|> -TITLE: Can we define the tensor product in the derived category $D^b_{\text{coh}}(X)$ just from $D^b_{\text{coh}}(X)$ in certain cases? -QUESTION [6 upvotes]: This question arise from the comparision of the reconstruction theorems of Bondal-Orlov and Balmer and is inspired by Shizhuo Zhang's mathoverflow question: How to unify various reconstruction theorems (Gabriel-Rosenberg, Tannaka,Balmers) -On the one hand, A. Bondal and D. Orlov in http://arxiv.org/abs/alg-geom/9712029 proved their celebrated reconstruction theorem: Let $X$ be a smooth projective variety such that the canonical bundle $\omega_X$ is either ample or anti-ample, and let $Y$ be any projective variety. If $D^b_{\text{coh}}(X)\cong D^b_{\text{coh}}(Y)$ as triangulated categories, then $X\cong Y$. -In fact, they reconstruct $X$ from $D^b_{\text{coh}}(X)$ and their construction heavily use the $\textit{Serre functor}$ S. In general, for a $k$-linear category $\mathcal{C}$, an equivalence $S: \mathcal{C}\rightarrow \mathcal{C}$ is called a Serre functor for $\mathcal{C}$ if their is a natural, bifunctorial isomorphisms -$$ -\varphi_{A,B}: \text{Hom}_{\mathcal{C}}(A,B)\xrightarrow{\sim} \text{Hom}_{\mathcal{C}}(B,SA)^{\vee} -$$ -for every $A,B$. Here $(\bullet)^\vee$ denotes the $k$-dual. When $\mathcal{C}=D^b_{\text{coh}}(X)$, the Serre functor $S_X$ is given by -$$ -S_X\mathcal{E}=\mathcal{E}\otimes \omega_X[\dim X] -$$ -i.e tensor product with the canonical bundle and shift by $\dim X$ in $D^b_{\text{coh}}(X)$. We can refer to Section 4 of A. Caldararu's lecture notes http://arxiv.org/abs/math/0501094 for an excellent introduction. We notice that Bondal-Orlov reconstruction does not involve the tensor structure of $D^b_{\text{coh}}(X)$ besides the definition of the Serre functor. -(As pointed out by Piotr and Qiaochu in the comments, the Serre functor is unique, hence is part of the data of $D^b_{\text{coh}}(X)$ and is not an extra data.) -On the other hand, P. Balmer in http://arxiv.org/abs/math/0111049 proved another reconstruction theorem: Let $X$ be a noetherian scheme, then we can reconstruct $X$ from the $\textit{tensor triangulated category}$ $(D^{\text{perf}}(X),\otimes^L_{\mathcal{O}_X})$. This reconstruction uses the extra structure of $D^{\text{perf}}(X)$: the tensor product, and it applies for more general $X$ than Bondal-Orlov. -Now we can look at the case when the two theorems overlap: If $X$ is smooth projective with $\omega_X$ ample or anti-ample, then $D^{\text{perf}}(X)\cong D^b_{\text{coh}}(X)$. Now from $D^b_{\text{coh}}(X)$ either using the Serre functor or using the tensor structure. It seems that there are some redundancy here. In fact we have -$$ -D^b_{\text{coh}}(X) \xrightarrow [\text{reconstruction}]{\text{Bondal-Orlov}}X\rightarrow (D^b_{\text{coh}}(X),\otimes^L_{\mathcal{O}_X})\xrightarrow [\text{reconstruction}]{\text{Balmer}} X\rightarrow D^b_{\text{coh}}(X) \ldots -$$ -Hence one may expect a direct construction of the tensor structure $(D^b_{\text{coh}}(X),\otimes^L_{\mathcal{O}_X})$ just from $D^b_{\text{coh}}(X)$ itself, considered as a triangulated category. -$\textbf{My question}$ is: If $X$ is smooth projective variety with $\omega_X$ ample or anti-ample, could we define the tensor product structure on $D^b_{\text{coh}}(X)$ just from the triangulated cateogry structure on $D^b_{\text{coh}}(X)$? -In the other direction we have $\textbf{a related question}$: What is the role of the Serre functor $S_X$ and the canonical bundle $\omega_X$ in the the tensor triangulated category $(D^b_{\text{coh}}(X),\otimes^L_{\mathcal{O}_X})$? - -REPLY [8 votes]: It seems like the answer to your question is no, at least without further clarification. If you could define the tensor product structure on $D^b(X)$ just from the triangulated structure and the Serre functor, then any $k$-linear derived autoequivalence $F:D^b(X)\rightarrow D^b(X)$, since it commutes with the Serre functor, would be a $\otimes$-autoequivalence as well. Now, consider $\mathbb{P}^1_k$ and the autoequivalences $F(n,k)(\mathscr{F})=\mathscr{F}(n)[k]$. These are not $\otimes$-autoequivalences unless $n=k=0$. More or less, they induce new $\otimes$-structures on $D^b(X)$ where the unit is $\mathscr{O}_{\mathbb{P}^1_k}(-n)[-k]$. So, I don't see how to make this work. -That being said, it's possible that you can reconstruct all of these $\otimes$-structures at once. Here is what I have in mind. Pick any invertible object $U$ in $D^b(X)$. Recall that the invertible objects are those complexes such that for every point-like object $P$ there exists $n_p\in\mathbb{Z}$ such that $Hom(U,P[i])=k(P)$ when $i=n_P$ and $0$ otherwise. Bondal and Orlov proved that the point-like objects are just the shifted sheaves $k(P)[n]$ for $P\in X$ (here on out I'll work over an algebraically closed field for simplicity). Then, they proved that the invertible objects are precisely the shifted line bundles $\mathscr{L}[n]$. So, our $U$ is equivalent to $\mathscr{L}[n]$ for some $\mathscr{L}$ and $n$. This can be ignored for time being; I'll come back to it later. -I claim that there is a tensor product $\otimes_U$ on $D^b(X)$ for which $U$ is the unit. Given two complexes $\mathscr{F}$ and $\mathscr{G}$, resolve $\mathscr{F}$ by terms of the form $S_X^n(U)$. The fact that we can do this is because the $S_X^n(U)$ form a spanning class of $D^b(X)$, which follows easily from the fact that the powers of $\omega_X$ form a spanning class, and the fact that $U=\mathscr{L}[n]$. Then, define $S_X^n(U)\otimes_U\mathscr{G}$ as $S_X^n(G)$. Then use these to define $\mathscr{F}\otimes_U\mathscr{G}$. Note that we need ampleness or anti-ampleness in order to perform this construction. It looks like the details work out: uniqueness follows from $S^n_X(U)\otimes_UU=S^n_X(U)$. -Now, recall that Bondal and Orlov also proved that the group of derived $k$-linear autoequivalences of $D^b(X)$ fits into an exact sequence $$0\rightarrow\mathbb{Z}\times Pic(X)\rightarrow Aut(D^b(X))\rightarrow Aut(X)\rightarrow 1.$$ We see that by fixing the invertible object $U$, i.e. the shifted line bundle $\mathscr{L}[n]$ we are picking exactly an element in $\mathbb{Z}\times Pic(X)$.<|endoftext|> -TITLE: Must the union of these two aspherical spaces be aspherical? -QUESTION [7 upvotes]: Let $X$ be a reasonably nice topological space (say, a connected CW complex), and let $Y$ and $Z$ be reasonably nice connected subspaces of $X$ such that $X = Y \cup Z$. -Suppose that $Y$, $Z$, and $Y\cap Z$ are aspherical, and the homomorphism $\pi_1(Y\cap Z)\to \pi_1(Y)$ induced by inclusion is injective. Does it follow that $X$ is also aspherical? -I think this is true, but I have been unable to construct a proof. If excision held for homotopy groups, then we would know that $\pi_n(X,Z) \cong \pi_n(Y,Y\cap Z) = 0$ for all $n \geq 2$, and it would follow from the long exact sequence for $(X,Z)$ that $\pi_n(X) = 0$ for all $n\geq 2$. -However, this does not appear to be a case to which excision applies. It seems that both $\pi_1(Y,Y\cap Z)$ and $\pi_2(Z,Y\cap Z)$ may be nonzero, so the pair $(Y,Y\cap Z)$ is potentially only $0$-connected, and the pair $(Z,Y\cap Z)$ is potentially only $1$-connected. -So, is it true that $X$ is always aspherical under these circumstances? - -REPLY [13 votes]: You also need the second inclusion-induced homomorphism to be injective, in order to reach the desired conclusion. That is: If $Y$, $Z$, and $Y\cap Z$ are connected, aspherical CW-complexes, and the inclusion-induced homomorphisms $\pi_1(Y\cap Z)\to \pi_1(Y)$ and $\pi_1(Y\cap Z)\to \pi_1(Z)$ are both injective, then the CW-complex $X=Y\cup Z$ is also aspherical. This is an old result of J. H. C. Whitehead: On the asphericity of regions in a 3-sphere, Fundamenta Mathematicae 32 (1939), -no. 1, 149-166.<|endoftext|> -TITLE: Generalizing the Notion of Nilpotent/Abelian/Cyclic Numbers -QUESTION [10 upvotes]: Let us call a number $n\in\mathbb{N}$ nilpotent if -$$n=p_1^{e_1}\cdots p_m^{e_m}$$ -with $p_i^k\not\equiv 1\mod p_j$ for $i,j\in\{1,\ldots,m\}$ and $1\leqslant k\leqslant e_i$. -A cute theorem says the following: - -Theorem: Every group of order $n$ is nilpotent, if and only if $n$ is a nilpotent number. - -But, the interesting thing is that the theorem doesn't stop there. It also says the following: - -Theorem: Every group of order $n$ is abelian, if and only if $n$ is a cubefree nilpotent number. -Theorem: Every group of order $n$ is cyclic, if and only if $n$ is a squarefree nilpotent number. - -For this reason squarefree and cubefree nilpotent numbers are called cyclic and abelian respectively. -The proofs of these theorems can be found here. -There is, of course, after hearing these last two theorems an obvious question: what does being a nilpotent number indivisible by an $\ell^{\text{th}}$ power, for $\ell\geqslant 4$ correspond to? In particular, one might guess that there is some filtration of classes of groups -$$C_2\subseteq C_3\subseteq C_4\subseteq\cdots\subseteq\{\text{nilpotent groups}\}$$ -where all groups of order $n$ are in $C_\ell$ if and only if $n$ is a nilpotent number not divisible by an $\ell^{\text{th}}$-power. -For example, we have already remarked that -$$C_1=\{\text{cyclic groups}\}\qquad C_2=\{\text{abelian groups}\}$$ -Also, if we can find such classes $C_j$, is there some unifying, governing statistic of groups naturally indexed by $\mathbb{N}$, for which $C_j$ is just the class of groups satisfying the $j^{\text{th}}$ statistic? -Thanks! - -REPLY [10 votes]: A nilpotent group is the direct product of its Sylow subgroups, hence the nilpotency class of a nilpotent group equals the maximum of the nilpotency classes of its Sylow subgroups. A group of order $p^n$ has nilpotency class at most n-1, since if $G/G'$ was cyclic, then $G$ itself would be cyclic. On the other hand, $p$-groups of maximal class do exist, so we obtain that for $n\geq 2$, $C_n$ consists precisely of the nilpotent groups of class $\leq n-1$, while $C_1$ is the class of cyclic groups.<|endoftext|> -TITLE: Central extension of the algebraic loop group -QUESTION [11 upvotes]: I'm doing some constructions with the universal central extension $\widehat{\Omega G}$ of the loop group $\Omega G$ (here $G$ is a matrix group), where a priori the loops involved are just smooth, but in fact turn out to be rational functions. This got me thinking to see if everything will in fact land inside the algebraic loop group $G\left(\mathbb{C}((t))\right)$. However, I would like to know how one constructs the central extension in that case, as for some reason I can't seem to find a decent discussion of this (I'm probably being stupid in my searching...). In particular, is the central extension something like an ind-affine algebraic group? Given the cocycle describing the extension, how do you get said extension? (the method I know gives it as a quotient of a split central extension of $P\Omega G$, see links above) -Now my intended aim is to package this into something like the crossed module $\widehat{\Omega G} \to PG$ representing the String 2-group, but using more algebraic ingredients. For instance, replace the Frechet manifold $PG$ of based paths in $G$ with the space of polynomial or rational connections on the trivial $G$-bundle on $\mathbb{C}^\times$. The part I don't know is the central extension as indicated above. -EDIT: let me add that I would be most interested in knowing whether the central extension of the loop group is something like an algebraic group or if it is in some sense 'inherently transcendental' (for instance, the cocycle one uses to build it uses a residue). - -REPLY [4 votes]: Since this recently got bumped by the community user, I thought I'd have a go at constructing an answer. -The central extension (of G(ℂ((t))) by ℂ×) is algebraic. In particular it is a group ind-scheme, which I believe is strict and ind-affine. -However, there does not seem to be a great place in the literature to find information about this central extension (so the OP is not being stupid in his searching). -What immediately came to mind for me is Beilinson and Drinfeld's unpublished manuscript "Quantization of Hitchin's Integrable System and Hecke Eigensheaves." I can't say with truth that I am an expert on the contents within, but I believe the answers sought are contained within. -Also worth looking at, as mentioned by abx, is the paper by Beauville and Laszlo, "Conformal blocks and generalized theta functions." -On the Kac-Moody side, there is the paper "Construction d’un groupe de Kac-Moody -et applications" by Mathieu. However once this is constructed, there is the question of comparing it to the loop group, which I don't know where to find an answer to. (On the other hand, Kumar's book doesn't seem to go as far as to construct the Kac-Moody groups as ind-schemes). -Since cocycles were brought up, let me point out that for central extensions of group objects in a category C, for cocyles to exist, there must be a splitting in C (not as groups) of the map to the quotient. There will be no cocyle within the cateogry of ind-schemes. However at the level of R-points for some R, you could reasonably expect to find a cocycle. The paper "Block-compatible metaplectic cocycles" by Banks, Levi and Sepanski may be useful to some readers wanting to try their hand at certain explicit cocycles (I've found it useful myself in the context of metaplectic groups over local fields). -Hopefully some intrepid MO reader will see this and be motivated to produce a thorough, reliable and accessible exposition of this field.<|endoftext|> -TITLE: Is there a continuous function $f$ satisfying the following Zygmund condition but not differentiable. -QUESTION [5 upvotes]: Suppose that a continuous function $f$ on the line and satisfies - $$ - |f(x+2h)−2f(x+h)+f(x)|\leq const \frac{|h|}{(\log\frac{1}{|h|})^{\beta}}\,\,\,\,\,\,\text{where}\,\,\,\, \beta \in(0, 1] - $$ - for all $x,h$ real. Is it true that $f$ is differentiable? - If not how can I prove it? - -REPLY [12 votes]: The Weierstrass-type function -$$ f(x) := \sum_{n=1}^\infty \frac{1}{2^n n^\beta} \sin(2^n x) $$ -will obey the hypotheses, yet fails to be differentiable at the origin. -In general, one should look to Weierstrass-type functions (perhaps weighted by power weights $|x|^{-\alpha}$ or variants such as $|x|^{-\alpha} \log(\frac{1}{|x|})^{-\gamma}$, in case some $L^p$ norms are involved for a finite $p$) as key test cases for these sorts of endpoint functional embedding problems. It is also quite clarifying to reformulate the hypotheses and conclusion in terms of Littlewood-Paley theory (either the classical theory using harmonic extensions, or the modern theory using smooth partitions of unity in frequency space; see e.g. Stein's "Singular integrals" for the former, or the appendix to my PDE book for the latter). For instance, the hypothesis here is basically equivalent to the Besov-type bound -$$ \| P_N f \|_\infty \ll N^{-1} \log^{-\beta} N$$ -for frequencies $N \gg 1$, where $P_N$ is a smooth Fourier projection to frequencies $|\xi| \sim N$, while the conclusion is roughly equivalent to the pointwise convergence of the series $\sum_N N P_N f(x)$ as $N$ ranges over dyadic integers. As $\sum_N \log^{-\beta} N$ diverges for $\beta \leq 1$, this indicates that the claim is false, and then guided by this analysis one can quickly come up with the aforementioned Weierstrass-type counterexample.<|endoftext|> -TITLE: About the ratio of the areas of a convex pentagon and the inner pentagon made by the five diagonals -QUESTION [34 upvotes]: Question : Letting $S{^\prime}$ be the area of the inner pentagon made by the five diagonals of a convex pentagon whose area is $S$, then find the max of $\frac{S^\prime}{S}$. -                   - -I've been interested in this simple question. It seems that a regular pentagon and its affine images would give the max, but I'm facing difficulty. -Remark : This question has been asked previously on math.SE without receiving any answers. - -REPLY [4 votes]: A complete solution is available at https://arxiv.org/abs/1812.07682 -On the polygon determined by the short diagonals of a convex polygon, -Jacqueline Cho, Dan Ismailescu, Yiwon Kim, Andrew Woojong Lee<|endoftext|> -TITLE: Dynamics of $3^x$ mod 1 -QUESTION [5 upvotes]: Consider the map $f(x)=3^x$ mod 1. Using the the iterated function system $T_{0}x=\log_{3}(x+1), T_{1}x=\log_{3}(x+2)$ we see that $f$ is dynamical conjugated to a full shift on two symbols. Moreover i think it follows from Lasota and York (Trans AMS, 1973) that there is an absolutely continuous ergodic measure for $f$. What is the density of this measure? Is this measure Bernoulli or Markov (an image of a Bernoulli or Markov measure under the conjugation)? What is the entropy of this measure? - -REPLY [9 votes]: As you say, it's an expanding map (min derivative at 0 is $\log 3$), so Lasota-Yorke and a bunch of other papers give that it has an absolutely continuous invariant measure. It's too much to hope that this measure is conjugate to a one-sided Bernoulli shift or a Markov chain. Instead, you can consider the natural extension, a canonically-define invertible map $\hat f$ that factors onto $f$. It is known that $\hat f$ is isomorphic to a Bernoulli shift. This is part of the industry dating back to the 1970's of studying expanding maps. Quite a good summary of everything is contained in Góra's 1994 paper in Ergodic Theory and Dynamical Systems.<|endoftext|> -TITLE: For G a Lie group, can I make sense of G/G as a derived manifold in a nice way? -QUESTION [9 upvotes]: The functor sending a smooth manifold $M$ to its de Rham algebra $\Omega^{\bullet}(M)$ does not send quotients by actions of Lie groups to invariant subalgebras. The example I have in mind is a connected Lie group $G$ acting on itself by left multiplication. The quotient is a point, which has de Rham algebra $\mathbb{R}$ concentrated in degree $0$, but the $G$-invariant subalgebra of $\Omega^{\bullet}(G)$ is much more interesting: it's the Chevalley-Eilenberg algebra $\Lambda^{\bullet}(\mathfrak{g}^{\ast})$ of the Lie algebra of $G$. -If I wanted to fix this, it seems like I ought to derive something, maybe the operation of taking quotients. Is there a nice category of derived manifolds in which the derived quotient $G/G$ has de Rham algebra the Chevalley-Eilenberg algebra (and vector fields given by $\mathfrak{g}$, and differential operators given by $U(\mathfrak{g})$, and so forth)? Or should I be thinking in terms of Lie algebroids, or what? -(I think I know what the category should be based on a talk I attended recently, but I know very little about it and would appreciate references. It should be the opposite of a category whose objects are something like commutative dg-algebras with degree $0$ part a smooth algebra.) -Edit: If what I said about quotients above is silly, let me ask a slightly different question: what I really want to know is what kind of smooth object, in some category of generalized smooth spaces, has de Rham algebra the Chevalley-Eilenberg algebra, vector fields $\mathfrak{g}$, differential operators $U(\mathfrak{g})$, etc. - -REPLY [5 votes]: The standard answer to: what has de Rham algebra the Chevalley-Eilenberg algebra, vector fields g, differential operators U(g), etc. is "the point with the Lie algebroid structure given by the Lie algebra $\mathfrak{g}$." All the notions you listed make sense for Lie algebroids, and you've listed the special cases for the one-point guy and the tangent Lie algebroid. (Funnily enough, I think usually I have to give this speech to explain in what sense differential operators are a universal enveloping algebra.) -I think Sam's answer is that in the modern world, we can think of Lie algebroids in terms of their quotients, which is surely true, but what you'll find good literature for is Lie algebroids.<|endoftext|> -TITLE: Sequential continuity of linear operators -QUESTION [6 upvotes]: Let $u\colon L\to M$ be a linear map of locally convex linear topological vector spaces. -Assume that $u$ is sequentually continuous, i.e. maps convergent sequences to convergent ones. -(This notion is formally weaker that the usual topological continuity in the case of non-metrizable spaces.) Let $L_0\subset L$ be a topologically dense linear subspace. Assume that -$u|_{L_0}\equiv 0$. -QUESTION: Does it follow that $u\equiv 0$? -I am interested in rather concrete examples of spaces: spaces of generalized functions on smooth manifolds (say $R ^n$) with the wave-front set contained in a given closed set. - -REPLY [8 votes]: Take $c(\Gamma)$ with $\Gamma$ uncountable under the topology of pointwise convergence. $c_0(\Gamma)$ is dense but not sequentially dense. Let $u$ be the linear functional that vanishes on $c_0(\Gamma)$ and is one at $1_\Gamma$.<|endoftext|> -TITLE: How does Tate cohomology fit into a derived categories framework? -QUESTION [7 upvotes]: I've read through one class field theory text after another, but there's something very non-intuitive for me about cohomology that makes it hard for me to understand why Tate cohomology was invented. -In order to make up for my lack of intuition regarding anything cohomological, I have gone through a few texts introducing the derived category framework. This approach has been very helpful for me, and has helped me gain intuition about cohomological methods in Algebraic Topology and Algebraic Geometry. My hope is that, with your help, it could also shed light on "Tate cohomology", which, at the moment, seems to me like a completely arbitrary definition that helps in mysterious and miraculous ways. -Is there a definition of Tate cohomology via derived categories that sheds light on why they are a natural thing to consider? What would be a reference of such a treatment? (If you think that I am looking in the wrong place for intuition and you have an alternative suggestion, I would be very happy to hear that as well!) - -REPLY [8 votes]: To any ring $R$, we can associate the bounded derived category $D^b(R)$. The full subcategory $D^{perf}(R)$ is spanned by bounded complexes of projectives. If $R$ is self-injective, e.g. $R=k[G]$ the group-algebra of a finite group $G$ over a field $k$, then the Verdier quotient $D^b(R)/D^{perf}(R)$ is equivalent to the stable module category $\underline{mod}(R)$. If $R=k[G]$, and $M$ is an $R$-module, the Tate cohomology groups are -$$H^n(G,M)=\underline{mod}(R)(k,M[n]),\quad n\in\mathbb Z,$$ -where $k$ is, as usual, the trivial $R$-module. The identification of the quotient with the stable module category is the non-trivial part of the story, and is due to Rickard ("Derived categories and stable equivalence"). - -REPLY [7 votes]: More of a comment to Fernando Muro's answer: Alexander Beilinson has nice handwritten lecture notes on class field theory which can be found here: http://www.math.lsa.umich.edu/~mityab/beilinson/. An expansion on the point of view explained by Fernando Muro can be found in Lecture 2 with many illuminating explanations, examples and applications.<|endoftext|> -TITLE: Galois Group of $x^n-2$ -QUESTION [22 upvotes]: Let $n \in \mathbb{N}$, then the order of the Galois Group -of $x^n-2$ coincide with $n \phi(n)$ for $n\in \{ 1 , \dots , 36 \}$ -except for $n=\{ 8, 16, 24, 32 \}$ where this order is $\frac{ n \phi (n)}{2}$ -and is easy to prove that for $p$ prime we have that the order -of the Galois Group of $x^p-2$ is $p(p-1)$. -What is the order of the Galois Group of $x^n-2 : n\in \mathbb{N}$ is general? -Thanks in advance. - -REPLY [35 votes]: The splitting field of $x^n-2$ over $\mathbb{Q}$ is $K.L$ where -$K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{2})$, so the order of the Galois group -is -$$ -[K.L:\mathbb{Q}] = \frac{[K:\mathbb{Q}]\cdot[L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]} -= \frac{n \phi(n)}{[K\cap L:\mathbb{Q}]}. -$$ -It remains to compute $m:=[K\cap L:\mathbb{Q}]$. -First show that $K\cap L = \mathbb{Q}(\sqrt[m]{2})$. For this, note that the norm -$N_{L/(K\cap L)}(\sqrt[n]{2})$ is in $K\cap L$. This norm is the product -of the $n/m$ conjugates of $\sqrt[n]{2}$ over $K$, so it is the product of $n/m$ of the conjugates of $\sqrt[n]{2}$ over $\mathbb{Q}$, and each of these conjugates has the form -$\zeta_n^i \sqrt[n]{2}$. Hence the norm has the form $\zeta_n^i \sqrt[n]{2}^{n/m} -= \zeta_n^i \sqrt[m]{2}$. -Since this is in $K\cap L$, and $\zeta_n^i\in K$, -it follows that $\sqrt[m]{2}\in K$, so $\sqrt[m]{2}\in K\cap L$. -But $[\mathbb{Q}(\sqrt[m]{2}):\mathbb{Q}]=m=[K\cap L:\mathbb{Q}]$, - so indeed $K\cap L=\mathbb{Q}(\sqrt[m]{2})$. -Next, since $\sqrt[m]{2}\in K$, and $K/\mathbb{Q}$ is abelian, it follows that $\mathbb{Q}(\sqrt[m]{2})/\mathbb{Q}$ is abelian and hence is Galois. Since $\zeta_m\sqrt[m]{2}$ is a conjugate of $\sqrt[m]{2}$ over $\mathbb{Q}$, it follows that $\zeta_m\in\mathbb{Q}(\sqrt[m]{2})$, so in particular $\zeta_m\in\mathbb{R}$, whence $m\le 2$. -The final step is to determine when $\sqrt{2}\in\mathbb{Q}(\zeta_n)$. For this, note -that $\sqrt{2}\in\mathbb{Q}(\zeta_8)$ but $\sqrt{2}\notin\mathbb{Q}(\zeta_4)$, and that -$\mathbb{Q}(\zeta_r)\cap\mathbb{Q}(\zeta_s)=\mathbb{Q}(\zeta_{(r,s)})$. Thus -$\sqrt{2}\in\mathbb{Q}(\zeta_n)$ if and only if $8\mid n$. So the conclusion is that -the splitting field of $x^n-2$ over $\mathbb{Q}$ has degree $n\phi(n)$ if $8\nmid n$, -and has degree $n\phi(n)/2$ if $8\mid n$. -Question: is there a way to do this without showing that $K\cap L=\mathbb{Q}(\sqrt[m]{2})$, by using from the start that $K\cap L$ is a subfield of $\mathbb{Q}(\sqrt[n]{2})$ which is Galois over $\mathbb{Q}$, and then somehow going directly to the final step?<|endoftext|> -TITLE: Geometry behind Rees algebra (deformation to the normal cone) -QUESTION [15 upvotes]: Let me start with the formal definition of Rees algebra. If $A$ is a commutative ring over some field $k$, $I \subset A$ is an ideal, then Rees algebra is by definition -$$ -R=\oplus_{i \in \mathbb{Z}} I^{i} t^{-i} \subset A[t, t^{-1}], -$$ -where $I^{i}=A$ for negative $i$. Basic fact about this algebra is that obvious map $k[t] \to R$ is flat and we get a flat family over an affine line, where fiber over $0$ is spec of associated graded algebra and all other fiber are isomorphic to the $\operatorname{spec}(A)$. -For example, if I take smooth hypersurface $S \subset \mathbb{A}^n$ then normal bundle is trivial and I get a family where fiber over $0$ is $S \times \mathbb{A}^1$, where this affine line "corresponds to normal bundle direction" and all other fibers are isomorphic to $\mathbb{A}^n$. This looks to me like an extremely weird family. -Intuitively, this construction suppose to be an analog of a tubular neighborhood theorem for differentiable manifolds, so in the example above is should be something like small neighborhood of $S$ in the total space of normal bundle is isomorphic to a small neighborhood of $S$ in $\mathbb{A}^n$. But in algebraic geometry we can't talk about "small neighborhoods" and naive version would be $S \times \mathbb{A}^1 \cong \mathbb{A}^n$ that fails in general, so instead this two spaces are fibers of a family. -Suppose that I don't know such algebraic construction and a want to construct geometrically such flat family over affine line. What train of geometric thoughts leads to such algebraic answer? - -REPLY [19 votes]: Let me try to trace a path from the picture in differential geometry to the algebraic formula you gave. -Starting in differential geometry, let $X$ be a manifold and $Z$ a submanifold with normal bundle $N$. Consider $X\times \Bbb R \to \Bbb R$. To construct the deformation to the normal bundle, the idea is that we want to replace the fiber at $0$ with a copy of $N$. The trick is in specifying the topology, i.e., given a sequence of points $(x,t)$ tending to the fiber at $0$ (so $t\to 0$), which ones do we declare to have a limit in $N$, and how do we describe that limit? -If we give ourselves a tubular neighborhood, this is easily done: we say that such a sequence $(x,t)$ as $t\to 0$ has a limit in $N$ if $x$ tends to a point in the tubular neighborhood, in which case the limit is that point, viewed as a point of $N$. (In simpler terms, the deformation to the normal bundle is just obtained by removing from $X\times \Bbb R$ the complement to the tubular neighborhood in the fiber at $0$.) -But we can accomplish the same thing more canonically without a tubular neighborhood. The idea is that we consider only sequences of points $(x,t)$ with $t\to 0$ for which $x$ actually tends to $Z$, and the limit in $N$ will be gotten by remembering the direction along which we've approached $Z\times 0$. -More precisely, consider the possible directions along which a point of $X\times \Bbb R$ can approach $Z\times 0$. This is the projective bundle of the normal bundle of $Z\times 0$ in $X\times \Bbb R$, or in other words the projective bundle of the direct sum of $N$ with a copy of $\Bbb R$. If we remove the section at infinity, we see just a copy of $N$. The section at infinity corresponds to points approaching $Z\times 0$ vertically, i.e. asymptotically along the fiber at $0$, $X\times 0$. So, truly, possible directions of points $(x,t)$ approaching $Z\times 0$ non-vertically correspond to points of the normal bundle $N$. This gives another description of the deformation to the normal bundle. -This description is also easy to translate into algebraic geometry. What we did is to take $X\times \Bbb A^1$, blow up $Z\times 0$ inside it, then remove the closure of the preimage of $(X-Z)\times 0$ in this blowup. So, we are more-or-less reduced to seeing how to describe blowups in terms of coordinate algebras. Maybe that's no less transparent than the original situation... but maybe I'll stop here for now and see if this was at all helpful. If so maybe we can try to figure out why the algebraic construction of the blow-up is what it is. - -Response to Dmitri's comments (this is dustin again, i just don't know how to log on anymore): yes, the part about removing the closure of the preimage of $(X-Z)\times 0$ might seem opaque from an algebraic perspective. But actually it's quite the opposite. Recall that, in general, one gets affine charts for a blowup by choosing generators for the ideal. If $J$ is the ideal in the ring $R$ and $f$ is an element of $J$, then the coordinate algebra of such a chart is $R[J/f]$. In our case one need only take $f$ to be the coordinate function $t$. In other words, the deformation to the normal cone is just the most obvious affine piece of the blowup $\operatorname{Bl}_{Z\times 0}(X\times \Bbb A^1)$. One can check chart-by-chart that this description agrees with the more geometric one, in terms of the preimage of $(X-Z)\times 0$ (this is also a specialization of a general fact about blowups).<|endoftext|> -TITLE: "'Category' was defined in order to define 'functor', which was defined in order to define 'natural transformation'" -QUESTION [9 upvotes]: I am looking for the source (and original version) of the above oft-repeated quotation. Mac Lane mentions it in Categories for the Working Mathematician, attributing it to Eilenberg-Mac Lane; however, I didn't see it while briefly skimming their paper General Theory of Natural Equivalences. - -REPLY [17 votes]: CW since some of the recent posts on MO have required little more than googling. - -Prior to the book you mentioned, MacLane attributed this saying to Peter Freyd in: -MacLane, S. (1965). Categorical algebra. Bulletin of the American Mathematical Society, 71(1), 40-106. -Relevant excerpt: (p. 48) - -With regard to the original language, Eric Wofsey points out that Freyd's Abelian Categories (1964) begins with this description:<|endoftext|> -TITLE: What interesting homotopy invariants can I write down using the universal property of homotopy types? -QUESTION [11 upvotes]: I've recently been led to believe some version of the following statement: - -Weak homotopy types, or equivalently $\infty$-groupoids (let me not commit myself to a particular model of these), are freely generated under homotopy colimits by a point in the sense that, if $C$ is an $(\infty, 1)$-category with small homotopy colimits, then the $(\infty, 1)$-category of homotopy colimit-preserving $(\infty, 1)$-functors $\infty\text{-Gpd} \to C$ should be equivalent to $C$, with the equivalence on objects being given by evaluating the functor at a point. - -This is the $(\infty, 1)$-categorical version of the $1$-categorical statement that $\text{Set}$ is freely generated under colimits by a point. -So we should be able to construct weak homotopy invariants by finding nice $(\infty, 1)$-categories and nice objects in them and seeing what the corresponding functor above produces. I think the following are examples, although the functors will be contravariant and I could have the details wrong: - -Take $C$ to be the ($(\infty, 1)$-category presented by) the dg-category of complexes of abelian groups, and take the object assigned to the point to be $\mathbb{Z}$. The corresponding weak homotopy invariant should be singular cochains on a space. This should be some version of Eilenberg-Steenrod. -Take $C$ to be the $(\infty, 1)$-category of dg-categories, and take the object assigned to the point to be the dg-category of complexes of abelian groups. The corresponding weak homotopy invariant should be the dg-category of $\infty$-local systems (with values in complexes of abelian groups) on a space. - - -What other interesting weak homotopy invariants can be described this way? - -REPLY [7 votes]: Here is a slight generalization of what you are thinking about, related to Thom spectra (disclaimer, I am still learning this stuff, so, someone correct me if I make a mistake): -Let $R$ be some $A_\infty$-ring spectrum. Let $R\mbox{-line}$ denote the connected space (infinity group) of self (homotopy) automorphisms of $R$ in the $(\infty,1)$-category of $R$-modules. Note that there is a canonical functor $$\theta:R\mbox{-line} \to R\mbox{-Mod}.$$ Also note that there is a canonical equivalence of $(\infty,1)$-categories $$\mbox{Fun}(R\mbox{-line}^{op},\infty Gpd) \simeq \infty Gpd/R\mbox{-line}.$$ An object on the right may be thought of as a generalized spherical fibration, since if $R$ is the sphere spectrum, (stable) spherical fibrations are classified by maps into $BGl_1 R = R\mbox{-line}.$ -The functor $\theta$ may be extended to a unique (homotopy) colimit preserving functor $${Th}_R:\mbox{Fun}(R\mbox{-line}^{op},\infty Gpd) \simeq \infty Gpd/R\mbox{-line} \to R\mbox{-Mod},$$ such that ${Th}_R$ restricted to the Yoneda embedded image of $R\mbox{-line}$ is just $\theta$. A simple calculation shows that in fact ${Th}_R$ sends a map (functor) $$f:X \to R\mbox{-line}$$ to the (homotopy) colimit of $\theta \circ f.$ This colimit is the generalized Thom spectrum of $f$. E.g., if $R$ is the sphere spectrum, and $M$ is a manifold with its canonical spherical fibration $$T_M:M \to R,$$ coming from the image of its tangent bundle under the $J$-homomorphism, then $Th_R\left(T_M\right)$ is the Thom spetrum of $M$. More generally, for the $J$-homomorphism itself $$J:BU \to BGL_1S,$$ $Th\left(J\right)$ is $MU.$<|endoftext|> -TITLE: Invariants of high-dimensional knots -QUESTION [10 upvotes]: In the study of knots in three dimensions, it can be shown that the fundamental group together with a specification of a meridian and longitude form a complete invariant for knots. What is known about the existence of complete or 'almost' complete algebraic invariants for codimension-2 embeddings of arbitrary manifolds $M^n$ into $S^{n+2}$? Are there any special classes for which such complete invariants exist? - -REPLY [8 votes]: I feel that I know both too much and too little about high-dimensional knots and links to answer the question properly. In particular, my 1998 book High-dimensional knot theory does not go beyond what was known by Levine in 1970, as far as complete algebraic invariants for codimension-2 embeddings of arbitrary manifolds $M^n$ into $S^{n+2}$ are concerned. My more recent collaboration with Maciej Borodzik and Andras Nemethi does concern invariants of such embeddings, e.g. proving in the preprint Codimension 2 embeddings, algebraic surgery and Seifert forms that the $S$-equivalence classes of Seifert matrices are isotopy invariants just as in the case $M=S^n$ ($n$ odd), but they are not complete invariants. And then there are boundary links, which deserve an answer by themselves.<|endoftext|> -TITLE: Rings in which every module has an injective image -QUESTION [8 upvotes]: Consider the class of rings $R$ with identity such that any left $R$-module has a non-zero injective homomorphic image. Any such ring is clearly a left V-ring. Is it true that any such ring must be semisimple (artinian) ? - -REPLY [7 votes]: Your class of rings is exactly the class of left V-rings: Let $R$ be a left V-ring and $M$ a non-zero left $R$-module. Let $0\neq m\in M$. Then $Rm$ has a maximal submodule and thus there is an epimorphism $\phi\colon Rm\to S$ with $S$ simple. Since $S$ is injective by assumption, the homomorphism $\phi$ extends from $Rm$ to $M$ and thus $S$ is a non-zero injective homomorphic image of $M$. -Since there are left V-rings that are not semisimple, the answer to your question is "No".<|endoftext|> -TITLE: Cubic forms and Hasse Principle -QUESTION [5 upvotes]: It's well-known that quadratic forms over the rational numbers $\mathbf{Q}$ satisfy the Hasse-Minkowski theorem. This is to say that they are isotropic over $\mathbf{Q}$ if and only if they are isotropic over $\mathbf{R}$ and over $\mathbf{Q}_p$ for all prime numbers $p$. -Of course the same statement does not hold for cubic forms, as demonstrated by Selmer, who showed that the ternary cubic form $3x^3 + 4y^3 + 5z^3$ has nontrivial zeros over $\mathbf{R}$ and over $\mathbf{Q}_p$ for all $p$. We say then that Selmer's example violates the Hasse Principle. -Now I saw a claim recently, which seems a little too good to be true, that if you take a ternary cubic form which violates the Hasse Principle then you can show that it must be diagonal. -Now I don't want to talk about approaches to this or potential proofs or anything of the sort. Merely, when I saw this I thought there ought to be some easy counterexample, but I couldn't immediately come up with one. If you're like me, when you think of Hasse Principle counterexamples, you first think of Selmer's example or of a hyperelliptic example, or maybe something that's just harder to write down. No obvious non-diagonal cubic forms lie in that bunch. As such, I pose the following question. - -Is there a non-diagonal ternary cubic form which violates the Hasse Principle? Can you write it down explicitly? - -REPLY [5 votes]: Continuing with Martin Bright's comment: if $F(X,Y,Z)$ is a ternary cubic form, say with integer coefficients and $M\in GL_3(\mathbf{Z})$ then $M$ acts on the variables $X,Y,$ and $Z$ in an obvious way and with this action, $F(MX, MY,MZ)$ is another ternary cubic form and any reasonable invariants of $F$ will also be true of this new cubic form. This of course includes anything about rational points and also the Jacobian as Prof. Voloch pointed out. -Of course this was not really the sort of example I was looking for. I was hoping to find something quantitatively different (say, they should not be equivalent to a diagonal under the action of $GL_3(\mathbf{Z})$ at least). And what I mean in my comment above about 182b3 is that using the search of Cremona's tables provided by LMFDB, we can find an example with $j$-invariant $-\frac{424962187484640625}{182}$. -Below are 4 ternary cubic forms, all of which are inequivalent under the action of $GL_3(\mathbf{Z})$ and all of which violate the Hasse Principle. - -[ - -x^3 + 17*x*y*z + 2*y^3 + 91*z^3, -2*x^3 + 17*x*y*z - 7*y^3 + 13*z^3, --x^3 + 17*x*y*z - 7*y^3 - 26*z^3, --x^3 - 17*x*y*z - 13*y^3 + 14*z^3 - ] - -These were generated with the magma commands -E := EllipticCurve("182b3"); -ThreeDescentByIsogeny(E);<|endoftext|> -TITLE: Existence of Morse functions on simply connected manifolds -QUESTION [13 upvotes]: Is it true that any simply connected closed manifold possesses a Morse functions that does not have critical points of index one? -If the dimension is at least 5, this is a consequence of the results from Milnor's "Lectures on the $h$-cobordism theorem", but what about dimensions 3 and 4? -If such a function does not exist in general in low dimensions, are there reasonable topological conditions for the existence of a Morse function without critical points of index one? - -REPLY [14 votes]: It is still an open and very interesting question in dimension 4. Akbulut (The Dolgachev surface. Disproving the Harer-Kas-Kirby conjecture. Comment. Math. Helv. 87 (2012), no. 1, 187–241) showed that the Dolgachev surface (and subsequently other elliptic surfaces in the same homotopy type) has a handle decomposition with no 1 or 3 handles. -In the opposite direction, Rasmussen gave an argument that an exotic homotopy $S^2 \times S^2$ detected by an Ozsvath-Szabo (or presumably Seiberg-Witten) invariant would require either 1 or 3-handles. That paper is currently withdrawn from the arxiv; quoting the abstract, `I am withdrawing the paper because it is unclear to me if such a manifold exists.'<|endoftext|> -TITLE: Spin manifolds with one parallel spinor -QUESTION [7 upvotes]: Are there any examples of D-dimensional Ricci-flat Riemannian (spin) manifolds of dimension D= 2,3,4,5 with the dimension of the space of parallel spinors equal to 1? And the same question for the pseudo-Euclidean case with dimensions D= 4,6. Here the manifolds are not assumed to be compact or complete, or simply connected. - -REPLY [5 votes]: In the split cases of interest to you ($D=4,6$ and of arbitrary signature), you can find a discussion of the local existence and generality in a 2000 paper of mine entitled Pseudo-Riemannian metrics with parallel spinor fields and vanishing Ricci tensor (http://arxiv.org/abs/math/0004073). -Note that, unlike the Riemannian case, a pseudo-Riemannian metric with a parallel spinor field need not be Ricci-flat; this usually imposes extra conditions. -In the discussion, I point out exactly what the possibilities are for holonomy groups, so that tells you when you can have the dimension of the space of parallel spinor fields equal to $1$. For example, with $D=4$, this can only happen with metrics of type $(2,2)$ and with $D=6$, this can only happen with metrics of type $(3,3)$. (In each case, such examples, even Ricci-flat ones, do exist.)<|endoftext|> -TITLE: Computing cotangent complex -QUESTION [6 upvotes]: I would like to know if one can compute all the cohomology sheaves of the cotangent complex of a subvariety of the affine space once a resolution of its ideal sheaf is given? -In my precise situation, I work with $X \subset \mathbb{A}^n$ Gorenstein of codimension 3 and I have a resolution: -$$0 \rightarrow \mathcal{O}_{\mathbb{A}^n} \rightarrow E^* \rightarrow E \rightarrow I_X \rightarrow 0$$ -Is it possible to compute the cotangent complex of the closed immersion $ X \hookrightarrow \mathbb{A}^n$ from this resolution? -Many thanks! -EDIT : I really want the cotangent complex of the closed immersion $X \hookrightarrow \mathbb{A}^n$ (and not of the map $X \rightarrow spec(k)$). - -REPLY [5 votes]: Your X is not a local complete intersection in A^n. Then the cotangent complex is going to have cohomology in infinitely many degrees. (Follows from a conjecture by Quillen proved by Avramov IIRC.) Your best bet would be to use Quillen's spectral sequence relating Tor_*(O/I, O/I) to the cohomology of the cotangent complex. You should read the original, but if you are in a hurry, take a look at Example Tag 08RG -of the Stacks project. If you only want to compute the first 3 terms, then I suggest looking at the method of Lichtenbaum-Schlessinger, see Section Tag 09AM.<|endoftext|> -TITLE: Conjecture of Spira on the zeros of $\zeta^\prime(s)$ -QUESTION [9 upvotes]: Let $N(T)$ be the number of complex zeros of $\zeta(s)$ with imaginary part between $0$ and $T$, and let $N_k(T)$ be the analogous counting function for the $k$th derivative $\zeta^{(k)}(s)$. Based on numerical evidence for $T<100$, Spira conjectured in 1965 ("Zero free regions of $\zeta^{(k)}(s)$, J. London. Math. Soc. v. 40 1965 pp. 677–682) that -$$ -N(T)=N_k(T)+[T\log(2)/(2\pi)]\pm 1. -$$ -Berndt later showed that -$$ -N(T)=N_k(T)+T\log(2)/(2\pi)+O(\log(T)). -$$ -Is Spira's original conjecture still open? (I don't expect this is true; finding a counterexample will be a nice project for an undergraduate.) - -REPLY [6 votes]: Shorokhodov gave an explicit counterexample to Spira's conjecture in "Pade approximates and numerical analysis of the Riemann zeta function", Computational Mathematics and Mathematical Physics, vol. 43 no. 9 (2003) pp. 1277-1298. He computed, for $T=1420$, 1000 zeros of $\zeta(s)$, 844 zeros of $\zeta^\prime(s)$, and noted $1420\log 2/2\pi\approx 156.65$.<|endoftext|> -TITLE: Does every manifold have a flat connection? -QUESTION [8 upvotes]: Suppose I have a manifold and a vector bundle over it, but not a connection or a metric. Can I always find a connection on it that has a Riemann curvature tensor that is identically zero? If so, can I always find a connection that has both Riemann curvature and torsion tensors identically zero? -I've attempted to simply for the Christoffel symbols, but couldn't make headway in the equations. - -REPLY [13 votes]: By Chern-Weil theory, the real Pontryagin classes $p_k \in H^{4k}(X, \mathbb{R})$ of a real vector bundle $V$ on a smooth manifold $X$ are determined by the curvature form of any connection on that bundle; in particular, if the curvature vanishes, then so do all of the $p_k$. Hence if any of the $p_k$ don't vanish, then $V$ does not admit a flat connection. (Note that all of the $p_k$ vanish if $\dim X \le 3$; Milnor's result regarding the case $\dim X = 2$ requires more difficult tools.) -If $V$ is taken to be the tangent bundle of $X$, then the first case where this happens is when $\dim X = 4$, where $p_1 \in H^4(X, \mathbb{R})$. If $X$ is closed and orientable then $p_1$ is nonzero iff $X$ has nonzero signature, by the Hirzebruch signature theorem. The simplest example of a $4$-manifold with nonzero signature is $\mathbb{CP}^2$; it follows that the tangent bundle of $\mathbb{CP}^2$ does not admit a flat connection.<|endoftext|> -TITLE: Can both G and BG be finite CW complexes? -QUESTION [23 upvotes]: Here's a question out of idle curiosity. Let $G$ be a topological group. Is it possible for both $G$ and (a model of) $BG$ to be finite CW complexes? (Apart from the obvious example of $G$ being [up to homotopy] the trivial group.) -A comment is that the fibration sequence $G \to EG \to BG$ shows that $\chi(G)\chi(BG) = 1$ in this case, so $\chi(G) = \pm 1$. This rules out plenty of examples, e.g. Lie groups, finite groups... - -REPLY [13 votes]: As stated in previous answers, the answer to the question is no. Here is another viewpoint, which gives additional information about the relationship between $G$ and $BG$: -A pair $(G,BG)$ where $G$ is a topological group homotopy equivalent to a finite CW complex, and $BG$ its classifying space, is called a finite loop space in the literature. So the question is if a finite loop space $G$ can have classifying space $BG$ homotopy equivalent to a finite CW complex, and the answer to this is no. -As pointed out in the previous answers, to get that $BG$ does not have a finite dimensional model, it is easy to reduce to the case where $G$ is connected (using that $BG$ is not finite for $G$ a finite group), so let's assume this. -Now, by the structure of Hopf algebras (Milnor-Moore) $H^*(G;{\mathbb Q})$ is an exterior algebra on a number $r$ of odd dimensional generators. The number $r$ is called the (rational) rank, and agrees with the usual notion of rank of compact Lie groups. Hence by a spectral sequence argument - - -$H^*(BG;{\mathbb Q})$ is a polynomial algebra on $r$ generators. - - -So the non-finiteness of $BG$ is a corollary of the following well-known fact: - - -Fact: A non-contractible connected finite loop space (or even H-space) has positive rank $r$. - - -Proof of Fact: The claim e.g., follows from a more general statement saying the rank is also equal to the number of odd degree generators for the mod $p$ cohomology (see Kane: Homology of Hopf Spaces Section 13-3, which uses the Bockstein spectral sequence to deduce this). To get the more limited statement of the rank being positive one can give a more pedestrian argument: Suppose that $G$ is non-contractible. Then (since its a simple space and a CW complex) $\tilde H^*(G;{\mathbb Z}) \neq 0$. If this cohomology is torsion free, then the rational cohomology is non-trivial, and we are done. So suppose there is torsion. If there is non-trivial $p$-torsion in $\tilde H^*(G;{\mathbb Z})$ then, by the universal coefficient theorem, $H^*(G;{\mathbb F_p})$ has cohomology in two consecutive degrees. So $H^*(G;{\mathbb F_p})$ has to have generators as a ring in odd degrees. But then the structure of Hopf algebras algebras shows that the Euler characteristic of $G$ has to be zero (since $H^*(G;{\mathbb F_p})$ contains an exterior tensor summand on a odd degree class, if $p$ is odd, and a truncated polynomial algebra on an odd degree class, truncated at $x^{2^k}$ for some $k$, if $p=2$). But $\chi(G) =0$ implies $\tilde H^*(G;{\mathbb Q}) \neq 0$ as wanted. -Additional info: There has been a lot of work on finite loop spaces, starting with the work of Hopf and Serre in the 1940's and 1950's, since they are generalizations of compact Lie groups. (Hopf looked at the weaker notion of H-spaces.) There is by now in fact a classification of (connected) finite loop spaces; see Section 3 of my survey paper http://www.math.ku.dk/~jg/papers/icm.pdf<|endoftext|> -TITLE: Simultaneous zero set of two equations in $\mathbb R^3$ -QUESTION [15 upvotes]: Can we have positive reals $x,y,z$ with -$$ x^{\left( y^z \right)} = y^{\left( z^x \right)} = z^{\left( x^y \right)} $$ in cyclic permutation, other than the line $x=y=z$? -I put this at https://math.stackexchange.com/questions/493739/this-is-stupid-but-i-have-a-bad-cold-with-cough and one poor guy has been hacking away at it. I still have no idea. -I would love to see some representation of the supposed surface $$ x^{\left( y^z \right)} = y^{\left( z^x \right)}, $$ which I feel really ought to be some ugly variant of the helicoid around the line $x=y=z$. -EDIT, Thursday: Is it true that the gradient of $$ \color{magenta}{ h(x,y,z) = x^{\left( y^z \right)} - y^{\left( z^x \right)}}, $$ is defined and a nonzero vector along the line, say, at the point $x=y=z=t$ for positive real $t?$ If so, the surface is orthogonal to that at the line...Further, if you switch to one of the other pairs as a difference, I expect the gradient vector to be rotated by $120^\circ;$ this gives a pretty good reason for there being no other points on all three such surfaces near the known line. - -REPLY [8 votes]: Maple found closed form for $z$ and parametrization over $\mathbb{C}$ -for $x^{(y^z)}=y^{(z^x)}$: -$$ z= {{\rm e}^{ \left( -x{\it LambertW} \left( -\ln \left( y \right) { {\rm e}^{\ln \left( {\frac {\ln \left( x \right) }{\ln \left( y \right) }} \right) {x}^{-1}}}{x}^{-1} \right) +\ln \left( {\frac { \ln \left( x \right) }{\ln \left( y \right) }} \right) \right) {x}^ {-1}}} \qquad (1) $$ -EDIT -Substituting $z$ in the other equation and solving, we are looking -for positive real solutions of -x^exp(ln(y)*exp((-x*LambertW(-ln(y)/x*exp(ln(ln(x)/ln(y))/x))+ln(ln(x)/ln(y)))/x))-exp((-x*LambertW(-ln(y)/x*exp(ln(ln(x)/ln(y))/x))+ln(ln(x)/ln(y)))/x)^(x^y) - -Complex solutions exist, e.g. -x1= (4.039760831390928810096443230122786460497 - 3.794516696260006267408154939005426227609j) -y1= (2.764005926356850384451036006794311106205 + 0.0j) -z1= (1.273990994985239416952018723391988170796 + 0.2830754921068781180566923204291057573632j) - -Things get complicated by the fact that with the principal branches -of $W$ and $\log$ $(1)$ need not be real and might have imaginary part. -Here is a plot of $(1)$ taking the principal branches and the real -part. The plot is wrong because it has some artifacts caused by -imaginary part. -Will try to make correct plot if I have the time.<|endoftext|> -TITLE: Separating pairs of points in R^n -QUESTION [6 upvotes]: Let $A$ be a set of $2k$ points in $\mathbb{R}^n$ such that no open set in $\mathbb{R}^n$ of diameter $2$ contains more than $k$ of these points. What is the largest possible distance $r_n>0$ one can guarantee so that for all $k$ we can pair up $2k$ points under the mentioned condition so that each pair of points is at least $r_n$ apart? It is not difficult to show that $1\leq r_n \leq 2$ ($2$ comes from just taking $|A|=2$ and proving that one can always have them at distance $1$, one can build a distance graph so that two point from $A$ are joined if they are at least $1$ apart - such graph has minimum degree at least $k$ and so has a Hamilton cycle). - -REPLY [6 votes]: Consider a regular $2k-1$-gon of diameter $2$, meaning the distance between the two most distant vertices is $2$. Then only $k-1$ vertices can be in any set of diameter less than two, becsuse any set of $k$ vertices contains a maximum-distance pair. So the set of vertices plus the center form a set satisfying your condition. As $k$ goes to $\infty$, the polygon approaches a circle of radius $1$, so $r_n=1$ for $n\geq 2$.<|endoftext|> -TITLE: Gamma Factors for Zeta Functions of Abelian Varieties -QUESTION [8 upvotes]: If $X$ is a scheme of finite type over $\mathbb{Z}$, and $X_0$ denotes its set of closed points, then one can define its zeta function on the half plane $Re(s)>\text{dim}(X)$: -\begin{equation} -\zeta_X(s)=\prod_{x\in X_0}\frac{1}{1-|k(x)|^{-s}}, -\end{equation} -where $k(x)$ denotes the (finite) residue field at $x$. -The zeta function has a completion, which includes the conductor and a gamma factor. I want to ask about the latter, in a special case, namely when $X$ is a (proper, regular) model of an abelian variety over a number field. -For example, if $X$ is a model of an elliptic curve $E$ over a number field $k$, then the zeta function of $X$ can be written: -\begin{equation} -\zeta_X(s)=n_{X}(s)\frac{\zeta_k(s)\zeta_k(s-1)}{L_E(s)}, -\end{equation} -where $n_X(s)$ depends only on the bad primes. Indeed, an identical expression holds for a model of any smooth projective curve over a number field. The quotient on the right hand side of the above expression is what some call the Hasse--Weil $\zeta$-function of $E$. This notion can be extended to any smooth projective variety $V$ of dimension $n$: -\begin{equation} -\frac{\prod_{i=0}^nL(H^{2i}_{et}(V),s)}{\prod_{i=0}^{n-1}L(H^{2i+1}_{et}(V),s)}. -\end{equation} -The gamma factor for the zeta function of a model of $V$ is the alternating product of the gamma factors for each Hasse--Weil L-factor (coming from the Hodge decomposition of each cohomology group). In the case of elliptic curves, direct computation shows that the gamma factor for a model is in fact a rational function. I have been lead to believe that this is true for any abelian variety, but I cannot prove it. -Firstly, is it true that the gamma factor of a model $\mathcal{A}\rightarrow\text{Spec}(\mathcal{O}_k)$ of an abelian variety $A$ over a number field $k$ is rational? Secondly, is there a profound reason that this should be true? - -REPLY [8 votes]: This amounts to an Euler characteristic argument. A ratio of products of Gamma functions such as -$$ \frac{\prod_{i=1}^r \Gamma(s - a_i)}{\prod_{j=1}^t \Gamma(s - b_j)} $$ -is rational if and only if $r = t$ (i.e. there are as many functions in the numerator as in the denominator). But the number of $\Gamma$ factors in the numerator of the zeta function is the sum of the dimensions of the even-degree cohomology groups, and the number of factors in the denominator is the sum of the odd-degree ones. So the ratio is a rational function of $s$ if and only if the Euler characteristic is 0. -However, all Abelian varieties have Euler characteristic 0 (since over $\mathbf{C}$ they are products of circles) and that does it.<|endoftext|> -TITLE: On the full rings of matrices -QUESTION [7 upvotes]: Let $R$ be a ring with elements $a,b$ such that $a^2 = 0 = b^2$ and $a+b$ is a unit. How to show that $R$ is a ring of $2\times2$ matrices? -COMMENT: The converse is clearly true, just take $a = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ and $b = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$. So it would be interesting if some one could generalize this to dimension $n$. - -REPLY [8 votes]: Let $c$ denote the inverse of $a+b$, i.e., $ac+bc=ca+cb=1$. Multiplying this by $a$ and $b$ from the left and from the right and using $a^2=b^2=0$, we obtain $abc=a$, $cba=a$, and $cab=b$. -Let us show that $R=aR\oplus bR$. It follows from $ac+bc=1$ that $R=aR+bR$. If $ax=by$, then $bax=0$, implying $0=cbax=ax$ in view of $cba=a$. -The $R$-modules $aR$ and $bR$ are isomorphic. Indeed, the rule $bcx\mapsto ax$, $x\in R$, defines an $R$-module homomorphism $bR=bcR\to aR$ because $bcx=0$ implies $ax=0$ due to $abc=a$. Its kernel vanishes: $ax=0$ implies $abcx=0$ because $abc=a$ and, hence, $0=cabcx=bcx$. -It remains to apply Propositions 5 and 6 on page 52 of "Structure of Rings" by N.Jacobson (1964) and to conclude that $R$ is a ring of $2\times2$ matrices. -Edit. By suggestion of Martin Brandenburg, I reproduce the Propositions quite literally: -Proposition 5 $\dots$ Conversely, if $\frak A$ is a ring with an identity $1$ and ${\frak A}={\frak I}_1\oplus\dots\oplus{\frak I}_n$ is a direct decomposition of $\frak A$ into right ideals which are isomorphic $\frak A$-modules, then there exists a set of matrix units $\{e_{ij}\mid i,j=1,\dots,n\}$ such that ${\frak I}_j=e_{jj}{\frak A}$. -Proposition 6. Let $\{e_{ij}\mid i,j=1,\dots,n\}$ be a set of matrix units in a ring $\frak A$ with identity $1$, and $\frak B$ the subring consisting of the elements which commute with the $e_{ij}$, $i,j=1,\dots,n$. Then every element of $\frak A$ can be written in one and only one way as -$\sum b_{ij}e_{ij}$ where $b_{ij}\in{\frak B}$ for all $i$ and $j$. Hence -${\frak A}\cong{\frak B}_n$. The ring $\frak B$ is isomorphic to -$e_{11}{\frak A}e_{11}$. -Second thought Edit. Those Propositions are an old stuff. It is probably better to prove them here in two lines. -For any $R$-module $M$, we have -$\text{End}_R(\underbrace{M\oplus\dots\oplus M}_n)\cong\text{Matr}_n(\text{End}_RM)$. -On the other hand, $\text{End}_RR\cong R$. -Question. The problem looks a bit artificial. Is it a known exercise?<|endoftext|> -TITLE: Complex structure of the Teichmüller space in terms of Fenchel-Nielsen coordinates -QUESTION [11 upvotes]: The Teichmüller space $T_g$ of genus $g$ Riemann surfaces can be parameterized in terms of Fenchel-Nielsen coordinates, taking values in $\mathbb{R}^{3g-3}\times \mathbb{R}_+^{3g-3}$. -The Teichmüller space $T_g$ also has a natural complex structure. -Could someone suggest a reference where the complex structure is given in terms of Fenchel-Nielsen coordinates? -A more rough question is: very naively I would combine $\mathbb{R}^{3g-3}\times \mathbb{R}_+^{3g-3}$ into $\mathbb{H}^{3g-3}$ where $\mathbb{H}$ is the upper half plane. In the natural complex structure, is $T_g$ the same as $\mathbb{H}^{3g-3}$ as complex manifolds? I guess it's not, but is it proved somewhere? - -REPLY [10 votes]: Although Fenchel-Nielsen coordinates do not give a complex-analytic parameterization of Teichmuller space, they do extend to give a complex-analytic parameterization of $T_g\times \overline{T}_g$ on a subset. This follows from Bers' simultaneous uniformization theorem, which parameterizes quasi-fuchsian groups by the Teichmuller spaces of the two components of the domain of discontinuity. In turn, (marked) quasi-fuchsian groups are parameterized by the character variety, which in turn is determined by a subset of complex Fenchel-Nielsen coordinates $\subset \mathbb{C}^{3g-3}\times \mathbb{C}^{* 3g-3}$. Bers then obtained a complex structure on Teichmuller space by fixing a conformal structure $x$ on one domain of discontinuity, and letting the other one vary, to get $T_g\times \{x\}\subset \mathbb{C}^{6g-6}$. He shows that as one varies $x$, the complex structure on $T_g$ does not change, and thus one obtains a well-defined complex structure on Teichmuller space, which is equivalent to the complex structure originally defined by Ahlfors.<|endoftext|> -TITLE: Why A. Weil considered elimination theory to be eliminated? -QUESTION [10 upvotes]: It is well known that André Weil declared, in the 1940's, that elimination theory must be eliminated from algebraic geometry. I would like to understand his mathematical reasons to adopt such an attitude against elimination theory. (N.B. Later Abhyankar answered back that eliminators of elimination theory have to be eliminated !. ). - -I was expecting a deep reason why André Weil , in his "Foundations of Algebraic Geometry (1946)", expressed his wish to see elimination theory completely eliminated from algebraic geometry (page 31). He writes himself that his book is first of all intended to give a "detailed and connected treatment of intersection-multiplicities" (For this, it is of course necessary to lay down "foundations of algebraic geometry"). Now, at these times (let us say between the two World Wars),the use of elimination theory in intersection theory led to contradictions .The main reason was lack of well established foundations (cf. the Introduction in Weil's book). André Weil did the job for intersection-multiplicities (and put the first foundations for our present algebraic geometry (opening the way to Grothendieck)).As for elimination theory, he quite simply hoped to see it eliminated ! (That avoids contradictions !) It is Jean-Pierre Jouanolou who did the work for elimination theory(during the 1980's). Thanks to him, elimination theory is now a full part of algebraic geometry, à la Grothendieck. - -REPLY [16 votes]: I think the usual interpretation is this (see S. Landsburg's comment): -The classical proof that $\mathbb{P}^n$ is proper uses elimination theory: we need to prove that for a ring $R$, $\mathbb{P}^n_R\to Spec(R)$ is closed. Say $R = k[t_1, \ldots, t_k]$, then a closed subset of $\mathbb{P}^n_R$ is given by equations in $t_1, \ldots, t_k, x_0, \ldots, x_{n}$, homogeneous in the $x_i$, and finding the image of this subset means eliminating the $x_i$. -Using Chevalley's valuation-theoretic approach ("valuative criterion of properness"), we can replace this rather elaborate argument by a very short and conceptual one. Hence we eliminated elimination theory.<|endoftext|> -TITLE: Hilbert's 10th problem and nilpotent groups -QUESTION [14 upvotes]: I am asking this question on behalf of a colleague of mine who does not have an MO account. Nevertheless I am also interested in the answer. -The question concerns relationships between Hilbert's 10th Problem (over $\mathbb{Z}$ and over $\mathbb{Q}$) and the Equations Problem (EP) in certain groups. The EP in a group $G$ is (apparently; I am not an expert here) to algorithmically decide whether an equation with parameters in a group has a solution. In other words, let $G$ be a countably infinite group and let $\{g_n\}_{n=1}^{\infty}$ be an enumeration of the elements of $G$. An equation in $G$ is a word in the $g_i$'s and in some formal variables $x_1,\ldots,x_N$ and their inverses, which is set equal to $1$. The EP is whether there is an algorithm which upon being given an equation, decides whether the equation has a solution in $G$, i.e., whether we can evaluate the indetermines to elements of $G$ so as to get a true identity. (Maybe this depends on the enumeration of the elements of $G$. If $G$ has a solvable word problem -- which I think is true for the groups which are coming -- then it shouldn't matter.) -Added: It seems I didn't get the formalism right. Let's restrict to finitely presented groups with solvable word problem, and enumerate the elements as distinct words in the generators with respect to some reasonable lexicographic ordering. Or, if you like, part of the question is to ask exactly what EP means in this context: certainly it means something, as it has been studied by many people. -More formally, in the language of groups augmented by a constant for each element $g_i$ of $G$, the EP is asking whether the positive existential theory of $G$ is decidable. -The questions concern a much-cited 1979 paper of Romankov: - -Romanʹkov, V. A. -Universal theory of nilpotent groups. (Russian) -Mat. Zametki 25 (1979), no. 4, 487–495, 635. - -A feature of the situation is that I haven't been able to get my hands on the entire paper, so but here is the MathReview by O.V. Belegradek: - -A finitely generated nilpotent group has a decidable theory if and only if it is abelian-by-finite [Ju. L. Eršov, Dokl. Akad. Nauk SSSR 203 (1972), 1240–1243; MR0297840 (45 #6892)]. The author gives an example of a finitely generated 4-step nilpotent group with undecidable universal theory. The proof depends on Matijasevič's undecidability result for the universal theory of the ring of integers. A. I. Malʹcev [Mat. Sb. (N.S.) 50 (92) (1960), 257–266; MR0118677 (22 #9448)] showed the undecidability of the theory of a free 2-step nilpotent group. The author proves that the decidability of the universal part of this theory is equivalent to the decidability of the universal theory of the field of rationals. {Reviewer's remark: A. M. Slobodskoĭ and È. I. Fridman [5th All-Union Conference on Mathematical Logic, p. 140, Akad. Nauk SSSR Sibirsk. Otdel., Inst. Mat., Novosibirsk, 1979] have announced the decidability of the universal theory of the field of rationals.} - -First Question: There are many free $2$-step nilpotent groups: one such group, call it $N(2,m)$ for each rank $m \geq 1$. Thus e.g. $N(2,2)$ is the standard Heisenberg group over $\mathbb{Z}$. Which group(s) does Romankov's decidability result refer to? -My guess on this: it should refer to $N(2,\infty)$, the free $2$-step nilpotent group of countably infinite rank. I also guess that it shouldn't matter much, in that the universal theories of $N(2,n)$ should be the same for all $2 \leq n \leq \infty$: I saw very similar results in the literature with solvable groups, and it seems very plausible. -Transition to the Second Question: There are many places in the literature where Romankov's result is characterized as: "Solving the equations problem in a free $2$-step nilpotent group is equivalent to Hilbert's 10th Problem over $\mathbb{Q}$." Here are some instances: -Page 1 of this arxiv preprint, which has since been published. -This 1997 paper -In the MathScinet reivew of a 1995 paper: MR1351615. The reviewer is Romankov. -But now I'm a little confused. Hilbert's 10th problem (over any ring) concerns the positive existential theory of that ring: it's about whether solutions exist to polynomial equations. Similarly for the EP. If we could omit the word "positive" then, sure, the full existential theory of any structure is decidable iff the full universal theory is, since -$\exists x P(x)$ is true exactly when $\forall x \ \neg P(x)$ is false. -So let's define E/IP to be the group theory problem with equations or inequations. So it seems to me that Romankov's result is rather that decidability of E/IP for $N(2,\infty)$ (and so perhaps also for $N(2,2)$ is my guess at Q1 is correct) is equivalent to the decidability of all polynomial equations and inequations over $\mathbb{Q}$. It seems to me though that "H10 over $\mathbb{Q}$" concerns equations only, so I wonder whether these inequations are actually necessary. -This question is equivalent to the definability of the set $R^{\bullet} = R \setminus \{0\}$ by a positive existential formula. Over $\mathbb{Z}$ this is well known to be the case: via Lagrange's Theorem that set is defined by -$\{y \mid \exists x_1,x_2,x_3,x_4 \ x_1^2 + x_2^2 + x_3^2 + x_4^2 +1= y^2\}$ -I found in a survey article of T. Pheidas that for $R = \mathbb{C}[[t]]$, $R^{\bullet}$ is not positively existentially definable. But what about over $\mathbb{Q}$? -Second Question: Is it really true that EP for $H(2,2)$ is equivalent to H10/$\mathbb{Q}$? Or just that E/IP is equivalent to the undecidability of polynomial equations and inequations over $\mathbb{Q}$? - -REPLY [6 votes]: For question 2: Over any field, every existential formula is equivalent to a positive existential one. Indeed, every negative basic formula $$\exists x_1,\dots,x_n[P(x_1,\dots,x_n) \neq 0]$$ is equivalent to the positive basic formula $$\exists x_0,x_1,\dots,x_n [x_0 \cdot P(x_1,\dots,x_n) - 1 = 0].$$ As you point out, the decidability of the existential and universal theories are equivalent problems, and with the above trick this is equivalent to the decidability of the positive existential theory ("H10") over a field.<|endoftext|> -TITLE: Isomorphism in category of finite automata -QUESTION [7 upvotes]: What does meanthat two finite automata is equivalent? I think that we must define category of finite automata, i.e. we must define $\mathrm{Hom}(A,B)$, where $A,B$ be an arbitrary finite automata. Hence two finite automata $A,B$ are equivalent if there exist isomorphism $f\in\mathrm{Hom}(A,B)$. So, how we can define category of finite automata? - -REPLY [2 votes]: The category of automata is discussed in Example 3.3. (3) in: -Adamek, Herrlich, Strecker - Abstract and concrete categories, the Joy of Cats, online. -It also appears in 4K, 5.2, 7.15, 13.13, 15.3, 20H in that book.<|endoftext|> -TITLE: CW complex and group action -QUESTION [6 upvotes]: This is a general question and any reference or related result will be extremely helpful. -Suppose $X$ is a Hausdorff topological space. Suppose G (a countable group) acts on it. Let $Y=X/G$ be the quotient space which is a CW complex. Can we say something about the cell structure of $X$ (if exists.)? If not in general is there some results which gives condition on the action (for example proper or cocompact) which gives us cell structures in $X$. In particular I am interested in the following question. -Under what condition on $X$ and on the group action $X$ will be a CW complex? - -REPLY [2 votes]: you may wish to check the following paper: -Illman, Sören -The equivariant triangulation theorem for actions of compact Lie groups. -Math. Ann. 262 (1983), no. 4, 487–501. -Let G be a compact Lie group. -Theorem 7.1: Let M be a smooth $G$-manifold with or without boundary. Then there exists an equivariant triangulation of $M$. -Corollary 7.2: Let M be a smooth $G$-manifold with or without boundary. Then $M$ can be given an equivariant CW complex structure.<|endoftext|> -TITLE: About elegant Reedy categories -QUESTION [5 upvotes]: I discovered today the notion of elegant Reedy category introduced in the paper Reedy categories and the $\Theta$-construction of Julia E. Bergner and Charles Rezk. An interesting property of such categories is the following (quoting nlab elegant Reedy category): -if $R$ is an elegant Reedy category and $M$ is a model category in which the cofibrations are exactly the monomorphisms, then the Reedy model structure and the injective model structure on $M^{R^{op}}$ coincide. -This is stated in Proposition 3.15 of the paper of Bergner and Rezk for $M$ being a simplicial set-valued presheaf. -I was wondering if it would be possible to obtain a kind of dual statement of this, something like -"if $R$ is a "coelegant" Reedy category and $M$ is a model category in which the fibrations are exactly the epimorphisms, then the Reedy model structure and the projective model structure on $M^{R^{op}}$ coincide". -At first sight it seems possible to do so, but I may have missed something. - -REPLY [5 votes]: Actually, I think the statement that you quoted from the nLab is wrong. It was copied from v1 of the Bergner-Rezk paper, but v2 corrected the statement to be only about simplicial presheaves on $R$ (although I think the proof would probably work for presheaves on $R$ with values in any other presheaf category). I've just fixed the nLab as well.<|endoftext|> -TITLE: Does positively curved sphere admit an isometric embedding as hypersurface in Euclidean space? -QUESTION [11 upvotes]: Let $(S^n, g)$ be an $n$-dimensional positively curved sphere. Assume the smoothness of the metric, does it admits an isometric embedding into $\mathbb R^{n+1}$? -for $n=2$ it is proved by A.D Alexandrov, also by H. Weyl - -REPLY [12 votes]: The answer of j.c. given prior to mine is of course correct but let me give a trivial reason why (in big dimensions) every Riemannian metric after an arbitrary small perturbation is not isometrically embeddable in $R^{n+1}$, even locally. -It is known that the curvature tensor of a metric induced from an imbedding in $R^{n+1}$ has the following form $$R_{ijkl} =h_{il} h_{jk} - h_{ik} h_{jl}, \ \ \ \ (*) $$ where $h$ is the second fundamental form. The formula $(*)$ is a very strong restriction on the curvature tensor, in big dimensions most $(0,4)$- tensors having the algebraic symmetries of the curvature tensor can not be represented in the form $(*)$. Now, by a arbitrary small perturbation of an arbitrary metric one can achive that - its curvature does not satisfies $(*)$; the existence of such a perturbation follows from the fact that at a point for any (0,4)-tensor $T$ satisfying the algebraic symmetries of the curvature tensor there exists a metric whose curvature tensor at this point coincides with $T$. Thus, one can slightly perturb an arbitrary metric such that the result is not imbeddable. -Since a small perturbation of a metric of positive sectional curvature again has positive sectional curvature, the answer on your question is negative. - -REPLY [11 votes]: When $n\geq 3$, the condition for a metric to be even locally isometrically embeddable in $R^{n+1}$ is nontrivial, so you may wish to add such an additional condition. Then this paper "A Priori Bounds for Co-dimension One Isometric Embeddings" by Yanyan Li and Gilbert Weinstein states as a conjecture: - -Conjecture. Let $g$ be a smooth metric of non-negative sectional curvature and positive scalar curvature on $S^n$ which is locally isometrically embeddable in - $R^{n+1}$. Then $(S^n,g)$ admits a smooth global isometric embedding $X:(S^n,g)\rightarrow R^{n+1}$. - -Section 5 of the paper discusses the positive sectional curvature case, and in particular Theorem 7 gives a result in the case $n=3$ for strictly positive curvature.<|endoftext|> -TITLE: About an embedding of abelian categories into categories of modules -QUESTION [7 upvotes]: Let $k$ be a field. Let $C$ be an abelian $k$-linear category with a symmetric tensor product $\otimes$ and internal homomorphisms, such that $\mathrm{End}(1)=k$. Let $M$ be another $k$-linear abelian category, and let -$$h:M\to C$$ -be an exact, faithful, $k$-linear functor. If $C$ has arbitrary limits and $M$ is small, we can understand the endomorphism ring of $h$ as a ring object (a monoid) in $C$, as follows. It is the equaliser -$$R := \ker \left( \prod_{m\in M}\mathrm{End}(h(m)) \xrightarrow{\quad u \quad} \prod_{m\to n}\mathrm{Hom}(h(m),h(n))\right)$$ -where $u$ is "precomposition minus postcomposition". The given functor $h$ factors then canonically over the category of $R$-modules in $C$ -$$M \xrightarrow{\quad\widetilde h \quad} R\mathrm{-Mod}_C \xrightarrow{\quad f \quad} C$$ -where $f$ is the forgetful functor. My question is: - -Is the functor $\widetilde h$ fully faithful? - -For example, if $C$ is the category of vector spaces over $k$, then the answer is Yes. For this, $k$ does not even have to be a field. On the other hand, $C$ could be much larger, for example, the category of sheaves of $k$-vector spaces on a connected topological space. This is the case i am ultimately interested in. -Remark: The existence of limits in $C$ was just for convenience, without them, one can still view $R$ as a pro-ring object and consider $R$-modules. The hypothesis $\mathrm{End}(1)=k$ is essential if one wishes for a positive answer, because the category of modules $R\mathrm{-Mod}_C$ is always $\mathrm{End}(1)$-linear. -Remark: Also, the whole story smells like it was some variant of the Freyd-Mitchell theorem, but i don't see a concrete way to link it to that. -Added: The proof that in the case $C = \mathrm{Vect}(k)$ the answer to my question is Yes goes roughly as follows: In this special case, we can view $M$, or at least the $\mathrm{Ind}M$, as a $C$-module, that is, for every vector space $V$ and object $m \in M$ we can give a sense to $\mathrm{Hom}(V,m)$ and $V\otimes m$ as objects in $M$. At this point, $\widetilde h$ is then even an equivalence. Define an object -$$X := \ker \left( \prod_{m\in M}\mathrm{Hom}(h(m),m) \xrightarrow{\quad u \quad} \prod_{m\to n}\mathrm{Hom}(h(m), n)\right)$$ -in $M$. Then $h(X)$ is $R$ as a right $R$-module, and given a left $R$-module $V$, we can define an object $Y$ of $M$ by -$$Y = X \otimes_R V = \mathrm{coker}(X \otimes R \otimes V \xrightarrow{\quad v \quad} X \otimes V)$$ -with $v$ = "left action on $V$ minus right action $X$". But then, $\widetilde h(Y)$ is isomorphic to the given $R$-module $V$, and one can treat morphisms in the same way: $V\to V'$ yields $X\otimes_RV \to X\otimes_RV'$. Note that if we only consider finite dimensional spaces, then the passage to ind-objects is superfluous, and $\widetilde h$ is an equivalence of categories. -The proof generalises for as far as $M$ is a $C$-module. But what is there to do if, for example, $M$ is "graded local systems" on a topological space and $C$="sheaves" and $h$=forget? - -REPLY [5 votes]: No. What follows appears to be a counterexample for $C = \text{Vect}$ (I don't understand where in your argument you prove fullness). -Let $M = \text{Vect}^{op}, C = \text{Vect}$, and let $h : \text{Vect}^{op} \to \text{Vect}$ be the contravariant functor $V \mapsto V^{\ast} \cong \text{Hom}(V, k)$. If smallness is important to you pretend that the first $\text{Vect}$ means vector spaces of at most countable dimension. The endomorphism ring of $h$ is $k$, so the lift $\tilde{h}$ is just $h$ again. -I claim that $h$ is not full. To see this, if $V$ is a countable-dimensional vector space, regarded as an object in $\text{Vect}^{op}$, then the induced map -$$\text{End}(V, V) \to \text{End}(V^{\ast}, V^{\ast})$$ -has the property that its source has dimension $\aleph_0^{\aleph_0} = 2^{\aleph_0}$, but its target has dimension at least $\left( 2^{\aleph_0} \right)^{2^{\aleph_0}}$. -(Taking the opposites of familiar abelian categories seems to be my new favorite trick! Note that vector space duality establishes that $\text{FinVect}$ is equivalent to its opposite. $\text{Vect}$ itself is the ind-completion of $\text{FinVect}$, so $\text{Vect}^{op}$ is the pro-completion; in other words, it's the category of profinite vector spaces. This category is also known as the category of linearly compact vector spaces: see this MO question and this n-cafe post for details. This gives some intuition for why $h$ is not full: it's the forgetful functor and it doesn't see the topology. -In the special case that $k = \mathbb{F}_p$ we can be a little more explicit: $\text{Vect}^{op}$ in this case is the full subcategory of profinite groups consisting of the ones whose finite quotients are all elementary abelian $p$-groups.)<|endoftext|> -TITLE: The symplectic geometry of cold coconuts -QUESTION [19 upvotes]: Consider the open set $M \subset \mathbb{C}^{2}$ given by the union of the unit ball $|z_1|^2 + |z_{2}|^2 < 1$ (the coconut) and the cylinder $|z_1| < \epsilon$, $0 < \epsilon < \! \!< 1$, (the straw, which in this case pierces the coconut through and through, but this is not important). -Fix a number $r$ strictly between $\epsilon$ and $1$. -Does there exist a strictly positive number $c$ so that the volume of the intersection of the unit ball with any symplectic image of the ball of radius $r$ lying wholly inside $M$ is greater than $c$? If so, is there a reasonable estimate for $c$ as a function of $r$ and $\epsilon$? -By Gromov's non-squezing theorem we know we cannot symplecticaly move the whole ball of radius $r$ up the straw, but it is not clear to me how much of its volume can we sip out of the coconut. -Motivation. -This problem is related to the comments I got on this question. Both questions arose from trying to understand a comment that V.I. Arnold made in one of his papers (I forgot where, but for some reason the statement came back to my mind after many years) saying or implying (I read this a long time ago ...) that despite the non-squeezing theorem and the symplectic camel theorem people in statistical mechanics still happily interchange regions of phase space that have the same volume. Brett's answer seems to imply that Arnold's criticism should not be taken too seriously and that the physicists are right: even if you cannot exchange the regions by canonical transformations, you can exchange all of their volume up to an arbitrarily small amount. -On the other hand, it is still possible that there exists a symplectic refinement of Poincaré recurrence. - -REPLY [7 votes]: Anton's answer is roughly what I meant by my comment above. In what follows, I'll just try to explain the intuition behind why thin straws can pretty much be as wiggly as you like. (You should be able to extend this argument to a full proof.) -It is easiest to visualize bending a symplectic cylinder when you consider it as a product of a small disk with an infinite strip, instead of as a product of a small disk with an unbounded plane. It is also a bit easier to imagine taking up all the volume with a straw with a square cross section, so I'll start with the following model for a thin straw. -$$ -2\epsilon< x_1<0, \ \ \ - \epsilon \epsilon^2$, equal to 0 when $x_1<0$ and which has small derivative close to 0. -In the region where $y_2<0$ or $x_1>\epsilon^2$, the effect of the flow of $H$ is to rotate $y_2$ into $y_1$, and $x_2$ into $x_1$. Running this flow will fold the straw back on itself until it looks roughly like a straw that has been bent almost 180 degrees. Using a further Hamiltonian flow, you can adjust this bent straw away from $0$ to look like the union of the upper half of the original straw with the upper half of the straw -$$ \epsilon^2< x_1<2\epsilon+\epsilon^2, \ \ \ - \epsilon -TITLE: Upper bound for class number of a real quadratic field -QUESTION [11 upvotes]: Unless I am mistaken, we know that an upper bound for the class number $h(D)$ of a real quadratic field $\mathbb{Q}(\sqrt{D})$ is $O(D^{1/2})$. Is the exponent of $1/2$ known to be the best possible? -Also, is there any better exponent known for the upper bound of -$$\liminf_{D \rightarrow \infty} h(D) \, ?$$ -Of course, the conjecture is that the exponent shoud be zero, but do we know any better than $1/2$? - -REPLY [18 votes]: The exponent $1/2$ is best possible. You can see this by varying $D$ along values of the form $n^2+4$ so that the regulator is only of size about $\log D$. Then the lower bounds for $L(1,\chi)$ (Siegel's theorem) and the class number formula give such a lower bound. This was worked out more precisely by Montgomery and Weinberger; see this paper of Duke which works out analogs for other number fields and gives the history: http://www.math.ucla.edu/~wdduke/preprints/number.pdf -Regarding the other question, I don't think we know much more than the regulator being larger than a power of $(\log D)$ infinitely often. So there is a big gap here. See the paper of Jacobson, Lukes and Williams on computational results towards this problem and which also discusses the known theoretical results: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.37.7045&rep=rep1&type=pdf<|endoftext|> -TITLE: Open coloring axiom vs. CH -QUESTION [15 upvotes]: Is there a simple, direct proof that the open coloring axiom contradicts CH (straight from the definitions, no machinery allowed)? The separable metric spaces version of OCA, if that helps. -Edit: probably it wasn't clear what I was asking. Let's say, hypothetically, that I'm writing a book on forcing that is meant to be as elementary and readable as possible. I've just introduced OCA and I have half a page to show the reader that it contradicts CH. Can I do this without quoting external theorems? - -REPLY [5 votes]: Another option is to use the same approach as Kunen in his book "Set Theory" (2011). In section V.6 he work with $SOCA$ (a weakening of $OCA_{T}$ that only ask for the existance of one of the homogeneous subsets, not for a covering). -In approximately two pages (364-366) he presents $SOCA$ and shows that the axiom implies $\neg CH$. The way he do it is with weakly Luzin sets. -His proof work as follows: -1) Define a set in ${R}^{n}$ ($R$ is the reals) as skinny iff the set of directions is not dense in the unit sphere. -2) Defines a weakly Luzin as a set whose all uncountable subsets are not skinny. -3) Define $A$ as an $\epsilon$-directed set iff there is a point in the unit sphere such that all the directions of $A$ have an angle at most of $\epsilon$ to that point. -4) Notice that every $\epsilon$-directed set is skinny. -5) Using induction and compactness show that $SOCA$ implies that every set is contains an $\epsilon$-directed set. -I was a litle vague because I left the reference. I can try to be more precise with definitions and proofs if needed.<|endoftext|> -TITLE: Proof-Theoretic Ordinal of ZFC or Consistent ZFC Extensions? -QUESTION [21 upvotes]: Let the proof theoretic ordinal $\alpha$ of a theory $T$ be the least recursive ordinal such that $T$ does not prove that $\alpha$ is well-founded. This ordinal is intended to quantify in some sense the complexity or power of a theory. -Does anyone know what is the proof theoretic ordinal of $ZFC$ or any non-trivial $ZFC$ extensions? Wikipedia says this is unknown for $ZFC$ as of 2008, but maybe there has been some recent progress? Thank you. - -REPLY [13 votes]: As Andres and I have stated in the comments, we are still nowhere near a proof-theoretic analysis of $ZFC$ or similar theories; even full second-order arithmetic remains well out of reach. -The paper "The Art of Ordinal Analysis" by Michael Rathjen does a good job of both describing ordinal analyses which have succeeded (e.g., $PA$), and showing how new difficulties arise as we climb towards higher and higher comprehension axioms (still around the level of $\Pi^1_2$-$CA_0$ by the end of the paper). I think this might be worth reading, if you want an explanation of why finding proof-theoretic ordinals of strong theories is "hard."<|endoftext|> -TITLE: What are the Possible Large Cardinals of $L[X]$? -QUESTION [5 upvotes]: I've been doing some basic reading in inner model theory, and I'm at the point where I've seen the definition of things like Martin-Steel and Mitchell-Steel inner models. I am wondering about the motivation for these constructions since they are much more complicated than the constructible universe $L$, or models like $L[X]$ built by adding a predicate to the construction of $L$. I am wondering, what makes a naive approach like taking $L[X]$ for the right sets fail for larger cardinals? Are there some results that pinpoint at what large cardinal strength more complicated constructions than $L[X]$ are needed? Thank you. - -REPLY [9 votes]: If $X$ is a set, then $L[X]$ does not have strong cardinals. In fact, $X^\sharp$ does not belong to $L[X]$, so any assumption that implies that $V$ is closed under sharps fails in these models. -Now, this is perhaps not the right example, as strong cardinals or supercompact cardinals essentially require a proper class of "witnesses". If you just want "local" versions, for example, having your model satisfy that there is an inaccessible $\kappa$ with $V_\kappa$ a model of "there is a proper class of strong cardinals" or more (say, having Woodin cardinals in your model), then models $L[X]$ are fine. In fact, contrary to your claim, the Mitchell-Steel models for Woodin cardinals are models of the form $L[X]$ for an appropriate set $X$. -So perhaps we need to refine your question to something much more specific, such as why we need complicated sequences of extenders (or measures) rather than just having $X$ be, say, an ultrafilter. This is easy to see when we try to model a degree of supercompactness, because nontrivial ultrafilters on $\mathcal P_\kappa(\lambda)$ concentrate on non-trivial sets (rather than sets of ordinals), so adding such an ultrafilter as a predicate $X$ ends up not doing much, since $L[X]$ cannot see any interesting measure $1$ sets, so $L[X]$ reduces to $L$ in this case. This has been known forever, it was noticed quickly after the $L[\mu]$ models for measurability began to be investigated. -Again, the above may not be the best example, since there are additional obstacles to developing fine-structure theory even at the level of $\kappa^+$-strong compactness, so we do not have "Mitchell-Steel" models for this assumption anyway. -Now, the complicated $L[\mathcal E]$ models are complicated because they are trying to do more than just modeling the large cardinal assumptions we are after. For example, to model that $\delta$ is Woodin, we need many "local" strong cardinals below $\delta$, so we need to add many ultrafilters to the model; this already indicates our constructions will have $\mathcal E$ be not a single measure, but rather a sequence of measures. -But the $L[\mathcal E]$ models do more: We want to add these ultrafilters in increasing order of "strength", which is measured by several parameters, such as their Mitchell order. This requires that we be able to organize the ultrafilters in some ordered fashion, avoiding clashes, so that if at some stage $\alpha$ we are to add a measure, there is exactly one candidate measure to add. This is useful to establish basic results such as condensation, the higher order versions of the result that elementary substructures of initial segments of $L$ are again initial segments of $L$. -We also want our models to be as canonical as possible, so in particular we must avoid coding sets by accident, which could happen if we are not careful on what measures we put in $\mathcal E$. There are additional requirements, all meant to ensure that the models we obtain can be compared with one another, carry no unwanted information, and add strength "from the ground up". To formalize this carefully ends up requiring some non-trivial amount of fine structure, and the verification that these fine structural assumptions indeed allow these constructions to succeed ends up being a careful inductive argument. This is the real source of the difficulties and technicalities associated with the theory of these models. -I suggest that you read the initial sections of the paper by Steel for the Handbook of set theory. They do a good job of spelling out some of these concerns and of describing how precisely the predicate $X$ that are used end up coding the ultrafilters witnessing the large cardinal assumptions one is after. -It may actually be useful to read first the appropriate sections in Zeman's book. The points above are illustrated there carefully, and one avoids the additional mathematical complications that come from iteration trees, which are inherent to the theory of Woodin cardinals. Truth is, when one first encounters iteration trees, they may appear rather complicated. It is natural that something complicated is needed, since we want to produce models that can code complicated sets of reals, and the complexity of the possible sets of reals of these $L[\mathcal E]$ models is closely tied up to the complexity of the comparison process. (This is nicely explained at the beginning of the Martin-Steel paper on Iteration trees.) -What may not seem so natural is the solution of using iteration trees rather than other coding devices when indexing the ultrafilters in the sequences. In fact, other approaches were suggested, by Baldwin and others, more or less generalizing directly the theory we get at the level of models for $o(\kappa)=\kappa^{++}$. But these suggestions were the ones that ended up being rather complicated and led to the theory being stuck until Steel had the idea of iteration trees. In fact, the theory was stuck well before the level of Woodin cardinals. Thanks to the development of the Mitchell-Steel machinery, these prior levels could then be developed as well.<|endoftext|> -TITLE: Higher degree generalizations of the Hard Lefschetz Theorem -QUESTION [7 upvotes]: Let $M$ be a $2d$-dimensional manifold. We say that $\omega \in H^2(M)$ has the Hard Lefschetz Property (HLP) if multiplication with $\omega^j$ is an isomorphism $H^{d-j} \to H^{d+j}$. This holds for the first Chern class of an ample line bundle on a smooth projective variety, this is the classical Hard Lefschetz Theorem. -My question is whether there is a generalization to classes in $H^4, H^6,$ ... etc. -My proposed definition is that $\eta \in H^k(M)$ has the k-HLP if multiplication by $\eta^j$ from $H^i$ to $H^{i+jk}$ is invective (surjective) if $|i-d|$ is greater (smaller) than $|i+jk-d|$. Then 2-HLP is HLP and if $\omega$ has the HLP then $\omega^j$ has the $2j$-HLP. -Are there other examples of classes with the $2j$-HLP? E.g. Chern classes of vector bundles satisfying an appropriate ampleness condition? - -REPLY [2 votes]: I guess I knew that I should've looked in part 2 of Lazarsfeld's book before posting this question, but it was checked out of the library... -Anyway, the speculation in the last paragraph is correct: if $X$ is a smooth projective variety and $E$ is an ample vector bundle on $X$ of rank $e$, then $c_e(E)$ has the $2e$-HLP. This is exactly the statement of Theorem 7.1.10 in Positivity in Algebraic Geometry II. (Lazarsfeld only considers multiplication by a single top Chern class, not powers of it, but this is equivalent by replacing $E$ by $E^{\oplus r}$).<|endoftext|> -TITLE: fpqc stackification -QUESTION [7 upvotes]: I am reading Lurie's Tannakian paper (http://www.math.harvard.edu/~lurie/papers/Tannaka.pdf) and I am confused about one point. -At the end of page 3 he defines a stack-hom in any topos, which is defined by stackification. Now, I know that there are problems with sheafification in the fpqc topology. Assume the topos is fpqc sheaves on Spec Z, shouldn't there be a problem with stackifying things? -Has this stackification nothing to do with fpqc sheafification? -Is it that this particular stackificaion does not cause any problems? -I would be glad if someone could clear this all up for me. - -REPLY [4 votes]: Summing up the comments: stackification works in any topos; fpqc sheaves do not form a topos (size does matter).<|endoftext|> -TITLE: Width of a random convex polygon -QUESTION [9 upvotes]: Consider a planar (2D) random walk comprised of N steps. -Consider the minimum convex polygon enclosing the N points visited by the random walker. -Assume the definition of the width of a convex polygon given in http://cgm.cs.mcgill.ca/~orm/width.html - -Is it possible to determine the probability density of the width of such a random convex polygon? - -REPLY [7 votes]: (Not an answer; rather an example.) -Here is a random walk with random $\pm 1$ $xy$-steps with equal probability (you didn't specify details), for $n=10^4$ steps, and its convex hull. -    -And here is the very same random walk extended to $n=10^5$ steps: -    - -Added 15Oct15. There is a new paper relevant to this question: - -Zakhar Kabluchko, Vladislav Vysotsky, and Dmitry Zaporozhets. - "Convex hulls of random walks, hyperplane arrangements, and Weyl chambers." - (arXiv abstract.) -"We give an explicit formula for the probability that the convex hull of an - $n$-step random walk in $\mathbb{R}^d$ with centrally symmetric density of increments - does not contain the origin." - -By "contain" here they mean "strictly contain in the interior of the hull." - -Update 1Apr2017. The same authors have revised their paper, -establishing a formula for the expected number of $k$-dimensional faces of -the convex hull of a random walk: - -Kabluchko, Zakhar, Vladislav Vysotsky, and Dmitry Zaporozhets. "Convex hulls of random walks: Expected number of faces and face probabilities." Revised 21 Aug 2017. (arXiv:1612.00249 Abs.)<|endoftext|> -TITLE: What's the relation between the heat kernel proof of the index theorem and deformation quantization? -QUESTION [7 upvotes]: In the book "Heat kernels and Dirac operators", I found the slogan by Quillen in the Introduction: "Dirac operators are a quantization of the theory of connections, and the supertrace of the heat kernel of the square of a Dirac operator is the quantization of the Chern character of the corresponding connection." In fact, Let $M$ be a manifold, $C(M)$ be the Clifford bundle over $M$, $\mathcal{E}$ a vector bundle with a $C(M)$ action, $D$ be the Dirac operator. -The strategy is first we proof that $e^{-tD^2}, t>0$ has a integral kernel $k_t(x,y)\in \Gamma(M\times M, \text{End}(\mathcal{E}))$. It is not difficult to prove that the integration of the super trace $\int_M\text{str}(k_t(x,x))|dx|$ is independent of $t$ and, when $t\rightarrow \infty$ the integration gives the analytic index of $D$ (Hence it gives the analytic index of $D$ for all $t>0$.) -The hard part is when $t\rightarrow 0$. For this we first get the asymptotic expansion -$$ -k_t(x,x)\sim(4\pi t)^{-n/2}\sum_{i=0}^{\infty}t^ik_i(x) -$$ -and then observe that $\text{End}(\mathcal{E})\cong C(M)\otimes \text{End}_{C(M)}(\mathcal{E})$. Moreover, the Clifford algebra $C(M)$ is a quantization of the exterior algebra $\mathcal{A}(M)$. Then we use the symbol map $\sigma$, which is the inverse of the quantization map, to get $\sigma(k)=\sum_{i=0}^{n/2}\sigma_{2i}(k_i)\in \mathcal{A}(M, \text{End}_{C(M)}(\mathcal{E}))$. After a careful rescaling we can proof that -$$ -\sigma(k)=\text{det}^{1/2}(\frac{R/2}{\sinh(R/2)})\exp(-F^{\mathcal{E}/S}) -$$ -which gives the local index theorem. For details of the proof see the first four chapters of "Heat kernels and Dirac operators", especially Chapter 4. -We see that involves the philosophy of quantization (Dirac operator, Clifford algebra, symbol map) as well as deformation (rescaling, $t\rightarrow 0$). On the other hand, the algebra of pseudo differential operators $\mathcal{D}(M)$ is a deformation quantization of the smooth functions $C^{\infty}(T^*M)$ (in fact, a subalgebra consists of functions satisfies some growing conditions)of the symplectic manifold $T^*M$. The $\textit{algebraic index theorem}$, developed by Fedosov, Nest-Tsygan, Dolgushev, ect., studies the general deformation quantization for any symplectic manifold. But their work is closer to Connes's approach to index theory, for example, method from cyclic homology are intensively used. (I'm not very familiar with algebraic index theorem, so please point out if I'm wrong.) -$\textbf{My question}$ is: do we have a more direct, clear relation between the heat kernel proof of the index theorem and deformation quantization? Do we also have a algebraic index theorem in this approach? Any references are very welcome! - -REPLY [2 votes]: I think you should have a look at the various papers of Louis Boutet de Monvel. But there is actually a construction of star-products on a symplectic manifold which makes use of the index theorem, due to Richard Melrose. -Last but not least, you might also want to have a look at Appendix B of this paper by Engeli and Felder, where they use heat kernal methods while proving a HRR formula for traces of holomorphic differential operators.<|endoftext|> -TITLE: Is $\varliminf_{n \rightarrow +\infty} |n \sin n| = 0$ correct, where $n$ is an integer? -QUESTION [13 upvotes]: Is it true that $\varliminf_{n \rightarrow +\infty} |n \sin n| = 0$, where $n$ -runs over the integers? -The existence of the limes inferior follows from Dirichlet's approximation theorem, -but the problem is to prove that it is $0$. - -REPLY [7 votes]: This is an open question. See the comments for more details.<|endoftext|> -TITLE: Is the Steiner ratio Gilbert–Pollak conjecture still open? -QUESTION [13 upvotes]: Gilbert-Pollak conjecture on the Steiner ratio: Consider a set $P$ of $n$ points on the euclidean plane. A shortest - network interconnecting $P$ must be a tree, which is called a Steiner minimum tree - and denoted by $SMT(P)$. A $SMT(P)$ may contain vertices not in $P$. Such vertices - are called Steiner points, while vertices in $P$ are called regular points. A spanning tree on $P$ is just a tree with vertex - set $P$. A shortest spanning tree on $P$ is also called a minimum spanning tree on $P$, - denoted by $MST(P)$. The Steiner ratio is defined to be - $$\rho=\inf\{L_s(P)/L_m(P)|P\},$$ - where $L_s(P)$ and $L_m(P)$ are lengths of $SMT(P)$ and $MST(P)$, respectively. Gilbert and Pollak conjectured that $p = \sqrt{3}/2$. - -I find recently that the proof of the Gilbert-Pollak conjecture given by D. Z. Du and F. K. Hwang in 1990 is disproved by N. Innami etc. (The Steiner Ratio Conjecture of Gilbert-Pollak May Still Be Open and The Steiner Ratio Gilbert–Pollak Conjecture Is Still Open). When I check related comments about this case on the Internet, I find there seems to be few people that care about this (very few information). -Is this conjecture still open? Is there anyone trying to sovle this problem in other ways or to fix the proof of D. Z. Du and F. K. Hwang? - -REPLY [11 votes]: The answer is yes. The basic concept used in Du and Hwang approach to the problem is so-called characteristic area constructed for a Steiner tree. But this object needs to have mutually exclusive properties. We put a short note in arXiv concerning these problems, see http://arxiv.org/abs/1402.6079. Some questions related to Gilbert-Pollack conjecture are discussed in http://arxiv.org/abs/1101.0106.<|endoftext|> -TITLE: Is every orientable circle bundle principal? -QUESTION [10 upvotes]: The only examples I found of nonprincipal circle bundle are nonorientable, like the Klein bottle that is an S^1 bundle over S^1 which is not principal and nontrivial. That makes me ask the question. -Is it true that every orientable circle bundle is principal? - -REPLY [17 votes]: Yes,it is true.You can find this result from "Geometry of differential forms" by Morita (Page 241) -PROPOSITION 6.15. Every oriented $S^1$ bundle admits the structure of principal $S^1$ bundle.<|endoftext|> -TITLE: Is there a notion of Skolemization for continuous logic? -QUESTION [8 upvotes]: Is there a notion of Skolemization for continuous logic (that of Ben Yaacov)? It seems to me that a Skolem function would be a minimizer, and minimizers are not continuous (from the parameters of the function being minimized). -If there isn't, is there a trick that gives the same effect (turning a formula into a universal one). - -REPLY [3 votes]: Many of the things you might want to do with Skolemization (such as producing Ehrenfeucht-Mostowski models) can be done by passing to a discretization of your theory and then Skolemizing there. Without having actually looked, I believe this is very similar to the technique Kaveh mentioned in their answer. (Although I should note that Henkin models can be constructed directly in continuous logic without too much difficulty.) -Of course, in some sense this requires modifying the underlying metric. As you point out, minimizing (or even approximately minimizing) is not continuous, even in compact structures, so a naïve definition of Skolemization is certainly not always possible. There is a weaker notion: - -A theory $T$ is weakly Skolemized if $\mathrm{dcl}(A)$ is a model of $T$ for every set of parameters $T$. - -This is equivalent to ordinary Skolemization in discrete first-order logic, but even with this it's unclear that you can always pass to an expansion which is weakly Skolemized without modifying the metric. -With regards to weakly Skolemizing theories, the best that I know at the moment is this: - -Every uniformly locally compact theory can be weakly Skolemized. - -Every ultrametric theory can be weakly Skolemized. - -If a theory $T$ has a weakly Skolemized expansion $T'$, then there is an intermediate expansion $T''$ (i.e., $T \subseteq T'' \subseteq T'$) such that the language of $T''$ is no larger than the language of $T$. (This is not a priori obvious.) - - -(Thinking about it now, I suspect that ultrametric theories can be Skolemized in a more direct sense. This is certainly true for ultrametric theories with discrete distance sets.) Beyond this I don't know very much. For instance, I don't know whether every expansion of Hilbert space can be weakly Skolemized. I gave a talk about this and also touch on it briefly in Remark 3.5.7 of my thesis. -If I interpret your broader question in what I would consider to be a 'non-trivially continuous' way, by which I mean, does every continuous theory $T$ have an expansion $T'$ with the same metric such that every formula is equivalent to a universal formula (or definable predicate, if you insist on considering those as distinct from formulas) over $T'$, then I believe it is currently unknown whether this always holds. (EDIT: I realized I was being dumb with this comment. Morleyization works fine in continuous logic and allows you to pass to an expansion with quantifier elimination, which is strictly stronger. That said, part of what Skolemization gives you is that the resulting theory is axiomatized by universal sentences. Since the standard semantic characterization of universal sentences works in continuous logic, I think this is equivalent to asking that the theory have a Skolemization in the strong sense, which I suspect is not always possible.) It is even unclear to me at the moment whether a theory $T$ having a weak Skolemization implies that it has an expansion with the property that every formula is logically equivalent to a universal formula, because, as I mention in my thesis, weak Skolemization is not witnessed by definable functions, but rather certain nice families of uniformly definable partial functions.<|endoftext|> -TITLE: Nagata's conjecture in positive characteristic -QUESTION [6 upvotes]: For a $\mathbb C P^2$ is known a result: if through the generic points $p_1,p_2,\dots p_n$ with multiplicities $m_1,m_2\dots, m_n$ correspondingly a degree $d$ irreducible reduced curve passes then $d^2\geq m_1^2+m_2^2+\dots+m_n^2-min (m_i)$. (see Lemma 1 in "Curves in $\mathbb P^2$ and symplectic packings" Geng Xu) -The idea of the proof is the following: move slightly a point $p_i$ and apply Bezout theorem; formally he gave a construction of a new curve via deformation of the old one. -I was wondering if the same proof works in positive characteristic ? -In the article it is written $\mathbb P^2$ without mentioning of characteristic, but the proof -doesn't work : as you know, a derivation of a polynomial in positive characteristic can drastically change its degree. -From the other hand it is very frequently when a result in complex geometry automatically translates to any characteristic by some logic reasons, it is the case here? - -REPLY [4 votes]: Xu's idea can be reproduced in arbitrary characteristic, without derivations, with the extra hypothesis that the singularities are ordinary (ie, $m_i$ distinct tangent directions at $p_i$), or at least each point has one direction of multiplicity one in the tangent cone. Without an assumption like this, I don't know if it can be done. EDIT: It can't be done. See below. Assume the base field $k$ is infinite; then the argument goes as follows. -Let $d$ and $m_1, \dots, m_n$ be such that for general $p_1, \dots, p_n$ there is an irreducible curve of degree $d$ with multiplicity $m_i$ at $p_i$. Fix such a set of points, and fix affine coordinates so that $p_1=(0,0)$ and $y=0$ is one of the tangent lines to the curve at $p_1$. Let $p(t)=(0,t)$ and for simplicity write $m=m_1$. By the assumption, for general $t\in k$ there is an irreducible curve of degree $d$ multiplicity $m$ at $p(t)$ and $m_i$ at $p_i$, $i=2,\dots, n$. Let $I\subset k[x,y]$ be the ideal of the points $p_2, \dots, p_n$ with their multiplicities, $I^e$ its extension to $k[x,y,t]$ and $J=I^e \cap (x,y-t)^{m}$. By hypothesis, there is an irreducible $F\in J$ of degree $d$ in $x$ and $y$. -Write $F=\sum_{i=0}^k t^iF_i$, where $F_i\in I$ are polynomials in $x,y$ of degree at most $d$, and $F_0$ has degree exactly $d$. Clearly $F\in J\subset(x,y,t)^m$ (a monomial ideal) -and it is easy to see that this implies $F_1 \in (x,y)^{m-1}$. If we show that $F_1\not\in (F_0)$, then the rest of Xu's argument works. -Since the multiplicity of $F_0$ at $p$ is exactly $m$, one can write $F=\sum_{j=0}^m G_j x^{m-j}(y-t)^{j}$, where $G_j$ are polynomials in $x,y,t$ and at least $G_1$ has nonvanishing constant term (by the choice of tangent line $y=0$). Then the summand for $j=1$ contributes a term $cx^{m-1}t$ to $F_1$ which can not be cancelled with any other term. This shows that the multiplicity at $p$ of $F_1$ is exactly $m-1$ and in particular that $F_1$ is neither zero nor equal to a scalar multiple of $F_0$. -EDIT (added Oct 11): Now consider the family of irreducible curves $C_{a,b}:(x-a)^2(y-b)^2=x^5+y^5$ which have a 4-fold point at the point $(a,b)$. For any parameterized curve $(a(t),b(t))$ in the plane $(a,b)$, expanding $F_t=(x-a(t))^2(y-b(t))^2-x^5-y^5=\sum_{i=0}^k t^iF_i$ gives $F_1=0$ in characteristic 2 and so the intersection-theoretic argument does not work (even after a linear change of variables, or for a germ of curve parameterized by two power series, the argument is the same). -Xu's argument is local in nature, and it has been used in other settings by many authors, always in characteristic zero (notably Lazarsfeld, see 5.2.3 in "Positivity in Algebraic Geometry I" and references there). I am quite sure that families of curves like the one above can be used to obtain counterexamples in positive characteristic on suitable surfaces. On the plane, however, Xu's conclusion is likely to be true, or at least counterexamples will be hard to construct (otherwise Nagata's conjecture would fail in positive characteristic, which is unknown and -as far as I am concerned- unexpected). In any case, Xu's argument can't be pushed to positive characteristic without some extra assumption on the singularities as above.<|endoftext|> -TITLE: Does existence of a proper class model imply the consistency? -QUESTION [9 upvotes]: The fundamental theorem of model theory says that: -Theorem: A first order theory is consistent if and only if it has a model. -In the above theorem we assume that the domain of any model is a non-empty "set". But in set theory sometimes we use proper class models of a theory. (For example $\langle L,\in\rangle \models ZFC$). Now a simple question is: -Question: Does existence of a proper class model for a first order theory (which is a weaker assumption than existence of a set model) imply its consistency? How can we formalize a similar theorem for proper class models? - -REPLY [13 votes]: $\newcommand{\ZFC}{\text{ZFC}}\newcommand{\KM}{\text{KM}}$ -The answer must of course be negative, since every model of a theory $T$ is a proper class model of $T$, from its own perspective, but this cannot imply $\text{Con}(T)$ because of the incompleteness theorem. -But the question is actually more problematic than this, since we cannot generally even express that a proper class model is a model of a given theory as a single statement; rather, it is generally a scheme. For example, in the case of the constructible universe, one often hears it said that the constructible universe $L$ is a model of $\ZFC+V=L$, but this is not a single assertion in the language of set theory. Rather, what one means is that one can prove in $\ZFC$ that any given axiom of $\ZFC$ holds also in $L$. Furthermore, because of Tarski's theorem on the non-definability of truth, we aren't really able to formulate the assertion "$L$ is a model of $\ZFC$" as a single assertion in the first-order language of set theory. In this sense, the answer to your question is no. -Meanwhile, however, if you work in a stronger theory, such as Kelley-Morse set theory, then you can recover an affirmative answer. This is because $\KM$ proves the existence of truth predicates for first-order truth relative to any given class. Thus, in $\KM$, if you have a proper class model of a theory $T$, then $\KM$ can build the truth predicate for first-order satisfaction in this model and see that it is consistent, so the answer turns to Yes. The assertion that $L\models\ZFC$, for example, can be formalized as a single assertion in Kelley-Morse set theory, and furthermore, this assertion will imply $\text{Con}(\ZFC)$, essentially by induction on proofs.<|endoftext|> -TITLE: What is the state of the art for algorithmic knot simplification? -QUESTION [17 upvotes]: Question: Given a `hard' diagram of a knot, with over a hundred crossings, what is the best algorithm and software tool to simplify it? Will it also simplify virtual knot diagrams, tangle diagrams, and link diagrams? - -The reason that I ask this question is that I have been reading: -Lewin, D., Gan O., Bruckstein A.M., -TRIVIAL OR KNOT: A SOFTWARE TOOL AND ALGORITHMS FOR KNOT SIMPLIFICATION, -CIS Report No 9605, Technion, IIT, Haifa, 1996. -This technical report is notable not only for its mathematical content, but also for its back-story. It was the undergraduate research project of Daniel M. Lewin, who was a few years later to found Akamai Technology which today manages 15-20% of the world's web traffic, to become a billionaire, and to be the first person to be murdered on 9-11. He is the subject of the 2013 biography No Better Time: The Brief, Remarkable Life of Danny Lewin, the Genius Who Transformed the Internet by Molly Knight Raskin, published by Da Capo Press. -The algorithm used by Bruckstein, Lewin, and Gan doesn't use 3-dimensional topology or normal surface theory, but instead relies on an algorithmic technique called simulated annealing. I suspect that this same technique could be used effectively for other problems in algorithmic topology. -To the best of my knowledge, Bruckstein, Lewin, and Gan's algorithm is unknown to the low-dimensional topology community, but it works very well. Despite it having been written a long time ago, I wonder (for reasons beyond mere curiousity) how far it is from being state of the art today. - -REPLY [3 votes]: A little Heegaard promotion. -Those of you interested in recognizing unknots, or the 3-sphere, or who are interested in simplifying general geometric 3-manifold presentations might want to try a new version of Heegaard. (Which may eventually replace the original outdated version posted on the website of Marc Culler and Nathan Dunfield at www.math.uic.edu/t3m.) -Heegaard accepts geometric presentations on up to 26 generators and 32 relators. So for knots, Wirtinger presentations, or geometric presentations obtained using over-crossing and under-crossing arcs will work provided the 26 generator restriction is satisfied. -For example, Heegaard easily determines each of the following examples is unknotted in only a few seconds. -Thistlethwaite's Unknot from Wikipedia, Culprit from Kaufmann-Lambropoulou, Fig5 from Kaufmann-Lambropoulou, H) from Henrich & Kauffman, J) from Henrich & Kauffman, G) Goeritz's unknot, Ochiai's unknot 1 from "Non-trivial projections of the trivial knot" URL http://hdl.handlenet/2433/99940, Ochiai's unknot 2 from the same paper, and Freedman's Twisted Unknot. -While the approach Heegaard uses isn't sexy, it does seem to work quite---perhaps unreasonably---well. -Those wishing to check such things for themselves can obtain a current copy of Heegaard from me by email request to jberge at charter 'dot' net. -Note Heegaard is a Unix executable, which runs under terminal on an apple macintosh. So you will probably need a mac. Though Ryan Budney has it running under linux.<|endoftext|> -TITLE: Bernstein-Sato polynomial (one variable) -QUESTION [11 upvotes]: Let $R = \mathbb{C}[x_1,...,x_n]$, $p \in R$. There exists a monic (of lowest degree) $b_p(x) \in \mathbb{C}[s]$ and a differential operator $D(s)$ such that -$$b_p(s) p^s = D(s)p^{s+1}.$$ -The polynomial $b_p(s)$ is called the Bernstein-Sato polynomial of $p$. The calculation of $b_p(s)$ is very complicated. -Consider $n=1$ and $R = \mathbb{C}[x]$. We have $p = (x-a_1)^{n_1}...(x-a_t)^{n_t}$ with $n_1 \leq n_2 \leq \cdots \leq n_t$. We have a formula for $b_p(s)$ in the following cases - -if $t=1$ we have $b_p(s) = (s+\frac{1}{n_1})\cdots (s+\frac{n_1-1}{n_1}) (s+1)$. -if $n_1 = \cdots = n_t=1$, then $b_p(s) = (s+1)$. - -Question. Does there exist a formula for $b_p(s)$ in term $(n_1,...,n_t)$? - -REPLY [5 votes]: It is known that the (global) Bernstein-Sato polynomial is the least common multiple of all local Bernstein-Sato polynomials. In the case at hand, the local polynomials are given by $b_{p,i}(s) = (s+\frac{1}{n_i})\cdots (s+\frac{n_i-1}{n_i}) (s+1)$ (localization at $a_i$) or 1 (localization somewhere else), which gives your desired formula. -Experimenting with these polynomials online is possible with Macauley2 at http://habanero.math.cornell.edu:3690/. For example, -needsPackage "Dmodules" - -R = QQ[x] - -f = x^3*(x-1)^5 - -b = globalBFunction f - -c = factor b - -returns - 1 -o9 = (s + 1)(3s + 1)(3s + 2)(5s + 1)(5s + 2)(5s + 3)(5s + 4)(----) - 5625<|endoftext|> -TITLE: Centralizers of elements in general linear group over Z mod prime power -QUESTION [6 upvotes]: I would like to know for which elements $x$ in $G:=Gl_n(\mathbb{Z}/\ell^e\mathbb{Z})$ their centralizers $C_G(x):=\{ y \in G \mid xy=yx\}$ are abelian groups. - -Here, $n$ is an integer $\geq 2$ and $\ell^e$ is a prime power. I am especially interested in the case $n=2$. -Of course, if $x$ is a multiple of the identity matrix, then $C_G(x)=G$, and thus is not abelian. -In case $e=1$, i.e. $\mathbb{Z}/\ell^e\mathbb{Z}$ is a field, then I have read a few times (but always without proof) that $C_G(x)$ is abelian, if $x$ is not a multiple of the identity matrix and $n=2$. To be precise, $C_G(x)$ will be either isomorphic to $\mathbb{F}_{\ell^2}^*$, $\mathbb{F}_\ell^* \times \mathbb{F}_\ell^*$, or $\mathbb{F}_\ell \times \mathbb{F}_\ell^*$. -So the question is what happens for arbitrary prime powers (and for arbitrary $n$)? -The question seems to me like a standard fact which should be contained in every text book about general linear groups. Can anybody give a good reference? -Thanks a lot! - -REPLY [9 votes]: Let $G_r=\mathrm{GL}_n(\mathbb{Z}/p^r)$. For $x\in G_r$ the centraliser $C_{G_{r}}(x)$ is abelian iff $x$ is regular iff the reduction mod $p$ of $x$ is regular. This is due to G. Hill, Regular elements and regular characters of $\mathrm{GL}_n(\mathcal{O})$, J. Algebra 174 (1995), no. 2, 610–635. The case $r=1$ is easy and was known earlier. -To add some details, note that Hill's Theorem 3.6 holds when the residue field is $\overline{\mathbb{F}}_p$, but I think his proof goes through for any algebraically closed residue field $\overline{k}$. If $k$ is any field and $C_{\mathrm{GL}_n(k)}(A)$ is abelian for $A\in \mathrm{GL}_n(k)$, then using for example the rational canonical form one can see that $A$ must be $\mathrm{GL}_n(k)$-conjugate to a companion matrix, and so $k^n$ is a cyclic $k[A]$-module. Extending scalars to $\overline{k}$ we get that $A$ is regular as an element of $\mathrm{GL}_n(\overline{k})$.<|endoftext|> -TITLE: character degree and solvability -QUESTION [6 upvotes]: There is an unsolved problem in Berkovich's book "Characters of Finite Groups Part 2" I state here: -Is $G$ solvable if $\chi(1)^2$ divides $|G|$ for all $\chi \in {\rm Irr}(G)$? -Can any one tell me some latest progresses for this? Maybe you can tell me some latest research papers. Thank you. - -REPLY [5 votes]: Steve Gagola and I have a paper where we address this very question. I am away from home and so I don't have the reference handy. It was in the Communications in Algebra in the late 90's. I think 1998. -Also, the example Marty gave was in that paper. Marty and I found that example while I was still working on my PhD.<|endoftext|> -TITLE: Characterisation of Q-rank 1 -QUESTION [7 upvotes]: I'm looking for a reference and/or the original source for the following fact: -An irreducible non-uniform lattice in a semisimple Lie group without compact factor has Q-rank 1 if and only if it does not contain a subgroup isomorphic to a finite index subgroup of $SL(3,{\Bbb Z})$ or $SO(2,3)_{\Bbb Z}$. - -REPLY [6 votes]: The proof of Kazhdan's property (T) for real simple Lie groups of real rank at least two as given in the old Bourbaki talk of Kirilllov and Delaroche involves showing (property (T) for $H=SL_3({\mathbb R}), Sp_2({\mathbb R})$ and then showing) that any such $G$ contains a subgroup locally isomorphic to $H$. -Exactly the same proof shows that any $\mathbb Q$-simple linear algebraic group $G$ of $\mathbb Q$ rank at least two contains a subgroup locally $\mathbb Q$-isomorphic to $SL_3$ or to $Sp_2$. -In detail, such a $G$ contains a subgroup $G_0$ which is $split$ over $\mathbb Q$ and of the same $\mathbb Q$-rank as $G$. By looking at the Dynkin diagram of $G_0$, one can extract a sub-diagram of type $A_2$ or $B_2$ except for $G_2$ where this (i.e. that $G_2$ contains $A_2$, the root system of short roots in the root system of $G_2$) can be proved directly by looking at its root system. -[Edit] Misha (in the comments) was right; this result that any $k$-simple group of $k$-rank at least two contains a subgroup locally isomorphic to $SL_3$ or $Sp_4$) is explicitly stated and proved in Margulis' book; see Proposition (1.6.2) of Margulis' book titled " Discrete subgroups of Semi-simple Lie groups (Ergbnisse tract, volume 17)". The result about arithmetic groups can be deduced from this by taking $k={\mathbb Q}$.<|endoftext|> -TITLE: If $\left(1^a+2^a+\cdots+n^{a}\right)^b=1^c+2^c+\cdots+n^c$ for some $n$, then $(a,b,c)=(1,2,3)$? -QUESTION [17 upvotes]: Question : Is the following conjecture true? -Conjecture : Let $a,b(\ge 2),c,n(\ge 2)$ be natural numbers. If $$\left(\sum_{k=1}^nk^a\right)^b=\sum_{k=1}^nk^c\ \ \ \ \ \cdots(\star)$$ -for some $n$, then $(a,b,c)=(1,2,3).$ -Remark : This question has been asked previously on math.SE without receiving any answers. -Motivation : This question comes from -$$\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3.$$ -This got me interested in $(\star)$. I've got the followings : -1. If $(\star)$ for any $n\in\mathbb N$, then $(a,b,c)=(1,2,3).$ -We can easily prove this by considering the limitation $n\to\infty$ of the both sides of -$$\frac{n^{(a+1)b}}{n^{c+1}}\left\{\sum_{k=1}^n\frac 1n\left(\frac kn\right)^a\right\}^b=\sum_{k=1}^n\frac 1n \left(\frac kn\right)^c.$$ -2. If $(\star)$ for $n=2$, then $(a,b,c)=(1,2,3).$ -3. If $(\star)$ for $n=3$, then $(a,b,c)=(1,2,3).$ -Since both 1 and 2 are easy to prove, I'm going to prove 3. -Proof : Supposing $c\le ab$, since $b\ge 2$, we get -$$(1+2^a+3^a)^b=1+2^{ab}+3^{ab}+\cdots\gt 1+2^c+3^c.$$ This is a contradiction. Hence, $c\gt ab$. Supposing $b\ge 3$, we get $c\gt ab\ge 3$. -Here, since $3^c+1\equiv 4,2$ (mod $8$) for any $c\in\mathbb N$, $3^c+1$ is not a multiple of $8$. By the way, since $1+2^a+3^a$ is even, $(1+2^a+3^a)^b$ is a multiple of $8$. Since $2^c$ is a multiple of $8$, this leads that $3^c+1$ is a multiple of $8$, which is a contradiction. Hence, $b=2, c\gt 2a$. -If $a\ge 3$, since -$$\left(\frac 23\right)^a+\left(\frac 13\right)^a\le\left(\frac 23\right)^3+\left(\frac 13\right)^3=\frac13,$$ -$2^a+1\le \frac{3^a}{3}.$ Hence, -$$3^c\lt 1+2^c+3^c=(1+2^a+3^a)^2\le \left(\frac{3^a}{3}+3^a\right)^2=3^{2a}\left(\frac 43\right)^2=3^{2a}\cdot\frac {16}{9}\lt 3^{2a+1}.$$ -$3^c\lt 3^{2a+1}$ leads $0\lt c-2a\lt 1$, which means that $c-2a$ is not an integer. This is a contradiction. Hence, we know $a=1$ or $a=2$. -The $(a,b)=(1,2)$ case leads $c=3$. -The $(a,b)=(2,2)$ case leads $c\ge5\Rightarrow 1+2^c+3^c\gt 196$, which is a contradiction. Now the proof is completed. -After getting these results, I reached the above conjecture. Can anyone help? - -REPLY [2 votes]: This is a partial solution. I've just been able to get the following theorem: -Theorem : If $(\star)$ for some $n=8k-5,8k-4\ (k\in\mathbb N)$, then $(a,b,c)=(1,2,3)$." -I wrote the proof for this theorem on MSE. -PS: This idea (using mod $8$) does not seem to work for the other $n$. Another idea would be needed.<|endoftext|> -TITLE: Is the inclusion version of Kunen inconsistency theorem true? -QUESTION [11 upvotes]: The relations $\in$ and $\subsetneq$ seem so similar in some sense. For example they are equal on ordinal numbers. So there is a natural question about their possible similar behaviors on the constructible universe or proper class of all sets for example in the case of Kunen inconsistency theorem. -Question (1): Is there a non-trivial elementary embedding $j:\langle V,\subsetneq\rangle \longrightarrow \langle V,\subsetneq\rangle$? -Question (2): Is there a non-trivial elementary embedding $j:\langle L,\subsetneq\rangle \longrightarrow \langle L,\subsetneq\rangle$? -Question (3): What is the consistency strength of existence of such embeddings relative to large cardinal axioms? Particularly what is the position of existence of a non-trivial elementary embedding from $\langle L,\subsetneq\rangle$ to itself relative to existence of $0^{\sharp}$? - -REPLY [14 votes]: $ \newcommand\ofnoteq{\subsetneq}$ -It is a very nice question! -The answer is that there are numerous definable automorphisms of $\langle V,\ofnoteq\rangle$. To see this, let $f:V\to V$ be any permutation of the universe, and define the induced function $\pi:V\to V$ by $\pi(x)=f[x]$, the image of $x$ under $f$. For example, $\pi$ maps the singleton $\{ a\}$ to the singleton $\{f(a)\}$. Using the fact that $f$ is a permutation, it is not difficult to see that $x\ofnoteq y\iff \pi(x)\ofnoteq\pi(y)$, and furthermore that $\pi$ is a bijection, and hence it is an automorphism of the universe with respect to $\ofnoteq$, and consequently an elementary embedding, which is nontrivial precisely when $f$ is. -So there is no large cardinal strength here to be found, and the lesson appears to be that $\ofnoteq$ does not capture much of the intended set-theoretic structure.<|endoftext|> -TITLE: Cohomology of Formal Groups -QUESTION [13 upvotes]: Lubin and Tate, in discussing moduli of 1-dimensional formal groups construct a cohomology theory of formal groups, at least in degrees 0,1 and 2. Does their result about deformations actually follow from the natural 3rd group of this cohomology being trivial? Has the obvious extension of this structure to a full cohomology theory been written down somewhere and had its properties studied? Moreover, Lubin and Tate's cohomology actually appears to be a special a case of what one might call the Hochschild cohomology of $\hat{\mathbb{G}}_a$ (using Demazure and Gabriel's notion of Hochschild cohomology of group schemes) with coordinates in some other formal group $\mathbb{G}$ (the one defining the group law of interest) with a trivial action of $\hat{\mathbb{G}}_a$ on $\mathbb{G}$. The same structure arises in Lazard's proof of the extensibility of formal group n-buds. Has this interpretation been written down anywhere? And more importantly, are there other interesting examples of applications of this sort of structure? -Looking at this question causes me to think I might find answers in the manuscript of Strickland's which is online, but I'm not sure yet. - -REPLY [4 votes]: Lazarev is doing calculations with this "cohomology theory" in his paper "Deformations of Formal Groups" -Among the thing he shows is that this cohomology is the E2 term of the bar spectral sequence from $E^*(K(\mathbb{Z},2)) \rightarrow E^*(K(\mathbb{Z},3) $<|endoftext|> -TITLE: Seeking conceptual explanation of these nice bijections on roots of unity -QUESTION [10 upvotes]: I proved the following facts by unenlightening calculations. Since the statements are quite clean, I think there should be a conceptual explanation for them, which my proof certainly is not. -Let $q$ be a prime power, and let $\mu_{q+1}$ be the set of $(q+1)$-th roots of unity in the finite field $\mathbf{F}_{q^2}$. If $b\in\mu_{q+1}$ and $c\in\mathbf{F}_{q^2}\setminus\mathbf{F}_q$ then -$$ -x\mapsto \frac{cx-bc^q}{x-b} -$$ -maps $\mu_{q+1}$ to $\mathbf{F}_q\cup\{\infty\}$. If $b\in\mu_{q+1}$ and $d\in\mathbf{F}_{q^2}\setminus\mu_{q+1}$ then -$$ -x\mapsto \frac{x-bd^q}{dx-b} -$$ -maps $\mu_{q+1}$ to itself. (It is also true that these are the only degree-one rational functions which map $\mu_{q+1}$ to either $\mathbf{F}_q\cup\{\infty\}$ or $\mu_{q+1}$, but I'm mainly interested in understanding the existence.) -I tagged this "group theory" because the first fact vaguely feels like a connection between orbits of a nonsplit torus and a split torus in $\textrm{PGL}_2(q)$. It's tempting to identify $\mathbf{F}_{q^2}$ with $\mathbf{F}_q\times\mathbf{F}_q$, and consider the resulting action of $\textrm{GL}_2(q)$ on $\mathbf{F}_{q^2}$, but I don't see how to go further in this way. -Any suggestions? - -REPLY [9 votes]: The question and the analog to the Cayley map in complex numbers pointed out in the comment by Jyrki Lahtonen is not only an analog, but in fact both are special cases of this more general observation: Let $z\mapsto\bar z$ be an involutory automorphism of a field $F$, and let $E$ be the subfield fixed by $\bar{\phantom{a}}$. Furthermore, set $S=\{z\in F\;|\;z\bar z=1\}$. Then, for $b\in S$, $c\in F\setminus E$, and $d\in F\setminus S$, -\begin{equation*} -z\mapsto \frac{cz-b\bar c}{z-b} -\end{equation*} -maps $S$ to $E\cup\{\infty\}$ and -\begin{equation*} -z\mapsto \frac{z-b\bar d}{dz-b} -\end{equation*} -maps $S$ to $S$.<|endoftext|> -TITLE: About sets with two mutually associative group structures -QUESTION [6 upvotes]: Consider the following: A set $S$ with two operations, $\ast$ and $\times,$ such that both $(S,\ast)$ and $(S,\times)$ are groups, and such that whenever one writes down a mixed product, e.g. $a\ast b\times c\times b \ast d \times a$ it doesn't matter where one puts the parentheses. My question is whether this structure has a name and whether it has been studied. - -REPLY [19 votes]: Let $e$ be an identity for $\cdot$ and $f$ for $*$. Then $e*e\cdot f=e$, i.e. $f^{-1}=e*e$. Further, -$$a*b=ae*(bb^{-1})b=a(e*e)b=af^{-1}b,$$ -so the second operation ($*$) is defined by the first one and by fixing an arbitrary element $f$.<|endoftext|> -TITLE: Rotation numbers for amenable group actions on the circle -QUESTION [7 upvotes]: Given an orientation-preserving homeomorphism $f: S^1 \to S^1$, one can define its rotation number $\rho(f) \in \mathbb{R}/\mathbb{Z}$, as $\rho(f) = (\lim_{n \to \infty} \tilde{f}^n(0)/n) + \mathbb{Z}$, where $\tilde{f}: \mathbb{R} \to \mathbb{R}$ is any lift of $f$. Intuitively, it measures the rate of circulation around the circle. -Now $\rho: Homeo_+(S^1) \to \mathbb{R}/\mathbb{Z}$ is not a homomorphism, but its restriction to any amenable subgroup of $Homeo_+(S^1)$ is. Thus if $G$ is amenable, and $\phi: G \to Homeo_+(S^1)$ is an action, then the composition $\rho \circ \phi: G \to \mathbb{R}/\mathbb{Z}$ is a homomorphism. -I am wondering, what homomorphisms $G \to \mathbb{R}/\mathbb{Z}$ arise as $\rho \circ \phi$ if $\phi$ is required to be 1-1? -(The requirement that $\phi$ be 1-1 is important, since otherwise we can realize any $\psi: G \to \mathbb{R}/\mathbb{Z}$ simply by making $G$ act on the circle via $\psi$.) -This question might be hard, since I don't even know which amenable groups act faithfully on the circle. But I'd even be curious about the case where $G$ is a finitely-generated, torsion-free nilpotent group. These guys do act on the circle faithfully, but the standard construction gives you something trivial in terms of rotation number. -I doubt it helps, but by Ghys and Matsumoto, for an amenable group $G$, two actions $\phi_1, \phi_2: G \to Homeo_+(S^1)$ are semi-conjugate if and only if $\rho(\phi_1(g)) = \rho(\phi_2(g))$ for every $g \in G$ (Matsumoto, "Numerical invariants for semiconjugacy of homeomorphisms of the circle"). - -REPLY [2 votes]: The answer to my question, if $G$ is assumed to be a finitely generated torsion-free nilpotent group, is that any homomorphism $\psi: G \to \mathbb{R}/\mathbb{Z}$ can be realized as $\rho \circ \phi$ for $\phi: G \to Homeo_+(S^1)$ 1-1. I'm sure this was already known, and the paper referenced by Dan Sălăjan definitely contains the right way to think about it. -Suppose we have $\psi: G \to \mathbb{R}/\mathbb{Z}$. Take the orbit of some $x \in S^1$ under $im(\psi)$, and blow up the points in this orbit so that they are intervals. We could call the circle with blown-up intervals $\tilde{S^1}$. Obviously, we can define an action of $im(\psi)$ on $\tilde{S^1}$, for instance by sending one interval to another affinely. -Now $G$ is a f.g. torsion-free nilpotent group, so it acts faithfully on the interval $[0, 1]$, and thus so does $\ker(\psi)$. If we take a copy of this action of $\ker(\psi)$ on each blown-up interval, together with the action of $im(\psi)$ that sends one interval to another, then altogether we have an action of the wreath product $\ker(\psi) \wr im(\psi)$ on $\tilde{S^1}$. -By the Kaloujnine-Krasner Theorem, $\ker(\psi) \wr im(\psi)$ contains a copy of every extension of $\ker(\psi)$ by $im(\psi)$, including $G$ itself. Indeed, under this embedding $\Phi\colon G \hookrightarrow \ker(\psi) \wr im(\psi)$, the projection of $\Phi(g)$ onto the $im(\psi)$ factor is precisely $\psi(g)$. (For more information on the Kaloujnine-Krasner Theorem, see e.g. Lectures on Finitely Generated Solvable Groups, in SpringerBriefs in Mathematics.) So this allows us to define our embedding $\phi\colon G \to Homeo_+(S^1)$ having the desired property.<|endoftext|> -TITLE: How do you define the strict infinity groupoids in Homotopy Type Theory? -QUESTION [17 upvotes]: In the setting of Homotopy Type Theory, how would you construct $\mathrm{isStrict} : U \rightarrow U$ which is inhabited exactly when the first type is (equivalent to?) a strict $\infty$-groupoid? -The "obvious" (to me, at least) approach is that if you have a strict $\infty$-groupoid you want associativity to be a judgemental equality and not just a propositional equality. However, there's a major problem with that approach, which is that composition itself has many different definitions which are all propositionally equal but not judgementally equal. Furthemore, by univalence, you shouldn't be able to distinguish between the types which are strict $\infty$-groupoids and the types which are equivalent to strict $\infty$-groupoids. (Similarly, isSet actually detects the property of being a $0$-type instead of being a set.) So is there an alternative approach? - -REPLY [27 votes]: I realize I'm dropping in long after all the excitement is over. However, I'd like to point out an implicit assumption which is (seemingly) inherent in the original question as well as some of the answers: namely, that "being strict" is a property of $\infty$-groupoids, rather than a structure put on an $\infty$-groupoid. -It is reasonable to suppose we have some kind of realization functor -$$ \Re: St\infty Gpd \to \infty Gpd,$$ -so that the homotopy groups of the space $\Re G$ naturally coincide with the homotopy groups of the strict infinty groupoid $G$. (This notion is due to Simpson, http://arxiv.org/abs/math/9810059). -Since there are notions of weak equivalence for both $\infty Gpd$ and $St\infty Gpd$, these both give (weak) $(\infty,1)$-categories, and $\Re$ is a functor between them. If $\Re$ happened to be fully faithful as a functor of $(\infty,1)$-categories, it would be reasonable to regard strictness as a property of $\infty$-groupoids. However, there no reason to think that is the case: being strict is not a property. -It is, of course, licit to ask "which $\infty$-groupoids are in the essential image of $\Re$", and to call that $\mathtt{isStrict}$. However, if $\Re$ is not fully faithful, then knowledge of $\mathtt{isStrict}$ is not at all the same as "defining strict infinity groupoids". (The same issue appears in Mike's variant involving the forgetful functor $H\mathbb{Z}Mod\to Spectra$.) -Actually, there is no need to speculate here. There are theorems: - -The category $St\infty Gpd$ is equivalent to the category $Crs$ of "crossed-complexes" (Brown-Higgins, http://groupoids.org.uk/pdffiles/x-comp.pdf). -There is a closed model structure on $Crs$, where the weak equivalences are what you expect (Brown-Golasinski, http://groupoids.org.uk/pdffiles/RB-golskyrev.pdf; see also Ara-Metayer, http://arxiv.org/abs/1010.2599). -There is an adjoint pair -$$ \pi\colon sSet \rightleftarrows Crs :N$$ -where the "nerve" functor $N$ is an example of a realization functor $\Re$. Furthermore, this adjunction is a Quillen pair relating the Brown-Golasinski model structure on $Crs$ with the Kan-Quillen model structure on $sSet$. (Brown-Higgins, http://groupoids.org.uk/pdffiles/crossedcomplexclass.pdf; I don't think they say "Quillen pair", but they prove what is needed in section 6.) -The derived functor of $N$ is evidently not fully-faithful. In fact (Ara, http://arxiv.org/abs/1206.2945), the homotopy category of simply connected strict $\infty$-groupoids is equivalent to $\mathcal{D}_{\geq2}(Ab)$, and any realization functor takes such objects to the evident products of Eilenberg-MacLane spaces. There are many more maps between Eilenberg-MacLane spaces than come from the derived category, so no realization functor can be fully faithful. - -It's still reasonable to ask if strictness is a structure on an $\infty$-groupoid (e.g., if $\Re$ is monadic in the $\infty$-categorical sense). For that matter, it's reasonable to ask if "weakness" is a structure on a strict $\infty$-groupoid (e.g., if the $\infty$-categorical left adjoint to $\Re$, modelled by $\pi$ in the Brown-Higgins paper, is comonadic). These seem like interesting questions. -Note: in the above, "strict $\infty$-groupoid" means a strict $\infty$-category in which all $k$-morphisms have inverses. There is also a more general notion of "quasistrict $\infty$-groupoid", in which $k$-morphisms have weak inverses. I gather less is known about these; see the paper by Ara referenced above.<|endoftext|> -TITLE: Does seeing beyond the course you teach matter? The case of linear algebra and matrices -QUESTION [33 upvotes]: This question is indeed very important for me. Thus I hope you bear with my subjective explanations for a few minutes. I am an "excellent" lecturer, at least according to course evaluation forms filled by students. More often than not, I use the so-called problem method in the courses I teach, and I advocate a particular philosophy of student-centered teaching. Yet, when I evaluate myself, something bothers me. As a professional mathematics educator, the best I can do is to help my students to learn the concepts and the techniques of the course internally, i.e. bounded to the syllabus of the course. -What if I could see beyond the course? What if I was an active mathematician who indeed works with those concepts and techniques, and knows a more advanced and perhaps more general version of those ideas? I was faced with these questions years ago when people started to compare my teaching with the teaching of a mathematician who is indeed an excellent "traditional" lecturer. To my own view, in a sense he could give to his students "more", since he could also see beyond the course. I had forgotten the whole issue until the current term; for the first time I am teaching a course in linear algebra and matrices for mathematics undergraduate students. That excellent colleague of mine is not around now (!), but the question is badly with me: - -If I could see beyond the course what ideas (concepts, techniques, - theorems, proofs, problems) would I stress more? - -To keep the question suitable for MO, please do not "argue", and just give one piece of concrete advice to a person who now teach to potentially some of your future colleagues! -PS. In this paper (Moore and Less; PRIMUS) you may find the story of the course that the comparison mentioned above started with. - -REPLY [23 votes]: In my opinion, what you should stress in a course on linear algebra depends more on what the particular students in your class want and/or need, and less on what you can "see beyond the course." However, since you asked this on MathOverflow, you are presumably asking for some insight into how professional mathematicians think about linear algebra, so I will try to address that question. -I would say that one the main hallmarks of those who have truly mastered linear algebra is that they can see how linear algebra is applicable in situations where the less well-trained do not. They are able to detect the presence of the "abstract structure" of linear algebra lying under the surface, even when it is not immediately evident from the statement of the problem. -Here are some examples. - -Sound waves can be decomposed into a weighted sum of pure tones. "Weighted sum" signals "linear algebra" to the cognoscenti. It doesn't matter that what you're adding together are functions and not finite sequences of numbers, and it doesn't matter that there are infinitely many possible pure tones. What matters is that you can take weighted sums, and that there is a precise sense in which different pure tones are "orthogonal" to each other. That means that linear algebra is applicable, and the concepts of eigenvalues and eigenvectors (or eigenfunctions) are applicable. -The Netflix Prize competition asked for an algorithm to recommend new movies based on your ratings of movies you've already seen. Where's the linear algebra? Start by writing down a large matrix with rows representing people, columns representing movies, and entries representing ratings. Experienced mathematicians know that the biggest singular values of this matrix capture most of the relevant information in it, and provide a good start to constructing the desired algorithm. -An old Putnam problem asked whether two matrices $A$ and $B$ with the property that $ABAB=0$ must also satisfy $BABA=0$. The obvious approach is to start playing around with examples, and there's nothing wrong with that. However, a more insightful approach is to build an abstract vector space with the basis $e_\emptyset, e_A, e_{BA}, e_{ABA}, e_{BABA}$ and define the linear transformations $Ae_S := e_{AS}$ and $Be_S := e_{BS}$, where $S$ is any string of $A$'s and $B$'s, $AS$ and $BS$ denote concatenation, and $e_{S} = 0$ if $S$ is not one of the strings $\emptyset$, $A$, $BA$, $ABA$, $BABA$. This is admittedly a very clever proof and even professional mathematicians might not think of it right away, but this example underlines the power of understanding that anything can be used as the basis of a vector space, even strings of symbols. -Suppose you have a large system of polynomial equations in $x$, $y$, and $z$, containing equations such as $xyz + 4x^2y - z^3 + 7 = 0$ and $y^2z^2 - xyz + 3 = 0$ and many others. At first glance we might think that linear algebra does not help here because we have variables multiplied together, and multiplication is nonlinear. However, if we have enough equations, and if the same terms appear often enough (in this example, $xyz$ appears in both equations), then we might be able to solve the system by using linear algebra, by treating each term as a separate variable and think of the system as a giant system of linear equations in a much larger number of variables. This might seem like a hopelessly optimistic approach, but in fact it is the basis for a general technique for solving systems of polynomial equations. Again, my point is that with a practiced eye, you can learn to see an entity such as $xyz$ not only as a product of three variables, but as a basis vector in a very large vector space. - -These examples may not translate directly into useful material for your teaching. However, I do believe that they give a good taste of how mathematicians think about linear algebra. They have internalized what "linear structure" means in the abstract and are able to detect it everywhere, to their advantage. Ideally, one would like to train students to think the same way. Of course, that may be more easily said than done.<|endoftext|> -TITLE: Sum of Difference of anti-diagonal matrix elements -QUESTION [7 upvotes]: Let $A \in \mathbb{R}^{n \times n}$, with elements $a_{ij}$ -What conditions on $A$ are required for the following to be true? -There exists some vector $x \in \mathbb{R}^n_+$, $x \neq 0$ such that for all $i=1\dots n$, -$$\sum_{j=1}^n x_j(a_{ij} - a_{ji}) \geq 0$$ -Obviously if $A$ is symmetric this is true. I believe it may be true for all $A$ but I haven't been able to prove it. -(sorry really wasn't sure what to title this post) -I've tried the usual Farkas' Lemma-style trick but the problem is self-dual - -REPLY [3 votes]: The original question asks if we can find a nonzero vector $x \ge 0$ such that $(A-A^T)x \ge 0$, of equivalently, for a skew-symmetric (aka antisymmetric) matrix $M$, the claim is that there always exists an $0 \neq x \ge 0$ such that $Mx \ge 0$. Chasing Gerry Myerson's suggestion above, here is a link to a proof of this claim by invoking Farkas' Lemma. -Lemma 4.5 here.<|endoftext|> -TITLE: The relationship between the dilogarithm and the golden ratio -QUESTION [6 upvotes]: Among the values for which the dilogarithm and its argument can both be given in closed form are the following four equations: -$Li_2( \frac{3 - \sqrt{5}}{2}) = \frac{\pi^2}{15} - log^2( \frac{1 +\sqrt{5}}{2} )$ (1) -$Li_2( \frac{-1 + \sqrt{5}}{2}) = \frac{\pi^2}{10} - log^2( \frac{1 +\sqrt{5}}{2} )$ (2) -$Li_2( \frac{1 - \sqrt{5}}{2}) = -\frac{\pi^2}{15} - log^2( \frac{1 +\sqrt{5}}{2} )$ (3) -$Li_2( \frac{-1 - \sqrt{5}}{2}) = -\frac{\pi^2}{10} - log^2( \frac{1 +\sqrt{5}}{2} )$ (4) -(from Zagier's The Remarkable Dilogarithm) -where the argument of the logarithm on the right hand side is the golden ratio $\phi$. The above equations all have this (loosely speaking) kind of duality, and almost-symmetry that gets broken by the fact that $Li_2(\phi)$ fails to make an appearance. Can anyone explain what is the significance of the fact that $\phi$ appears on the right, but not on the left? Immediately one can see that the arguments on the lefthand side of (2)-(4) are related to $\phi$ as roots of a polynomial, but what other meaning does this structure have? - -REPLY [2 votes]: Let $K$ be the field of real algebraic numbers. Denote by $B_2(K)_\mathbb Q$ the pre-Bloch group of raional numbers. It is quotient of free vector generated by $\{x\}_2, x\in K\backslash\{0,1\}$ by Abel five tern relation. There is a map -$$\delta_2\colon B_2(K)_\mathbb Q\to \Lambda^2 K^\times\otimes\mathbb Q$$ -which is given by the formula $\{x\}_2\mapsto x\wedge (1-x)$. This map is well-defined. -It follows from some number of results(Borel theorem, Suslin theorem etc.). That this map is isomorphism. From the other side there is a conjecture that all the relations between values of dilogarithms comes from Abel five-term relations. So practical application is as follows: If you have some element of the form $x=\sum\limits_{i=1}^nn_iLi_2(x_i)$ and you want to know whether this element can be expessed via elementary functions, you need to apply $\delta_2$ to the element $\widetilde x=\sum\limits_{i=1}^n n_i\{x_i\}_2$ and to see whether it zero or not. -Now in your case $x=Li_2\left(\dfrac{3-\sqrt{5}}{2}\right)$ then $\delta_2(\widetilde x)=\left(\dfrac{3-\sqrt{5}}{2}\right)\wedge \left(\dfrac{\sqrt{5}-1}{2}\right)$. -But this is equal to zero because $\left(\dfrac{\sqrt{5}-1}{2}\right)^2=\left(\dfrac{3-\sqrt{5}}{2}\right)$. -So we see that behind your identity there is some formula from algebraic number theory. I think the same computation can be done with other formulas.<|endoftext|> -TITLE: Where does Segal's category come from? -QUESTION [8 upvotes]: Segal's category $\Gamma$ is the skeleton of the category $\text{FinSet}_{\ast}$ of pointed finite sets. It is used to write down $\Gamma$-spaces, which are functors $\Gamma \to \text{Top}$ satisfying some conditions, and which model infinite loop spaces. I would like to be able to tell myself a story about this category which would explain in some sense why one might have come up with it as a candidate to be part of a delooping machine. -For comparison, here is the analogous story about $\text{FinSet}$: equipped with disjoint union, it is the free symmetric monoidal category on a commutative monoid. (I guess all of my stories are universal properties.) Since infinite loop spaces are in particular supposed to be like homotopy coherent commutative monoids I can see how one might have come up with $\text{FinSet}$ as a candidate to model infinite loop spaces, but not $\text{FinSet}_{\ast}$. -Riffing off of the above, it seems like $\text{FinSet}_{\ast}$, equipped with wedge sum, is the free symmetric monoidal category on something like a "copointed" commutative monoid; that is, a commutative monoid together with a map $\varepsilon : M \to 1$. The idea is that $\text{FinSet}_{\ast}$ can equivalently be thought of as the category of sets and partial functions, and throwing in a map $\varepsilon : M \to 1$ lets us model partial functions by using $\varepsilon$ to throw away the points at which a partial function isn't defined. -Why is this, and not $\text{FinSet}$, a reasonable category to use to model infinite loop spaces? (The inclusion of $\varepsilon : M \to 1$ is particularly strange because in any cartesian monoidal category, such as $\text{Top}$, it is unique because $1$ is the terminal object.) I guess $\varepsilon$ is needed to get an inclusion of $\Delta^{op}$ into $\Gamma$ so we can define the geometric realization of a $\Gamma$-space, but now I don't understand why there should be such an inclusion; the universal properties don't suggest it. The augmented simplex category $\Delta_a$ is the free monoidal category on a monoid, so the universal properties suggest instead a monoidal functor $\Delta_a \to \Gamma$. - -REPLY [4 votes]: If we can agree that it is natural to use $\Delta$ to encode categories and their homotopical generalisations, then I don't think it is farfetched to view $\Gamma$ as the analogous gadget to encode commutative monoids and their homotopical generalisations. -So let's think a little about $\Delta$ first. The category $\Delta$ has objects in bijection to $\mathbb{N}$, so that a presheaf $C$ on $\Delta$ is a graded set; for every $n \in \mathbb{N}$ we want to think of $C(n)$ as a set of chains of composable morphisms of length $n$. The morphisms in $\Delta$ are supposed to encode things like composition and associativity; so for example there should be some morphism $[1] \to [2]$ which gives composition of pairs of morphisms. But how do we know that we have all necessary morphisms in $\Delta$? For every $n \in \mathbb{N}$ denote by $\Delta_n$ the free category generated by a graph consisting of a chain of $n$ directed arrows. For a small category $\mathscr{C}$ the set of chains of morphisms of length $n$ is canonically isomorphic to $\mathbf{Cat}(\Delta_n, \mathscr{C})$. We could now hope that there is a functor $\Delta \to \mathbf{Cat}, \; [n] \mapsto \Delta_n$ which induces the fully faithful functor $\mathbf{Cat} \hookrightarrow \widehat{\Delta}, \; \mathscr{C} \mapsto ([n] \mapsto \mathbf{Cat}(\Delta_n,\mathscr{C}))$. Indeed, if we choose this functor to be fully faithful, which completely determines the structure of $\Delta$, then we obtain such a functor $\mathbf{Cat} \hookrightarrow \widehat{\Delta}$, the classical nerve functor. -If we play the same game with monoids we obtain the $\Gamma$ category. If we view any commutative monoid as a category, then we see that we might again suppose that $\Gamma$ has objects in bijection to $\mathbb{N}$; for any number $n \in \mathbb{N}$ we call the corresponding object $(n)$. Now a commutative monoid has exactly one object, so without knowing anything else, we may already assume that $(0)$ gets mapped to $\{*\}$. Like for $\Delta$, let's try to view $\Gamma$ as a subcategory of $\mathbf{AbMon}$, the category of commutative monoids, in order to again obtain a nerve functor. The objects have to be $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$ for all $n \in \mathbb{N}$. The morphisms of $\Gamma$ are then completely determined by where the generators of $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$ are sent. To get an appropriate subcategory of $\mathbf{AbMon}$ we will only consider those morphisms which send generators to elements, which themselves are sums of distinct generators (this will be important below). If we label and keep track of only the generators, we rediscover Segal's original description of $\Gamma$. The induced nerve functor is again fully faithful, and it is now straightforward how to view any commutative monoid as a $\Gamma$-set. Furthermore, we note that there is a unique morphism $(n) \to (0)$ for all $n \in \mathbb{N}$, so that any presheaf on $\Gamma$ taking $(0)$ to $\{ * \}$ factors through the category of pointed sets $\mathbf{Set}_*$, and we might as well consider contravariant functors $\Gamma^{\mathrm{op}} \to \mathbf{Set}_*$. -Finally, let us show that there is a canonical functor $\Delta \to \Gamma$, so that every $\Gamma$-set has an underlying simplicial set. This is simple: by again viewing any commutative monoid as a category, we simply send the generators of $\Delta_n$ to the generators of $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$ for every $n \in \mathbb{N}$. In this last step we are implicitly using that we have labelled the generators of $\mathbb{N} \oplus \underbrace{\cdots}_{n \times} \oplus \mathbb{N}$, but this is not a problem: Let us denote the nerve functor $\mathbf{Cat} \hookrightarrow \widehat{\Delta}$ by $N$, and the nerve functor $\mathbf{AbMon} \hookrightarrow \mathbf{Cat}(\Gamma^{\mathrm{op}}, \mathbf{Set}_*)$ by $N_\Gamma$. Then for any monoid $M$, any functor $\Delta \to \Gamma$ induced by any labelling will take $N_\Gamma(M)$ to $N(M)$ (viewing $\mathbf{AbMon}$ as a subcategory of $\mathbf{Cat}$). I don't, however, know whether arbitrary presheaves $\Gamma^{\mathrm{op}} \to \mathbf{Set}_*$ get mapped to the same simplicial set under the different functors induced by different labellings.<|endoftext|> -TITLE: What is Tropicalization, and how is it applied -QUESTION [20 upvotes]: My question is: -What is Tropicalization, how is it done, and what are some basic applications of it? -motivation -I am interested especially in how questions about enumerative algebraic geometry translate into the tropical world and how information obtained in the tropicalization reflects back on the original question. -This rather general question is similar to earlier questions I asked about integrable systems and about categorification. There are some relevant Wikipedia articles but while interesting I don't find them sufficient, and several point of views from MO participants can be enlightening and lead to a good resource. -A related MO question Why tropical geometry? - -REPLY [5 votes]: One may also consult the book-in-progress by Diane Maclagan and Bernd Sturmfels -at this link: -Introduction to Tropical Geometry. The first five chapters are available as of -this summer.<|endoftext|> -TITLE: Absent 2nd order terms in deformation quantization of Poisson manifolds -QUESTION [5 upvotes]: I am reading Kontsevich' famous paper on deformation quantization of Poisson manifolds. In section 1.4.2 on page 4 he gives the general formula for the star product associated to a Poisson structure on $\mathbb{R}^n$ up to and including order 2. Now I have noticed that in the second order term, terms of the form (for example) $\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})\partial_j(f)\partial_l(g)$, and I do not understand why. This term would correspond to the following graph: - -I.e., there is an edge going from 1 to 2 and an edge going from 2 to 1; all such terms would be of the form $\partial_j(\alpha^{i\cdot})\partial_i(\alpha^{j\cdot})\times\text{derivatives of $f$ and $g$}$, and none of them occur in his formula in section 1.4.2. As far as I can see, such graphs would be admissible under the rules on page 5 in the paper. I have yet to find a way to evaluate the weight associated to this kind of graphs, but at least the integrand of the weight for the graph above is nonzero. So my question is: - -Why do terms of this form not contribute to the second order term of the star product? - -REPLY [6 votes]: Using a remark in this recent paper by Thomas Willwacher as a hint, I found out. The problem is solved by the following: there exists a gauge transformation $D$ that transform the extra term away. To see this, we first need the second order term of a gauge transformed star product: If $\star = I + \hbar^nB_n$ and $D = I + \hbar^nD_n$, and $f\star'g = D^{-1}(Df\star Dg)$ then -$$ -B_2'(f,g) = B_2(f,g)+B_1(f,D_1g)+B_1(D_1f,g)-D_1(B_1(f,g))+D_1(f)D_1(g)- - D_1(g D_1f)-D_1(f D_1g)+D_1(D_1(f g))+g D_2f+fD_2g-D_2(f g), -$$ -where, of course, $B_2'$ is the second order term of $\star'$. -Setting $D_1 = 0$ reduces this to -$$ -B_2'(f,g) = B_2(f,g)+g D_2f+f D_2g-D_2(f g) = B_2(f,g) + d^H(D_2)(f\otimes g) -$$ -where $d^H$ is the Hochschild differential. (Also, $B_1' = B_1$.) -Now comes the crucial part: the term in the question $\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})\partial_j\otimes\partial_l =: A$ is $d^H$-exact: indeed, $A = d^H\left(-\frac12\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})\partial_j\partial_l\right)$. So if we set $D_2$ to minus the expression in the brackets, then, $A$ no longer occurs in $B_2'$ while the other terms of $B_2$ occur unmodified in $B_2'$. -Finally, expressions of the form $\partial_k(\alpha^{ij})\partial_i(\alpha^{kl})f\partial_j\partial_l(g)$ (corresponding to the diagram in the question but with both downwards arrows now pointing to $g$) also cannot occur in $B_2$ because $B_n$ is zero for any $n$ when one of its arguments is constant, and such a term would violate this rule. -Thus, the star product in section 1.4.2 in Kontsevich's paper does not equal his general formula up to order 2; instead, they are equivalent. -(By the way, the weight of the term corresponding to the diagram is, including multiplicity from similar graphs, equal to $-\frac16$ according to this paper by Giuseppe Dito. In particular, it is nonzero.)<|endoftext|> -TITLE: Conjugacy classes of PGL(3,Z) -QUESTION [9 upvotes]: We know that every $2\times 2$ matrix in $PGL(2, \mathbb{Z})$ of order $3$ is conjugate to the matrix $$ \left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \end{array} \right) $$. -I am interested in finding out to what extent this holds for $3\times 3$ integer invertible matrices. -In other words how many conjugacy classes of order 3 matrices in $PGL(3, \mathbb{Z})$ are there? - -REPLY [8 votes]: The finite subgroups of $GL_3(\mathbb{Z})$ are known in the literature: -$\qquad$ Tahara: On the finite subgroups of $GL(3,\mathbb{Z})$. Nagoya Math. J. 41(1971), 169-209. -In particular Proposition 3 states that there are exactly two non-conjugate subgroups of order three. Representants are -$$U_1=\langle \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}\rangle, \qquad U_2=\langle\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}\rangle$$ -Added: In $PGL_3(\mathbb{Z})=GL_3(\mathbb{Z})/\langle -I\rangle$ there are also exactly two conjugacy classes of subgroups of order 3 and representants are $\bar{U}_1,\bar{U}_2$. -For, let $V_i=\langle x_i\rangle \le G := GL_3(\mathbb{Z}),i=1,2$ be subgroups of order 3. If $\bar{V}_i$ are conjugate in $\bar{G} :=PGL_3(\mathbb{Z})$ then there is $g \in G$ s.t. $x_2=(\pm I)gx_1^kg^{-1} (k=1,2)$ and hence $(\pm I)^3=I$ and $x_2=gx_1^kg^{-1}$, i.e. the $V_i$ are conjugated in $G$. -Conversely, let $\bar{V}$ be a subgroup of $\bar{G}$ of order three. It's preimage in $G$ has order 6. Hence there is $V \le G$ of order three that maps to $\bar{V}$ and by the above $V$ is conjugated to some $U_i$.<|endoftext|> -TITLE: Condition for two matrices to share at least one eigenvector? -QUESTION [21 upvotes]: Suppose that I have two matrices $A$ and $B$, and I want them to share a common eigenvector $x$. For simplicity let's just assume that the eigenvalue associated with $x$ is $1$ for both matrices, so $Ax=x$ and $Bx=x$. Is there a simple condition on $A$ and $B$ which is both necessary and sufficient for this to occur? -Edit: loup blanc's answer covers the case where the eigenvalues are not known, which is generally much more interesting than the case I was asking about, which is when both eigenvalues are 1. The solution to my case is just that $\ker(A-I) \cap \ker(B-I) \ne 0$. I would still be interested if someone found an even simpler condition which is equivalent to this, though. - -REPLY [10 votes]: I'm not sure if this is what you're looking for, but Donu Arapura's website contains notes on algebraic geometry. Starting on page 24 of the notes, he proves that the set of pairs of matrices $(A,B)\in M_n(k)\times M_n(k)$ (with $k$ algebraically closed) having a common eigenvector is Zariski closed in $M_n(k)\times M_n(k)$. -In particular, for each $n$, there is a system of polynomials in the entries of two arbitrary $n\times n$ matrices with the property that all of the polynomials vanish iff the matrices share a common eigenvector. So, once you know these polynomials, you have an "simple, easily computable" way to check whether or not a pair of matrices shares a common eigenvector. -Unfortunately, he mentions that starting with $n=3$, the computation of what these polynomials is "painfully slow" on his computer. (When $n=2$, is turns out the polynomial is $\det(AB-BA)$). He also mentions, and later proves, that for $n > 2$, the system of polynomials must consist of more than one polynomial. Thus, $\det(AB-BA)=0$ is necessary for sharing a common eigenvalue, but not sufficient as other answers have shown.<|endoftext|> -TITLE: Exponential tail bounds without the moment generating function -QUESTION [7 upvotes]: As an exercise, I thought I would try to prove some classical Chernoff bounds without ever using the moment generating function, but then found myself getting stuck in certain places. Before I state my question, let me give some background. -$\mathbf{Background}$: -Consider for example the fact that there exist constants $c,c'$ such that for any $\lambda > 0$ -\begin{equation} -(*)\ \mathop{\mathbb{P}}_\sigma\left(\sum_i \sigma_i x_i > \lambda\right) \le c \cdot e^{-c' \lambda^2/\|x\|_2^2} -\end{equation} -where $\sigma_1,\sigma_2,\ldots$ are independent Rademachers. One way to prove (*) for small $\lambda$ is to say -$$ -\mathop{\mathbb{P}}_\sigma\left(\sum_i \sigma_i x_i > \lambda\right) = \mathop{\mathbb{P}}_\sigma\left(e^{t\sum_i \sigma_i x_i} > e^{t\lambda}\right) < e^{-t\lambda} \cdot \mathop{\mathbb{E}}_\sigma e^{t\sum_i \sigma_i x_i} = e^{-t\lambda}\cdot \prod_i \mathop{\mathbb{E}}_{\sigma_i} e^{t\sigma_i x_i} = e^{-t\lambda}\cdot \prod_i \frac 12(e^{-tx_i} + e^{tx_i}) \le e^{-t\lambda} \prod_i \frac 12(1 + e^{tx_i}t^2x_i^2/2) \le e^{-t\lambda} \prod_i e^{e^{tx_i}t^2x_i^2/2} \le e^{c\cdot t^2\|x\|_2^2/2 - t\lambda} -$$ -as long as $t$ (which will depend on $\lambda$) is small enough so that the $e^{tx_i}$ term is at most $c$. Then one optimizes and sets something like $t \sim \lambda/\|x\|_2^2$. This proof uses the moment generating function (MGF), i.e. we talk about $\mathbb{E} e^{tX}$ in the proof for our random variable $X = \sum_i \sigma_i x_i$. At various points in the proof we also use analytic properties of the exponential function, e.g. based on Taylor's theorem. -A proof that avoids the MGF and Taylor's theorem is to work with moments. Define $\|X\|_p = (\mathbb{E}|X|^p)^{1/p}$ as usual and let $g_1,g_2,\ldots$ be i.i.d. gaussians of mean $0$ and variance $1$. Then -$$ -\|\sum_i \sigma_i x_i\|_p = \sqrt{\frac{\pi}2} \|\mathop{\mathbb{E}}_g \sum_i \sigma_i |g|_i x_i\|_p \le \sqrt{\frac{\pi}2} \|\sum_i \sigma_i |g_i| x_i\|_p = \sqrt{\frac{\pi}2} \|\sum_i g_i x_i\|_p \lesssim \|x\|_2\cdot \sqrt{p} -$$ -where the last inequality used $2$-stability of the gaussian (also the $\sqrt{\pi/2}$ can be avoided but never mind about that). Then we use that -$$ -\mathop{\mathbb{P}}_\sigma\left(\sum_i \sigma_i x_i > \lambda\right) \le \mathop{\mathbb{P}}_\sigma\left(|\sum_i \sigma_i x_i|^p > \lambda^p\right) < \lambda^{-p} \cdot \|\sum_i \sigma_i x_i\|_p^p -$$ -and set $p\sim \lambda^2 / \|x\|_2^2$. This proof is of the form I'm looking for: no MGF, no analytic tricks/Taylor's theorem. (It also works for all $\lambda$.) -$\mathbf{My\ question}$: -One formulation of the Chernoff bound is the following. There is some constant $c>0$ such that the following holds. Say $X_1,\ldots,X_n$ are independent and each bounded by $K$ in magnitude almost surely. Let $X = \sum_i X_i$ with $\mu = \mathbb{E} X$ and $\sigma^2 = \mathbb{E}(X - \mathbb{E}X)^2$. Then for all $\lambda > 0$, -$$ -(**)\ \mathop{\mathbb{P}}\left(|X - \mu| > \lambda\right) \lesssim \max\left\{e^{-c\lambda^2/\sigma^2}, \left(\frac{\sigma^2}{\lambda K}\right)^{c\lambda/K}\right\} . -$$ -One way to prove the above is to use the moment generating function and analytic tricks (see Exercise 3 at http://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/). My question is, similar to the first example, can we prove (**) simply using moment methods while avoiding the MGF and Taylor's theorem/analytic tricks? e.g. using combinations of Jensen's inequality, symmetrization, triangle inequality on $\|\cdot \|_p$, and other such things. -Note that (**) seems to be equivalent to the statement that for all $p \ge 1$, -$$ -\|X - \mu\|_p \lesssim \sigma\sqrt{p} + K\frac{p}{\ln(epK^2/\sigma^2)} . -$$ - -REPLY [3 votes]: Ok, I figured out the answer to my own question, though without the $\ln(ep K^2/\sigma^2)$ in the denominator. I'll see whether I can get that later (or maybe someone else sees it?). We can assume without loss of generality that $\mathbb{E} X_i = 0$ for each $i$ (otherwise apply what follows with the random variables replaced by $X_i - \mathbb{E} X_i$). The trick is to use that $\|\langle g,x \rangle\|_p$ is monotonically increasing as a function of $\|x\|_2$ (this follows by 2-stability of the gaussian). Let $\sigma_i$ be i.i.d. Rademacher and $g_i$ be i.i.d. normal. -\begin{align} -\|\sum_i X_i\|_p &= \|\sum_i X_i - \mathbb{E} X_i\|_p\\ -&\le \|\sum_i (X_i - X_i')\|_p\\ -&\le 2\|\sum_i \sigma_i X_i\|_p\\ -&= 2\sqrt{\frac{\pi}{2}}\|\mathbb{E}_g\sum_i \sigma_i |g_i| X_i\|_p\\ -&\lesssim \|\sum_i \sigma_i |g_i| X_i\|_p\\ -&= \|\sum_i g_i X_i\|_p (1)\\ -&\lesssim \sqrt{p} \|(\sum_i X_i^2)^{1/2}\|_p\\ -&\le \sqrt{p} \|\sum_i X_i^2\|_p^{1/2}\\ -&= \sqrt{p} \|\sum_i X_i^2 - \mathbb{E}\sum_i X_i^2 + \sum_i X_i^2\|_p^{1/2}\\ -&\le \sigma\sqrt{p} + \sqrt{p}\|\sum_i X_i^2 - \mathbb{E}\sum_i X_i^2\|_p^{1/2}\\ -&\lesssim \sigma\sqrt{p} + \sqrt{p}\|\sum_i g_i X_i^2\|_p^{1/2}\\ -&= \sigma\sqrt{p} + \sqrt{p}\left(\mathbb{E}_X\|X\|_\infty^p\left(\mathbb{E}_g|\sum_i g_i X_i \frac{X_i}{\|X\|_\infty}|^p\right)\right)^{1/2p}\\ -&\le \sigma\sqrt{p} + \sqrt{pK}\left(\mathbb{E}_X\mathbb{E}_g|\sum_i g_i X_i \frac{X_i}{\|X\|_\infty}|^p\right)^{1/2p}\\ -&\lesssim \sigma\sqrt{p} + \sqrt{pK}\left(\mathbb{E}_X\mathbb{E}_g|\sum_i g_i X_i|^p\right)^{1/2p} (2)\\ -&= \sigma\sqrt{p} + \sqrt{pK}\|\sum_i g_i X_i\|_p^{1/2} (3) -\end{align} -Inequality (2) used that $\|\langle g,x\rangle\|_p$ is monotonically increasing as a function of $\|x\|_2$, as mentioned above. Now notice we have bounded (1) in terms of its square root in (3). Thus $\|\sum_i g_i X_i\|_p^{1/2}$ is at most the larger root of the associated quadratic equation, and this gives us the bound -$$ -\|\sum_i g_i X_i\|_p \lesssim \sigma\sqrt{p} + Kp -$$ -as desired. -I tried to be careful and write all the details to make it clear, but much of the above argument can be compressed once you're familiar with the right bag of tricks (symmetrization, gaussian comparison principle, and triangle inequality in the right places).<|endoftext|> -TITLE: Does this cross-product norm inequality hold? -QUESTION [6 upvotes]: I asked this on MSE over a month ago, but the one answer I got doesn't seem to work. - -Let $\times$ denote the cross-product. $\;$ Is it the case that - - -For all unit vectors $\:\mathbf{x}\hspace{.01 in},\hspace{-0.03 in}\mathbf{y}\hspace{-0.03 in},\hspace{-0.02 in}\mathbf{z}\:$ in $\mathbf{R}^3$, $\;\;\;\; \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{z}\hspace{.02 in}\right|\right| \:\: \leq \:\: \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{y}\hspace{.02 in}\right|\right| \hspace{.02 in}+\hspace{.02 in}\left|\left|\hspace{.03 in}\mathbf{y} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{z}\hspace{.02 in}\right|\right| \;\;\;\;\;$. - -? - -(If yes, then $\;\;\; \langle [\mathbf{x}]\hspace{.01 in},\hspace{-0.03 in}[\hspace{.02 in}\mathbf{y}] \rangle \: \mapsto \: \left|\left|\hspace{.03 in}\mathbf{x} \hspace{-0.03 in}\times \hspace{-0.03 in}\mathbf{y}\hspace{.02 in}\right|\right| \;\;\;$ defines a nice metric on the projective plane.) - -REPLY [6 votes]: The alleged inequality is true. -Using the definition of the (vector) cross product, for unit vectors $x, y \in R^3$ the original claim boils down to showing that -\begin{equation*} - d(x,y) = \sin(\cos^{-1}(x^Ty)) -\end{equation*} -is a distance. But it turns out that $d(\cdot,\cdot)$ is actually is a distance on unit vectors in $R^n$. -Showing the above function to be a distance is equivalent to showing that -\begin{equation*} - \sqrt{1 - (x^Ty)^2} -\end{equation*} -is a distance over the set of unit vectors in $R^n$. But this is a known result of Wang and Zhang, "A trace inequality for unitary matrices", AMM, 1994, pp. 453-455. -EDIT: Also worth comparing with the recent determinant based distance described in this MO question (which suggests that actually after suitable normalization, other symmetric matrix functions should also generate distances, e.g., $d(X,Y) := [1-\text{trace}(\wedge^k(X^TY))]^{1/2}$.<|endoftext|> -TITLE: Is this error in this paper of Langlands fixable? -QUESTION [14 upvotes]: The FQS criterion for the Virasoro algebra was discovered by Friedan, Qiu and Shenker (1), but the mathematicians found their proof insufficient, so that, FQS (2) and Langlands (3), published in the same time a complete proof. -There is an error in the paper of Langlands : (3) lemma 7b page 148 (see also here page 7) : - -$p=2$, $q=1$, $m=2$, $h_{p,q}(m)= \frac{5}{8}$, $M=4$ -$p=4$, $q=1$, $m=3$, $h_{p,q}(m)= \frac{7}{2}$, $M=13$ -... - -yield case $(B)$, but $(p,q) \ne (1,1)$ and $m \ngtr q+p-1$. -Remark : In fact, we need to distinguish between $q \ne 1$ and $q=1$, not between -$(p,q) \ne (1,1)$ and $(p,q)=(1,1)$). -This lemma is used in the rest of the paper. - -Question: Is there a way to fix the rest of the paper ? - -Remark : This way was used by Sauvageot ((4) lemma 2 (ii) p 648), without fixing. -References : - (1) D. Friedan, Z. Qiu, S. Shenker, Conformal invariance, unitarity, and critical exponents in two dimensions. Phys. Rev. Lett. 52 (1984), no. 18, 1575--1578. -(2) D. Friedan, Z. Qiu, S. Shenker, Details of the nonunitarity proof for highest weight representations of the Virasoro algebra. Comm. Math. Phys. 107 (1986), no. 4, 535--542. - -(3) R. P. Langlands, On unitary representations of the Virasoro - algebra. Infinite-dimensional Lie algebras and their applications - (Montreal, PQ, 1986), 141--159, World Sci. Publ., Teaneck, NJ, 1988. - -(4) F. Sauvageot, Représentations unitaires des super-algèbres de Ramond et de Neveu-Schwarz. Comm. Math. Phys. 121 (1989), no. 4, 639--657. - -REPLY [8 votes]: This morning I just discovered these corrigenda of K. Iohara and Y. Koga (in which my name is cited in acknowledgement). In fact, three years ago, I have contacted K. Iohara (author, with Y. Koga, of the book Representation Theory of the Virasoro Algebra) about this error (but I didn't know they fixed it). -I do not yet read these corrigenda into details, but I guess it's ok (I hope). -Another unfixed error for the Ramond algebra : -As I said in remark, F. Sauvageot gave a proof 'à la Langlands' of the FQS criterion for the superVirasoro algebras ($N=1$): the Neveu-Schwarz algebra and the Ramond algebra. -Seven years ago, during my PhD, I have discovered this error in this paper of Langlands, reproduced in this paper of Sauvageot, so I decided to write a proof "à la FQS" of the FQS criterion for the Neveu-Schwarz and Ramond algebras. -In fact I discovered that this criterion runs for the Neveu-Schwarz case, -but not for the Ramond case : -We can prove lemma 4.19 p 22 of this paper (Neveu-Schwarz case) -thanks to the curves $h=h^{m}_{pp}$, but the Ramond case doesn't have these curves ! -So it's ok for the Neveu-Schwarz case (and I guess Sauvageot's paper is fixable in this case by using the corrigenda above), but for the Ramond case, -the FQS criterion gives the discrete series plus some representations of -charge $c_{m}$ with $m$ non-interger ! -Sketch of fixing : -For excluding these last representations we can use an argument of -fusion: -$$ (R) \boxtimes (R) \to (NS) $$ -It's known that the fusion of two representations of the Ramond algebra (R), in the discrete series at central charge $c_{m}$, give a (discrete series) representation of the Neveu-Schwarz algebra (NS) at the same central charge $c_{m}$ (the Ramond algebra is given by a twisted vertex module over the vertex operator algebra of the Neveu-Schwarz algebra). But we know -that the Neveu-Schwarz case doesn't contain such representations at central -charge $c_{m}$ with $m$ non-interger, the result follows.<|endoftext|> -TITLE: Physical Disturbances to Computations -QUESTION [6 upvotes]: In this paper, page 7 (160 of the Journal), Fig 3, there is a particularly amusing (not to the authors!) caption: -"... On April 1 of year 2 in the $S_0$ experiment, the computer was hit by a cosmic ray or some other disturbance which caused improper numbers to be stored in the ground temperature array." -Note that the temperature array they are referring to is NOT a physical probe; the entire experiment was done numerically on what's probably a room sized computer for the duration of at least a year. The authors even plot a dashed corrected curve in the figure to try and account for this. -Are there any other documented cases of numerical errors arising from physical phenomena affecting older computers? I'm aware of stuff like the Pentium FDIV bug and the collection from this question but, I was wondering of examples more in line with the above caption from seemingly ridiculous sources like cosmic rays? - -REPLY [3 votes]: Yes. This problem is particularly salient for satellite systems, which are less shielded than regular terrestrial electronics. As transistors get smaller, particles are ever more likely to cause logical errors. -You have to design HW that is robust to these sorts of particles: Wikipedia:Radiation Hardening -I have friends that have gotten PhDs in this field: http://www.isde.vanderbilt.edu/wp/tag/radiation-effects/ -I think it's pretty cool stuff.<|endoftext|> -TITLE: Who is Petrov of the Petrov-Galerkin method? -QUESTION [14 upvotes]: I was not able to find the origin of the name Petrov in the Petrov-Galerkin method for the numerical approximation of PDEs. -Wikipedia refers to a certain Alexander G. Petrov, but it is still not clear who he was: on the Internet I found two persons called Alexander G. Petrov, both born during the 40's, but I think that the Petrov I am looking for should be born before, since Boris Galerkin lived in 1871-1945. -Can you suggest me any bibliographic resource to consult? -[EDIT: The Wikipedia page I was referring to now contains the correct name of Petrov] - -REPLY [3 votes]: I've recently found a very interesting SIAM review article [1] regarding the history of the Petrov-Galerkin method. The figure of G.I. Petrov seems to have contributed with a simple, but significant improvement to the Galerkin method: - -In Russia, Petrov [2] proposed to use approximation spaces different from the test spaces, which led to the now-called Petrov–Galerkin family of methods. - -I highly suggest the reading of [1], since the history of the Petrov-Galerkin method is particularly thrilling! - -[1] Gander, M.J., and Wanner, G., "From Euler, Ritz, and Galerkin to Modern Computing", SIAM Review 54.4 (2012), pp. 627-666. -[2] G. I. Petrov, "Application of Galerkin’s method to the problem of stability of flow of a viscous fluid", J. Appl. Math. Mech., 4 (1940), pp. 3–12 (in Russian).<|endoftext|> -TITLE: Abelian groups injective over their endomorphism -QUESTION [6 upvotes]: Let $M$ be an abelian group and let $R = \mbox{End}_\Bbb{Z}(M)$. Under what conditions (on $M$), $_RM$ is injective!? - -REPLY [7 votes]: It seems that such modules where classified by Fred Richman and Elbert A. Walker, in Modules over PIDs that are injective over their endomorphism rings, Ring Theory (Proc. Conf., Park City, Utah, 1971), p.363-372, Academic Press, N.-Y., 1972. -The paper is available on E. A. Walker's website.<|endoftext|> -TITLE: Tetrahedron insphere iteration -QUESTION [24 upvotes]: I know that iterating the following incircle construction approaches an equilateral triangle in the limit: -      - -Starting with any triangle $T$, one forms $T'$ by connecting the three points -of tangency of the circle inscribed inside $T$. -Does the analogous process for tetrahedra approach a regular tetrahedron? -And the same question may be asked for simplices in $\mathbb{R}^d$. -A reference would be appreciated—Thanks! - -REPLY [11 votes]: There is likely a theorem here that, for almost every start tetrahedron (left column below), -very quickly the iterates form an approximate $2$-cycle. Below shows three random -tetrahedra and five iterates each (rescaled): - - -Below I plot the distance between the incenter and the circumcenter of $25$ random tetrahedra, as the process is iterated and rescaled at each step. This strongly supports -fedja's conjecture. If there are exceptions, they are not common. - - -Here is a sample of what the inscribed and circumscribed spheres look like (with red & green centers converging):<|endoftext|> -TITLE: Is this inverted integral transform valid? -QUESTION [5 upvotes]: I have the following transform: -$$F(y) = \int_{0}^{\infty} y\exp{\left[-\frac{1}{2}(y^2 + x^2)\right]} I_0\left(xy\right)f(x)\;\mathrm{d}x$$ -with the following conditions: - -$f(x)$ and $F(y)$ must be real and positive (they are continuous probability distributions) -$x,y$ must be real and positive (they are magnitudes) - -@Carlo suggested the following approach. Re-arrange such that we have: -$$\frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} = \int_{0}^{\infty} \left(\frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]}\right) I_0\left(xy\right)x\;\mathrm{d}x$$ -Now define: -$$ G(y) = \frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} \quad\quad g(x) = \frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]}$$ -We can now write: -$$G(y) = \int_{0}^{\infty} g(x) I_0\left(xy\right)x\;\mathrm{d}x$$ -Make the substitution $y = iq$ along with the fact that $I_0(ixq) = J_0(-xq) = J_0(xq)$ and we have: -$$G(iq) = \int_{0}^{\infty} g(x) J_0\left(xq\right)x\;\mathrm{d}x$$ -This is a Hankel transform, which has an inverse: -$$g(x)=\int_{0}^{\infty}G(iq)J_{0}(xq)q\;\mathrm{d}q$$ -Substituting in $q = \frac{1}{i}y$ and $\mathrm{d}q = \frac{1}{i}\mathrm{d}y$ we obtain: -$$g(x)= -\int_{0}^{i\infty}G(y)I_{0}(xy)y\;\mathrm{d}y$$ -Substituting out $G(y), \;g(x)$: -$$\frac{f(x)}{x}\exp{\left[-\frac{1}{2} x^2\right]} = -\int_{0}^{i\infty} \frac{F(y)}{y}\exp{\left[\frac{1}{2}y^2\right]} I_{0}(xy)y\;\mathrm{d}y$$ -and finally solve for $f(x)$: -$$f(x) = -\int_{0}^{i\infty} x\exp{\left[\frac{1}{2}(y^2 + x^2)\right]} I_0\left(xy\right)F(y)\;\mathrm{d}y$$ - -Is this inversion valid given the aforementioned conditions? -How can I perform this integral numerically? - -I'm directly measuring $F(y)$ (by normalising the histogram of a large set of values) and so I need to calculate $f(x)$ by doing this integral numerically. However I don't understand how to do this when the data is all real but the integral is over the imaginary axis. Can anyone help? -Thank you! - -REPLY [6 votes]: define $G(y)=y^{-1}\exp(y^2/2)(Tf)(y)$ and $g(x)=x^{-1}\exp(-x^2/2)f(x)$, then you seek the solution to the integral equation -$$G(iq)=\int_{0}^{\infty}g(x)J_{0}(xq)xdx$$ -This is a Hankel transform. The inverse is -$$g(x)=\int_{0}^{\infty}G(iq)J_{0}(xq)qdq$$<|endoftext|> -TITLE: $SL_2(\mathbf{Z},8\mathbf{Z})$ differs from $E_2(\mathbf{Z},8\mathbf{Z})$. Has this result appeared in the literature? -QUESTION [9 upvotes]: I know a proof that the congruence subgroup $SL_2(\mathbf{Z},8\mathbf{Z})$ differs from its subgroup $E_2(\mathbf{Z},8\mathbf{Z})$, but can't find this fact in the literature. Does anyone know a citation for this? - -REPLY [11 votes]: In fact, this holds for any $N$ such that $X=\mathbb{H}^2/PSL_2(\mathbb{Z},N\mathbb{Z})$ has genus $>0$. The point is that the normal subgroup generated by parabolics in $SL_2(\mathbb{Z},N\mathbb{Z})$ will generate the kernel of the map obtained by filling in the cusps of the Riemann surface $X$ to get a closed Riemann surface $\hat{X}$. If $\hat{X}$ has genus $\geq 1$, then the kernel will have infinite index (more generally, if there's torsion, the filling will be an orbifold, which will have non-trivial fundamental group if it has genus $>0$ or more than two orbifold cone points, or two cone points of the same order). In fact, $X$ has genus $=5$. -Actually, the group I've described will contain $E_2(\mathbb{Z},N\mathbb{Z})$, and is the normalizer in $SL_2(\mathbb{Z})$ of a primitive upper triangular element in $SL_2(\mathbb{Z},N\mathbb{Z})$ . If you take the normalizer in $SL_2(\mathbb{Z},N\mathbb{Z})$, you get an even small subgroup obtained by the kernel of filling in a single cusp of $X$. In this case, one need only show that there are at least 3 cusps, or the genus is $>0$, or there is some combination of elliptic points and cusps which makes the filling have non-trivial fundamental group.<|endoftext|> -TITLE: Kontsevich integral : state of the art -QUESTION [13 upvotes]: The Kontsevich integral is known to be a universal Vassiliev invariant. -It is still an open question whether it is a complete knot invariant, i.e. whether it distinguishes a given knot from all other knots up to ambient isotopy and mirror image. -I am curious to know if there are some references giving the state of the art for this important question. Have some clues for a possible answer emerged in the last few years? - -REPLY [10 votes]: I don't think that there has been a tremendous amount of progress in understanding the Kontsevich Invariant of a knot in the last decade or so. It appears that essential new ideas may be needed in order to answer the fundamental questions. - Quantum Invariants: A study of knots, $3$-manifolds, and their sets, T. Ohtsuki, World Scientific 2002. This is still more-or-less current, I think, and is a basic reference. Also T. Ohtsuki's Problems on invariants of knots and 3–manifolds. -A more recent textbook reference is: Introduction to the Vassiliev Knot Invariants by Chmutov, Duzhin, and Mostovoy, Cambridge University Press, 2012. It contains a detailed discussion of the Kontsevich Invariant of a knot, including developments of the last decade such as work that has been done on the loop expansion and on the 2-loop polynomial in particular. -With regard to clues for a final answer, not to toot my own horn, but Ohtsuki and I showed that 3 loop Vassiliev invariants do not distinguish a knot from its reverse (this is trivial for lower loop degree): - -Vanishing of 3-loop Jacobi diagrams of odd degree, D. Moskovich and T. Ohtsuki, J. Comb. Theory, Ser. A 114(5): 919-930 (2007). -If this were true for general loop degree, then that would imply that there exist prime unoriented knots which cannot be distinguished by the Kontsevich invariant (this is a result of Kuperberg). My personal suspicion (without solid support) is that this is false for general loop degree, and indeed that the Kontsevich invariant does separate (unoriented) knot types.<|endoftext|> -TITLE: Does every set of reals contain a measure-zero set of the same cardinality? Does it contain a meager set of the same cardinality? -QUESTION [27 upvotes]: This question arises from an issue in my post on Ashutosh's excellent question on Restrictions of the null/meager ideal. -Question 1. Does every set of reals contain a measure-zero subset -of the same cardinality? -In other words, if $A\subset\mathbb{R}$, is there a measure-zero set $B\subset A$ with $|B|=|A|$? Is this assertion at least consistent? Does it follow from the -continuum hypothesis? Does it follow from some other cardinal characteristic hypothesis? -In the intended application, what is needed is that the assertion is consistent with the additivity number for measure being equal to the continuum. Is this consistent? Can anyone prove the consistency of the failure of the property? -Similarly, in the case of category rather than measure: -Question 2. Does every set of reals contain a meager subset of the same cardinality? -And similarly, is this statement consistent? Does it follow from CH or other cardinal characteristic hypotheses? Is it consistent with the additivity number for the meager ideal being large? Can anyone show the consistency of the failure of the property? -The questions arise in my post on Ashutosh's question, where I had proposed as a solution idea the strategy of a back-and-forth construction of length continuum, where the domain and target remain measure-zero during the course -of the construction. But in order for this strategy to succeed, we seem to need to know in the context there that one may extend a given -measure-zero set inside another non-measure-zero set to a larger -measure-zero set with the same cardinality (and the same with meagerness). I had thought at first -that this should be easy, but upon reflection I am less sure about it, and so I ask these questions here. - -REPLY [4 votes]: I think that I have a compelete answer (but see PS below). -Let $\mathcal{M}$ (repectively, $\mathcal{N}$) be the ideal of meager (respectively, Lebesgue null) -sets in $\mathbb{R}$. -Let $\kappa$ be an infinite cardinal, and $\mathcal{I}$ be an ideal of sets in $\mathbb{R}$. -A set $X\subseteq\mathbb{R}$ is $\kappa$-$\mathcal{I}$-Luzin -if $|X|\ge\kappa$ but $|X\cap I|<\kappa$ for every set $I\in\mathcal{I}$. -In other words, if it has no subset of cardinality $\kappa$ in $\mathcal{I}$. -A set $X\subseteq\mathbb{R}$ is Luzin (respectively, Sierpi\'nski) -if it is $\aleph_1$-$\mathcal{M}$-Luzin (respectively, $\aleph_1$-$\mathcal{N}$-Luzin). -cov$(\mathcal{I})$ is the minimal cardinality of a cover of the real line by elements of -the ideal $\mathcal{I}$. The following folklore result is easy. -Theorem: If cov$(\mathcal{I})={}$cof$(\mathcal{I})$ then there is a -cov$(\mathcal{I})$-$\mathcal{I}$-Luzin set. -Thus, the answer to Question 1 is "No" if cov$(\mathcal{N})={}$cof$(\mathcal{N})$, -and similarly for Question 2. Both hypotheses follow from CH, MA, etc. -Lemma. Assume that there is a $\kappa$-$\mathcal{I}$-Luzin set $X$. -Then cf$(\kappa)\le{}$cov$(\mathcal{I})$. -Indeed, by moving to a subset of $X$ we may assume that $|X|=\kappa$. -Let $\{I_\alpha : \alpha<\text{cov}(\mathcal{I})\}$ be a family in $\mathcal{I}$ covering the real line. -Then $$X=\bigcup_{\alpha<\text{cov}(\mathcal{I})} X\cap I_\alpha.$$ -Thus, $\kappa$ is a union of $\text{cov}(\mathcal{I})$ sets of cardinality smaller than $\kappa$. -Theorem (Judah, Shelah): It is consistent that non$(\mathcal{M})={}$non$(\mathcal{N})=\aleph_1$ -and there are no Luzin or Sierpi\'nski sets. -Moreover, they prove that a Miller real kills the ground model Luzin and Sierpi\'nski set, -and iterate with countable support iteration. -The consistency result you hoped for follows. -Theorem. It is consistent that every set of reals contains a measure-zero subset and -a Lebesgue null subset of the same cardinality. -Indeed, in Miller's model, non$(\mathcal{M})={}$non$(\mathcal{N})=\aleph_1$ and the continuum is $\aleph_2$. It follows from the Cihon diagram that $\text{cov}(\mathcal{M})=\text{cov}(\mathcal{N})=\aleph_1$ -there. By the above lemma, there are no $\aleph_2$-$\mathcal{M}$-Luzin or $\aleph_2$-$\mathcal{N}$-Luzin -sets there. By the Judah--Shelah Theorem, there are also no $\aleph_1$-$\mathcal{M}$-Luzin or $\aleph_1$-$\mathcal{N}$-Luzin sets there. -PS. I am writing this after an all-night work, I hope I do not mess up things.<|endoftext|> -TITLE: Quotient of $Z[x_1,...,x_n]$ by a maximal ideal is a finite field -QUESTION [5 upvotes]: I am seeing the proof of the Ax-Groethendieck theorem from commutative algebra and I have a problem. How can I prove that if $x_1,...,x_n$ are complex numbers and $I$ is a maximal ideal of $\mathbb{Z}[x_1,...,x_n]$, the quotient $\mathbb{Z}[x_1,...,x_n]/I$ is a finite field? -Thanks. -Infinite fields, finite fields, and the Ax-Grothendieck theorem - -REPLY [7 votes]: Use that ${\mathbb Z}$ is a Jacobson ring, so that according to the generalized Nullstellensatz, the inverse image of the maximal ideal $(0)$ in $k={\mathbb Z}[x_1,\ldots, x_n]/I$ is also a maximal ideal, i.e. an ideal of the form $(p)$ (with $p\neq 0$). This means the image of ${\mathbb Z}$ in $k$ is finite, so $k$ is a finite extension of a finite field, hence finite. - -REPLY [5 votes]: Let $R$ be a finitely generated integral domain (over $\mathbb Z$) and let $I$ be a maximal ideal of $R$. We show $R/I$ is a finite field. Let $K$ be an algebraic closure of $R/I$. Let $p$ be the characteristic of $K$. Suppose $n$ elements generate $R$. Then we can write $R/I= \mathbb Z[x_1,\ldots x_n]/(f_1,\ldots, f_m)$. Therefore, the first order sentence $\phi=\exists y_1,\ldots, y_n[f_1(y_1,\ldots,y_n)=0\wedge\cdots \wedge f_m(y_1,\ldots,y_n)=0]$ -is true in $K$. There are two cases. -If $p>0$, then since the first order theory of algebraically closed fields of characteristic $p$ is complete we have $\overline {\mathbb F_p}\models \phi$. It follows that $R/I$ embeds in $\overline {\mathbb F_p}$ and hence is finite being finitely generated. -Next suppose $p=0$. By completeness the theory of an algebraically closed field of characteristic $0$ models $\phi$. It is a standard consequence of the compactness theorem of first order logic that there is an algebraically closed field of prime characteristic that models $\phi$. The previous paragraph now shows $R/I$ is a finite field.<|endoftext|> -TITLE: Existence of CM Newforms in Level p -QUESTION [9 upvotes]: If $p$ is a prime and $k \geq 2$ is an even integer, what can we say about the existence of CM forms in the space $S_k^\text{new}(\Gamma_0(p))$? If it helps at all, I'm specifically interested in the primes $2,3,5$ and $7$, and I'm looking at spaces without a character. - -REPLY [15 votes]: No such newform exists. -If $\psi$ is a Groessencharacter of an imaginary quadratic field $K$, the level of the associated newform is $N_{K/\mathbf{Q}}(\mathfrak{f}) \cdot \operatorname{disc}(K/\mathbf{Q})$, where $\mathfrak{f}$ is the conductor of $\psi$. So a CM-type newform of prime level would have to come from an imaginary quadratic field $K$ of prime discriminant (which automatically rules out levels 2 and 5), and a Grossencharacter of $K$ with conductor 1, and infinity-type $(1-k, 0)$. Since you insist on small primes, $K$ will have class number 1, so such a Grossencharacter is unique if it exists and sends a fractional ideal to the $(k-1)$-st power of a generator. So we are in trouble if $k-1$ is not a multiple of the order of the unit group of $K$. But the unit group always contains $\pm 1$, so $k$ must be odd. -This also proves that an $f$ of level $\Gamma_1(p)$ exists when $p=3$ and $k = 1 \bmod 6$, or when $p = 7$ and $k$ is odd (and its Nebentypus is the quadratic character modulo $p$). I'll leave you to work out what the story is for larger prime levels.<|endoftext|> -TITLE: primes represented by an indefinite binary quadratic form -QUESTION [13 upvotes]: Suppose I have a form $$ f(x,y) = a x^2 + b x y + c y^2, $$ with $a,b,c$ integers, $\gcd(a,b,c)=1$ and $\Delta = b^2 - 4 a c > 0,$ but $\Delta \neq n^2$ for any integer $n.$ -Do there exist (positive) primes $p,q$ such that $f$ integrally represents $p$ and $-q?$ -I have most books on quadratic forms of which I've ever heard, but I do not see this. I will keep checking. For positive forms we have Chebotarev density and infinitely many primes. And, of course, this may also follow from Chebotarev density; if so, is there a cheaper way as well? -I would like to allow odd $b,$ but I suppose it does not really matter: $f$ represents a superset of the numbers represented by $f(x,2y), f(2x,y), f(x-y,x+y),$ one of which is primitive. -I WILL FIND OUT - -REPLY [8 votes]: Meyer (Über einen Satz von Dirichlet, Crelle 103 (1888)) proved that a primitive -binary quadratic form with nonsquare discriminant represents infinitely many primes that -lie in any given compatible residue class modulo a given integer N. He did so only for forms with even middle coefficient, and did not address the signs of primes, but both of these require only minor modifications. -This answer makes me wonder whether someone actually proved, with Dirichlet's methods, that a -finite number of forms with coprime discriminants simultaneously represent infinitely many primes - by Chebotarev, this has to be true.<|endoftext|> -TITLE: Can transcedence degree be defined for arbitrary ring homomorphism? -QUESTION [8 upvotes]: Fix a homomorphism $f:A\rightarrow B$. -Choose $\{b_1,\dots,b_n\}$, $\{b'_1,\dots,b'_m\}$ subsets of elements in $B$. Suppose that $B$ is algebraic over $f(A)[b_1,\dots,b_n]$ and $\{b_1,\dots,b_n\}$ are algebraically independent over $f(A)$. Suppose that $\{b'_1,\dots,b'_m\}$ satisfies the same condition. Does it imply $m=n$? -I checked the proof for an abstract dependence relation in Jacobson's Basic Algebra II (see 3.6). The condition $x m$. -Let $B'$ be the $A$-subalgebra of $B$ generated by the $b_i$ and the $b_j'$. Then, $A$, $C$ and $C'$ are subrings of $B'$. Moreover, the ring extension $B' / C$ is integral (since a subring of a ring integral over $C$ is still integral over $C$), and similarly the ring extension $B' / C'$ is integral. Hence, we can replace the ring $B$ by $B'$ without anything else changing. So let us WLOG assume that $B = B'$. Thus, $B$ is the $A$-subalgebra of $B$ generated by the $b_i$ and the $b_j'$. Hence, $B$ is a finitely generated $C'$-algebra and integral over $C'$, therefore a finitely generated $C'$-module. Choose any finite generating set of the $C'$-module $B$, and throw in the element $1$. The resulting set is finite and generates $B$ as a $C'$-module. Denote this set by $S$. Let $U$ be the $A$-submodule of $B$ spanned by $S$. -For every polynomial algebra $\mathfrak C$ over $A$ (such as $C$ and $C'$) and any nonnegative integer $i$, let $\mathfrak C_{\leq i}$ denote the $A$-submodule of $\mathfrak C$ consisting of polynomials of degree $\leq i$. Recall that -(1) $\mathfrak C_{\leq i}$ is a free $A$-module of rank $\dbinom{i+r}{r}$, where $r$ is the number of indeterminates of the polynomial algebra $\mathfrak C$. -Also, $\mathfrak C_{\leq a+b} = \mathfrak C_{\leq a} \mathfrak C_{\leq b}$ for all nonnegative integers $a$ and $b$. -We have $C' = \bigcup\limits_{p\geq 0} C'_{\leq p}$ and thus $C' U = \bigcup\limits_{p\geq 0} C'_{\leq p} U$ (since $C'_{\leq 0} \subseteq C'_{\leq 1} \subseteq C'_{\leq 2} \subseteq \cdots$). -Since $B$ is generated by $S$ as a $C'$-module, while $U$ is the $A$-linear span of $S$, we have $B = C' U = \bigcup\limits_{p\geq 0} C'_{\leq p} U$. -Let $\ell = \left|S\right|$ (we know that $S$ is finite) and $S = \left\lbrace s_1, s_2, ..., s_{\ell} \right\rbrace$. Then, the $A$-module $U$ is spanned by the $s_j$ with $j$ ranging over $\left\lbrace 1, 2, ..., \ell \right\rbrace$ (since the $A$-module $U$ is spanned by $S$). -There are only finitely many products $b_i s_j$ with $i \in \left\lbrace 1, 2, ..., n\right\rbrace$ and $j \in \left\lbrace 1, 2, ..., \ell \right\rbrace$, and thus there exists some nonnegative integer $p$ such that these products all lie in $C'_{\leq p} U$ (since $B = \bigcup\limits_{p\geq 0} C'_{\leq p} U$). Fix such a $p$. Then, -(2) $b_i U \subseteq C'_{\leq p} U$ for all $i \in \left\lbrace 1, 2, ..., n\right\rbrace$ -(because the $A$-module $U$ is spanned by the $s_j$ with $j$ ranging over $\left\lbrace 1, 2, ..., \ell \right\rbrace$, so that the $A$-module $b_i U$ is spanned by the $b_i s_j$ with $j$ ranging over $\left\lbrace 1, 2, ..., \ell \right\rbrace$). -Now, every nonnegative integer $N$ satisfies -(3) $C_{\leq N} U \subseteq C'_{\leq pN} U$. -Why is this so? Indeed, in order to prove (3), we need to show that $b_{i_1} b_{i_2} ... b_{i_k} U \subseteq C'_{\leq pN} U$ for every $k\leq N$ and any $k$-tuple $\left(i_1,i_2,...,i_k\right) \in \left\lbrace 1,2,...,n\right\rbrace^k$ (because such products $b_{i_1} b_{i_2} ... b_{i_k}$ span $C_{\leq N}$ as an $A$-module). In order to prove this, it is clearly enough to show that $b_{i_1} b_{i_2} ... b_{i_k} U \subseteq C'_{\leq pk} U$ for every $k\leq N$ and any $k$-tuple $\left(i_1,i_2,...,i_k\right) \in \left\lbrace 1,2,...,n\right\rbrace^k$ (because if $k\leq N$, then $pk \leq pN$ an thus $C'_{\leq pk} U \subseteq C'_{\leq pN} U$). This is shown by induction over $k$, using the induction step -$b_{i_1} b_{i_2} ... b_{i_k} U = b_{i_1} b_{i_2} ... b_{i_{k-1}} \underbrace{b_{i_k} U}_{\subseteq C'_{\leq p} U\ \text{(by (2))}} \subseteq b_{i_1} b_{i_2} ... b_{i_{k-1}} C'_{\leq p} U$ -$= C'_{\leq p} \underbrace{b_{i_1} b_{i_2} ... b_{i_{k-1}} U}_{\subseteq C'_{\leq p(k-1)} U\ \text{(by induction hypothesis)}} \subseteq \underbrace{C'_{\leq p} C'_{\leq p(k-1)}}_{= C'_{\leq p + p(k-1)} = C'_{\leq pk}} U = C'_{\leq pk} U$. -Armed with (3), we can close in for the kill. Recall that $1 \in S \subseteq U$. Hence, every nonnegative integer $N$ satisfies -(4) $C_{\leq N} \subseteq C_{\leq N} U \subseteq C'_{\leq pN} U$ (by (3)). -Since $C_{\leq N}$ is a free $A$-module of rank $\dbinom{N + n}{n}$ (by (1)), there exists an $A$-module isomorphism from $A^{\dbinom{N + n}{n}}$ to $C_{\leq N}$. Due to (4), this yields that there exists an $A$-module injection from $A^{\dbinom{N + n}{n}}$ to $C'_{\leq pN} U$. Denote this injection by $\phi$. -But $U$ is generated by $S = \left\lbrace s_1, s_2, ..., s_{\ell} \right\rbrace$ as $A$-module. In other words, $U = \sum\limits_{q=1}^{\ell} s_q A$. Hence, $C'_{\leq pN} U = \sum\limits_{q=1}^{\ell} C'_{\leq pN} s_q$ is a quotient of the $A$-module $\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$. Since $C'_{\leq pN}$ is a free $A$-module of rank $\dbinom{pN + m}{m}$ (by (1)), the direct sum $\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$ is a free $A$-module of rank $\ell \dbinom{pN + m}{m}$. In other words, there exists an $A$-module isomorphism from $A^{\ell \dbinom{pN + m}{m}}$ to the direct sum $\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$. Hence, there exists an $A$-module surjection from $A^{\ell \dbinom{pN + m}{m}}$ to the $A$-module $C'_{\leq pN} U$ (because the $A$-module $C'_{\leq pN} U$ is a quotient of this direct sum). Denote this surjection by $\psi$. -The $A$-module $A^{\dbinom{N + n}{n}}$ is free and thus projective. Hence, the $A$-module injection $\phi : A^{\dbinom{N + n}{n}} \to C'_{\leq pN} U$ lifts (through the $A$-module surjection $\psi : A^{\ell \dbinom{pN + m}{m}} \to C'_{\leq pN} U$) to an $A$-module map $\chi : A^{\dbinom{N + n}{n}} \to A^{\ell \dbinom{pN + m}{m}}$ satisfying $\phi = \psi \circ \chi$. This $A$-module map $\chi$ is clearly injective again. -Since $n > m$, the polynomial $\dbinom{x + n}{n} \in \mathbb Q\left[x\right]$ has a higher degree than the polynomial $\ell \dbinom{px + m}{m} \in \mathbb Q\left[x\right]$. Hence, the former polynomial (having a positive leading coefficient) grows faster than the latter. Thus, there exists a nonnegative integer $N$ such that $\dbinom{N + n}{n} > \ell \dbinom{pN + m}{m}$. Fix this $N$. -But there is a well-known fact saying that if $a$ and $b$ are two nonnegative integers satisfying $a > b$, and if there is an injective $A$-module map $\gamma : A^a \to A^b$, then $A$ is the trivial ring. (This is equivalent to theorem (2) in Fred Richman's Nontrivial uses of trivial rings, Proceedings of the American Mathematical Society, vol. 103, no. 4, 1988, pp. 1012-1014, and part of Corollary 5.11 in Keith Conrad's Exterior powers.) Applying this fact to $a = \dbinom{N + n}{n}$, $b = \ell \dbinom{pN + m}{m}$ and $\gamma = \chi$, we conclude that $A$ is the trivial ring, qed. $\blacksquare$ -Remark. We can use the above argument to prove a slightly stronger result: - -Theorem 2. Let $A$ be a subring of a commutative ring $B$. Let $n$ and $m$ be two nonnegative integers such that $n > m$. Assume that $C = A\left[b_1, b_2, ..., b_n\right]$ and $C' = A\left[b'_1, b'_2, ..., b'_m\right]$ be two subrings of $B$ (with all $b_i$ and all $b'_j$ lying in $B$, obviously) such that $b_1$, $b_2$, ..., $b_n$ are algebraically independent over $A$ (that is, $C$ is the polynomial ring in $b_1$, $b_2$, ..., $b_n$ up to isomorphism). Assume that the ring extension $B / C'$ is integral. Then, $A$ is the trivial ring (that is, $A=0$). - -Proof of Theorem 2. This proof is similar to the proof of Theorem 1 above, but the following changes need to be made: - -We no longer need to WLOG assume that $n > m$, because $n > m$ is already given by the assumptions. -Our definition of $\mathfrak{C}_{\leq i}$ no longer gives us $A$-modules $C'_{\leq i}$, because $C'$ is not necessarily a polynomial algebra. We thus need to define $C'_{\leq i}$ differently. For any nonnegative integer $i$, we define $C'_{\leq i}$ to be the $A$-submodule of $C'$ consisting of all elements that can be written as polynomials of degree $\leq i$ in the generators $b'_1, b'_2, \ldots, b'_m$. This $A$-module $C'_{\leq i}$ might not be free, but - -(5) it is finitely generated with $\dbinom{i+m}{m}$ generators -(namely, the $\dbinom{i+m}{m}$ monomials of degree $\leq i$ in the generators $b'_1, b'_2, \ldots, b'_m$). We can easily see that $C'_{\leq a+b} = C'_{\leq a} C'_{\leq b}$ for all nonnegative integers $a$ and $b$. Again, $C' = \bigcup\limits_{p\geq 0} C'_{\leq p}$ holds. - -Our definition of $\psi$ needs to be modified from the place on where I claim that $C'_{\leq pN}$ is a free $A$-module of rank $\dbinom{pN + m}{m}$. In fact, the $A$-module $C'_{\leq pN}$ is not necessarily free of rank $\dbinom{pN + m}{m}$ anymore. But it is finitely generated with $\dbinom{i+m}{m}$ generators (according to (5)), and thus is a quotient of a free $A$-module of rank $\dbinom{pN + m}{m}$. Hence, the direct sum $\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$ is a quotient of a free $A$-module of rank $\ell \dbinom{pN + m}{m}$. In other words, there exists an $A$-module surjection from $A^{\ell \dbinom{pN + m}{m}}$ to the direct sum $\bigoplus\limits_{q=1}^{\ell} C'_{\leq pN}$. Hence, there exists an $A$-module surjection from $A^{\ell \dbinom{pN + m}{m}}$ to the $A$-module $C'_{\leq pN} U$ (because the $A$-module $C'_{\leq pN} U$ is a quotient of this direct sum). Denote this surjection by $\psi$. $\blacksquare$ - -Theorem 2 has a well-known corollary: - -Corollary 3. Let $A$ be a subring of a commutative ring $B$. Let $n$ and $m$ be two nonnegative integers such that $n > m$. Let $b_1, b_2, \ldots, b_n$ be $n$ algebraically independent elements of $B$. Furthermore, let $b'_1, b'_2, \ldots, b'_m$ be $m$ elements of $B$ such that $B = A\left[b'_1, b'_2, ..., b'_m\right]$. Then, $A$ is the trivial ring (that is, $A=0$). - -Proof of Corollary 3. Define a subring $C$ of $B$ by $C = A\left[b_1, b_2, ..., b_n\right]$. The ring extension $B / B$ is clearly integral. Hence, Theorem 2 (applied to $C' = B$) yields that $A$ is the trivial ring. This proves Corollary 3. $\blacksquare$<|endoftext|> -TITLE: Nontrivially nontrivial automorphisms of $P(\omega_1)/$fin -QUESTION [14 upvotes]: Velickovic proved (Theorem 4.1 of OCA and automorphisms of $\mathcal{P}(\omega)/\mathrm{fin}$) that, assuming - -OCA (Open Coloring Axiom) and -$\rm MA_{\aleph_1}$, - -every (Boolean algebra) automorphism of $\mathcal{P}(\omega_1)/\mathrm{ fin}$ is trivial, i.e. for every such automorphism $\varphi$ there is a function $e : \omega_1\to \omega_1$ such that for all $a\subseteq \omega_1$, $\varphi[a] = [e''a]$. -On the other hand, if there is a nontrivial automorphism $\psi$ of $\mathcal{P}(\omega)/\mathrm{fin}$ (as there are in any model of CH), one can easily construct a nontrivial automorphism $\varphi$ of $\mathcal{P}(\omega_1)/\mathrm{fin}$ by just copying $\psi$ on $\omega$ and the identity on $\omega_1\setminus \omega$: -$$ \varphi[x] = \psi[x\cap \omega]\vee [x\setminus \omega] $$ -Of course one can replace $\omega$ with any countable set $a$, and the identity with any trivial automorphism of $\mathcal{P}(\omega_1 \setminus a)/\mathrm{fin}$. But this is somewhat unsatisfying; all of these automorphisms seem to be nontrivial "for trivial reasons." Hence the following question: - -Is there, consistently, an automorphism of $\mathcal{P}(\omega_1)/\mathrm{fin}$ which is nontrivial on every cocountable set? - -REPLY [4 votes]: EDIT: The following does not work, as pointed out in the comments. I do not know how to make $\varphi$ a homomorphism. - -The answer is yes, but for trivial reasons. Identify $\omega_1$ with $\omega_1\times \omega$. Every automorphism $\psi$ of $P(\omega)/fin$ induces an automorphism of $P(\omega_1\times \omega)/fin$ as follows: - -If $ A=\bigcup_{i\in \omega_1} \{i\}\times A_i$, then let $\varphi(A) = \bigcup_{i\in \omega_1} \{i\}\times \psi(A_i)$. - -Note that $\varphi$ is well-defined on $P(\omega_1\times \omega)/fin$, and is nontrivial on any co-countable set.<|endoftext|> -TITLE: Crossed modules and degree-3 group cohomology -QUESTION [16 upvotes]: It is well known (see e.g. K. Brown, "Cohomology of groups") that a degree-3 cohomology class of a group G with coefficients in a module A can be thought of as an equivalence class of crossed modules, in the sense that every crossed module gives a double extension of G by A. Now, it is also well known that degree-3 cohomology of Z/n with coefficients in R/Z (with a trivial action of Z/n) is isomorphic to Z/n. My question is, how does one actually construct a crossed module corresponding to an element of Z/n? - -REPLY [15 votes]: Here is a general procedure to construct a crossed module $\mathcal{G} = (G_1 \stackrel{d}{\to} G_0)$ from a normalized 3-cocycle $\alpha \in Z^3(G; \mathbb{R}/\mathbb{Z})$ which is additive in the third variable. -First, we get a cocycle $\omega \in Z^2(G; G^{\ast})$, where $G^{\ast}$ is the Pontryagin dual of $G$ equipped with the trivial $G$-action, by $\omega(a,b)(c) = \alpha(a,b,c)$. We use this cocycle to get an extension -$$0 \to G^{\ast} \to G_0 \to G \to 0.$$ -We will construct the $G_0$-module $G_1$ as an extension of $G_0$-modules -$$0 \to \mathbb{R}/\mathbb{Z} \to G_1 \to G^{\ast} \to 0.$$ -Since $\mathbb{R} / \mathbb{Z}$ is divisible, this sequence splits as an exact sequence of abelian groups. Thus, $G_1 \cong \mathbb{R} / \mathbb{Z} \times G^{\ast}$ as abelian groups. The action of $G_0$ factors through an action of $G$: -$$a: G \times (\mathbb{R} / \mathbb{Z} \times G^{\ast}) \to \mathbb{R} / \mathbb{Z} \times G^{\ast}$$ -$$a(g, (\theta, \gamma)) = (\theta + \gamma(g), \gamma)$$ -The map $G_1 \stackrel{d}{\to} G_0$ is just the composite $G_1 \to G^{\ast} \to G_0$. -OK, let's specialize now to the case $G = \mathbb{Z} / n\mathbb{Z}$. We can realize the class in $H^3(\mathbb{Z} / n\mathbb{Z}; \mathbb{R}/\mathbb{Z})$ corresponding to $m \in \mathbb{Z} / n\mathbb{Z}$ by the normalized cocycle -$$\alpha(a, b, c) = \begin{cases} -0 & \text{if $a + b < n$,}\\ -cm & \text{otherwise.} -\end{cases}$$ (Here we are taking the image of the cocycle as living in $\mathbb{Z} / n\mathbb{Z}$ as a subgroup of $\mathbb{R}/\mathbb{Z}$, and we will likewise identify $\mathbb{Z} / n\mathbb{Z}$ with its Pontryagin dual by sending $a \in \mathbb{Z} / n\mathbb{Z}$ to $\frac{a \cdot -}{n} \in (\mathbb{Z} / n\mathbb{Z})^\ast$.) So explicitly, the group $G_0$ is the set $\mathbb{Z} / n\mathbb{Z} \times \mathbb{Z} / n\mathbb{Z}$ equipped with the operation -$$(a, b) \oplus (c, d) = \begin{cases} -(a + c, b + d) & \text{if $b + d < n$,}\\ -(a + c + m, b + d) & \text{otherwise.} -\end{cases}$$ -The group $G_1$ is just $\mathbb{R} / \mathbb{Z} \times \mathbb{Z} / n\mathbb{Z}$. The action of $G_0$ on $G_1$ is given by -$$(a, b) \rhd (\theta, c) = (\theta + \frac{bc}{n}, c).$$ -The map $G_1 \stackrel{d}{\to} G_0$ is the composite of projection onto the second factor followed by inclusion into the first.<|endoftext|> -TITLE: If G is a sequential topological group, must G x G be sequential? -QUESTION [13 upvotes]: Using standard definitions, the topological space $Y$ is sequential if for each nonclosed $A \subset Y$, there exists a convergent sequence $a_{1}$ , $a_{2}$,...$\rightarrow b$ -so that $a_{n} \in A$ but $b \notin A$. -Working in the topological category TOP, we assume the group $G$ is a sequential space, inversion is continuous, and group multiplication is continuous with respect to the standard product topology on $G \times G$. -Must $G \times G$ be a sequential space? -A `yes' answer would provide a sharp dividing line between sequential topological groups in TOP, and sequential topological groups in SEQ\TOP. -(SEQ is a category in which the standard product topology of $G \times G$ is refined to ensure $G \times G$ is sequential). - -REPLY [2 votes]: A simple example of sequential topological groups $G,H$ with non-sequential product $G\times H$: $G=\mathbb R^\omega$ and $H=\mathbb R^\infty=\lim \mathbb R^n$ be the direct limit of finite-dimensional Euclidean spaces. The group $G$ is metrizable and $H$ is a sequential $k_\omega$-space. The non-sequentiality of $G\times H$ follows from a result [Banakh, Taras; Zdomskyĭ, Lubomyr. The topological structure of (homogeneous) spaces and groups with countable $cs^*$-character. Appl. Gen. Topol. 5 (2004), no. 1, 25--48; MR2087279] saying that each sequential topological group $G$ of countable $cs^*$-character is either metrizable or contains an open $k_\omega$-subgroup. -We say that a topological space $G$ has countable $cs^*$-character if for every point $x\in X$ there is a countable family $\mathcal N$ of subsets of $X$ such that for every neighborhood $O_x\subset X$ and every sequence $(x_n)$ convergent to $x$ in $X$ there is a set $N\subset O$ in the family $\mathcal N$ containing infinitely many points of the sequence $(x_n)$. -The mentioned result of Banakh and Zdomskyy implies that for any sequential topological group $G$ of countable $cs^*$-character the square $G\times G$ is sequential (more precisely, $G\times G$ is metrizable or contains an open $k_\omega$-subgroup).<|endoftext|> -TITLE: Can we always make a strictly functorial choice of pullbacks/re-indexing? -QUESTION [10 upvotes]: $\newcommand{\C}{\mathbf{C}} \newcommand{\D}{\mathbf{D}}$ Let $\C$ be a category with pullbacks. Taking any choice of pullbacks gives us re-indexing functors $f^* \colon \C /Y \to \C/X$, and these will be functorial in $f$ up to natural isomorphism, in that $g^* \cdot f^* \cong (f \cdot g)^*$. However, these will usually not be strictly functorial in $f$; that is, $g^* \cdot f^*$ and $(f \cdot g)^*$ will not be literally equal. Strict functoriality also requires that $1_X^* = 1_{\C/X}$; while this typically does hold on the nose, it’s still not automatic. -My main question: Is there always some choice of pullbacks that make re-indexing strictly functorial? I believe the answer should be “no”, but I don’t know any counterexample. Even in the case of $\mathbf{Set}$, it’s not obvious whether there’s a choice that works. -An equivalent phrasing of the question is: can the codomain fibration $\mathrm{cod} \colon \C^\rightarrow \to \C$ be equipped with a splitting? It can always be replaced by an equivalent split fibration over $\C$; but splitting the codomain fibration itself seems hard. - -REPLY [3 votes]: Here's an argument (assuming global choice) that the codomain fibration will always be split in "natural" cases. Perhaps it's well-known, or perhaps it has an important shortcoming (such as the use of choice?). Or perhaps there's an error -- strictification is tricky! - -Claim: Let $C$, $D$ be categories with pullbacks, and suppose there is a functor $p: C \to D$ with the following properties: - -$p$ preserves pullbacks. -$p$ is an isofibration, and induces an isofibration on all slices (perhaps this is redundant?). -$p$ reflects identities (i.e. if $f: c \to c$ is an endomorphism in $C$, and $pf: pc \to pc$ is an identity, then so is $f$). -There is a cardinal $\kappa$ such that for every $f: x \to c \in C^{[1]}$, the isomorphism class $[f]$ of $f$ in the fiber category $p^{-1}(pf) \subseteq C/c$ has cardinality $\kappa$. - -The claim is that if $D$ admits strict reindexing, then so does $C$, and $p$ preserves the strict reindexing. - -Upshot: Most "natural" categories $C$ admit such a functor $p: C \to \mathsf{Set}$, with $\kappa$ being the cardinality of the universe (perhaps with an exception for objects over $\emptyset$, which often doesn't matter because there are few maps into $\emptyset$). I'm assuming that $\mathsf{Set}$ admits strict reindexing (right?). So most "natural" categories do as well. -EDIT: $\mathsf{Set}$ does indeed admit strict reindexing. To see this, as in the comment below, note that the fibration $\mathsf{Set}^{(-)} \to \mathsf{Set}$ admits a splitting, and the fibration $\mathsf{Set}/(-) \to \mathsf{Set}$ is isomorphic as a fibered caegory -- the fact that it's equivalent as a fibered category is standard, and each isomorphism class in each fiber category has the cardinality of the universe, in both fibrations, so an isomorphism of fibrations exists. So $\mathsf{Set}/(-) \to \mathsf{Set}$ also admits a splitting as desired. -So indeed, most "natural" categories admit a strict reindexing. Of course, it's completely undecideable to compute it because the choice function must solve the isomorphism problem! - -Proof of Claim: Let $()^\ast$ be a system of strict reindexing in $D$. Fix enumerations of the isomorphism classes of the fibers of $p$ by $\kappa$ (using condition (4) and choice). Now, if $\gamma : c' \to c$ is a morphism in $C$ and $f: x \to c \in C/c$, take an arbitrary pullback of $f$ along $\gamma$. Because $p$ is an isofibration (condition (2))and $p$ preserves pullbacks (condition (1)), we can correct this choice to lie over the canonical pullback $(p\gamma)^\ast(pf)$. Moreover, we can choose a representative which has the same index under our enumeration as $f$ does; let's call it $\gamma^\ast(f)$. It remains only to nail down the lift of the other leg of the pullback square, but any two choices differ by an automorphism of $\gamma^\ast(f)$ which lies over the identity on $(p\gamma)^\ast(pf)$, so by condition (3) this choice is uniquely determined. -It's now obvious that the lifted operation $()^\ast$ is strictly functorial, because it preserves the indexing by our enumerations.<|endoftext|> -TITLE: Two-sided bar construction -QUESTION [5 upvotes]: On page 4 of this paper by H. Abbaspour, the author defines the two-sided bar construction -$$B(A,A,A):=A\otimes T(s\bar{A})\otimes A$$ -of a differential graded algebra $(A,d_A)$ (over a field). -The definition of the differential $d=d_0+d_1$ on $B(A,A,A)$ is unclear to me. While $d_1$ seems to lower the wordlength on $T(s\bar{A})$ by $1$, $d_0$ seems to raise the degree (as a tensor product) of an element in $A\otimes T(s\bar{A})\otimes A$ by $1$. -As far as I understand, $(B(A,A,A),d)$ should be a chain complex, in fact it should give a free resolution of $(A,d_A)$ as an $(A\otimes A^{op},d_A\otimes 1+1\otimes d_A)$-module. -What is the grading on $B(A,A,A)$? Why does $d$ lower the degree by $1$? -Thanks to anyone who can shed some light on the two-sided bar construction. - -REPLY [6 votes]: I haven't looked at Abbaspour's paper, but here is what is going on, -in a bit greater generality. Let $N$ be a right, $M$ a left DG -$A$-module. Then $B=B(N,A,M)$ is defined and it is bigraded. The -grading with differentials that raise degree, which you apparently -have in mind, is a bit awkward, so regrade by $A_n = A^{-n}$ and -similarly for $M$ and $N$. Then we have $B_{p,q}$, where $p$ is -the homological degree (via $N\otimes \bar{A}^{p}\otimes M$ and $q$ -is the internal degree (add up the degrees of $n$, the $a_i$, and $m$ -of an element $n[a_1,\cdots,a_p]m$). There is a horizontal (or internal) -differential $d^h\colon B_{p,q} \to B_{p,q-1}$ given by the differentials on $N$, $A$, -and $M$ and there is a vertical (or homological) differential $d^v: B_{p,q}\to B_{p-1,q}$. -These commute. Now regrade by total degree, $B_n = \sum_{p+q=n} B_{p,q}$. Then the -differential is given by $d = d^h + (-1)^p d^v$ on the summand $B_{p,q}$.<|endoftext|> -TITLE: About non-stationary sets of $\omega_1$ -QUESTION [5 upvotes]: Suppose $A$ is a non stationary set of $\omega_1$. Define by induction the following sequence of sets:\ - -$A_0 = A$ -$A_{\alpha+1} = A_{\alpha}'$ [$X'$ is the subset of $X$, of all points the are limits of sequences from $X$] -For limit stage we take the intersection. - -Is it true that for some $\alpha < \omega_1$, $A_{\alpha}$ is null ? - -REPLY [10 votes]: No. Take a fast growing continuous $h:\omega_1 \to \omega_1$ such that $\operatorname{otp}(h(\alpha+1)-h(\alpha)) \geq \omega^\alpha$ for each $\alpha$. Consider the non-stationary set $X = \omega_1 - \{h(\alpha) : \alpha \lt \omega_1\}$. Since it takes $\alpha$ derivatives to exhaust $\omega^\alpha$, the interval $[h(\alpha)+1,h(\alpha+1)) \subseteq X$ cannot be exhausted in fewer than $\alpha$ steps.<|endoftext|> -TITLE: Categorical Construction of Quotient Topology? -QUESTION [12 upvotes]: The product topology is the categorical product, and the disjoint union topology is the categorical coproduct. But the arrows in the characteristic diagrams for the subspace and quotient topologies point the same way as in the diagrams for the product and disjoint union topologies, respectively (but there are different conditions on the "constructor" arrow). This leads me to wonder: - -Are there categorical constructions that generalize the subspace and quotient topologies? - -REPLY [3 votes]: Nobody gave this reference so I give it : http://www.tac.mta.ca/tac/reprints/articles/17/tr17abs.html, "The joy of cats", especially chapter 21 p350. The notions of initial and final topologies are generalized. The point is that the underlying set functor from general topological spaces to sets is topological.<|endoftext|> -TITLE: Janelidze's Galois theory -QUESTION [23 upvotes]: I am interested in learning about categorical Galois theory, as developed by Janelidze. I am a graduate student who has good familiarity with category theory, but not in the level of doing research on it. -First of all, I would like to ask where is the best place to study this theory in the greatest generality developed until now? Also, I know there are many papers, in different directions, dealing with Galois theory in category and topos theory - is Janelidze's theory "included" in some other theory, in the way Grothendieck's Galois theory is "included" in Janelidze's? -I have started reading the book Galois Theories. In section 1 of Chapter 5, the section that presents the theorem, Borceux and Janelidze use definitions that I haven't seen before, like "admissible class of morphisms" and "relatively admissible adjunction". Are these other concepts in disguise? And if they are notations created by the authors, do they have actual use anywhere else except from the development of the theory? I am asking all these questions because I am not sure what I should learn first before moving on to this generalization of Galois theory. I have tried reading other papers by Janelidze on this topic, e.g Pure Galois theory in Categories but the notation there seems very technical and the theorems are not proved. -Thanks for any help. - -REPLY [17 votes]: I only began to understand Galois Theory when I picked up Janelidzes book. -In the first chapter he covers the usual Galois Theory and mentions of course that an adjunction between posets is a Galois Connection. This is categorified up a categorical level as described here in nlab. They mention: - -It is the particular case where the 2-relation $R$ is the hom-functor $S^{op}×S\rightarrow Set$; the corresponding adjunction is something which Lawvere calls conjugation. - -In the second chapter, Janelidze covers the Galois Theory of Grothendieck, but not in: - -in its full generality, that is in the context of Schemes: this would require a long technical introduction. But the spirit of Grothendiecks approach is applied to the context of fields. - -These notes by Lenstra do. He also mentions Galois Categories (as Janelidze does not)and this links up with the Tannaka-Krein Reconstruction. -In the fourth chapter, besides mentioning the Pierce Spectrum, which is an interesting variation on the Zariski Spectrum and links up to Stone Spaces, he mentions effective monadic descent - -In its more specific meaning descent is the study of generalizations of the sheaf condition on presheaves to presheaves with values in higher categories. - -The descent is monadic when the pseudo-functor, aka the 2-presheaf, in its fibered category avatar (by the Grothendieck construction) are bifibrations. This allows the use of monads. He also mentions internal presheafs which are a concept in Enriched Category Theory. -In Chapter five, Janelidze relativises these concepts in the context of slice categories in preparation of proving the Abstract Categorical Galois Theorem. -This is extended in Chapter seven to 'Non-Galoisian Galois Theorem' by removing the Galois condition on effective descent, and by way of the Joyal-Tierney Theorem places it in the context of Grothendieck Toposes. -NLab goes on to say: - -For $K$ a field let $Et(K)$ be its small étale site. And let $\bar{E}:=Sh(Et(K))$ be the sheaf topos over it. This topos is a - -local topos -locally connected topos -connected topos - -Then Galois extensions of $K$ correspond precisely to the locally constant objects in $\bar{E}$. The full subcategory on them is the Galois topos $Gal(\bar{E})\rightarrow \bar{E}$. -The Galois group is the fundamental group of the topos. - -Hence, they conclude: - -Accordingly in topos theory Galois theory is generally about the classification of locally constant sheaves. The Galois group corresponds to the fundamental group of the topos. - -This can then be established in higher Topos Theory where a cohesive structure on the higher topos is used to make the constructions go through. - -Proposition. For $H$, a ∞-topos that is: - -locally ∞-connected -∞-connected, - -We have a natural equivalence $LConst(X)\backsimeq ∞Grpd[Π(X),∞Grpd_{\kappa}]$ -of locally constant ∞-stacks on $X$ with ∞-permutation representations of the fundamental ∞-groupoid of $X$. - -One, ought to here make the connection with the usual classification of topological covering spaces: - -Let $X$ be a topological space, that is: - -locally connected - -connected - - -then, $Cov(X)\backsimeq Top[\pi (X),Set]$<|endoftext|> -TITLE: Exact sequences of the cohomology induced by fiber bundle -QUESTION [6 upvotes]: I'm reading section 2.1 of Lawson's book, Spin Geometry. The book states the following fact. Let $X$ be a manifold and $E$ a vector bundle over it. Equip $E$ with a Riemannian structure. Let $P_O$ be the bundle of orthonormal frames in $E$ which is a principal $O_n$ bundle. The fibration $O_n \rightarrow P_O(E) \rightarrow X$ gives an exact sequence $0 \rightarrow H^{0}(X;\mathbb{Z}_2) \rightarrow H^{0}(P_O(E);\mathbb{Z}_2) \rightarrow H^{0}(O_n;\mathbb{Z}_2) \rightarrow H^{1}(X;\mathbb{Z}_2) $ and the fibration $SO_n \rightarrow P_{SO}(E) \rightarrow X$ gives another exact sequence $0 \rightarrow H^{1}(X;\mathbb{Z}_2) \rightarrow H^{1}(P_{SO}(E);\mathbb{Z}_2) \rightarrow H^{1}(SO_n;\mathbb{Z}_2) \rightarrow H^{2}(X;\mathbb{Z}_2) $. Lawson only says that we can deduce them from Serre spectral sequence but I don't know how. Could someone give an explicit recipe? (By the way, we are around page 79 to page 81.) -Thank you. - -REPLY [8 votes]: What you are asking about is a consequence of a more general statement about fibrations. Let $p: E \to B$ be a fibration with $B$ path connected and based. Set $F = p^{-1}(*)$. -Assume $B$ is $r$-connected and $p$ is $s$-connected. Then there's a exact sequence -$$ -0 \to H^0(B) \to H^0(E) \to H^0(F) \to H^1(B) \to \cdots \to H^{r+s}(F) \to H^{r+s+1}(B) -$$ -One way to prove this is to show that the evident map $E \cup CF \to B$, whose domain is the mapping cone of $F\to E$, is $(r+s+2)$-connected. There are a variety of ways to show this, one of which is called the "dual Blakers-Massey Theorem." Then the long exact cohomology sequence of the cofiber sequence $F \to E \to E\cup CF$ combined with the connectivity statement gives what you want.<|endoftext|> -TITLE: Smooth affine algebraic subgroups as complete intersections -QUESTION [7 upvotes]: Let's fix an algebraically closed field $k$ of arbitrary characteristic and a connected nonsingular affine algebraic $k$-group $G$. Under what conditions can I assume that a connected nonsingular affine algebraic subgroup $H \subseteq G$ is a complete intersection in $G$? Or are such subgroups always complete intersections? (I would imagine not, but I don't know any counterexamples.) If it helps I'm happy to assume that $G$ is semisimple. -EDIT: I should probably say "smooth k-group scheme" here instead of "nonsingular." - -REPLY [5 votes]: This answer is just to record that Will Sawin's example works: The Borel subgroup of $PGL_2$ is not a complete intersection in $PGL_2$. Recall that the coordinate ring of $GL_2$ is $k[w,x,y,z,\Delta^{-1}]$ where $\Delta=wz-xy$. We are thinking of $w$, $x$, $y$, $z$ as entries of the matrix $\left( \begin{smallmatrix} w & x \\ y & z \end{smallmatrix} \right)$. The coordinate ring of $PGL_2$ is the subring of homogenous elements, where $w$, $x$, $y$ and $z$ are graded in degree $1$ and hence $\Delta^{-1}$ has degree $-2$. -The Borel is given by the equations $wy/\Delta=xy/\Delta=y^2/\Delta=yz/\Delta=0$. Conceptually, the defining equation is $y=0$, but the element $y$ isn't in the coordinate ring of $PGL_2$. Since the Borel is a hypersurface, if it were a complete intersection, it would be defined by the vanishing of a single equation, say $g(w,x,y,z)/\Delta^k$, where $g$ is a homogeneous polynomial of degree $2k$. We may also assume that $\Delta$ does not divide $g$. Since $g$ has even degree, it must either vanish to even order along $y=0$, or else vanish along some other hypersurface besides $y=0$. Either way, $g/\Delta^k$ does not define the ideal of the Borel.<|endoftext|> -TITLE: Ultrafilter theorem and translation invariant measures -QUESTION [13 upvotes]: The usual Vitali construction of a non-Lebesgue measurable set generalizes to a proof that there are no (non-trivial) translation invariant measures on $\mathcal P\mathbb R$. -On the other hand, there are many proofs of the existence of non-Lebesgue measurable sets just relying on the ultrafilter theorem instead of AC, but I can't see they can be generalized this way. They use that Lebesgue measure is determined by the measure of the intervals, or that the Haar measure on the Cantor cube is determined by the measure of the clopen sets. -Does the ultrafilter theorem imply that there are no translation invariant measures on $\mathcal P\mathbb R$? Of course, I ask for non-trivial ones, i.e., $\sigma$-finite and with finite sets having measure zero. - -REPLY [3 votes]: Theorem. If there exists a free ultrafilter $\mathcal U$ on $\omega$, then there exists a non-measurable subset of the real line. -Proof. First observe that the free ultrafilter $\mathcal U$ is a non-measurable subset of the Cantor cube $\{0,1\}^\omega$ with respect to the standard product measure on $\{0,1\}^\omega$ (here we identify subsets of $\omega$ with their characteristic functions, which are elements of the Cantor cube $\{0,1\}^\omega$). -Next, consider the standard Cantor ladder map $$f:\{0,1\}^\omega\to[0,1]\subset\mathbb R,\;\;f:(x_n)_{n\in\omega}\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$ -and observe that $f$ is measure-preserving in the sense that for any measurable subset $B\subset [0,1]$ the set $f^{-1}(B)$ is measurable in $\{0,1\}^\omega$ and has product measure equal to the Lebesgue measure of $B$. -Since the symmetric difference $\mathcal U\triangle f^{-1}(f(\mathcal U))$ is at most countable, the set $f^{-1}(f(\mathcal U))$ is not measurable in $\{0,1\}^\omega$ and hence its image $f(\mathcal U)$ is not measurable in $[0,1]$.<|endoftext|> -TITLE: Elementary Submodels in Partitions Theorems -QUESTION [8 upvotes]: I'am reading the paper Elementary Submodels in Infinite Combinatorics from Soukup (http://eprints.renyi.hu/45/1/elementary_submodels_revised.pdf) and there are a lot of proofs using elementary submodels, such as the proof of Delta-System lemma and partitions theorems. However, I don't take the intuition and I would like more examples of the applications of elementary submodels. Anyone knows goods references for it in infinite combinatorics, specially in Partition Theory? -Thanks. - -REPLY [10 votes]: Decent introductions can be found in Discovering modern set theory. II, by Weese and Just, and in Introduction to cardinal arithmetic, by Holz, Steffens, and Weitz. Most likely there are a few other presentations at a comparable level. -The technique is so useful that is employed in many arguments, in a large variety of ways, and I fear I would leave out too many references even if I just tried to list relevant examples covering the key applications. -That said, you definitely want to examine arguments related to negative partition relations, that is, arguments establishing the existence of colorings where large sets not only fail to be monochromatic, but actually take as many colors as possible. There are three papers I suggest you look at in this regard: Consider the following theorem: - -Suppose $\lambda$ is an uncountable regular cardinal and there is some non-reflecting stationary subset of $\lambda$. Then $\lambda\not\to[\lambda]^2_\lambda$. That is, there is a function $f:[\lambda]^2\to\lambda$ such that whenever $X$ is a subset of $\lambda$ of size $\lambda$, we have that $f[[X]^2]=\lambda$. - -This result was proved by Shelah using Todorcevic's method of walks, and it is an elegant elementary substructure argument. See - -Saharon Shelah. Was Sierpiński right? I. Israel J. Math., 62 (3), (1988), 355–380. MR0955139 (89m:03037). - -The method of walks itself is central to many modern results in infinitary combinatorics. Several key properties (of the so-called $\rho$-functions) are established by elementary substructure arguments. Todorcevic's introduction, where $\omega_1\not\to[\omega_1]^2_{\omega_1}$ is first established, is highly recommended: - -Stevo Todorcevic. Partitioning pairs of countable ordinals. Acta Math., 159 (3-4), (1987), 261–294. MR0908147 (88i:04002). - -Todorcevic's result was later extended by Moore, showing that we can in fact ensure the existence of an $f:[\omega_1]^2\to\omega_1$ such that for any uncountable $A,B\subseteq\omega_1$, $f[A\otimes B]=\omega_1$, that is, for any $\xi<\omega_1$ there are $\alpha<\beta$, with $\alpha\in A$ and $\beta\in B$ such that $f(\alpha,\beta)=\xi$. This is a key component of his solution to the $L$-space problem: - -Justin T. Moore. A solution to the $L$ space problem. J. Amer. Math. Soc., 19 (3), (2006), 717–736. MR2220104 (2008c:54022). - -Even here, the list above is severely outdated: For example, there are fairly recent results of Rinot and Rinot-Todorcevic on the "rectangular square-bracket operation", and Stevo has written a beautiful book on the technique of walks, Walks on ordinals and their characteristics. Also, Eisworth-Shelah have extensions to coloring theorems at successors of singular cardinals. -The method of elementary substructures is also used to establish positive partition relations. A very nice paper, fairly readable and a good way of acquiring some intuition on the technique, is - -James E. Baumgartner, András Hajnal, Stevo Todorcevic. Extensions of the Erdős-Rado theorem. In Finite and infinite combinatorics in sets and logic (Banff, AB, 1991), pp. 1–17, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 411, Kluwer Acad. Publ., Dordrecht, 1993. MR1261193 (95c:03111). - -A direct continuation of the methods used in this paper, but now to establish polarized relations, is in - -James E. Baumgartner, András Hajnal. Polarized partition relations, J. Symbolic Logic, 66 (2), (2001), 811–821. MR1833480 (2002k:03068). - -A good survey of these methods can also be seen in the Hajnal-Larson paper for the Handbook. - -I expect the above directly addresses your question, but I would be remiss if I did not mention that the technique is also used in many forcing arguments. For example, to establish chain conditions. There is also Todorcevic's technique of forcing with models as side conditions, which recently has been used by Friedman and Neeman, independently, to obtain an essentially different proof of the consistency of $\mathsf{PFA}$ relative to a supercompact cardinal, in the process answering a question of mine and others on the reals of inner models of models of $\mathsf{PFA}$. -Again, it would be futile to even attempt to give a reasonably complete list of suggestions here. But I expect the above should give you a good starting point anyway. Let me add that the technique is particularly useful when studying proper and semiproper forcing notions (this is emphasized in the Martin's maximum paper by Foreman, Magidor, and Shelah, for example). -I still hope one day (soon...) to finish a book Boban Velickovic and I started a long while ago. For now, I have posted on my site a short excerpt showing how the techniques of elementary substructures can be used even at a very basic level. -(A more sophisticated application by Boban and I can be seen in this paper.) - -REPLY [9 votes]: Complementing Andres's excellent answer, let me simply try to help build your intuition for elementary submodels. -The basic situation is just like the familiar fact that if you -have finitely many group elements $g_0,\ldots,g_n$ in a large -group $G$, then they generate a countable subgroup of $G$. One -simply starts with the finite set, and applies the group operation and inverses -in all possible ways to what one has generated so far, arriving in countably many steps in a countable set that is closed under inverses and any -further applications of the group operation. -More generally, and for essentially similar reasons, if one has -countably many functions $f:N^{k}\to N$ on a set $N$, then any countable subset $A\subset N$ is included -in a countable set $M\subset N$ that is closed under all the -operations $f$. One simply applies the functions in all possible -ways to what one has generated so far, and there are only -countably many ways to do this. -In particular, consider the case where $N$ is a structure of some -kind, in a countable first order language. Perhaps we have a -formula $\varphi$, such that for some $\vec a$ in $N$, there is a -witness element $x$ for which $\varphi(x,\vec a)$. In this case, -let $f_\varphi(\vec a)=x$ be the function that selects such a -witness when there is one. Such a function $f_\varphi$ is called a -Skolem function for $\varphi$, a function that selects -existential witnesses when they exist. The observation of the -previous paragraph shows that any countable set $A\subset N$ can -be extended to a countable set $M\subset N$ that is closed under -$f_\varphi$ for every $\varphi$. But notice that this means that -whenever you have some $\vec a$ in $M$, and have some property -$\varphi(x,\vec b)$ with a witness $x$ in $N$, then you can find -such an $x$ already inside the small set $M$. This feature is -called the Tarski-Vaught property, and it is equivalent to -saying that $M$ is an elementary substructure of $N$, written -$M\prec N$. Closure under witnesses in this way implies that $M$ -and $N$ satisfy exactly the same assertions about objects in $M$ -that are expressible in the language of $N$. Thus, we've -essentially proved the Downward -Löwenheim-Skolem theorem, which says that every first-order -structure in a countable language has a countable elementary -substructure. -The basic theory of elementary substructures is covered in any -introductory logic text. -In the example of the Soukop paper to which you link, the author -uses as $N$ the entire universe $V$ of all sets. In this case, -there are certain meta-mathematical difficulties with getting a -fully elementary substructure, and so they make themselves happy -with a substructure $M$ that is elementary with respect to some -fixed finite collection $\Sigma$ of first-order properties. Thus, -one has $M\prec_\Sigma V$, which means that $M$ agrees with $V$ on -the truth of any property in $\Sigma$. Such a way of proceeding is -an essential application of the Reflection theorem. -The examples of that paper are quite nice. For example, in theorem -4.1 there, he proves the classic $\Delta$-system lemma via -elementary substructures: if $\cal{A}$ is an uncountable -collection of finite sets, then it contains an uncountable -$\Delta$-system. To see this, find a countable elementary -substructure $M\prec_\Sigma V$, using a suitably large collection -of formulas $\Sigma$, with ${\cal A}\in M$, with $M$ closed under -finite sequences. Let $D=M\cap A$ for some $A\in {\cal A}-M$. In -$V$, there is a maximal $\Delta$-system ${\cal B}\subset{\cal A}$ -with root $D$. By elementarity, there is such a $\cal{B}$ inside -$M$. But now $\cal{B}$ cannot be countable, for then it could be -enumerated in $V$ and $M$ would have to have such an enumeration, -and hence also have every element of $\cal{B}$, so ${\cal B}\subset -M$. But in this case, ${\cal B}\cup\{A\}$ would be a strictly -larger subfamily of $\cal{A}$ that is a $\Delta$-system with root -$D$, contrary to the maximality of $\cal{B}$. So it is -uncountable, and we are done. QED -That way of proving the $\Delta$-system lemma is very different -from the other proofs with which I am familiar, and it seems to be -a powerful method. -The method of elementary substructures is used pervasively in set -theory, and there are numerous other applications. For example, -almost any use of proper forcing involves elementary -substructures of some kind.<|endoftext|> -TITLE: Non continuous Linear form on $E=C([0,1],\mathbb{R})$ without AC -QUESTION [8 upvotes]: Let's note $E=C([0,1],\mathbb{R})$ the Banach space of real continuous funtions from the [0,1] interval with the uniform norm. -Is it possible to show a non-continuous linear form on $E$ exists without using a basis, i.e. without AC? - -REPLY [19 votes]: No. It's impossible. -In certain models of $\sf ZF+DC$ there is a property known as "automatic continuity" for Banach spaces, that means that every linear operator to a normed space is continuous. -Such models are, for example, Solovay's model where all sets of reals are Lebesgue measurable, and have the Baire property; and Shelah's model where all sets of reals have the Baire property. Note that Solovay's model require the existence of an inaccessible cardinal, a statement which is unprovable from the usual axioms of set theory, whereas Shelah's model does not require assumptions beyond those of $\sf ZF$ (and was in fact used to show that the consistency of the statement "all sets of reals have the Baire property" requires no additional hypothesis). - -You can find a nice exposition to Solovay's model in Aki Kanamori's "The Higher Infinite", as well Solovay's original paper "A model of set-theory in which every set of reals is Lebesgue measurable". -Shelah's result is much more difficult (but more rewarding too, as it omits the requirement of an inaccessible cardinal), and can be found in the seventh section of Shelah's celebrated paper "Can you take Solovay's inaccessible away?". The paper is long and requires quite an understanding of forcing and set theory in order to full appreciate its content. -To see why either one of these results imply the answer above, see Kechris' "Classical Descriptive Set Theory", there he proves Pettis theorem stating that if a group homomorphism between Polish groups is Baire measurable then it is continuous. The proof goes through in $\sf ZF+DC$, and while it does not immediately imply the automatic continuity for all Banach spaces, the one in the question is indeed separable and so forms a Polish group. -However a nice argument suggested to me by Harvey Friedman is that in $\sf ZF+DC$ continuity and sequential continuity are equivalent. If $T$ was a discontinuous linear operator on a non-separable space, then we could have found a sequence which witnesses that, and the restriction of $T$ to the separable space generated by that sequence has to be discontinuous as well. -Other authors have dealt with the question of automatic continuity (for Banach spaces) in such models. Amongst them are Garnir, Wright and Brunner. You can also find some information in Eric Schechter's "Handbook of Analysis and Its Foundations".<|endoftext|> -TITLE: Why do we need a $G$-universe? -QUESTION [13 upvotes]: Let $G$ be a compact Lie group. Before defining $G$-prespectra, we have to define a $G$-universe $\mathcal U$. -Question: Why do we need a $G$-universe? -A $G$-universe is defined to be a countably infinite-dimensional (real) representation of $G$ with an inner product such that - -$\mathcal U$ contains the trivial representation. -$\mathcal U$ contains countably many copies of each finite-dimensional subrepresentation. - -After fixing a $G$-universe, we can define a $G$-prespectrum (indexed on $\mathcal U$): a collection $\lbrace EV \rbrace_V$ of $G$-spaces indexed by finite-dimensional subrepresentations $V$ of $\mathcal U$ together with $G$-maps $\sigma_{V,W}:\Sigma^{W-V}EV \to EW$ for pairs $V \subset W$. -(For details and relevant papers, see "Basic notions in equivariant stable homotopy theory".) -A $G$-universe seems to be used only for taking (co)limits and to be regarded as a source of finite-dimensional (orthogonal) representations of $G$. I feel that it is very artificial to use a $G$-universe for the purposes. -For example, what is wrong if we deal with the category of (equivalent classes of) finite dimensional (orthogonal) representations of $G$? This is a small category, so we do not have any problem to consider (co)limits. - -REPLY [14 votes]: Tyler, you are too fast: didn't give me a chance to answer first! -Of course, I agree with everything you say. I wrote the following -before seeing your answer (except for the last paragraph). -Since I introduced this choice, let me explain. But first, echoing -André, taking equivalence classes would be a wrong choice -even if it gave a category, which it doesn't. In fact, the word -“representation” in this context is a convenient lie: -we are not doing representation theory here, and we must not think -at all in terms of equivalence classes. For example, isomorphisms -between “representations” control signs in equivariant -cohomology theory. -One point is to obviate set theoretic nonsense. It has become -unfashionable, perhaps, to pay attention to this, but of course -the collection of all finite dimensional representations is not -a set, and for many purposes, such as taking colimits as you -say, one does want a set. -A mathematical point is that different universes give different -categories of G-spectra, and that matters enormously: change of -universe plays an essential role in equivariant stable homotopy -theory. This could be dealt with in other ways, but use of -universes is convenient. -Actually, how essential a universe is, depends on which choice -of a category of $G$-spectra one has in mind. For all choices, -it is very convenient to work with $G$-vector spaces with a -fixed given $G$-inner product. For orthogonal $G$-spectra, the fact -that the category $\mathcal I$ of such $G$-inner product spaces -is essentially small (equivalent to a small category) allows us to -use it without actually specifying a universe, although one does -obtain a different $\mathcal I$ for each choice of a set of irreducible -representations (the complete universe, allowing all, being the -most important). -For $G$-spectra in the sense of Gaunce Lewis and myself, and therefore -for the $S$-modules of EKMM (Elmendorf–Kriz–Mandell–May) use of a -universe is truly essential: $G$-spectra are obtained from $G$-prespectra -as colimits over inclusions of sub $G$-inner product spaces of a universe. -Such colimits make no sense without use of some device to ensure smallness. -In this line of development, use of a universe seems truly essential. The -linear isometries $G$-operad $\mathcal L$ is central to the construction of the -smash product (and to lots of work in equivariant infinite loop space -theory), and $\mathcal L(j)$ is the $G$-space of linear isometries $U^j\to U$, -where $U$ is the universe in which one is working. It would be ludicrous -to try to make sense of that without working in a universe. -As a philosophical point, it is essential to be eclectic in this area and to -allow use of different categories of $G$-spectra, such as orthogonal and Lewis–May -or EKMM, since there are many things that one can readily prove with one and not -the other. For a comparison of these two and discussion of change of universe, see -for example Mandell–May, Equivariant orthogonal spectra and $S$-modules. -That source explains how, in orthogonal $G$-spectra, one can actually work with one -fixed universe, even the trivial one, and obtain equivalent categories as Tyler -says. Hill–Hopkins–Ravenel took that observation from Mandell–May and ran with it. -To be honest, I sometimes regret we made that observation; as Tyler notes, it can -be a source of confusion, and it can sometimes obscure the mathematics. Here again -it is wise to be eclectic and think in terms of both physical change of universe -and “phantom” change of universe in terms of that observation.<|endoftext|> -TITLE: "MultiCatalan numbers" -QUESTION [7 upvotes]: Could anyone provide a reference for the following (sort of) generalization of Catalan numbers: the multinomial coefficient -$$ -\binom{2k_1+3k_2+4k_3+...}{k_1+2k_2+3k_3+...,k_1,k_2,k_3,...} -$$ -is divisible by $k_1+2k_2+3k_3+...+1$. -Denoting the quotient by $C(k_1,k_2,k_3,...)$, one may call these the multiCatalan numbers. They definitely must appear somewhere in combinatorics, but I could not find any reference. -The reason I am sure these numbers must be known is that, for example: $(-1)^{k_1+k_2+...}C(k_1,k_2,...)$ is the coefficient at $x_1^{k_1}x_2^{k_2}\cdots$ of the composition inverse of the formal power series $t+x_1t^2+x_2t^3+...$; -$C(k_1,k_2,...)$ is the number of faces of the Stasheff polytope $S_{k_1+k_2+...}$ of shape $S_1^{k_1}\times S_2^{k_2}\times\cdots$ (here for convenience I have redenoted by $S_n$ the standard $K_{n+1}$; so $S_1$ is a point, $S_2$ a segment, $S_3$ a pentagon, etc.); -hence they also enumerate certain kinds of trees, etc., etc. ... - -REPLY [2 votes]: If you reduce the partition arrays associated with these numbers, you get the face polynomials of the Stasheff associahedra or their simplicial duals, depending on the ordering. These are the vintage Kirkman-Cayley numbers of the late 1800's: -$$ K(n,k) = \frac{1}{k+1} \binom{n-3}{k} \binom{n+k-1}{k}\;.$$ -Kirkman asserted they were the number of dissections of convex polygons in 1857, and Cayley gave the correct proof in 1890. See G. Gaiffi.<|endoftext|> -TITLE: On extended Riemann Hypothesis and coefficients of Selberg Class L-functions -QUESTION [6 upvotes]: There is the conjecture that Selberg Class L-functions satisfy RH. -So that an L-function needs to have its coefficient multiplicatives (plus other conditions: functional equation,...) in order to satisfy RH. But I would like to know if for a L-function of Selberg class it is possible to find another L-function having the same non trivial zeros (strictly in the critical strip) but having "the sum of its coefficient" bounded instead of "multiplicative coefficients". -Meaning that for a L-function of Selberg class $L(s)= \sum a_n n^s$ with multiplicative coefficient $a_n$ there is another L-function $L_1$ (of Selberg class or not) having same non trivial zeros as $L(s)$ such that $L_1(s)= \sum b_n n^s$ with $\exists K, \forall N, |\sum_{n=1}^{n=N} b_n| \le K$ -For primitive Dirichlet L-functions the function is itself satisfying the condition. But for the Zeta function we need to multiply it by $(1-2^{1-s})$ in order to obtain the Dirichlet Eta function which has the same zeros (strictly in the critical strip) as Zeta with the sum of its coefficient bounded. (And no more multiplicative property for its coefficients) -Any existing results on this question ? May be it is obvious that such equivalence does not always exists? - -REPLY [5 votes]: Here's an argument that I think demonstrates why no such tweak is possible if the degree is -strictly bigger than $1$. For simplicity let us just consider the case when $L(s,f)$ has no pole. -Suppose there is some tweaked Dirichlet series $H(s,f)=\sum_{n=1}^{\infty} b(n)n^{-s}$ with coefficients - that have bounded partial sum and such that $H(s,f)$ and $L(s,f)$ have the same zeros in $\sigma >0$. - Then $L(s,f)/H(s,f)$ is an analytic function in $\sigma >0$, and in some the half-plane $\sigma >2$ say - this function is bounded. Now we can use a complex function theory argument (as in Riemann hypothesis implies - Lindel{\" o}f) to show that $L(s,f)/H(s,f)$ is bounded by $(1+|s|)^{\epsilon}$ for all $s$ with Re$(s)\ge \delta >0$ - for any fixed positive $\delta$. -Next, we note that the Plancherel formula can be used to show that - $$ - \int_0^{\infty} \Big| \sum_{n\le e^{x} } b(n) \Big|^2 e^{-2\sigma x} dx - = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{|H(\sigma+it)|^2}{|\sigma+it|^2}dt, - $$ - and the formula holds for all $\sigma >0$ since we are assuming that the partial sums - of $b(n)$ are bounded. However the right hand side is, in view of our previous observation, - $$ - \gg \int_{-\infty}^{\infty} \frac{|L(\sigma+it,f)|^2}{|\sigma+it|^{2+\epsilon}} dt, - $$ - if $\sigma \ge \delta >0$. Now the idea is to show that the functional equation for - $L(s,f)$ will prevent this from happening. Basically the functional equation shows - that for $0<\sigma <1/2$, $L(s,f)$ roughly is of size $|s|^{d(1/2-\sigma)}$ times $L(1-s,f)$ - (which in mean square can be estimated). In other words, $|L(\sigma+it,f)|^2$ will - roughly grow like $|s|^{d(1-2\sigma)}$ and if $d$ is $>1$ for $\sigma$ suitably close - to zero it will not be possible for - $$ - \int_{-\infty}^{\infty} - \frac{|L(\sigma+it,f)|^2}{|\sigma+it|^{2+\epsilon}} dt - $$ - to converge. -This last part was sketched quickly, but it is a standard way of producing $\Omega$ results - for arithmetic functions. What it shows is that the partial sums of the coefficients of - the actual $L$-function must be $\Omega(x^{(d-1)/(2d)-\epsilon})$ which is the kind of result mentioned in rlo's - answer.<|endoftext|> -TITLE: Maximal $p$-subgroups in nilpotent groups -QUESTION [5 upvotes]: By using Zorn's Lemma one can establish the existence of maximal $p$-subgroups in any group, even infinite. Using this existence, exactly as in the finite case, it is easy to show that in a nilpotent group the set of $p$-elements form a subgroup (take the normalizer of the normalizer etc.) But then the maximal $p$-subgroup is unique and therefore one should not need to use Zorn's Lemma to prove that the set of $p$-elements in a nilpotent group form a subgroup. In fact there are proofs of this fact that do not use Zorn's Lemma. But none is easy. I am looking for an easy proof, using induction on the nilpotency class for example, or using some commutator calculus. Does anyone have an accessible (by undergraduates) proof of this fact that does not use Zorn's Lemma? Thanks a lot. - -REPLY [2 votes]: Very nice. Of course... Thanks a lot. In the mean time I found another proof. -Let $P$ be the set of prime elements of the group $G$. Let $P_1=P\cap G'$. By induction on the nilpotency class of $G$ we may assume that $P_1 \leq G$. -To prove that $P$ is a subgroup I proceed by induction on the nilpotency class of $P_1$. The trick is to divide the group $G$ by $Z(P_1)$, which is of course a normal subgroup of $G$. The inductive step is easy. The initial case where $P_1=1$ is handled as follows: It is easy to show that if $G/Z(G)$ has an element of order $p$, then $G'\cap Z(G)$ has an element of order $p$. Therefore $G/Z(G)$ does not have elements of order $p$, meaning that $P\subseteq Z(G)$. Therefore $P\leq G$. Thanks again.<|endoftext|> -TITLE: What is the status of (universal) algebra in type theory? -QUESTION [13 upvotes]: With the recent interest in homotopy type theory as a foundation for mathematics, it seems natural to develop algebra within the framework of type theory. So far, I can't find much literature regarding this. -I know of Danielsson and Coquand's result that isomorphism implies equality, and work by Spitters and van der Weegen using type classes (without univalence), but has any other work been done to develop algebra within dependent type theory? Is this even a worthwhile task? - -REPLY [9 votes]: It largely depends on how general you want to make your algebra; in particular, do you want to look just at structures on $n$-types, for some finite $n$, or consider algebraic structures on all types? -The universal algebra of 0-types should look much like classical universal algebra; this is what the Danielsson–Coquand result you mention talks about, for instance, and as far as I know no general work beyond that has been done yet. The most novel aspects of this, I guess, would be in giving more exploration of working with Ahrens–Kapulkin–Shulman categories (and related structures) than anyone’s done so far. -The universal algebra of 1-types is wide open, and should be reasonably approachable. I don’t know of any existing work in this direction; and I also don’t know quite what to expect it to look like — possibly like classical 2-categorical algebra (in the 2-monad sense), or possibly nicer, if AKS–style (2-)categories give a simplification of the language? Algebra on $n$-types, for fixed $n>1$, is also open, but I guess this would be a subsequent project to the 1-types case. -Algebra on arbitrary types is open, but probably difficult. Several of us at the IAS last year spent some time trying to crack this (not for general u.a., just for specific algebraic structures), and all ran up against the barrier that Urs alludes to in comments. Essentially, classical approaches to coherently homotopy-algebraic structures seem to all sooner or later use on-the-nose equality in some way that’s not available in HoTT (e.g. the axioms of an operad action). This is a known open problem, and a nice one, but not easy, I think!<|endoftext|> -TITLE: Reference for Automorphisms of K3 surfaces -QUESTION [5 upvotes]: I am looking for some introductory reference concerning Automorphisms (of finite order) on K3 surfaces. Any suggestion? - -REPLY [7 votes]: You can start with -Hans Sterk -Finiteness results for algebraic K3 surfaces -Mathematische Zeitschrift, 1985, Volume 189, Issue 4, pp 507-513 -The aim of this paper is to prove the following theorem: - -Theorem. Let $X$ be an algebraic K3 surface over the complex numbers. - Then - a) the group $\mathrm{Aut}(X)$ of (biholomorphic) automorphisms is finitely generated, and - b) for every even integer $d\geq -2$, the number of $\mathrm{Aut}(X)$-orbits in the collection of complete linear systems which contain an irreducible curve of selfintersection $d$ is finite. - -If this does not give you enough, look at the references that seem promising and especially at the papers that refer to this one. Very likely anything newer that's relevant will refer to this. -Also look at the many papers of Nikulin on K3's. He has made lots of computations in the K3 lattice about automorphisms. -I believe Oguiso also has a paper where he studies finite automorphisms of Calabi-Yau's, but this is only a vague memory, so I could be off. Someone might correct me on this....<|endoftext|> -TITLE: What are good English-language sources for reading about the Luzin affair? -QUESTION [12 upvotes]: What are good English-language sources for reading about the Luzin affair? - -I'm interested in the subject and am wondering about good historical sources. - -REPLY [5 votes]: Wikipedia article on Luzin mentions several English resources in the section on Luzin affair. -(Here is a link to the recent revision of the article, just in case there will be some substantial changes.) -References in English language which are mentioned there, are: - -Levin, A. E. (1990). "Anatomy of a public campaign: "Academician Luzin`s case" in Soviet political history.". Slavic Review (Slavic Review, Vol. 49, No. 1) 49 (1): 90–108. doi:10.2307/2500418. JSTOR 2500418. Maybe you will find something interesting also in the publications citing this paper. -The book Graham L., Kantor J.-M. Naming Infinity: A True Story of Religious Mysticism and Mathematical Creativity. Belknap Press, Cambridge and London (2009). Google Books link. This book has already been mentioned in another answer. -The book Demidov, Sergei S.; Ford, Charles E. (1996). N. N. Luzin and the affair of the "National Fascist Center". San Diego, CA: Academic Press. pp. 137–148. ISBN 5-88812-103-7. MR 1388788.. In History of mathematics: states of the art, edited by Joseph W. Dauben MR 1388780, Google Books link. -S.S. Kutateladze, An Epilog to the Luzin Case, Siberian Electronic Mathematical Reports, Vol.10 (2013),A.1-6. The same author also has an article called Roots of Luzin's Case, doi: 10.1134/S1990478907030015. - -If you search in Google Books for other books mentioning Luzin affair, you might find some other useful information. And, of course, Google Scholar is worth checking, too. - -REPLY [3 votes]: See also The Tragedy of Mathematics in Russia<|endoftext|> -TITLE: Hamiltonian circuit -QUESTION [9 upvotes]: Let us consider a disk with one labelled point on the boundary and $n$ labelled points in the interior. -Let T be a triangulation of the whole disk with vertices on the labelled points such that T contains no self-loops except the boundary of the disk, no multiple edges between two points in the interior of T, arcs of T can be curved. -1) Does T have an Hamiltonian circuit ? -2) If R is another triangulation of the same time, how many flips (as a function of $ n $ should one perform at most in order to get from R to T ? - -REPLY [2 votes]: Not necessarily. -See the famous paper of Sleator/Tarjan/Thurston (Journal of the AMS, vol 1, issue 1, I believe).<|endoftext|> -TITLE: A conjecture about odd path and odd cycle -QUESTION [5 upvotes]: Let $k$ be a positive integer and $G=(V,E)$ be a $2$-connected simple graph.Suppose $v\in V(G)$ satisfy: -$(1)$there exists at least one vertex $u\in V(G)\backslash\{v\}$ such that $u$ is not adjacent with $v$; -$(2)$for any $u\in V(G)\backslash\{v\}$ such that $u$ is not adjacent with $v$,there is a $u$-$v$ path in $G$ which has an odd length $\geq2k+1$. -I want to ask is it ture that there must exists a odd cycle in $G$ whose length$\geq2k+1$. - -REPLY [5 votes]: False. As we want a counterexample, we naturally start with the Petersen graph, P. Note that for any vertex v of P and non-adjacent edge uw of P there is a Hamiltonian path from v to w that does not use the uw edge. On the other hand, there is no Hamiltonian cycle in P. -Our graph G will have 10t+15 vertices (where t is some large number) and is obtained from P by subdividing each edge with a vertex and replacing each original vertex by t vertices, connected to each other and to the subdividing vertices which the original vertex was connected to. Let v be a vertex that was not obtained by subdividing an edge and let 2k+1=9t+16 (or +17, whichever is odd). Using the above mentioned properties of P, it is easy to see that G satisfies the requirements if t is big enough. -This graph is even 3-connected and can be generalized to give even stronger counterexamples. -Note that my original construction used t=2, which is not sufficient for the above reasoning, as was noted in the comment by Dani.<|endoftext|> -TITLE: On mathematical aspects of the most recent Nobel Prize in economics winners' work -QUESTION [16 upvotes]: Can somebody briefly introduce the mathematical aspects, in particular, those related to mathematical finance, of the three economists who were just awarded this year's Nobel Memorial Prize in Economic Sciences? -According to the New York Times: - -The three economists, who worked independently, were described as collectively illuminating the workings of financial markets by showing that stock and bond prices move unpredictably in the short term but with greater predictability over longer periods. - -This seems very related to mathematical finance: all kinds of random noises, yet in the long run, may not deviate too much from, say, their rational means. Many thanks for the comments, introductions and ideas! - -REPLY [8 votes]: Just to add a little something to arsmath's very good answer: The mathematics in Fama's main idea that returns are impredictable are indeed simple, and moreover, not due to him. What Fama did is a huge empirical study to support that claim. For the mathematical argument itself, which is simple but not absolutely trivial, in other words which has some mathematical content, -I recommend any interested reader this very nice article of Samuelson (1965), Proof that properly anticipated prices fluctuate randomly.<|endoftext|> -TITLE: Equivariant normalization? -QUESTION [10 upvotes]: Let $G=\mathrm{Gl}_n\mathbb C$ and let $X$ be an affine $G$-variety. Let $\phi:\tilde X\to X$ be the normalization of $X$, i.e. the spectrum of the integral closure of $\mathbb C[X]$ in its fraction field. Can $\tilde X$ be given the structure of a $\tilde G$-variety such that $\phi$ is equivariant? - -REPLY [7 votes]: Here is the sort of example I think Jason Starr was raising. (I looked at Brian's webpage, but it wasn't obvious which paper to read.) Take $k$ to be a perfect field of characteristic $p$, with $p \neq 0$, $2$. Let $A = k[x,y]/(y^2-x^p)$. The normalization of $A$ is $\tilde{A} = k[t]$, with $y=t^p$ and $x=t^2$. -Let $G$ be the group scheme with underlying space $k[\epsilon]/\epsilon^p$ and multiplication given by the map $\epsilon \to \epsilon \otimes 1 + 1 \otimes \epsilon$. If, like me, you prefer to think in terms of functors of points, $G(R) = \{ \epsilon \in R : \epsilon^p =0 \}$ and the multiplication map $G(R) \times G(R) \to G(R)$ is $(\epsilon_1, \epsilon_2) \mapsto \epsilon_1 + \epsilon_2$. -Let $G(R)$ act on $A(R)$ by $(\epsilon, (x,y)) \mapsto (x+2 \epsilon, y)$. If you prefer maps of algebras, $x \mapsto 1 \otimes x + 2 \epsilon \otimes 1$, $y \mapsto 1 \otimes y$. I claim that this action does not lift to $\tilde{A}$. Suppose, to the contrary, that $t \mapsto 1 \otimes t_0 + \epsilon \otimes t_1 + \cdots + \epsilon^{p-1} \otimes t_{p-1}$, with the $t_i \in k[t]$. Writing out that the action must preserve the relation $t x^{(p-1)/2} = y$ gives -$$ \left( 1 \otimes t_0 + \epsilon \otimes t_1 + \cdots \right) (1 \otimes t^2 + 2 \epsilon \otimes 1)^{(p-1)/2} = 1 \otimes t^p $$ -Equating the coefficients of $1$ and $\epsilon$ gives $t_0 t^{p-1} = t^p$ and $t_1 t^{p-1} + (p-1) t_0 t^{p-3} = 0$. So $t_0=t$ and $t_1 = 1/t$. But $1/t$ isn't in $k[t]$. -Morally, the action wants to be $(\epsilon, t) \mapsto t \sqrt{1+2\epsilon t^{-2}}=1+\epsilon t^{-1} - (1/2) \epsilon^2 t^{-3} + \cdots $. The trouble is that $\left( t \sqrt{1+2\epsilon t^{-2}} \right)^k$ is in $k[t, \epsilon]/\epsilon^p$ when $k$ is even or is $\geq p$, but not in general. - -This can never occur when $G$ is normal. Over a field of characteristic zero, all group schemes are regular, and regular implies normal, so there are no examples over a field of characteristic zero. -Recall the universal property of normalization: For any normal variety $Y$, the induced map $\mathrm{DomHom}(Y, \tilde{X}) \to \mathrm{DomHom}(Y,X)$ is bijective, where $\mathrm{DomHom}$ is the dominant homomorphisms. -Proof: If $G$ is normal, then $G \times \tilde{X}$ is normal. We have a map $G \times \tilde{X} \to G \times X \to X$, where the first map is $\mathrm{Id} \times \phi$ and the second map is the group action. By the universal property of the normalization, there is a map $G \times \tilde{X} \to \tilde{X}$ making the obvious diagram commute. We claim that this map gives an action of $G$ on $\tilde{X}$. -Consider the two maps $G \times G \times \tilde{X} \to \tilde{X}$. We must show they are equal. Again, by the universal property of normalization, it is enough to show that the two compositions $G \times G \times \tilde{X} \to X$ are equal. But these are the same as $G \times G \times \tilde{X} \to G \times G \times X$, follwed by the two maps $G \times G \times X \to X$, and these are equal because $G$ acts on $X$.<|endoftext|> -TITLE: Norms of B-spline coefficients -QUESTION [7 upvotes]: In Shumaker's book (Spline Functions: Basic Theory), we know that the $l^\infty$-norm of B-spline coefficients is bounded above and below by the $L^\infty$-norm of the spline itself. Are there similar results about other norms? Such as $l^1$-norm of the coefficients? - -REPLY [5 votes]: This question is closely related to the so-called "condition number" of the B-spline basis. Basically, for a spline $f$ of some degree $p$ with a coefficient vector $c=(c_i)$, you generally have for any $q \in [1,\infty]$ that -$$ - A_{p,q} \|c\|_{\ell_q} \le \| f \|_{L_q} \le B_{p,q} \|c\|_{\ell_q}, -$$ -and the smallest possible ratio $\kappa_{p,q} = B_{p,q} / A_{p,q}$ is called the condition number of the basis. -Unfortunately, although this type of estimate seems to be "folklore" in the spline community, it's surprisingly hard to find a good reference for it. I will point you back to Schumaker's book, but to Theorem 9.27, equation (9.68), which states essentially this. The caveat is that this equation occurs in a chapter on Chebysheffian splines, and I don't yet fully understand if it fully applies to the standard B-spline situation. I have seen more knowledgeable people cite this equation for this purpose, though. -Update: I have since found a more palatable reference, namely De Boor, Splines as linear combinations of B-splines. A Survey (1976). Theorem 5.2 states what you need. I also forgot to mention above that, for $q \ne \infty$, the knot spacing enters the inequality, but with the same order for the lower and upper bound.<|endoftext|> -TITLE: How do you find maximal orders in quaternion algebras? -QUESTION [5 upvotes]: Let A be the 4-dimensional algebra over Q with basis 1,i,j,k, and multiplication table -$$ -i^2 = -1 \quad j^2 = -11 \quad k^2 = -11 \quad ij = k \quad jk = 11i \quad ki = j -$$ -So, A is the unique "definite quaternion algebra of discriminant 11" over the rational numbers. I'm led to believe that there are two conjugacy classes of maximal orders in A, but I had a hard time finding them explicitly. -A has elements of multiplicative order 4 (e.g. $i$) and $6$, (e.g. $1/2 + i/4 + j/4$). I presume then that $(1,i)$ can be extended to a basis for a maximal order, and that $(1,1/2 + i/4 + j/4)$ can be extended to a basis for a non-conjugate maximal order. What are these bases? -I'd appreciate an answer even just for discriminant 11. Of course any information about how to answer this kind of question systematically would be a nice bonus for me. -PS. Thanks Aeryk and Aurel. One of the two orders, say $O_1$, has basis -$$1, \quad i,\quad 1/2+j/2, \quad i/2 + k/2$$ -Can anyone tell me the other one? Are the Ivanyos/Ronyai/Voight algorithms good for finding all maximal orders, or would they output something conjugate to $O_1$ again? - -REPLY [6 votes]: For semisimple algebras over $\mathbb{Q}$, there is a general algorithm due to Gabor Ivanyos and Lajos Ronyai, described in Finding maximal orders in semisimple algebras over $\mathbb{Q}$ and implemented in the computer algebra system Pari/gp. For quaternion algebras, there is a dedicated algorithm due to John Voight, described in Identifying the matrix ring: algorithms for quaternion algebras and quadratic forms and implemented in the computer algebra system Magma. - -REPLY [5 votes]: See Example B on page 9 of these notes: http://www.math.polytechnique.fr/~chenevier/coursIHP/chenevier_lecture6.pdf -Let $D =\left(\frac{−1,−11}{\mathbb{Q}}\right)$ be the quaternion algebra with discrimimant 11. -A discriminant computation shows that a maximal order $O$ is given by $\mathbb{Z}[z] + i\mathbb{Z}[z]$ where $z = \frac{1+j}{2}$. -Sorry I couldn't give a more general answer.<|endoftext|> -TITLE: Automorphism group of a finite group -QUESTION [8 upvotes]: I would like to ask if there exists an explicit description of $\mathrm{Aut}(G)$, the group of automorphisms of a finite group $G$, in particular, when $G$ is abelian. E.g., if $G = \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, where $m$ is a positive integer, how can we describe $\mathrm{Aut}(G)$? Which relation do we have between it and $GL_m(\mathbb{Z})$? If $m$ is prime then $\mathrm{Aut}(G) \cong GL_m(\mathbb{Z})$, but what happens for $m$ general? E.g., if $m=4$, I find that the cardinality of $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z})$ is $8$, by seeing to where the generators $(0,1)$ and $(1,0)$ are sent. But I have the feeling that $GL_4(\mathbb{Z})$ should be contained in $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z})$, so I should have a problem with the cardinalities. -More general, is it sufficient to see where a minimal set of generators is sent? Maybe someone could indicate me a paper where I could find a good description of such automorphism groups? - -REPLY [13 votes]: A nice set of generators for the automorphism group of a finite abelian group is described by Garrett Birkhoff in his paper titled "Subgroups of abelian groups", Proc. London Math. Soc., s2-38(1):385-401, 1935 (MR). -Note that each finite abelian group is the product of its $p$-primary subgroups. An automorphism preserves the primary parts. So it is sufficient to consider $p$-abelian groups. -It is convenient to think of automorphisms of finite abelian groups as integer matrices. Expressing the group $A = \mathbf Z/p^{\lambda_1}\oplus \dotsb \oplus \mathbf Z/p^{\lambda_n}$ as a quotient of the free abelian group $\mathbf Z^n$, lift an automorphism $\phi$ of $A$ to an automorphism $\tilde\phi$ of $\mathbf Z^n$: -$$ -\begin{matrix} -\mathbf Z^n & \xrightarrow{\tilde \phi} & \mathbf Z^n\\ -\downarrow & &\downarrow\\ -A & \xrightarrow{\phi} & A -\end{matrix} -$$ -The matrix $(\phi_{ij})$ representing $\tilde\phi$ is an invertible integer matrix. -As far as the automorphism $\phi$ is concerned, the its entries in the $i$th row are in $\mathbf Z/p^{\lambda_i}\mathbf Z$. Also $\phi_{ij}$ is divisible by $p^{\max(0, \lambda_j-\lambda_i)}$. With this matrix representation, it is easy to do calculations. For example, composition is matrix multiplication. -As for your query concerning cardinalities: if $\lambda_1>\lambda_2>\dotsb>\lambda_n$ and -$$ A = (\mathbf Z/p^{\lambda_1}\mathbf Z)^{\oplus m_1}\oplus\dotsc \oplus (\mathbf Z/p^{\lambda_n}\mathbf Z)^{\oplus m_n}, -$$ -then it is possible to deduce from the above description of the automorphism group that -$$ -\lvert\mathrm{Aut}(A)\rvert = q^{\sum_{i,j}m_im_j\min(\lambda_i,\lambda_j)}\prod_{k=1}^n \prod_{l=1}^{m_k} (1 - q^{-l}). -$$ -The group $\prod_{k=1}^n GL_{m_k}(\mathbf Z/p\mathbf Z)$ is a quotient of $\mathrm{Aut}(A)$ by a $p$-group.<|endoftext|> -TITLE: Orbifolds vs. branched covers -QUESTION [11 upvotes]: Forgive me if this is a basic question. I'm just learning about orbifolds, and covering spaces are my happy place for thinking about group actions. -If $M$ is a manifold and $G$ is a group acting properly discontinuously on $M$, then $M/G$ is an orbifold. It seems to me like $M$ should also be a branched cover of $M/G$. Is this always true? If not, is there an illuminating example? -Right now when I think of orbifold, I'm secretly thinking of quotients of manifolds by group actions. And when I think of quotients of manifolds by group actions, I'm secretly thinking about covering spaces. If anyone could help me out by showing me some spots where either of these bridges break down I'd be very appreciative. - -REPLY [2 votes]: I will not presume that you asked about what a good cover is. Thinking about orbifolds the way you described is useful: a group $G$ acts on $M$ properly and discontinuously but because the action is not free the topological quotient $X = G\backslash M$ may be missing some information. In particular the map $M \twoheadrightarrow G\backslash M=X$ is not a covering map. -One definition of an orbifold is the space $X$ equipped with an atlas $\mathcal A$ such that each chart contains some extra information (i.e. a local group action.) This is the extra information that is lost, for example, in passing from $M$ to $X=G\backslash M$. With this information, you can recover $M$ from $X$ and $G$, should $M$ actually exist (this has to do with "goodness"). So in this sense an orbifold is a quotient of a proper discontinuous group action. -A covering space of $Y$ can be obtained by taking a open cover $\mathcal{U}=\{U_i\}$ of $X$ and glueing together copies of the $U_i$, one of the criteria for being a covering space though is that two distinct lifts of some $U_i \in \mathcal U$ are disjoint. -If you want to make an orbifold cover, i.e. get $M$ from $G\backslash M$, the problem is that distinct lifts of the same chart in $\mathcal A$ can not be made to be disjoint. -Suppose first that the non-trivial fixed sets of the $G$ action on $M$ are points (and that $M$ is a surface). Let $X = G\backslash M$ and let $\mathcal A$ be the orbifold atlas on $X$. Then $M$ can be obtained by taking copies of the charts $V_i$ in $\mathcal A$ (which are all discs), but sometimes you will need to slice some of the $V_i$ open (to get something that looks like a slice of watermelon), you can then get $M$ by glueing together possibly sliced charts from $\mathcal A$ in a manner analogous to building a covering space. Moreover, these "slices" in charts are exactly the cuts in branched covers. So in this case the orbifold cover is a branched cover. -However, as Andy Putman commented, not all actions of $G$ on $M$ have fixed sets that are points. For example a reflection will fix an arc, and the example that Andy gave ($\mathbb R^2$ modulo reflection) will not correspond to a branched cover. So to answer your question, branched covers are a way to construct special (but important) cases of orbifold covers. The special case, as Andy commented, is when all the non trivial $G$-fixed sets in $M$ are points (provided $M$ is a surface).<|endoftext|> -TITLE: Jacquet module for Lie algebras? -QUESTION [5 upvotes]: Let $G$ be a reductive group and let $P$ be a parabolic subgroup with Levi $L$ and unipotent radical $N$. Let $F$ be a finite or local non-Archimedean field. The Jacquet Functor is a functor from $G(F)$-modules to $L(F)$-modules defined by -$V \mapsto V_N:=V/\langle n\cdot v-v\rangle$, where $\langle n\cdot v-v\rangle$ denotes the vector subspace of $V$ generated by expression of the form $n\cdot v-v$, where $n\in N(F)$ and $v\in V$. The Jacquet functor satisfies many nice properties: it is exact, it is adjoint to parabolic induction, etc. It plays a fundamental role in representation theory of finite and $p$-adic groups. -Now suppose $\mathfrak{g}$ is a reductive Lie algebra over $\mathbb{C}$, and let $\mathfrak{p}$ be a parabolic with Levi $\mathfrak{l}$ and nilpotent radical $\mathfrak{n}$. I guess one can define a functor from $\mathfrak{g}$ modules to $\mathfrak{l}$ modules by $V\mapsto V_N:=V/(\mathfrak{n}\cdot V)$. Is this a reasonable thing to do? Is there a reference where basic properties of this functor are spelled out? -The properties I am interested in is in relation to "parabolic induction" in Lie algebras. For instance, is it true that this functor is the adjoint of the functor of parabolic induction? Can one hope for an analogue of second adjointness theorem? - -REPLY [5 votes]: Yes, this is indeed a reasonable thing to do. -The space $V / \mathfrak{n}V$ is called space of $\mathfrak{n}$-coinvariants and its $p$th left derived functor is the $p$th Lie algebra homology $H_p(\mathfrak{n},V)$. -The most comprehensive reference would be Knapp, Vogan: Cohomological Induction and Unitary Representations. -Edited: -By superficially skimming over the book by Knapp & Vogan, I would be inclined to say that the second adjointness theorem is true. I am certainly far from being an expert in these matters. If this book is too big for you I can recommend Dirac Operators in Representation Theory by Pandzic and Huang and its two chapters on cohomological induction.<|endoftext|> -TITLE: Random vs Unknown -QUESTION [20 upvotes]: Is there any distinction at all between a random quantity and an unknown quantity or is it impossible to distinguish? -Example: 5 minutes in the future, I plan to roll a die the the number of the die roll at that point in time is called X. Suppose the number X will happen to be 4 , although there is no way of knowing that at the current time (pretend I didn't say it was going to be 4, or that I retroactively inserted a 4 into the post after I witnessed it, but let's continue this discussion as if it were unknown). -Is X a deterministic constant (in reality equal to 4), which is simply unknown (once again, pretend I didn't say it was 4)... or is it a random variable uniformly distributed from 1 to 6 (discrete)? -Does there exist any situation where we can say that a number is definitely a random variable and definitely not an unknown deterministic constant or vice versa (definitely an unknown deterministic constant and definitely not random)? -If there is no way to distinguish these two things, then is it possible they are in fact descriptions of the same thing i.e. everything we call random in the universe really just originates from lack of knowledge? Does there exist objective randomness which is not a function of the knowledge any specific observer has? -Thanks! - -REPLY [3 votes]: In "Theory of Probability and Random Processes" by V.V.Tutubalin (it exists, alas, only in Russian: http://padabum.com/d.php?id=10681), the second half of the book is devoted to applications of probability to real world problems. Unlike most textbooks, it contains not an optimistic list of successful applications, but case studies and analysis of arising difficulties. After reading this book you may agree with the author that successful applications of probabilistic models to practice should often be viewed as miracles. -I like the book a lot. Retelling it is not for a MO comment, so let me give you just one quote. After saying that there is no accurate, systematic, and consistent way to apply mathematics to physics and engineering, Tutubalin writes (p.317): This is especially true for applications of probabilistic models due to the fundamental haziness of the notion of statistical ensemble.<|endoftext|> -TITLE: No big clique minor but a big grid minor -QUESTION [6 upvotes]: I was wondering if the following result is known (or if there's a nice short proof without treewidth/brenchwidth related theorems): as the title says, suppose you have a graph without a big clique minor but which has a big grid minor; then, show that it either has an induced wall or the line graph of an induced wall as an induced subgraph. -Any input welcome, thanks! -EDIT: Here's a paper I googled randomly that includes all the definitions from above (and also a picture of what a wall graph is - see page 6): http://www.automata.rwth-aachen.de/~grohe/pub/grokawree13.pdf. - -REPLY [5 votes]: The structure of graphs with big tree-width but no big clique minor is partially described by the Flat-wall theorem (Theorem 9.8 of GM XIII). With Ken Kawarabayashi and Robin Thomas, we recently posted a new proof of this result - http://arxiv.org/abs/1207.6927. -I don't know of your proposed theorem having appeared anywhere else, and I don't know of any short proof avoiding the usual graph minor technicalities. Here, though, is an outline of a proof using the standard graph minors tools. -By the Flat-wall Theorem, if there is no $K_t$ minor and the tree-width is sufficiently big, then there exists an induced subgraph H containing an $r'$-wall $W$ such that $H$ has a decomposition up to 3-separations capturing the wall. -This means that there exist pairwise edge disjoint subgraphs $H_0, H_1, \dots, H_s$ such that: -a) $\bigcup_0^s H_i = H$, -b) $|H_i \cap H_0| \le 3$ for $i \ge 1$ and $H_i \cap H_j \subseteq H_0$ for $i, j \ge 1$, $i \neq j$, and -c) $H_i - H_0$ contains at most one vertex of $W$ of $deg_W = 3$. -d) Let $\bar{H}_0$ be the graph obtained from $H_0$ by adding an edge to every pair of non-adjacent vertices $x, y$ for which there exists an index $i$ with $x, y \in H_i \cap H_0$. Then $\bar{H}_0$ is planar and for all $i \ge 1$, there exists a face of $\bar{H}_0$ containing $V(H_i) \cap V(H_0)$. -Consider the subgraph of $H$ induced by the wall $W$. It again has a decomposition satisfying a) - d) above (just take the subgraphs $H_i \cap W$ for $i = 0, \dots, s$.) We first do some analysis on the subgraphs $H_i$ for $i \ge 1$. By picking the wall and decomposition together to minimize the number of vertices in the wall, it is not too hard to show that each $H_i$ can be embedded in the plane using using the rural societies theorem of Robertson and Seymour. The conclusion is that we may assume that the subgraph induced by the wall is in fact planar. -To prove the theorem, one now must only show it for a wall with some edges added in a planar way. This should be possible by first taking every second vertical and every second horizontal path of the wall so that the only edges not contained in the wall should form triangles with two edges incident a vertex of degree 3 in the wall.<|endoftext|> -TITLE: Irreducibililty tests for cubic and quartic polynomials over finite fields -QUESTION [6 upvotes]: The unpublished preprint: -D. G. A. Jackson, The irreducibility of a cubic over $\mathbb{F}_q$, Research Report 98-17, Univ. of Sydney (1998) -gives necessary and sufficient conditions (when ${\rm char}({\mathbb F}_q)\neq2,3$) for a cubic polynomial over $\mathbb{F}_q$ to be irreducible. -I have conditions (below) which work for cubics when ${\rm char}(\mathbb{ F}_q)=2,3$, and also quartics. While I guess that such results have been published, I have been unable to find references. I would be most grateful if someone knows precise reference (if they exist). -Theorem A. Suppose $q=2^e$ and $c(x)=x^3+c_2x+c_3$ where $c_2c_3\neq0$. Then $c(x)$ is irreducible over $\mathbb{F}_q$ if and only if ${\rm Tr}_{\mathbb{F}_q/\mathbb{F}_2}(c_2^3/c_3^2)\equiv e\pmod2$, and a root $\beta$ -of $b(x)=x^2+c_3x+c_2^3$ is a noncube. -Theorem B. Suppose $q=3^e$ and $c(x)=x^3+c_2x+c_3$. Then $c(x)$ is irreducible over $\mathbb{F}_q$ if and only if $-c_2$ is a square in $\mathbb{F}_q^\times,$ and ${\rm Tr}_{\mathbb{F}_q/\mathbb{F}_3}(c_3/\eta^3)\neq0$ where $\eta^2=-c_2$. -The factorization of a quartic should depend on two square roots and one cube root, at least when ${\rm char}(\mathbb{F}_q)\neq2,3$. Given that 1/4 of quartic -polynomials are irreducible (in the limit as $q\to\infty$) one may guess that there is an irreducibility test of the form: -$x^4+ax^3+bx^2+cx+d$ is irreducible over $\mathbb{F}_q$ if and only if $f$ or $g$ is a square, and $h$ is a cube; where $f,g,h$ are given rational functions of $a,b,c,d$. [Thus irreducibility should occur with limiting probability $(1-(1/2)^2)\times 1/3=1/4$, as $q\to\infty$; agreeing with the correct limiting probability.] -Question. Are expressions for $f,g,h$ known? What if ${\rm char}(\mathbb{F}_q)=2,3$? [I should say: "f is a square or g is a nonsquare". The probabilistic argument still works.] -Stephen Glasby - -REPLY [10 votes]: These results have been known for many years. Here are some references and further developments; full bibliographic details are at the end of this answer. -Reducibility criteria for cubics over finite fields of characteristic at least $5$ were given already in 1906 by Dickson. In fact he gave conditions for a cubic to have each of the possible factorization types. Reducibility criteria for cubics in characteristic $2$ were given in 1966 by Berlekamp, Rumsey and Solomon (see also their 1967 paper). Those criteria are slightly different than your Theorem A. Your Theorems A and B appear to have been published for the first time in 1975 by Williams, who also gave criteria for each of the possible factorization types. -Reducibility criteria for quartics over prime fields of odd order were given by Skolem in 1952. In 1969, Leonard gave a different proof of Skolem's result, which extends at once to non-prime fields of odd order. Finally, reducibility criteria for quartics in characteristic $2$ were given by Leonard and Williams in 1972. -Your Theorems A and B can be generalized in various ways. Bluher's 2004 paper generalizes your Theorem A to polynomials of the form $x^{p^k+1}+ax+b$. In a series of papers culminating in 1980, Agou determined all instances when $f(L(x))$ is irreducible over $\mathbf{F}_{p^s}$, where $f,L\in\mathbf{F}_{p^s}[x]$ and all terms of $L(x)$ have degree $p^i$ with $i\ge 0$. The case $f=x+c_3$ and $L=x^3+c_2 x$ with $p=3$ yields your Theorem B. Subsequently, a simpler proof of Agou's results was given by Cohen in 1982; following two related papers by Moreno, Cohen gave an even simpler proof in 1989. -References: -S. Agou, Irréducibilité des polynômes $f(\sum_{i=0}^m a_i X^{p^{ri}})$ sur un corps fini $\mathbf{F}_{p^s}$, Canadian Mathematical Bulletin 23 (1980), 207-212 -E. R. Berlekamp, H. Rumsey and G. Solomon, Solutions of algebraic equations in field of characteristic 2, Jet Propulsion Lab. Space Programs Summary No. 4 (1966), 37-39 -E. R. Berlekamp, H. Rumsey and G. Solomon, On the solution of algebraic equations over finite fields, Information and Control 10 (1967), 553-564. -A. W. Bluher, On $x^{q+1}+ax+b$, Finite Fields and their Applications 10 (2004), 285-305 -S. D. Cohen, The irreducibility of compositions of linear polynomials over a finite field, Compositio Mathematica 47 (1982), 149-152 -S. D. Cohen, The reducibility theorem for linearised polynomials over finite fields, Bulletin of the Australian Mathematical Society 40 (1989), 407-412 -L. E. Dickson, Criteria for the irreducibility of functions in a finite field, Bulletin of the American Mathematical Society 13 (1906), 1-8 -P. A. Leonard, On factoring quartics (mod $p$), Journal of Number Theory 1 (1969), 113-115 -P. A. Leonard and K. S. Williams, Quartics over GF($2^n$), Proceedings of the American Mathematical Society 36 (1972), 347-350 -Th. Skolem, The general congruence of 4th degree modulo $p$, $p$ prime, Nordisk matematisk tidskrift 34 (1952), 73-80 -K. S. Williams, Note on cubics over GF($2^n$) and GF($3^n$), Journal of Number Theory 7 (1975), 361-365<|endoftext|> -TITLE: Explicit description of the stack associated to a groupoid -QUESTION [5 upvotes]: Let $\{X_1 \rightrightarrows X_0\}$ be a smooth groupoid object in the category of affine schemes ($X_0 \to X_1$, $X_1 \to X_1$ and $X_1 \times_{X_0} X_1 \to X_1$ also belong to the datum). Equivalently, we have a smooth commutative Hopf algebroid. Let $X$ be the associated algebraic stack. Thus, we have a presentation $X_0 \twoheadrightarrow X$ (which is smooth, surjective and affine) with $X_0 \times_X X_0 = X_1$, and $X$ is geometric. Every geometric stack arises this way. -Question. If $Y$ is a scheme, how can we describe $X(Y)$ explicitly in terms of $Y$ and the $X_i$? -This should be somewhere in the literature? For example, if $X_0 = \mathrm{Spec}(\mathbb{Z})$, then $X_1$ is a group scheme, $X$ is its classifying stack, so that $X(Y)$ consists of $X_1$-torsors on $Y$, right? -In general, I expect that the answer will be some kind of "torsors under the groupoid". A morphism $Y \to X$ may be pulled back to $Y_0 \to X_0$, where $Y_0 \to Y$ is smooth, surjective and affine. Conversely, if $Y_0 \to Y$ is smooth, surjective and affine, and $Y_0 \to X_0$ is a morphism, then an extension to $Y \to X$ corresponds to a descent datum of $Y_0 \to X_0 \to X$ with respect to the fpqc cover $Y_0 \to Y$. But this still involves $X$. How can we get rid of $X$ in the description of morphisms $Y \to X$? - -REPLY [3 votes]: Niels already links to references that answer the question, but I'll briefly summarize them. -If $X$ is a stack presented by a groupoid $X_1 \rightrightarrows X_0$ then a map from a scheme $S$ into $X$ induces a groupoid presentation $S \mathop{\times}_X X_1 \rightrightarrows S \mathop{\times}_X X_0$ of $S$. Conversely, suppose that we have a groupoid presentation $S_1 \rightrightarrows S_0$ of $S$ and compatible maps $S_i \rightarrow X_i$ (such that the two maps $S_1 \rightarrow S_0$ are induced by pullback from the two maps $X_1 \rightarrow X_0$) then there is an induced map $S \rightarrow X$. -The associated stack of the groupoid $X_\bullet$ is therefore the fibered category $X$ for which $X(S)$ is the category of groupoid presentations of $S$ with a map to $X_\bullet$ satisfying the above parenthetical condition.<|endoftext|> -TITLE: When does a group action on a k-algebra induce an algebraic action on the spectrum? -QUESTION [5 upvotes]: This question arose from my last question, which I considered answered - from the comments, however, it is obvious that the answer is only complete in characteristic zero, and I am trying to understand why. -Let $A$ be a $\Bbbk$-algebra1, where $\Bbbk$ is some algebraically closed field. Let $G$ be a reductive algebraic group. If $G$ acts algebraically on $X=\newcommand{\Spec}{\mathrm{Spec}}\Spec(A)$, then it induces an action of $G$ on $A$. -On the other hand, assume that $G$ acts on $A$ by $\Bbbk$-algebra automorphisms. For any maximal ideal $\mathfrak m\subset A$ and any $g\in G$, the image $g.\mathfrak m$ is again a maximal ideal. This defines an action of $G$ on the closed points of $X$ and for any $f\in A$, note that the image of $f$ in $A/g.\mathfrak m=\Bbbk$ is the same as the image of $g^{-1}.f$ in $A/\mathfrak m=\Bbbk$, -$$\begin{matrix} -g^{-1}.f & \in & A & \xrightarrow{\quad g\quad} & A & \ni & f \\ -&& \downarrow && \downarrow && \\ -g^{-1}.f+\mathfrak m & \in & A/\mathfrak m & \xrightarrow{\quad \sim\quad} & A/g.\mathfrak m & \ni & f+g.\mathfrak m -\end{matrix} -$$ -so if $\mathfrak m$ is viewed as a closed point of $X$, we have the familiar formula $(g^{-1}.f)(\mathfrak m) = f(g.\mathfrak m)$. This is good, but I forgot to ask myself (and now I am asking you): -When is this action algebraic? -By this, I mean that there is a morphism $G\times X\to X$ of $\Bbbk$-schemes (or varieties) which gives the above action on closed points. -The first assumption should probably be that each $f\in A$ is contained in a finite-dimensional $G$-module. because this property holds when the action on $A$ comes from an algebraic action on $X$. On the other hand, I suspect that one will need at least characteristic zero (or more generally, some separability condition). However, I don't know exactly how to put this together. -1 You may assume $A$ finitely generated over $\Bbbk$ and reduced, or even a domain, but I have a feeling that it won't matter much whether we deal with varieties or $\Bbbk$-schemes. - -REPLY [2 votes]: If I understand the question correctly, you have a map of schemes $G \times X \to X$, and the corresponding map of $k$-points is a group action (meaning that the obvious two maps $G(k) \times G(k) \times X(k) \to X(k)$ coincide), but you are not sure that it is a group action in the category of schemes. In other words, you fear that you may have two maps $G \times G \times X \to X$ which coincide on $k$ points but not as maps of schemes. -This certainly can't happen if $G$ and $X$ are reduced. So, if you are talking about varieties, there is no issue. It's not obvious to me what happens when $G$ is reduced (which is automatic in characteristic zero) but $X$ isn't. - -Based on comments below, and on reading the motivating question, I didn't understand right. The question is, given an action $G(k) \times X(k) \to X(k)$, so that $g \times X(k) \to X(k)$ is algebraic for every $k \in G(k)$, can we conclude that it comes from an algebraic map $G \times X \to X$. But I don't think there is any good way to force this. For example, suppose that $k$ has a nontrivial automorphism $\sigma$ and $G$ is defined over the fixed field of $\sigma$. (Think of complex conjugation.) Then $\sigma$ induces an automorphism of $G(k)$ as an abstract group. Take any algebraic action $G \times X \to X$ and compose with the automorphism of $k$ to get a very nonalgebraic action of $G(k)$ on $X(k)$.<|endoftext|> -TITLE: Word length in the symmetric group -QUESTION [6 upvotes]: Let $n \geq 1$ and let $H_n$ be a 2-Sylow subgroup of the symmetric group $\mathrm{Sym}(2^n)$. Let also consider the cycle $\gamma_n = (1, \ldots, 2^n)$ of order $2^n$. -If we assume moreover that $H_n$ contains the transposition $(1,2)$ then $\Sigma_n = \{{\gamma_n^{\pm1}\}} \cup H_n$ is a generating set of $\mathrm{Sym}(2^n)$, and we can consider the word length on $\mathrm{Sym}(2^n)$ associated to $\Sigma_n$, defined by $$ |g|_{\Sigma_n} = \min \{ k \geq 0 : \exists s_1, \ldots, s_k \in \Sigma_n; \sigma = s_1 \cdots s_k \},$$ for $g \in \mathrm{Sym}(2^n)$. -Now let $m_n$ be the maximum of the $|g|_{\Sigma_n}$ when $g$ ranges over $\mathrm{Sym}(2^n)$. -Question 1: is the sequence $(m_n)_n$ bounded ? (this may not depend on the choice of the Sylow subgroup $H_n$) -(EDIT: the answer to Question 1 is no according to Peter Mueller's answer below) -If the answer to Question 1 is no, then -Question 2: are there known bounds for $m_n$ ? -(EDIT: I am interested in an upper bound, i.e. an upper bound of the diameter of the Cayley graph $\mathrm{Cay}(\mathrm{Sym}(2^n), \Sigma_n)$ of the symmetric group $\mathrm{Sym}(2^n)$.) - -REPLY [3 votes]: I think we can get all transpositions with words at length at most 5 over $\Sigma_n$, which will give an upper bound of $5(2^n-1)$. This can easily be improved slightly - for example, you can use elements of $H_n$ for the first and last terms of the product of length $(2^n-1)$, but I don't know whether we can do better than $O(2^n)$ for the upper bound. It seems likely that you could. -To show how to get the transpositions, we can assume that $H_n$ contains $(1,2)$, $(3,4)$, $(5,6)$, etc, and by conjugating these by $\gamma_n$, we get $(2,3)$, $(4,5)$, $(6,7)$, etc. -I claim that every transposition $(i,j)$ in $S_n$ is a conjugate to one of $(1,2)$, $(2,3)$, $(3,4)$, by an element of $H_n$, which will give the required bound. Use induction $n$. We can assume that $\{1,2,\ldots,2^{n-1}\}$ and $\{2^{n-1}+1,\ldots,2^n\}$ are blocks of imprimtivity for $H_n$. If $i$ and $j$ are in the same block, then the claim follows by induction. Otherwise, since the stabilizer of the block system in $H_n$ is $H_{n-1} \times H_{n-1}$, where $H_{n-1}$ acts transitively on the block it fixes, we can conjugate $(2^{n-1},2^{n-1}+1)$ to$(i,j)$ by and element of $H_{n-1} \times H_{n-1}$.<|endoftext|> -TITLE: About two 'negative' continued fractions whose sum equals $1$ -QUESTION [5 upvotes]: Letting $a_1,a_2,\cdots,a_r$ be integers which are larger than or equal to $2$, let us define -$$[a_1,a_2,\cdots,a_r]=\cfrac{1}{a_1-\cfrac{1}{a_2-\cfrac{1}{\ddots-\cfrac{1}{a_r}}}}$$ -(Note that the negative signs are used) -Also, let $X, Y, Z$ be positive integers which satisfy -$$Z\lt X+Y,\ Z\gt X,\ Z\gt Y$$ -and let -$$\frac XZ=[a_1,a_2,\cdots,a_r],\ \frac YZ=[b_1,b_2,\cdots,b_s].$$ -Then, here is my question. -Question : Is the following true? -"There exist $r^{\prime}\le r, s^{\prime}\le s$ such that -$$[a_1,a_2,\cdots,a_{r^{\prime}}]+[b_1,b_2,\cdots,b_{s^{\prime}}]=1$$ -for any $(X,Y,Z)$." -Remark : Observing the initial numbers is not sufficient because the nearer to $1$ the value $\frac XZ+\frac YZ$ is, the harder it is to find the answer (see example 2). -This question has been asked previously on math.SE without receiving any answers. -Examples : - -$\frac XZ=\frac 37=[3,2,2]$ and $\frac YZ=\frac 57=[2,2,3]$ leads $[3]+[2,2]=\frac 13+\frac 23=1$ where $\frac 37+\frac 57=\frac 87\approx 1.143$ -$\frac XZ=\frac{901}{2067}=[3,2,2,4,2]$ and $\frac YZ=\frac{1170}{2067}=[2,5,2,2,3]$ leads $[3,2,2,4]+[2,5,2,2]=\frac{10}{23}+\frac{13}{23}=1$ where $\frac XZ+\frac YZ=\frac{2071}{2067}\approx 1.002.$ - -Motivation : I've got an algorithm to find $b_1,b_2,\cdots,b_s$ such that -$$1-x=[b_1,b_2,\cdots,b_s]$$ -for any given $x=[a_1,a_2,\cdots,a_r]$. -Algorithm : Supposing that $2^r$ represents $r$-consecutive $2$s, I'm going to write -$$[a_1,a_2,\cdots,a_r]=[2^{q_1},p_1,2^{q_2},p_2,\cdots,2^{q_s},p_s,2^{q_{s+1}}]$$ -where $p_i\ge 3\in \mathbb N, q_i\ge 0\in \mathbb Z$. For example, $[2,2,5,3,2,4]=[2^2,5,2^0,3,2^1,4,2^0]$. -Then, the algorithm is -$$1-[2^{q_1},p_1,2^{q_2},p_2,\cdots,2^{q_s},p_s,2^{q_{s+1}}]$$ -$$=[(q_1+2),2^{(p_1-3)},(q_2+3),2^{(p_2-3)},(q_3+3),2^{(p_3-3)},\cdots,(q_s+3),2^{(p_s-3)},(q_{s+1}+2)].$$ -After getting this algorithm, I reached the above expectation. I can neither find any counterexample even by using computer nor prove that the expectation is true. Can anyone help? - -REPLY [4 votes]: $\let\ds\displaystyle$Here is a different proof of the fact proven by Alexey Ustinov; it was found by Alexey Volostnov. -Assume that $[a_1,\dots,a_r]=\frac ab$, $[b_1,\dots,b_s]=\frac cd$, $\frac ab+\frac cd>1$, and $d\leq b$. We will prove that if we delete $a_r$ the resulting fractions will sum up to at least one; the conclusion follows. Notice that the case $r=1$ is impossible: otherwise we would have $\frac ab=\frac1{a_r}$ and $\frac cd\leq 1-\frac 1d\leq 1-\frac 1b=1-\frac ab$. -Let $\frac pq=[a_1,\dots,a_{r-1}]$, and assume that $\frac pq+\frac cd<1$. Then we have $\frac pq+\frac cd\leq 1-\frac 1{qd}$, $\frac ab+\frac cd\geq 1+\frac1{bd}$, and $aq-bp=1$. This implies -$$ - \frac 1{bq}=\frac ab-\frac pq\geq \frac1{bd}+\frac1{qd}, -$$ -in particular, $bq -TITLE: Where to find digitized old papers on the internet? -QUESTION [9 upvotes]: Following Murphy's law for the published material: -"The paper you need is too old to be in the arXiv, it is not in any online database which your institution has subscription to, and... it not even in the library!!!" -I really need -Z. Ran. On Subvarieties of Abelian Varieties. $Invent. Math.$ 62 (1981) p. 459--479. -Is there anyone that could help? - -REPLY [2 votes]: Perhaps it is worth mentioning that several important journals have been digitized by GDZ. It is one of reasonable places where to look for older papers. (If you can't find it through the subscription of your institution.) -Here is list of mathematical journals available at GDZ. -In fact, Google Scholar search finds several versions of the paper mentioned in the question. One of them is eudml link which links further to GDZ. -Several other reasonable possibilities are mentioned in the answers to this question: Finding a paper.<|endoftext|> -TITLE: Topological Problems Solved by Lattice Duality -QUESTION [6 upvotes]: It is well known the success of lattice dualities (as Pontryagin duality for abelian groups, Stone duality for Boolean algebras and Priestley duality for distributive lattices) to solve algebraic problems using their representation in the correspondent topological space. -Natural duality theory provides a general framework that encompasses many important and well-known representation theorems in algebra. -In fact, when dualities are useful, the duals are simpler structures: the dual of a free algebra is a direct exponent. -In Pontryagin duality, the structure of the dual space is the same when adding the topology. As the semilattice duality Hofmann-Mislove-Starlka, the Pontryagin duality is for algebras for which the basic operations are homomorphisms. The same happens with vector spaces. -Gelfand Duality is a representation theorem for commutative Banach algebras describing them as algebras of continuous functions. Moreover, for commutative $C^{*}$-algebras, this representation is an isometric isomorphism. It is also a classical example of the contributions of lattice duality to topology. -A famous example is the proof of Tychonoff theorem without using the axiom of choice: if you pass from topological spaces over to the algebraic side and use locales, then one can show that a product of compact locales is compact, without choice (cf. Peter Johnstone's book Stone spaces). -Otherwise, the main use of lattice dualities seems to be rather one-sided, i.e., (citing "Natural Dualities for the Working Algebraist") to translate algebraic problems, usually stated in an abstract symbolic language, into dual, topological problems, where geometric intuition comes to our help. -Are there other clear examples of problems of topological nature solved by lattice duality? Thank you in advance. - -REPLY [4 votes]: The idea that these dualities are only used in the direction of proving algebraic results using topological spaces is not correct. Any sort of completion or compactification process (that I can think of at least) can be constructed using special knids of filters or ultrafilters on lattices. These constructions include the Stone-Cech compactification, the Hewitt realcompactification, the completion of a uniform space, the Smirnov compactification of a proximity space, and the completion of the hyperspace of a uniform space and other constructions. There are countless examples of using ultrafilters and filters on Boolean algebras and lattices to prove results about topology. -For example, one can obtain the characterization of weakly compact cardinals given in [1] using a Boolean algebraic characterization of when a uniform space generated by equivalence relations is supercomplete. This result says that a cardinal $\kappa$ is weakly compact if and only if the space $2^{\kappa}$ with an appropriate uniformity is supercomplete. Furthermore, one can use the same technique to obtain characterizations of strongly compact cardinals. - -Artico, Giuliano (I-PADV-PM); Marconi, Umberto (I-PADV-PM); Pelant, Jan (CZ-AOS) -On supercomplete $\omega_\mu$-metric spaces. -Bull. Polish Acad. Sci. Math. 44 (1996), no. 3, 299–310.<|endoftext|> -TITLE: Poincare recurrence theorem and convergence on compact metric spaces -QUESTION [6 upvotes]: I am looking for a proof (or a reference to a proof) of the following theorem: -Let $X$ be a compact metric space with metric $d$, endow $X$ with the Borel $\sigma$-algebra and a probability measure $\mu$. Let $T\colon X\to X$ be a continuous map which is $\mu$-preserving. Then for $\mu$-almost every $x\in X$ there is a sequence $n_k\to\infty$ in $\mathbb{N}$ such that $T^{n_k}x\to x$ as $k\to\infty$. -I tried already for some time and looked for references, unfortunately unsuccessful. It seems that one needs to find the right formulation to be able to use the Poincare recurrence theorem. Any ideas? - -REPLY [3 votes]: This is part of the statement of Proposition 4.1.18 in Introduction to the Modern Theory of Dynamical Systems by Katok and Hasselblatt. As far as I am aware there is essentially only one proof, which is that given by Anthony. It generalises to the following result which I have found useful once or twice but which is less frequently stated: -Proposition: Let $T$ be a measure-preserving transformation of the probability space $(X,\mathcal{F},\mu)$, $Y$ a second-countable topological space, and $f \colon X \to Y$ measurable. Then $$\mu\left(\left\{x \in X \colon f(x) \in \overline{\left\{f(T^nx) \colon n \geq 1\right\}}\right\}\right)=1.$$ -Proof: Let $\{U_k\colon k \geq 1\}$ be a countable base for the topology of $Y$. For each $k \geq 1$ we have $$\mu\left(\left\{x \in f^{-1}(U_k) \colon f(T^nx ) \in U_k \text{ infinitely often}\right\}\right)=\mu\left(f^{-1}(U_k)\right)$$ -by the Poincare recurrence theorem. Taking an appropriate intersection over all $k$ we find that there is a set of full measure on which every $x$ returns infinitely often to every $f^{-1}(U_k)$ to which $x$ originally belonged. Since $\{U_k\}$ is a base the result follows. -(Your question corresponds to the case $Y:=X$ and $f:=\mathrm{id}$.)<|endoftext|> -TITLE: The non-simplicity of $SO(4)$ and $A_4$ -QUESTION [25 upvotes]: It is well known that the alternating group $A_n$ is simple unless $n=4$. It is likewise well known that the special orthogonal group $SO(n)$ is essentially simple unless $n=4$ (specifically, the group $SO(n)$ is simple for odd $n$ and the group $SO(n)/\{\pm I\}$ is simple for even $n\neq 4$). -My question is: are these two facts equivalent? The non-simplicity of $SO(4)$ can be proved by observing that the double-cover of $SO(4)$ is $SU(2)\times SU(2)$ which, being a direct product, is very much not simple. This double-cover is closely related to properties of the quaternions (see Stillwell's Naive Lie Theory). Is there an analogous proof of the non-simplicity of $A_4$ based on a geometric structure related to the quaternions? -P.S. This relationship is an example of the "field of one element" heuristic. Can that be formalized? - -REPLY [24 votes]: First, the non-simplicity of $A_4$ has a very beautiful proof, which I heard summarized by Gromov as : $2+2=4$, or rather $4=2+2$. -Or rather, there are 3 ways to pair 4 objects 2 by 2. The action of $S_4$ on the 4 objects therefore induces an action on the 3 pairings, hence a non-trivial morphism $S_4\to S_3$, whose kernel intersects $A_4$ in a non-trivial simple subgroup. -The sequel is a bit rough, but when looking at $SO(4)$, you can try to reinterpret the same proof, using the elements of a basis instead of the permuted objects. $SO(4)$ naturally acts on the set of direct orthonormal bases of $\mathbb{R}^4$; with each basis comes 3 decompositions of $\mathbb{R}^4$ into pairs of orthogonal planes (which should correspond in some sense to 3 complex structures satisfying the quaternionic relations, probably using that the planes are endowed with particular bases). So $SO(4)$ acts on such triples of complex structures, which if I remember well identifies with $SO(3)$, the unit tangent bundle $S^2$ (choosing the first complex structure $I$ is picking a point on the unit sphere in purely quaternionic numbers, then you have left to choose an orthogonal pure unit quaternion). Considering dimension you get a relatively large non-simple subgroup. - -REPLY [10 votes]: Here is an argument using the finite quaternion group. -Let $Q_8$ be the quaternion group of order $8$, namely $$Q_8= \{{\pm 1, \, \pm i, \, \pm j, \, \pm k\}}.$$ -It is well known that $\textrm{Aut}(Q_8)=S_4$, and this is usually proven by constructing an explicit isomorphism between $\textrm{Aut}(Q_8)$ and the symmetry group of the cube, see for instance here. -On the other hand, since $\textrm{Z}(Q_8)=\{\pm 1\}$, one also has $$\textrm{Inn}(Q_8)=Q_8/\textrm{Z}(Q_8)=V_4,$$ -where $V_4$ denotes the Klein group of order $4$, which is isomorphic to $C_2 \times C_2$. -Finally, the inner automorphism group is always a normal subgroup of the full automorphism group, so the argument above shows that $S_4$ contains a normal subgroup isomorphic to $V_4$. Using again the identification of $\textrm{Aut}(Q_8)$ with the symmetry group of the cube, it is no difficult to show that such a normal $V_4$ consists of even permutations, in other words it is contained in $A_4$. -This shows that $A_4$ is not simple. - -REPLY [8 votes]: Here are some general remarks which hold for arbitrary semisimple Lie algebras ${\mathfrak g}$ relating its algebraic properties to that of its Weyl group $W$. - -${\mathfrak g}$ is simple if and only if the standard linear action of its Weyl group $W$ is irreducible. However, $W$ itself might still split nontrivially as a direct product (this happens in few cases); let's call such $W$ "reducible". This was analyzed by Luis Paris in http://arxiv.org/pdf/math/0412214v2.pdf. The main focus of his paper was on infinite Coxeter groups, but he also classified reducible finite Coxeter groups whose root system is irreducible (section 7). In all cases, the factor of $W$ by its center is irreducible, i.e., does not split nontrivially as a direct product. Therefore, the statement is: ${\mathfrak g}$ is simple iff $W/Z(W)$ is irreducible. -As for simplicity of $W$ itself, in the case of "classical" root systems, $W$ always has the form of semidirect product of a permutation group and a finite abelian group. Thus, if one is willing to divide by the abelian normal subgroup and then pass to the alternating group, then in the classical case one does obtains that simplicity of ${\mathfrak g}$ is equivalent to simplicity of a certain "subquotient" of $W$. I do not know enough about exceptional Coxeter groups to make a similar conclusion in general.<|endoftext|> -TITLE: On Turan's theorem -QUESTION [6 upvotes]: Turan's theorem provides minimum number of edges of a graph on $n$ vertices to surely contain a clique of a prescribed size. This has been generalized to regular graphs. -What additional specializations have been made in the literature if the graph is regular and contains additional algebraic structure? - -REPLY [6 votes]: There are versions of Turan's theorem that add spectral assumptions to regularity. -This one is due to Sudakov, Szabó, & Vu (CiteSeer link): -A graph is called an $(n, d, \lambda)$-graph if it has $n$ vertices, is $d$-regular, -and $\max_{i \ge 2} |\lambda_i| \le \lambda$, where $\lambda_i$ are the -adjaceny-matrix eigenvalues, -largest to smallest. -If for some $r \ge 2$, $d^r \gg \lambda n^{r-1}$, -then every $(n, d, \lambda)$-graph contains a clique of size $r+1$. -They establish bounds on the size of the largest $K_{r+1}$-free subgraph. -In some sense this requires that the 2nd-largest eigenvalue is sufficiently small.<|endoftext|> -TITLE: Quantitative lower bounds related to Zhang's theorem on bounded gaps -QUESTION [20 upvotes]: Let $\mathcal{H}=\left\{ h_{i}\right\} _{i=1}^{k}$ be an admissible set, and define $$\pi_{\mathcal{H}}(x)=\left|\left\{ n\leq x\ :\ \exists\ i,j\leq k,\ i\neq j\ \text{such that both }n+h_{i},\ n+h_{j}\ \text{are prime}\right\} \right|.$$ The Hardy-Littlewood $k$-tuple conjecture implies that $$\pi_{\mathcal{H}}(x)\sim C_{\mathcal{H}}\frac{x}{\log^{2}x},$$ where $C_{\mathcal{H}}$ is a constant depending on $\mathcal{H},$ and the Selberg sieve can be used to prove that $$\pi_{\mathcal{H}}(x)\ll_{\mathcal{H}}\frac{x}{\log^{2}x}.$$ Yitang Zhang recently proved that for any sufficiently large admissible set $\mathcal{H}$, we have $$\lim_{x\rightarrow\infty}\pi_{\mathcal{H}}(x)\rightarrow\infty.$$ (the current lower bound on $\mathcal{H}$ can be found here) - -Question: Does Zhang's work give a lower bound on $\pi_{\mathcal{H}}$ with the correct order of magnitude? That is, can we prove that $$\pi_{\mathcal{H}}(x)\gg_{\mathcal{H}}\frac{x}{\log^{2}x}?$$ - -REPLY [18 votes]: In the main theorem of this recent paper of Pintz, Zhang's method is used to show that (for $k$ large enough), there are $\gg_{\mathcal H} \frac{x}{\log^k x}$ values of $n \le x$ such that two of the $n+h_i$ are prime and the remaining $n+h_j$ are almost prime (have $O_k(1)$ prime factors, all of which are $\gg x^{c_k}$ for some $c_k>0$ independent of $x$). This is the correct asymptotic (up to multiplicative constants) if one enforces the almost primality conditions, but has the wrong power of log in the denominator if those conditions are removed. -In a slightly different direction, in this paper of Goldston, Pintz, and Yildirim (a followup to their most famous paper), it is shown that for any fixed $\varepsilon>0$, there are $\gg_\varepsilon \frac{x}{\log x}$ primes $p_n$ less than $x$ such that $p_{n+1}-p_n \leq \varepsilon \log p_n$. Again, this is the correct asymptotic up to multiplicative constants. -I have a graduate student who has recently started looking at improvements to these results, with some encouraging preliminary findings, but the work is not yet completed. It is indeed unlikely that one will be able to obtain the optimal asymptotic here unconditionally, but some improvement upon the partial results given above look plausible.<|endoftext|> -TITLE: Work of plenary speakers at ICM 2014 -QUESTION [89 upvotes]: The next International Congress of Mathematicians (ICM) will take place in 2014 in Seoul, Korea. The present question is meant to gather brief overviews of the work of the plenary speakers for the ICM 2014. -More precisely, anybody who feels qualified to give a short description of the work of one of the plenary speakers at ICM 2014 is invited to put that in an answer below, or to add to the existing answers. Ideally, there would be a single answer dedicated fully to each speaker. Answers which summarize this thread are also very welcome. -Most importantly, this thread is meant to be informative and educational for anyone in the mathematical community. Therefore, please strive to make the answers broadly accessible. -Background: A similar, very successful MathOverflow question was asked a few years ago concerning the work of the plenary speakers for the International Congress of Mathematicians in 2010. Please look there to get an idea of what could be achieved with the present question. -List of plenary speakers at ICM 2014 -For completeness, here is a list of the scheduled plenary speakers at ICM 2014, (copied from here): - -Ian Agol, University of California, Berkeley, USA -James Arthur, University of Toronto, Canada -Manjul Bhargava, Princeton University, USA -Alexei Borodin, Massachusetts Institute of Technology, USA -Franco Brezzi, IUSS, Pavia, Italy -Emmanuel Candes, Stanford University, USA -Demetrios Christodoulou, ETH-Zürich, Switzerland -Alan Frieze, Carnegie Mellon University, USA -Jean-François Le Gall, Université Paris-Sud, France -Ben Green, University of Oxford, UK -Jun Muk Hwang, Korea Institute for Advanced Study, Korea -János Kollár, Princeton University, USA -Mikhail Lyubich, SUNY Stony Brook, USA -Fernando Codá Marques, IMPA, Brazil -Frank Merle, Université de Cergy-Pontoise/IHES, France -Maryam Mirzakhani, Stanford University, USA -Takuro Mochizuki, Kyoto University, Japan -Benoit Perthame, Université Pierre et Marie Curie, France -Jonathan Pila, University of Oxford, UK -Vojtech Rödl, Emory University, USA -Vera Serganova, University of California, Berkeley, USA - -REPLY [12 votes]: Fernando Coda-Marques -Together with Andre Neves he proved the Willmore conjecture. This states that the Clifford torus in the 3-sphere minimizes the Willmore energy, with energy $2\pi^2$. To prove this conjecture, they show that for 5-parameter sweepouts of $S^3$ by surfaces, with certain boundary conditions, the min-max area is $2\pi^2$. This theorem has other consequences besides the Willmore conjecture.<|endoftext|> -TITLE: Approximation of infinite set in generic extension -QUESTION [5 upvotes]: Suppose $M$ is a c.t.m and suppose $P$ is $Fn(I,2)$ where $I$ is infinite. Now suppose $G$ is $P$-generic, and $A \in M[G]$ is infinite set. -Is it guaranteed that the exist $B \in M$ such that $B \subset A$, and $B$ is infinite ? -If not, is there any 'reasonable' condition that we can add to $M$ or $P$ to get the desired result ? (Like CH is $M$, or large enough $I$) - -REPLY [5 votes]: Theorem: Let $\kappa$ be a regular cardinal in $M$. Let $\mathbb{P} \in M$ be a separative partial order of size $\kappa$, and let $G$ be $\mathbb{P}$-generic over $M$. Then the following are equivalent: -(1) There is some $p \in G$ such that in $M$, $\mathbb{P} \restriction p$ has a dense subset of size $< \kappa$. -(2) For all sets of ordinals $A \in M[G]$ of size $\kappa$, there is $B \subseteq A$ such that $B \in M$ and $|B| = \kappa$. - -Per Mohammad's request, I will give a proof. I think this is a nice theorem because it characterizes a certain combinatorial property of partial orders in terms of model comparison with generic extensions, like the distributivity properties. I have found some use of it in comparing submodels of generic extensions. -For a partial order $P$, let $d(P)$ be the least size of a dense subset of $P$. For a given $P$, and $q \leq p$ in $P$, $d(P \restriction q) \leq d(P \restriction p)$. Call an element of $P$ "$d$-stable" if $(\forall q \leq p) d(P \restriction q) = d(P \restriction p)$. By well-foundedness, the $d$-stable elements are dense. -Let $G$ be $P$-generic over $M$. Suppose there is $p \in G$ with $d(P \restriction p) < \kappa$, and let $D \subseteq P \restriction p$ witness this. Then $p \Vdash \kappa$ is regular. Let $A \in M[G]$ be a set of ordinals of size $\kappa$, and let $\dot{A}$ be a name for it. $A = \{ \alpha : (\exists q \in D \cap G) q \Vdash \check{\alpha} \in \dot{A} \}$, so for some $q \in D \cap G$, $B = \{ \alpha : q \Vdash \check{\alpha} \in \dot{A} \}$ has size $\kappa$. -Suppose now that $p \in G$ is $d$-stable, and $d(P \restriction p) = \kappa$. We will find a set of ordinals $A \in M[G]$ unbounded in $\kappa$ that contains no unbounded set from $M$. This suffices because, although $\kappa$ may no longer be regular, the function $f : \kappa \to A$ defined by $f(\alpha) =$ the least $\beta \in A$ above $\alpha$, is an object of size $|\kappa|$, and if it had a size-$\kappa$ subset $g \in M$, then the range of $g$ would be an unbounded subset of $A$ from $M$. -Let $D \subseteq P \restriction p$ be dense, and recursively construct $D' \subseteq D$ also dense, and with the property that for all $q \in D'$, $| \{ r \in D' : r \geq q \}| < \kappa$ (exercise). Let $D' = \langle p_\alpha : \alpha < \kappa \rangle$, and in $M[G]$ consider $A = \{ \alpha : p_\alpha \in D' \cap G \}$. $A$ is unbounded in $\kappa$ (exercise). If we had $q \in D'$ and an unbounded $B \in M$ such that $q \Vdash \check{B} \subseteq \dot{A}$, then $\{ p_\alpha : \alpha \in B \} \subseteq D'$, and by separativity, $q \leq p_\alpha$ for all $\alpha \in B$, which contradicts the property of $D'$.<|endoftext|> -TITLE: Why are optimization problems often called "programs"? -QUESTION [33 upvotes]: Why are optimization problems often called programs? - -linear programming -geometric programming -convex programming -Integer programming -... - -REPLY [2 votes]: Solving an optimization problem is not programming in any sense. However, the results of the optimization are then used as key factors in the making of decision related to resources or strategy. And that is the "programming" part. -So this is a case of metonymy: naming something after something else which has an associated meaning. -Although linear programming does precede computer programming, the term program as a "list of things to be done" precedes linear programming! -For instance, a symphony orchestra's formal concert performance has a program. This is very much like a computer program. First we play this, then we play that, then there is an intermission, and so forth. A copy of the program is usually available to members of the audience. -The word program is made up of "pro" (beforehand) and "gram" (write): literally writing something down before doing it with the intent of closely sticking with what is written. -Once we have solved the optimization problem, then we chart a "program" for our organization to take specific steps with specific resources.<|endoftext|> -TITLE: Euclidean real quadratic fields -QUESTION [10 upvotes]: It is known that, under GRH, a real quadratic field is Euclidean iff it is a UFD. So, assuming the conjecture of Gauss and GRH, we expect that there are infinitely many Euclidean real quadratic fields. -I want to know if there are any approaches in showing that there are infinitely many Euclidean real quadratic fields by explicitly constructing the Euclidean functions. Can we hope for a single (or finitely many) Euclidean function(s) that will do the job or we need infinitely many different Euclidean functions? Can these Euclidean functions be obtained by modifying the norm function? -Since very few non-norm Euclidean quadratic fields are known, can we at all hope for showing the infinitude of real UFD quadratic fields by finding infinitely many Euclidean quadratic fields? - -REPLY [5 votes]: The problem of finding explicit Euclidean functions is even more complicated. -Usually the euclidean property is proved indirectly, using sieve-theoretic methods (in particular, Wilson's large sieve over number fields and the Gupta-Murty lemma). For this kind of argument see Harper's work on $\mathbb{Q}(\sqrt{14})$ and Harper-Murty. -I think that the only known non-trivial example (that is, for fields which are not norm-Euclidean) of a real quadratic field with an Euclidean function is $\mathbb{Q}(\sqrt{69})$: - -D. A. Clark, "A quadratic field which is euclidean but not norm-euclidean" (1994) - -"Since very few non-norm Euclidean quadratic fields are known" I don't think this is true, depending on the meaning of "few" here. Malcolm Harper showed in his thesis that all real quadratic fields with class number $1$ and discriminant $\leq 500$ are Euclidean, but I believe this was never published.<|endoftext|> -TITLE: geometric interpretation of "Euclidean domain" -QUESTION [5 upvotes]: I cite from Wikipedia: - Commutative rings ⊃ integral domains ⊃ integrally closed domains ⊃ unique factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields -All of these properties have a well-known geometric interpretation, except perhaps "(norm) Euclidean domain". Do you know one? - -REPLY [3 votes]: Although this isn't "if and only if", one property of Euclidean domains $R$ which is not true of all PID's is that the special Whitehead group $SK_1(R)$ is trivial. -This is Theorem 2.3.2 of Rosenberg's book "Algebraic K-theory and its applications". -I leave it to others to decide whether this property is geometric.<|endoftext|> -TITLE: Deligne Weil II -QUESTION [14 upvotes]: Deligne's Weil I has been published under the title "La conjecture de Weil: I" in 1974, and Weil II in 1980. So did Deligne know in 1974 that there would be a Weil II, and can one explain the period between the two publications? - -REPLY [26 votes]: To complete Carlo's answer, I think that one thing that can explain the long -gap (in addition of the amount of difficult material in Weil II) is that Deligne -felt the need to consolidate his result of Weil I before going further. -It should be reminded that Weil I was criticized from various directions for relying on results that were not yet formally published. Those results were essentially the theory of Grothendieck of application of Etale Cohomology to L-functions (SGA 5) and the theory of Lefschetz as generalized to schemes by Grothendieck (SGA 7). At the time of the publishing of Weil I, Grothendieck had left the IHES and was not working anymore on the SGA's (nor the EGA's), and his students and colleagues were left with notes of his talks in various -states of redaction. Serre tells, for example, in a letter to Grothendieck (published by the SMF in the volume "correspondence Serre-Grothendieck"), -that Illusie, who was in charge to prepare for publication some crucial parts of SGA 5, told him he was not able to check the commutativity of certain diagrams, -commutativity considered as obvious in the notes. -So something that occupied Deligne for quite a while between 1974 and 1980 was this huge work of publishing/completing the work that Grothendieck left abruptly in 1970 (how much this work was just cleaning, proofreading and publishing, and how much it involved original mathematical work is the subject of polemics in Grothendieck's "Récoltes et Semailles"). For instance, Deligne (with others) published SGA 4+1/2 in 1977 with the stated intent to be a partial substitute for the then missing SGA 5. Deligne also worked with Katz and the second part of SGA 7, etc.<|endoftext|> -TITLE: Algebraic K-theory of odd-dimensional spheres -QUESTION [8 upvotes]: Let $A(X)$ denote the Waldhausen's algebraic K-theory of a space $X$, and let $n$ be odd. - -Are the rational homotopy groups of $A(S^n)$ known? -Is the group $\pi_{2k}(A(S^n))$ finite for all positive $k\ll n$? - -A reference (or proof sketch) would be appreciated. -EDIT: I found that the answers are stated (without reference or proof) on page 1 in "Homological stability of diffeomorphism groups" by Alexander Berglund and Ib Madsen, http://arxiv.org/abs/1203.4161. Namely, the answers to both questions is yes. - -REPLY [10 votes]: Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead. -The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in -G. Carlsson, R. Cohen, T. Goodwillie, and W. Hsiang, The free loop space and the algebraic K-theory of spaces. -K-Theory 1 (1987), no. 1, 53–82. -(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor: -$$ -\prod_{n\ge 1} Q(Y^{[n]}_{h\Bbb Z_n}) -$$ -where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z_n$ acts by cyclic permutation and ${({-})}_{h\Bbb Z_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}_{h\Bbb Z_n})$ coincide the homology groups of $Y^{[n]}_{\Bbb Z_n}$. -When $Y$ is a $j$-sphere, the rational homology is not hard to compute. -Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(*)$ by a version of Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.<|endoftext|> -TITLE: The Dissertation of F. J. van der Linden -QUESTION [9 upvotes]: Does anyone have access to the 1984 dissertation of Franciscus Jozef van der Linden under Hendrik Lenstra? It is called Euclidean Rings with two infinite primes. The theory is that this has the details on Lenstra's three non-principal norm-Euclidean classes for real quadratic fields. In case it is a matter of copying a few pages, apparently the discussion takes place in Section 5.5; this is according to the 2012 dissertation of Pierre Lezowski: -http://genealogy.math.ndsu.nodak.edu/id.php?id=170068 and -http://www.math.u-bordeaux1.fr/~plezowsk/index.php -Note that I have written to both Pierre Lezowski and Hendrik Lenstra, and something may come of that in time. On the other hand, sometimes email gets shunted to spam folders, or people lack adequate time to reply. -http://genealogy.math.ndsu.nodak.edu/id.php?id=47922 - -See also: -"The Defenestration of Ermintrude Inch" -"The Love Song of J. Alfred Prufrock" -"The Prime of Miss Jean Brodie" -"The Picture of Dorian Gray" -"The Injudicious Prayers of Pombo the Idolater" -"The Coronation of Mr Thomas Shap" - -REPLY [10 votes]: The thesis can be found in full here. -Section 5.5 begins on pdf 84/216 (p. 72). - -FYI: Because the thesis was published through Centrum voor Wiskunde en Informatica, you can simply search for it through their website: https://repository.cwi.nl/<|endoftext|> -TITLE: How general is the convergence of the exponential function's power series? -QUESTION [6 upvotes]: I asked essentially this over two weeks ago on MSE, and nothing was else was posted to that question. - -Let $\mathbf{V}$ be a Fréchet space whose underlying set is $V$. - -Let $\;\; \beta \: : \: V\times V \: \to \: V \;\;$ be a continuous bilinear map - -that has an identity element and is power-associative. - -For vectors $v$ and non-negative integers $n$, define $\hspace{.02 in}v^{\hspace{.02 in}n}\hspace{.02 in}$ in the obvious way. -Does it follow that for all vectors $\hspace{.02 in}v$, $\;\;\; \displaystyle\sum_{n=0}^{\infty} \; \left(\hspace{-0.03 in}\frac1{n!}\hspace{-0.05 in}\cdot \hspace{-0.02 in}v^{\hspace{.02 in}n}\hspace{-0.05 in}\right) \;\;\;$ exists? -If no, what if we additionally assume that $\hspace{.02 in}\beta\hspace{.02 in}$ is associative - -and/or commutative and/or every vector has an inverse? - -REPLY [6 votes]: Let me give you a counter-example with an associative $\beta$, i.e. a Frechet algebra for which the exponentials do not exist in general: The main reason is that a Frechet algebra needs not to be locally multiplicatively convex. On the Weyl algebra with two generators $Q$ and $P$ subject to the commutation relations $[Q, P] = 1$ there are several locally convex topologies possible such that the completion yields a Frechet algebra. Now it is well-known that none of them can be locally multiplicatively convex. In fact, one can show that e.g. the exponentials of quadratic expressions in the generators like $Q^2$ do not converge. -If of course you have a Frechet algebra which is locally multiplicatively convex then you have an entire calculus and hence in particular exponentials of all elements. -Edit: here is just simple argument that in an algebra with elements satisfying ccr one can not have submultiplicative seminorms: you compute $[Q, \cdots [Q, P^n] \cdot]$ directly (with $n$ commutators) and get essentially $n!$ times the identity of the algebra. On the other hand, if there would be a submultiplicative seminorm then using the submultiplicativity and a simple counting of all terms gives that the seminorm on this expression grows like $2^n$ times the seminorm of the identity. Thus the seminorm has to vanish on the identity and hence it is identically zero. It is essentially the same proof as the one to show that there is no normed algebra with canonical commutation relations (ccr). -In case you are interested in, I have a recent preprint on the arXiv where several locally convex topologies for the Weyl algebra are discussed, none of course mutliplicatively convex ;)<|endoftext|> -TITLE: On the solvable octic $x^8-x^7+29x^2+29=0$ -QUESTION [17 upvotes]: The irreducible but solvable octic, -$$x^8-x^7+29x^2+29=0\tag{1}$$ -was first mentioned by Igor Schein in this 1999 sci.math post. This does not factor over a quadratic or quartic extension, but over a 7th deg one. It can also be nicely solved using the $29th$ root of unity. Let $\omega = \exp(2\pi i /29)$ then define, -$$y = y_k = \omega^{k}+\omega^{12k}+\omega^{17k}+\omega^{28k}\tag{2}$$ -$$z_k = 4(y^3+y^2-9y-4)(y^2-2)(y-1)+9\tag{3}$$ -then I found a pair of roots of $(1)$ as, -$$x = \frac{1\color{red}{-}\sqrt{z_{1}}+\sqrt{z_{2}}+\sqrt{z_{4}}+\sqrt{z_{8}}+\sqrt{z_{16}}+\sqrt{z_{32}}+\sqrt{z_{64}}}{8} \approx 1.79106+0.8286\,i\dots$$ -$$x = \frac{1+\sqrt{z_{1}}\color{red}{-}\sqrt{z_{2}}\color{red}{-}\sqrt{z_{4}}+\sqrt{z_{8}}+\sqrt{z_{16}}\color{red}{-}\sqrt{z_{32}}+\sqrt{z_{64}}}{8} \approx 1.79106-0.8286\,i\dots$$ -and the other pairs using appropriate signs of the square roots. -Note: Of course, $y_k$ and $z_k$ are roots of two different 7th-deg eqns with integer coefficients, while $(3)$ is the 6th-deg Tschirnhausen transformation between them. (In an earlier edit, I used an alternative expression for $z_k$ by P. Montgomery found in the sci.math link, but I like this one better.) -Question: Does anyone know why $(1)$ has such a simple form, and if we can find other similar irreducible but solvable octics involving a $p$th root of unity for other prime $p$? (For some reason, this does not appear in the Kluener's database of number fields for 8T25.) - -REPLY [8 votes]: To answer your second question, there are soluble octics with the same Galois group involving -other $p$th roots of unity. Take $p\equiv 1\mod 7$, and $K={\mathbb Q}(\alpha)$ -the unique degree 7 extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta_p)$. E.g. take -$\alpha=\sum_i \zeta_p^i$ where $i$ ranges over all seventh powers in ${\mathbb F_p}$. -If $f(x)$ is the minimal polynomial of $\alpha$ (degree 7), then $f(x^2)$ is the minimal -polynomial of $\sqrt\alpha$, which defines, generally, a 'random' quadratic extension of $K$. -That is, its Galois group over ${\mathbb Q}$ is -$$ - G=C_2\wr C_7\cong C_2^7:C_7. -$$ -Viewing $C_2^7$ as a 7-dimensional representation -of $C_7$ over ${\mathbb F}_2$, it decomposes as a 1-dimensional (trivial) representation plus -two distinct 3-dimensional ones. (The reason for this is that $2^3\equiv 1\mod 7$.) Factor out -$C_2^4\triangleleft G$, which is one of those -plus the trivial one. This gives a -Galois group $C_2^3:C_7$ that you are after, -and a subgroup $C_7$ in it cuts out the required octic field. -Here is a Magma code that can be used in the -Magma online calculator -to get such an octic: -p:=43; // or some other p = 1 mod 7 - -K:=CyclotomicField(p); -alpha:=&+[z^i: i in [1..p] | IsPower(GF(p)!i,7)]; - -R:=PolynomialRing(Rationals()); -f:=Evaluate(MinimalPolynomial(alpha),x^2); -K:=NumberField(f); -assert exists(a){a: a in ArtinRepresentations(K) | #Kernel(Character(a)) eq 16}; -F:=Field(Minimize(a)); -DefiningPolynomial(F); - -You can also stick in a Tschirnhaus transformation, say, -alpha:=alpha^3+alpha+1; - -in the 5th line to vary the generator of $K$ - in this way you get all possible -$C_2^3:C_7$-extensions involving $p$th roots of unity. -For your questions in the comments, the roots might be real or complex, and the constant term -may or may not be a square - this depends on whether $\alpha$ is chosen to be totally real, -and on the way 'Minimize' works; you can always use an additional Tschirnhaus transformation -to modify the final output or Pari's 'polredabs' function to try and get the coefficients smaller. -I do not know the reason why for $p=29$ there is such an elegant octic, this is very curious. It is a bit like the Trinks polynomial $x^7-7x+3$ with Galois group PSL(2,7), and I wonder whether simple polynomials having interesting Galois group is such a statistical blip, or there is a reason behind it.<|endoftext|> -TITLE: Homotopy theory of acyclic categories -QUESTION [6 upvotes]: Homotopy theory of category of posets is well-developed and explained in various places. My interest is in acyclic categories. Recall that in acyclic categories only invertible morphisms are the identity morphisms. A poset is an acyclic category; here there is at most $1$ morphism between any two objects. -The category $\mathbf{Ac}$ of acyclic categories lies (strictly) between the category $\mathbf{Pos}$ of posets and the category $\mathbf{Cat}$. -I am looking for references where categorical and homotopy theoretic properties of $\mathbf{Ac}$ are discussed. -A few references that I am aware of are the following: - -Combinatorial algebraic topology by D. Kozlov. Chapter 10 deals with basic constructions involving acyclic categories, their subdivisions, category of intervals and Mobius inversion. Chapter 14 is about group actions on posets and quotients in $\mathbf{Ac}$. Finally, in Chapter 15 he gives an explicit construction of homotopy colimit of diagram of spaces indexed over an acyclic category. -Homotopy limits and colimits by R. Vogt. The paper mainly deals with topological categories (i.e., small categories whose morphism sets can be topologized), however, some constructions can be translated to $\mathbf{Ac}$. -Cellular stratified spaces I and II by Dai Tamaki. This is an extensive account of various useful constructions involving (generalized) cell complexes whose face categories are acyclic. He describes a range of applications of acyclic categories; from configuration spaces to toric topology. - -I am interested in knowing answers to the following: - -Homotopy category of $\mathbf{Ac}$. -Model structures on $\mathbf{Ac}$. -Grothendieck construction for diagrams in $\mathbf{Ac}$. -Quillen's theorems A and B. -The category of $\mathbf{Ac}$-diagrams. - -REPLY [4 votes]: Here is a cool new (and very readable) preprint which uses the second barycentric subdivision (as discussed in Zhen Lin, Fernando Muro and Peter May's comments) to construct a cofibrantly generated model structure on $\textbf{Ac}$ which is Quillen-equivalent to the Thomason model structure on $\textbf{Cat}$. - -R Bruckner. A Model Structure On The Category Of Small Acyclic Categories, arXiv:1508.00992 [math.AT] - -Caveat: this only applies to small acyclic categories.<|endoftext|> -TITLE: A name for matrices with only simple eigenvalues? -QUESTION [6 upvotes]: I am constantly working with hermitian matrices without multiplicity in their spectrum. Since this hypothesis appear in several important problems, for instance perturbation theory, I looked in the literature for an accepted terminology but found nothing. Does anyone know a reference where these matrices, or their set, have been given a name ? I am considering calling them "simple matrices" but it is a bit ambiguous... - -REPLY [6 votes]: Apparently, there is something I don't understand in this discussion. Why should one reinvent the wheel? Such matrices have always been known as Hermitian matrices with simple spectrum. Just look at this wiktionary entry or Terry Tao's blog out of thousands of other examples.<|endoftext|> -TITLE: Is the mapping $f: \mathbb{R} \rightarrow [0,1], \ x \mapsto \sum_{n=1}^\infty \frac{\lfloor x^n \rfloor \mod 2}{2^n}$ surjective? -QUESTION [7 upvotes]: Is the mapping -$$ - f: \mathbb{R} \rightarrow [0,1], \ x \mapsto \sum_{n=1}^\infty \frac{\lfloor x^n \rfloor \mod 2}{2^n} -$$ -surjective? - -If not, what is its image? -If yes, what can be said about images of intervals, besides the obvious $f([-1,1]) = \{0,\frac{2}{3},1\}$? - -REPLY [3 votes]: I agree with Johan - it should be surjective. Here is a proof. I claim that $f([4,6])=[0,1]$. Let $[x,y]\subset [4,6]$ be such that $x^k=m$, and $y^k=m+1$. Then $y^{k+1}-x^{k+1}>x(y^k-x^k)=x\geq 4$. Which means $[x,y]$ contains inside itself two intervals $[x',y']$, and $[y',z']$ such that $(x')^{k+1}=m'$, $(y')^{k+1}=m'+1$, $(z')^{k+1}=m'+2$. Then one proceed recursively.<|endoftext|> -TITLE: Writing Mathematics : Linking words -QUESTION [23 upvotes]: I'm trying to write mathematics in English and I'm clearly missing something : linking words. I'm writing "so, we get", "Observe that" too many times and I'm afraid to use some expressions : "it implies" sound weird for me (and I don't know if I'm right here...) for example. -So I'm looking for good AND bad linkings words for writing Mathematics in order to diversify my mathematical language. -Thank you - -REPLY [4 votes]: If you are Russian, everything was done for us by A.B. Sosinsky in his great book "How to write a mathematical article in English". Here is a link (http://www.ega-math.narod.ru/Quant/ABS.htm). The book is of course in Russian. It helped me many times in my life. -Other suggestions: read other articles or find some English speaking friends to reread them after. If these friends know mathematics, even better. The friend who helps me out was studying English literature and now is doing her MD career. But her mathematical English is much better than mine!<|endoftext|> -TITLE: Limit of pushforward measures of random variables is "represented" by a random variable -QUESTION [10 upvotes]: Suppose we have an arbitrary probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a sequence of real random variables $X_n:\Omega\to\mathbb{R}$ such that the pushforward measures $(X_n)_*(\mathbb{P})$ converge weakly to a probability $\tilde{\mathbb{P}}$. -Can we always construct a random variable $X:\Omega\to\mathbb{R}$ such that $\tilde{\mathbb{P}}=X_*(\mathbb{P})$? (i.e. such that $X_n\to X$ in distribution?) -(I know this is not the kind of questions that should be asked here, but I received no helpful answers on StackExchange..) - -REPLY [7 votes]: Yes I think so. We may assume that $\mathcal F$ is the $\sigma$-algebra generated by sets of the form $X_n^{-1}(a,\infty)$ as $n$ runs over the positive integers and $a$ runs over the reals as $\mathcal F$ already contains these sets. In so doing, we might be making $\mathcal F$ smaller, but that only makes the problem harder. -Let $\mathcal P_1,\mathcal P_2,\mathcal P_3,\ldots$ be a sequence of refining finite partitions of $\mathbb R$ such that the intersection of an element of $\mathcal P_n$ for each $n$ consists of at most one point. For example, $\mathcal P_n$ could consist of intervals $[n,\infty)$, $(-\infty,-n)$ and half-open dyadic intervals of length $1/2^n$ between $-n$ and $n$. -Then let $\mathcal F_n$ be the finite sub-algebra of $\mathcal F$ generated by -the sets $X_i^{-1}A$ for $1\le i\le n$ and $A$ running over $\mathcal P_n$. -Write $\mathcal F_n=\{B^n_1,\ldots B^n_{k_n}\}$. Now there is a measurable map $\Phi$ from $\Omega$ to $\Xi=\prod_{n=1}^\infty\{1,\ldots,k_n\}$ sending $\omega$ to sequence of partition elements that it lies in. -Equip $\Xi$ with the lexicographic ordering. We can then check that $\mathbb P$ induces a measure $\mu$ on $\Xi$. There are at most countably many atoms in $\Omega$ with respect to $\mathcal F$. Let the set of atoms be $A$. We can map each of these atoms disjointly to atoms of the same mass (at least) in the limit measure $\tilde {\mathbb P}$ (some justification needed here, but I'm pretty confident). -First we couple the atoms in both $\tilde{\mathbb P}$ and $\Omega$. That is we define $X$ restricted to the atoms. -To finish, use quantile coupling to couple what's left. Given an $\omega$, set $t(\omega)=\mu(\{\omega'\in \Omega\setminus A\colon \Phi(\omega')\le \Phi(\omega)\})$ and finally, define for $\omega$ not in an atom, $X(\omega)=\inf\{s\colon \big(\tilde{\mathbb P}(-\infty,s])-\sum\text{(atoms of $\tilde{\mathbb P}\le s$)} \big)\ge t\}$.<|endoftext|> -TITLE: Analogy between Lagrange's Theorem and Rank-Nullity Theorem? -QUESTION [10 upvotes]: One can view view Lagrange's Theorem $$|G/H|=|G|/|H|$$ and the Rank-Nullity Theorem $$\dim(V/U)=\dim(V)-\dim(U)$$ as directly analogous. Does anyone know a high-level explanation of this analogy? I suspect it may involve the word "valuation" or the word "Artinian". - -REPLY [4 votes]: As Igor Rivin suggests, the explanation involves the word "logarithm". In fact, if $\mathfrak C$ is the category of groups, you can define an invariant of the category -\begin{align*} -\log\lvert{-}\rvert:Ob(\mathfrak C)&\to \mathbb R_{\geq 0}\cup\{\infty\}\\ -M&\mapsto \log \lvert M\rvert -\end{align*} -This invariant is additive, in the sense that, given a surjective morphism $\phi:M\to N$ in the category, $\log\lvert M\rvert=\log\lvert\ker(\phi)\rvert+\log\lvert N\rvert$. -In any category you can give analogously the definition of additive invariant. If your category is Abelian the usual thing is to say that an invariant $i:Ob(\mathfrak C)\to \mathbb R_{\geq 0}\cup\{\infty\}$ is additive provided $i(B)=i(A)+i(C)$ for any short exact sequence $0\to A\to B\to C\to 0$. -The dimension of vector spaces and the composition length of modules are examples of such additive invariants. -There is another property that the above invariants satisfy, that is, they commute with direct unions, that is, if -$$M=\underset{\to}{\bigcup}M_\alpha$$ -then, $i(M)=\sup_\alpha i(M_\alpha)$. Such an invariant is called upper continuous. In general, additive and upper continuous invariants are called length functions (as they generalize the usual composition length). -There are some attempts in the literature to classify in some sense all the additive invariants of some category, typically a category of modules. See for example: -Northcott, D. G.; Reufel, M. A generalization of the concept of length. -Quart. J. Math. Oxford Ser. (2) 16 (1965) 297-–321. MR195905 -Vámos, P. Additive functions and duality over Noetherian rings. Quart. J. Math. Oxford Ser. (2) 19 (1968) 43–-55. MR223434<|endoftext|> -TITLE: Category of motivic spectra -QUESTION [9 upvotes]: When the survey Axiomatic Stable Homotopy, Neil Strickland, 2004 was written the category of motivic spectra was not investigated from the point of view of axiomatic stable homotopy, as considered e.g. in Hovey, Palmieri and Strickland, Axiomatic Stable Homotopy Theory, 1997. -The question is this: was any work done in the last ten years in this direction? - -REPLY [18 votes]: Here's a direct link to the book by Hovey–Palmieri–Strickland. -The category of motivic spectra is known to satisfy the axioms of Definition 1.1.4 in the book when the base is a countable field of characteristic zero. Axioms (c) and (e) are problematic in general: - -Motivic spectra are generated by strongly dualizable objects when the base is a field of characteristic zero. The proof uses Hironaka's resolutions of singularities, see Röndigs–Østvær, Modules over motivic cohomology. -The representability of cohomology functors holds if the base is covered by finitely many spectra of countable commutative rings. See Naumann–Spitzweck, Brown representability in A1-homotopy theory. - -My 2 cents: axiom (c) sounds like a reasonable conjecture over general base schemes, but it seems very unlikely that axiom (e) would hold beyond the known case. -ETA In case someone stumbles upon this answer, let me take back one cent and say that axiom (c) should not hold over any scheme of positive dimension. The correct expectation is that it holds over zero-dimensional schemes (this being only known in characteristic zero for now). The idea is that the subcategory generated by dualizable object is the category of locally constant motivic spectra. Here is a proof that (c) does not hold for $SH(S)$ assuming $S$ admits a regular codimension 1 point $s$ (eg $S$ is normal): take an open $U$ whose restriction to $\mathrm{Spec}(\mathcal{O}_{S,s})$ is the generic point. Then $\Sigma^\infty_T U_+\in SH(S)$ is not generated by dualizable objects; in fact its rational motive isn't: localization and absolute purity imply easily that pullback to the generic point $DM_{\mathbb Q}(\mathcal{O}_{S,s})\to DM_{\mathbb Q}(\eta)$ is conservative on locally constant motivic sheaves.<|endoftext|> -TITLE: Dual of the space of continuous functions -QUESTION [5 upvotes]: Let $T \subseteq \mathbb R$ be a closed set of real numbers. Let $X := C(T, \mathbb R)$ denote the Fréchet space of continuous real-valued functions on $T$. The topology on $X$ is generated by seminorms $\|x\|_K := \sup_{t \in K} |x(t)|$ for any compact $K \subseteq T$. -Let $X^*$ denote the dual space of continuous linear functionals on $X$. Is there a nice characterization of the dual space using the Riesz representation theorem? -Let $e_t : X \to \mathbb R$ denote the evaluation functional, defined by $e_t[x] := x(t)$ for all $t \in T$. Are the evaluation functionals dense in $X^*$? - -REPLY [6 votes]: It is the space of compactly supported Radon measures. See Nicolas Bourbaki, Intégration, chapter 4, page 156 in Springer’s 2007-edition. -It seems to me that the space spanned by evaluation functionals is dense in the weak-*-topology (given any finite set of continuous functions choose—using compactness of the support—a finite open cover of the support such that in each of these open sets the given functions do not vary a lot, then use an evaluation functional for each of the open sets, at a point in the respective set), but not in the usual norm topology.<|endoftext|> -TITLE: Maximal geometric mean of distances between points on an interval -QUESTION [5 upvotes]: Suppose I had T points in the interval $[0,1]$. Call them $e_1, \dots, e_T$. -Question 1: -What is a good nontrivial bound on the geometric mean of $$\{|e_i - e_j| : 1 \leq i < j \leq T \}, $$ as a function of $T$, independent of our choice of $e_i$? -Question 2: -Suppose we can choose $i$ that minimizes the geometric mean of $$E_i = \{ |e_i-e_j| : 1 \leq j \leq T, j \neq i \}.$$What is a good nontrivial upper bound on the geometric mean of $E_i$ as a function of $T$? -Some context: -Suppose I were given an arithmetic circuit that computes $$f(z) = \sum_{i=1}^t c_tz^{e_t},$$ with bounds $D$ on the degree and $T$ on the number of terms, and we would like to test whether $f$ is zero. One way to test this for $f$ over an arbitrary field is to compute images $f \bmod (z^p-1)$ for a set comprised of the smallest primes $p$ whose product exceeds $$\prod_{j \neq 1} (e_1 - e_j),$$ -though one could replace $e_1$ with any of the $e_i$'s. A naive upper bound on this product is $D^T$. An answer to either of the questions above could improve this bound. - -REPLY [5 votes]: The extremal $e_i$ for Question 1 are probably well-known: -they are $0$, $1$, and $(1+r)/2$ where $r$ ranges over the roots of the -Gegenbauer - polynomial $C_{T-2}^{(3/2)}$. -The product of the $|e_i-e_j|$ is then a power of $2$ times -the square root of $\mathop{\rm disc}\bigl((x^2-1)C_{T-2}^{(3/2)}(x) / c_{T-2}\bigr)$, -where $c_n$ the leading coefficient $(2n+1)!/2^n n!^2$ of $C_n^{(3/2)}$. -This discriminant can in turn be computed from the values of $C_{T-2}^{(3/2)}$ -at $\pm 1$ and the discriminant of $C_{T-2}^{(3/2)} / c_{T-2}$, -which is known but somewhat complicated $-$ experimentally it seems to be -$$ -\prod_{m=1}^{T-2} \frac{(m+2)^{m-2}m^m}{(2m+1)^{2m-3}}. -$$ -To prove that these are the optimal $e_i$: let $P(x) = \prod_{i=1}^T (x-e_i)$ and -suppose $\{e_i\}$ maximizes $\prod_{i -TITLE: Nondeterministic Turing machines and the recursion theorem -QUESTION [5 upvotes]: This is almost certainly a silly question, but: -I am currently reading Moschovakis' article "Kleene's amazing second recursion theorem" (http://www.math.ucla.edu/~ynm/papers/1602-002-1.pdf) and there is a footnote in it which confuses me. -In footnote 10, on page 195, Moschovakis writes: - -. . . [T]he operation $$\Phi(p)=\begin{cases} -1, & \text{if $p(0)\downarrow$ or $p(1)\downarrow$,}\\ -\perp, & \text{otherwise}\\ -\end{cases}$$ is effective but not computable by a deterministic Turing machine. - -Here "operations" are functions on indices of programs which take the same value on indices representing the same partial computable function (bottom of pg. 194): so in more conventional terminology, $\Phi(e)=1$ if $\varphi_e(0)\downarrow$ or $\varphi_e(1)\downarrow$, and is undefined otherwise. -The problem I'm having is that this seems completely false. Deterministic Turing machines compute exactly the same functions as nondeterministic Turing machines. -But of course Moschovakis knows this, so my question is: - -What is Moschovakis actually saying? - -I am sure the answer is quite simple, but I don't see it. - -REPLY [7 votes]: Moschovakis by "computable" means: can be computed "by values" (see the last paragraph on the previous page). Where "computed by values" refers to the fact that we do not have access to the definition of a function, but can only ask for values of the function on its arguments. -There is, however, much deeper context of the footnote. The operation: -$$\Phi(p)=\begin{cases} -1, & \text{if $p(0)\downarrow$ or $p(1)\downarrow$,}\\ -\perp, & \text{otherwise}\\ -\end{cases}$$ -is a variant of "parallel or" function: -\begin{array} -\\ -1 &\mathit{or}& {-} &=& 1 \\ -{-} &\mathit{or}& 1 &=& 1 \\ -0 &\mathit{or}& 0 &=& 0 \\ -\end{array} -highly studied in computer science in the context of denotational semantics. The undefinability of "parallel or" in "sequential" calculi was the main obstacle to have a fully abstract continuous (domain-theoretic) semantics. This crucial observation led to some famous works (mostly by Plotkin and Milner) on the full abstraction theorem for PCF (a typed lambda calculus with a fixed-point operation).<|endoftext|> -TITLE: How to solve this quadratic matrix equation? -QUESTION [14 upvotes]: I would like to solve for $X$ in the matrix equation -$$ -XCX + AX = I -$$ -where all the matrices are $n\times n$, have real components, $X$ is positive semidefinite and $C$ is symmetric. My (possibly optimistic) hunch is that $X$ will be unique because of the positive semidefinite requirement; if not I only care about finding a single solution. -My searches have turned up a lot of work on solving non-symmetric riccati equations, unfortunately I don't meet the requirements for any of them. For example, everything I've seen requires that the entries of $C$ all be nonnegative, which is not true in my case. - -REPLY [6 votes]: In fact, your problem reduces to a Riccati equation in dimenion $n-1$ that is easy to solve. Here, if I correctly understood your problem, $A=I_n+aa^T, C=(cc^T)(cc^T)=uu^T$ (since $rank(C)=1$), where $a,u$ are known vectors. $A,X$ are symmetric $>0$ and $C$ is symmetric $\geq 0$, then $AX=XA$. We may assume $A=diag(1+\alpha,I_{n-1})$ where $\alpha\geq 0$. Then $X=diag(\beta,S_{n-1})$ where $\beta>0$ and $S$ is symmetric $>0$. Put $u=[v,w_{n-1}]^T$. $\alpha,v,w$ are known and $\beta,S$ are unknown. The equation can be rewritten: $\begin{pmatrix}\beta^2v^2&\beta vw^TS\\v\beta Sw&Sww^TS\end{pmatrix}+diag((1+\alpha)\beta,S)=I_n$. That implies $\beta^2v^2+(1+\alpha)\beta=1,vSw=0,Sww^TS+S-I_{n-1}=0$. -Case 1. (easy) $v\not=0,Sw=0$. Then $\beta^2v^2+(1+\alpha)\beta-1=0,S=I_{n-1}$. thus necessarily $w=0$ and for every $v$, there is a unique solution $\beta>0$. -Case 2. (more interesting) $v=0$. Then $(1+\alpha)\beta=1,Sww^TS+S-I_{n-1}=0$. You obtain $\beta>0$ if $\alpha>-1$ and finally you must to solve a standard symmetric Riccati equation in dimension $n-1$. As above, we may assume $ww^T=diag(\gamma,0_{n-1})$ where $\gamma\geq 0$. Then put $S=\begin{pmatrix}e&f^T\\f&H_{n-1}\end{pmatrix}$, where $e>0$ and $H$ is symmetric $>0$. The equation can be rewritten $(\gamma e+1)e=1,(\gamma e+1)f=0,H+ff^T=I_{n-1}$. Thus $f=0$ and $H=I_{n-1}$. For every value of $\gamma\geq 0$, there is a unique solution $e>0$. -EDIT: in fact, I assumed $\alpha>0$. It remains to study the case $\alpha=0$.<|endoftext|> -TITLE: Adding large sets not containing countable ground model sets -QUESTION [5 upvotes]: The question is motivated by Toni's question "Approximation of infinite set in generic extension" (see Approximation of infinite set in generic extension). -Before I state the question, let me add some remarks. In what follows, it is always assumed that $V$ is our ground model: -Remark 1. If we force to add $\lambda-$many Cohen reals by $Add(\omega, \lambda),$ then we get a cardinal preserving extension $W$ in which there is a set $C \subset \lambda$ of size $\lambda,$ such that $C$ contains no countable set in $V$. But note that in $W, 2^{\aleph_0} \geq \lambda,$ and so $GCH$ may fail in it (if $\lambda\geq \aleph_2$). -Remark 2. If we allow collapsing cardinals (by collapsing $\aleph_1$ to $\aleph_0$ or forcing with Namba forcing), then for any regular cardinal $\lambda,$ we can find a $GCH$ preserving extension $W$ of $V$ such that in $W$ there is a club $C\subset \lambda$ which avoids points of countable $V-$cofinality. This $C$ contains no countable set in $V$ and has finite intersection with any countable set in $V$. -Remark 3. If there are $\lambda-$many measurable cardinals, then there is a cardinal and $GCH$ preserving extension $W$ of $V$ with the same reals as $V$ such that $W$ contains a set $C$ of ordinals of size $\lambda$ which contains no countable set in $V$ and has finite intersection with any countable set in $V$. -Also note that if we require such a set $C$ in a cardinal preserving and not adding new reals extension, then some large cardinals are needed. -Now my questions are as follows: -Question 1. Suppose $V$ satisfies $GCH$ and contains no inner models with measurable cardinals. Is there a $GCH$ and cardinal preserving extension $W$ of $V$ such that in $W$ there is a set $C\subset \lambda$ of size $\lambda,$ for $\lambda\geq \aleph_3,$ such that $C$ has finite intersection with any countable ground model set? -Question 2. Suppose $V$ satisfies $GCH$ and contains no inner models with measurable cardinals. Is there a $GCH$ and cardinal preserving extension $W$ of $V$ such that in $W$ there is a set $C\subset \lambda$ of size $\lambda,$ for $\lambda\geq \aleph_3,$ such that $C$ contains no countable set from $V$. -Update. -Regarding Prof. Hamkins answer, I would like to add a few more comments (both of them are joint work with M. Gitik). -A. Assuming the existence of enough measurable cardinals, there is a pair $(V_1, V_2)$ of models of $ZFC$ with the same cardinals and reals, such that if $\kappa$ is the first fixed point of the $\aleph-$function in them, then in $V_2$, then there is a splitting $(S_\sigma: \sigma<\kappa)$ of $\kappa$ into sets of size $\kappa,$ such that any $S_\sigma$ has finite intersection with any countable set in $V_1$. This shows that Hamkins argument does not extend to the first fixed point of the $\aleph-$function. -B. Suppose $V \subset V_1$ have the same cardinals and reals and $\delta$ is less than the first fixed point of the $\aleph-$function. if $X \subset \aleph_\delta, X\in V_1$ and $|X|\geq \delta^+$ then $X$ has a countable subset which is in $V$. -Our proof of B is essentially the same as Hamkins argument and is by induction on $\delta$. Now Hamkins argument suggests that if we require $X$ has finite intersection with any countable set in $V$, then we do not require $V$ and $V_1$ to have the same reals (of course if $V$ and $V_1$ have the same reals then the statements "$X$ does not contain a countable set in $V$" and "$X$ has finite intersection with any countable set in $V$" are equivalent). - -REPLY [4 votes]: The answer to your original question (now question 1) is no, this is impossible, and the GCH and measurable cardinals are not involved. -Theorem. There is no cardinal-preserving forcing extension $V[G]$ with a set $C\subset\aleph_3$ having finite intersection with every countable set in $V$. -Proof. Suppose that there is such an extension $V[G]$ with such a set $C$. Let $\alpha$ be the supremum of the first $\omega$-many elements of $C$. So $\alpha\lt\aleph_3$ and $C\cap\alpha$ is a subset of $\alpha$ with finite intersection with every ground model countable set. Since $\alpha$ has size $\aleph_2$ in $V$, we may apply a bijection $\pi:\alpha\to\aleph_2$ in $V$ to get a set $B\subset\aleph_2$, namely $B=\pi[C]$, such that $B$ has finite intersection with every ground model set. So we have reduced to $\aleph_2$. Now, let $\beta$ be the supremum of the first $\omega$ many elements of $B$. So $\beta\lt\aleph_2$ and $B\cap\beta$ is a subset of $\beta$ having finite intersection with every countable ground model set. There is a bijection in $V$ of $\beta$ with $\aleph_1$, so $B\cap\beta$ is isomorphic to a set $A\subset\aleph_1$, by an isomorphism in the ground model, which has finite intersection with every ground model set. Let $\gamma$ be the supremum of the first $\omega$ many elements of $A$. So $\gamma$ is a countable ordinal in $V$, but $A\cap\gamma$ is infinite, a contradiction. QED -Clearly, the argument can be generalized beyond $\aleph_3$. -Your revised question (question 2) seems to be trivialized by the case of simply adding a Cohen real. This adds a set $C\subset\omega$, which is therefore also a subset of $\aleph_3$, but it contains no infinite ground model set as explained in the answers to Toni's question Approximmation of infinite set in generic extension. I suppose you intend to add a cofinal subset to $\aleph_3$?<|endoftext|> -TITLE: A theorem of Markov about completely regular spaces and topological groups -QUESTION [5 upvotes]: In Pontriaguin's classic book Grupos continuos (in English Continuous Groups), says that A. Markov proved that: - -There are topological groups that are not normal. - -Furthermore, he says it is deduced from a deeper result of Markov that says: - -Every completely regular space can be embedded as a closed subspace of some topological group. - -When looking at the references of the book, the article of Markov is cited in russian -which I am unable to transcript-. Nevertheless, a translation to spanish in my copy of the book has in brackets "Sobre grupos topológicos libres", i.e. "On free topological groups". However, I haven't been able to find neither Markov's article (a trasnlation into English) nor any related article to this result. -Does anyone know any reference (in English or Spanish) that exposes this results of Markov? - -REPLY [3 votes]: A simple proof can be found in the book: A. Arhangel'skii, -M.Tkachenko: Topological Groups and Related Structures, Atlantis Press 2008 (pp. -81-82)<|endoftext|> -TITLE: New trends in Applied Graph Theory -QUESTION [5 upvotes]: What are current trends in Applied Graph Theory? I am interested mainly in non-algorithmical problems. Maybe even in applications of graphs to other mathematical disciplines. For example, abstract algebra naturally produces the notion of non-commuting graph of a group and prime graph of a ring. -Also, what about modern Chemical Graph Theory? Is it still bounded by finding upper and lower bounds on graph indexes like Wiener's, Zagreb's, Randic's and other? -Finally, what graph-theoretical problems arose say from biological networks or from the study of social graphs? - -Clearly I am not specific on this question. Sorry for that! However, I have strong interest in graphs applications, so I had to ask mathoverflowers something like that! Also, thanks for your answers! - -REPLY [5 votes]: The book "Networks, Crowds, and Markets: Reasoning About a Highly Connected World" by David Easley and Jon Kleinberg contains amazing material and it is very readable. One can download a copy: -http://www.cs.cornell.edu/home/kleinber/networks-book/ - -REPLY [3 votes]: One relatively new area is studying the spread of infectuous diseases via graph models, with arcs representing contacts. -For example: - -"Using network properties to predict disease dynamics on human contact networks." - Proc. Royal Society B. 2011. (journal link)<|endoftext|> -TITLE: Does a left coset inclusion in a right coset imply they are equal? -QUESTION [10 upvotes]: Let $H$ be a subgroup of a group $G$ and $x$ be an element of $G$. -If $G$ is finite, $xH\subset Hx$ implies $xH=Hx$ since $xH$ and $Hx$ have the same finite cardinality (that of $H$). -What happens when $G$ and $H$ are infinite? Can a left coset be strictly included in a right coset? - -REPLY [13 votes]: In the free group on two generators $x, y$, if we take $H$ to be the subgroup generated by elements of the form $x^n y^m x^{-n}$ for $n \geq 0$, then $x H x^{-1}$ is strictly contained in $H$ because $y \notin x H x^{-1}$.<|endoftext|> -TITLE: Algebraic Groups in Characteristic p -QUESTION [18 upvotes]: It is well-known that Lie groups are, under nice conditions, essentially determined by their Lie-algebras. What's the corresponding statement for algebraic groups over fields of finite characteristic? - -REPLY [15 votes]: Since the question is somewhat open-ended, it may be useful to add further comments to what Dietrich and Marguax have said. [Also, it's important to correct Dietrich's first sentence: While any finite dimensional Lie algebra over $\mathbb{C}$ is the Lie algebra of some Lie group, it need not be the Lie algebra of an algebraic group. Some references are given in my answer to an earlier question here.] -1) In the setting of (real or complex) Lie groups, it's always essential to focus on connected groups, since by definition the Lie algebra only depends on the identity component of the group. Moreover, the classical correspondence between Lie groups and their Lie algebras isn't quite bijective: semisimple groups with the same root system have isomorphic Lie algebras even though the groups may vary through an isogeny class from simply connected to adjoint type. Apart from such qualifications, the exponential and logarithm maps allow one to pass back and forth between the analytic and the algebraic theories pretty effectively. -2) In the 1950s Chevalley extended the work on linear algebraic groups begun by Kolchin and Borel, imitating the Lie algebra correspondence to some extent. Here the Lie algebra is geometrically the tangent space to the group at the identity, so again one wants to consider mainly the connected groups (equivalent to being irreducible in the Zariski topology). Over an algebraically closed field of characteristic 0, there is even a partial substitute for analytic methods in his use of formal exponential power series. But obviously this has its limits, and in prime characteristic it breaks down badly. Even in characteristic 0, the Jordan-Chevalley decomposition in the group allows one to distinguish the additive and multiplicative groups (as algebraic groups) but not their Lie algebras. -3) For the crucial study (including classification) of semisimple groups, Chevalley abandoned the Lie algebra but was still able to imitate almost exactly the Cartan-Killing classification by root data. Here again it's essential to focus on simply connected groups, since isogenous groups often have isomorphic Lie algebras (though in characteristic $p$ that's not always true). -4) For semisimple (connected and simply connected) $G$, the question asked amounts to this: If the Lie algebras of two such groups are isomorphic over an algebraically closed field of prime characteristic, are the groups isomorphic as algebraic groups? To this the answer turns out to be YES, but it seems to require case-by-case study using the Chevalley classification. For some $p$, the Lie algebra of a simple $G$ is not simple, which complicates matters. Probably the most thorough study was done by G.M.D. Hogeweij in his Utrecht thesis: see his papers in Indag. Math. 44 (1982). -5) The books by Chevalley, Demazure-Gabriel, and myself all contain various details about the characteristic 0 correspondence for algebraic groups and their Lie algebras, which concerns mainly centers, centralizers, etc. But examples show clearly how much breaks down in characteristic $p$. So one always has to be cautious. -P.S. Besides the further insights into algebraic groups and their Lie algebras given by scheme theory and by formal groups, the hyperalgebra approach is quite useful. It's developed thoroughly in the book Representations of Algebraic Groups by J.C. Jantzen (second ed., AMS, 2003), where the complications in characteristic $p$ representation theory are explored.<|endoftext|> -TITLE: Integral Identity Involving Bell Numbers -QUESTION [7 upvotes]: Is the following identity true ? -$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad \overset{?}{=} \quad \int_0^\infty \frac{x!}{x^x} dx$$ -where -$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad \text{and} \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$ -NOTE: A short sketch of the demonstration proving the convergence of the integral on the left can be found here. Also, the numerical value of the integral on the right is about 2.5179+. Furthermore, if the position of $x$ and $n$ in the numerator of each sum were reversed, and both sums were to start at n = 0, we would have the following identity: - -$$\int_0^\infty \frac{E(x)}{e^x} dx \quad = \quad \sum_{n=0}^\infty \frac{n!}{n^n}$$ -where $\lim_{n \to 0} n^n = 1,$ and -$$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad \text{and} \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$ - -REPLY [8 votes]: It seems that the integral on the left-hand side exceeds $2.57$, -so it's not even close to the numerical value $2.5179\ldots$ of the -integral on the right-hand side. -I told gp: -b(x) = suminf(n=1,n^x/n^n) -B(x) = suminf(n=1,n^x/n!) -r(x) = b(x)/B(x) -intnum(x=0,25,r(x)) - -and got $2.5793$+. Since the integrand $r(x)=b(x)/B(x)$ is positive but -apparently decreasing, the Riemann sum $.01 \sum_{n=1}^{2500} r(n/100)$ -should be a lower bound on $\int_0^{25} r(x)\,dx$, -and thus on $\int_0^\infty r(x)\,dx$; and already that lower bound -exceeds $2.57$: replacing the last gp command above by -sum(n=1,2500,.01*r(.01*n)) - -returns $2.5755599998001798\ldots > 2.57$.<|endoftext|> -TITLE: Relationship between fractal dimension and Hurst exponent -QUESTION [10 upvotes]: For many basic random processes (like fractional Brownian motion) Hurst exponent "H" and fractal dimensions ( Hausdorf = Minkoswki typically) "D" -are realated by simple formula D = 2-H. -I want to clarify what are exact statemetns known about this relationships. -Consider some function y = f(t) (not a random process, but just function). -Assume that fractal dimensions of its graph equals to "D". - -Is it true that Hurst exponent is somehow correctly defined and equals to H=2-D ? - -There possibly can be some subtleties defining the Hurst exponent. -Let me try to give the following precise definition. Assume function f(t) is defined on the interval [0, 1]. Let us look on the values of function at points t=k/n, k=0...n. -So we obtain discrete time series f_0, ...,f_n. -And use the standard way of calculating the rescaled range $(R/S)_{n}$, -and existence of the Hurst exponent means that $(R/S)_{n}$ ~ $C n^H$. -I am not quite sure this definition is the correct one, but I think it can be suitable modified. - -REPLY [10 votes]: In principle, fractal dimension and Hurst exponent are independent of - each other: fractal dimension is a local property, while the - long-memory dependence characterized by the Hurst exponent is a global - characteristic. For self-affine processes, the local properties are - reflected in the global ones, resulting in the celebrated relationship - $D+H=n+1$ between fractal dimension $D$ and Hurst exponent $H$ for a - self-affine surface in $n$-dimensional space. -More generally, local and global behavior are decoupled. Stochastic models of the socalled Cauchy class separate fractal dimension and Hurst exponent and allow for any combination of the two parameters, without any linear relation. - -Stochastic Models That Separate Fractal Dimension and Hurst Effect, Tilmann Gneiting and Martin Schlather (2001).<|endoftext|> -TITLE: Chevalley restriction theorem for exterior algebras -QUESTION [8 upvotes]: Suppose $G$ is semisimple Lie group, $\mathfrak{g}$ is its Lie algebra, $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, and $W$ is the correspondent Weyl group. -Chevalley restriction theorem says that there is isomorphism of algebras of invariants $\mathbb{C}[\mathfrak{g}]^{G}\simeq \mathbb{C}[\mathfrak{h}]^{W}$ given by restriction. -Here $\mathbb{C}[\mathfrak{g}]$ and $\mathbb{C}[\mathfrak{h}]$ denote the algebras of polynomial functions on correspondent algebras viewed as vector spaces. In other words, $\mathbb{C}[\mathfrak{g}]=Sym(\mathfrak{g}^*)$ is a symmetric algebra. -My question is: is there a similar result for algebra of $G$-invariants of the exterior algebra $\bigwedge(\mathfrak{g}^*)$? -Thank you very much! - -REPLY [6 votes]: See: arXiv:dg-ga/9406006 and arXiv:math.DG/9506223. - -Edit: - -See lemma 3.3 in the first paper: $C^\infty$ and polynomial (by Solomon, cited in the paper) -$W$-invariant differential forms on $\mathfrak h$ correspond to horizontal $G$-invariant forms on $\mathfrak g$. -I just noted: You are asking for $G$-invariants of $\bigwedge \mathfrak g^\star$, i.e., constant differential forms. -These describe the de Rham cohomology of a compact form of $G$, by the theorem of Chevalley and Eilenberg, and can be described as the set of primitive elements. See books, for example, - -MR0400275 (53 #4110) Reviewed -Greub, Werner; Halperin, Stephen; Vanstone, Ray -Connections, curvature, and cohomology. -Volume III: Cohomology of principal bundles and homogeneous spaces. Pure and Applied Mathematics, Vol. 47-III. Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1976. xxi+593 pp. -MR1379333 (97j:57057) Reviewed -Onishchik, A. L.(RS-YAR) -\cyr Topologiya tranzitivnykh grupp preobrazovaniĭ. (Russian. English, Russian summary) [Topology of transitive transformation groups] Fizmatlit ``Nauka'', Moscow, 1995. 384 pp. ISBN: 5-02-014724-9 - -There is a translation into English which cannot be found in MathRev: - -Onishchik, Arkadi L. -Topology of transitive transformation groups. (English) Zbl 0796.57001 -Leipzig: Johann Ambrosius Barth. xv, 300 p. (1994).<|endoftext|> -TITLE: Are there locally jammed arrangements of spheres of zero density? -QUESTION [16 upvotes]: I know of a remarkable result from a paper of -Matthew Kahle (PDF download), that there are arbitrarily low-density -jammed packings of congruent disks in $\mathbb{R}^2$: - -In 1964 Böröczky used - a subtle disk packing construction to disprove a conjecture of Fejes Tóth, that a - locally jammed arrangement of disks in the plane, one where each disk is held in - place by its neighbors, must have positive density. -Böröczky, K. "Über stabile Kreis- und Kugelsysteme." Ann. Univ. Sci. Budapest, Eötvös Sect. Math. 7, 79-82, 1964. - -Surprisingly (to me), this natural candidate is not the lowest density: -      - -My question is whether or not Böröczky's construction extends to $\mathbb{R}^3$ -and higher dimensions. (I do not yet have access to Böröczky's 1964 paper.) -Addendum. (I add this prompted by user j.c.) -My question is answered: Yes. First, through the help of -Wlodek Kuperberg and other commenters, we learned what exactly is Böröczky's -1964 zero-density packing. Indeed it is "subtle" as Matthew Kahle said. -Second, simply layering identical copies of that packing of spheres in $\mathbb{R}^3$ -results in a jammed packing of zero density, as confirmed by Benoît Kloeckner. -What remains is to correct Wikipedia's misleading entries on the topic, as detailed by -Gerry Myerson. - -REPLY [12 votes]: Wlodek Kuperberg sent me this -'diagram from the "Lagerungen"' -of Böröczky's packing, over which I overlaid several congruent disks: -  -(Yes, small cracks or overlays are visible between disks that should be tangent. -The diagram is not 100% metrically accurate.) - -Added: The vertical red lines suggest a slight concavity.<|endoftext|> -TITLE: Subgroups of S_n consisting of elements having each less than 2 fixed points -QUESTION [5 upvotes]: I would like to know what is already known in literature about the following problem: -For what $n>0$ is it possible to find subgroup $H$ of $S_n$ having exactly $n(n-1)$ elements -with the property that every element of $H$ is identity or has less than 2 fixed points as permutation? -Proving it has no more than $n(n-1)$ elements is quite easy. I can show big class of $n$ for which there is $H$ we ask for. -More generally, let's ask: -Let $n>0$ and let's say that subgroup $H$ of $S_n$ is $k$-free if every element of $H$ as permutation has less than $k$ fixed points or is identity. For what $n$ is it possible to find such $H$ having exactly $n(n-1)\ldots (n-k+1)$ elements? Is it unique up to conjugation? -Edited off-by-one error -Edited the error in the title too - -REPLY [9 votes]: I've been staring at Zassenhaus's table, and it looks to me like there is a uniform way to describe all the exceptional near fields. Let $\Gamma$ be a finite subgroup of the quaternions, with entries in number field $K$. Let $\pi$ be a prime of $\mathcal{O}_K$ which does not appear in the denominator of any of the entries of $\Gamma$ and let $p$ be the norm of $\pi$. Assume $p$ is odd. Then we can reduce the entries of $\Gamma$ modulo $\pi$ and get a subgroup of $\mathbb{F}_p\langle i,j \rangle / (i^2=j^2=-1,\ ij=-ji)$. This ring is known to be isomorphic to $\mathrm{Mat}_{2 \times 2}(\mathbb{F}_p)$, so we get an action of $\Gamma$ on $\mathbb{F}_p^2$. In all of the examples that occur, we can make this more explicit: In each case $\pi$ is principal and we can choose a generator $\alpha$ which is a sum $a^2+b^2+c^2+d^2$ of four squares in $K$. So we can realize the vector space $\mathbb{F}_p^2$ explicitly as the quotient of the quaternions by the right ideal generated by $g:=a+bi+cj+dk$. -All of Zassenhaus's examples are of the form $\Gamma \times C$ acting on $\mathbb{F}_p^2$, where $C$ is a subgroup of $\mathbb{F}_p^{\ast}$ acting by scalars and chosen such that $p^2-1 = |\Gamma| \times |C|$. In terms of the original question, we then take $n=p^2$ and $G = (|\Gamma| \times |C|) \ltimes (\mathbb{Z}/p)^2$. -The tetrahedral cases Take $\Gamma$ to be the double cover of the rotational symmetries of the tetrahedron. This can be embedded in the quaternions with rational coefficients, as the group $\{ \pm 1, \pm i, \pm j, \pm k, (\pm 1 \pm i \pm j \pm k)/2 \}$. So $|\Gamma|=24$. -We can take $p=5$, $g=2+i$, $|C|=1$ or $p=11$, $g=3+i+j$, $|C|=5$. -The octahedral cases Take $\Gamma$ to be the double cover of the rotational symmetries of the octahedron. This can be embedded in the quaternions with coefficients in $\mathbb{Q}(\sqrt{2})$. So $|\Gamma| = 48$. -We can take $p=7$, $\pi = (3 + \sqrt{2})$ and $g=(1+\sqrt{2}/2) + i+j/2+k/2$ and $|C|=1$. We can also take $p=23$, $\pi = (5+\sqrt{2})$ and $g=(1+\sqrt{2}/2) + 3i/2+j+k/2$ and $|C|=11$. -The icosehedral cases Take $\Gamma$ to be the double cover of the rotational symmetries of the icosahedron. This can be embedded in the quaternions with coefficients in $\mathbb{Q}(\sqrt{5})$. So $|\Gamma| = 120$. -We can take $p=11$, $\pi = (4+\sqrt{5})$ and $g = (1 + \sqrt{5})/2+(1 + \sqrt{5})/2 i + j$ and $|C|=1$. We can also take $p=29$, $\pi = (7+2 \sqrt{5})$, $g=(1+\sqrt{5}) + i$ and $|C|=7$. Finally, we can take $p=59$, $\pi = (8+\sqrt{5})$, $g=(1 + \sqrt{5})/2+(1 + \sqrt{5})/2 i + 2j+k$ and $|C|=29$.<|endoftext|> -TITLE: Measurable and definable sets -QUESTION [5 upvotes]: Is there a model of set theory such that: - -AC holds, -Every ordinal definable set is measurable, -Every ordinal definable set of sets of $\mathbb{R}^2$ whose projection on the first (or second axis) is ordinal definable contains a definable member. - -REPLY [4 votes]: The answer is no, there is no such model. And we don't even need AC. -To see this, note first that statement 3 is equivalent to the assertion that every real number is ordinal definable. The reason is that the set of reals that are ordinal definable is itself definable, and so the complement is also definable, but contains no OD member. -But from this, we can use the HOD order, which is definable, to find an ordinal-definable well-ordering of the reals. And from any such ordering, we can define a non-measurable set by the Vitali argument, which would be a definable violation of statement 2.<|endoftext|> -TITLE: Dual Banach space of $B(X,Y)$ when $X$ is finite dimensional -QUESTION [8 upvotes]: Denote $B(X,Y)$ the Banach space of bounded operators between Banach spaces $X$ and $Y$. -When $X$ and $Y$ are both finite dimensional, it follows from the formula -$$\|u\|_{B(X,Y)} = \sup_{\|x\|_X < 1,\|\xi\|_{Y^*}< 1} \xi(u(x))$$ -and the Hahn-Banach separation theorem that that the dual Banach space of $B(X,Y)$ is the projective tensor product $X \hat \otimes Y^*$ (= $X \otimes Y^*$ as a vector space with unit ball the convex hull of the $x \otimes \xi$ for $\|x\|_X < 1,\|\xi\|_{Y^*}< 1$). -When only $X$ is finite dimensional, this identification of $B(X,Y)^*$ with $X \hat \otimes Y^*$ still holds isometrically. For a long time I thought that this was as elementary as the case when both spaces are finite dimensional, but I recently realized that I could not find an elementary proof. Can anybody help me, or provide me with a reference? (I think the only reference I know is Grothendieck's memoire). -Some elementary facts~: - -The dual of $B(X,Y)$ is $X \otimes Y^*$ as a vector space. -The formula above and Hahn-Banach tell me that the closed unit ball of $B(X,Y)^*$ corresponds to the weak-$*$ closure of the unit ball of $X \hat \otimes Y^*$. So the question is why is the closed unit ball of $X \hat \otimes Y^*$ weak-$*$ closed? - -REPLY [7 votes]: If $X=\ell_1^N$, then $B(X,Y)\equiv \ell_\infty^N(Y)$ and the result is easy. -If the unit ball of $X$ has $N$ extreme points, then $X$ is a quotient of $\ell_1^N$ and you can deduce what you want from (1). -The general case follows from (2) by approximation. - -Notice that this is basically Dean's proof of local reflexivity (minus the steps needed to deduce local reflexivity from what you want), and is, to my way of thinking, the "right" approach.<|endoftext|> -TITLE: On the prime number theorem in arithmetic progression -QUESTION [8 upvotes]: The prime number theorem tells us that , if $\pi\left(x\right)$ denotes the number of primes less than or equal to $x$, we have $$\pi\left(x\right)\sim\frac{x}{\log x}.$$ -In a similar manner considered $1\leq a \leq q$ with $(a,q)=1$ and defined $\pi\left(x,a,q\right)$ the number of primes less than or equal to $x$ congruous $a\,\textrm{mod}\, q$ and $\phi\left(n\right)$ the number of minor numbers and coprime with $n$, we have $$\pi(x,a,q)\thicksim\frac{1}{\phi(q)}\frac{x}{\log x}.$$ -If $q$ is "small" you have asymptotic formulas for $\pi\left(x,a,q\right)$ (see the Siegel - Walfisz theorem). For any $q$ we have the estimate -$$\pi(x,a,q)\gg\frac{1}{\phi(q)}\frac{x}{\log x}.$$ -I would like to know if there is an estimate of the type $$\pi(x,a,q)\ll\frac{1}{\phi(q)}\frac{x}{\log x}$$ for any $q$. -I hope I was clear! Sorry for my bad english! - -REPLY [11 votes]: For $x\leq\phi(q)$ the estimate $\pi(x,a,q)\ll\frac{1}{\phi(q)}\frac{x}{\log x}$ would imply $\pi(x,a,q)\ll\frac{1}{\log x}$, i.e. $\pi(x,a,q)=0$ for large $x$ which is clearly false. So a bound you envision can only hold for $x$ slightly above $\phi(q)$. On the other hand, for any $\epsilon>0$, the Brun-Titchmarsh inequality implies -$$\pi(x,a,q)\ll_\epsilon\frac{1}{\phi(q)}\frac{x}{\log x},\qquad x>q^{1+\epsilon}.$$<|endoftext|> -TITLE: Ordering on words -QUESTION [5 upvotes]: What are the known computation-friendly well-orderings on words from $A^*$, where $A$ is a finite alphabet, except the standard weightlex and syllable-order? - -REPLY [3 votes]: Here are three relevant references for partial wqo on words: -[1] F. D'Alessandro, S. Varricchio, Well quasi-orders, unavoidable sets, and derivation systems, RAIRO - Theoretical Informatics and Applications 40 (2006) 407-426, DOI -[2] A. Ehrenfeucht, D. Haussler, G. Rozenberg, On regularity of context-free languages, Theoret. Comput. Sci. 27 (1983) 311–332. -[3] M. Kunc, Regular solutions of language inequalities and well quasi-orders, Theoret. Comput. Sci. 348 (2005) 277–293, ISSN 0304-3975, DOI.<|endoftext|> -TITLE: What does a theoretical mathematician do? -QUESTION [22 upvotes]: I'm 12, and really like mathematics and physics. I was just wondering what does a 'theoretical mathematician' do? - -REPLY [30 votes]: There are several things that mathematicians do: - -teachers of mathematics teach math, and you surely know some of those, -applied mathematicians use their knowledge of mathematics to help engineers, physicists, chemists, and other scientists solve math problems. For example, they help engineers calculate how water is going to flow around a ship that the engineer designed. Or they might help doctors calculate how a flu is going to spread and how many people will get sick. -theoretical mathematicians figure out new math that nobody has done before. For example, they solve open math problems that nobody has been able to solve. Or they think of new interesting math that nobody has ever done before. Sometimes they invent new math that allows physicists and other scientists solve physics problems that they could not solve with old math. - -Most mathematicians do a bit of everything. For instance, I teach at a university, I do theoretical math, but I also do applied math because my research is about math and computers. So I often figure out how to program computers so that they can do math by themselves, which means that I need to write programs as well as invent new math that allows me to write even smarter programs. If you would like to know examples of this, just ask.<|endoftext|> -TITLE: What is the source of this E̶r̶d̶ő̶s̶ quote? -QUESTION [14 upvotes]: Namely, the following one - -"All problems appeared once in the [American Mathematical] Monthly." - -I remember reading it several years ago... When I first posed the question, I believed that I had read it somewhere in Krantz' Mathematical Apocrypha but according to Carlo Beenakker the quote is nowhere to be found in the said recollection of stories and anecdotes of Krantz. I gather that Erdős might have expressed it in relation with the history of what is nowadays known as the Sylvester-Gallai theorem but it'd be perfectly possible for me to be wrong in this respect too. -I thank you in advance for your replies. - -REPLY [22 votes]: In Index to Mathematical Problems, 1980-1984 - Page xi by Stanley Rabinowitz this quote is attributed to Léo Sauvé: - -Léo Sauvé, former editor of Crux Mathematicorum, an international problems journal, once quipped that it seemed like all problems "had once been published in the Monthly"<|endoftext|> -TITLE: On the irrationality measure of $\sum_{n=1}^\infty a^{-b^n}$ -QUESTION [8 upvotes]: Pick integers $a, b \ge 2$ and let $\xi_{a,b}$ be the sum of the series $\sum_{n=1}^\infty a^{-b^n}$. It is known that $\xi_{2,2}$ is transcendental: I learned a proof of this from notes by M. Filaseta (who attributes the basic idea to P. Erdős, but does not provide a reference), and I'm confident that the argument can be made to work at least in the more general case when $a = b$, but I have contrasting feelings about the case $a \ne b$ (and no much time to look closer at this right now). Then, my first question is: - - -Q1. Could you kindly provide me with a reference to Erdős' original work on the subject (if ever published) or any published work addressing the more general question of the transcendence of $\xi_{a,b}$? - - -While I've my own proof for the general result in the above, I've also read that $\xi_{2,2}$ is known not to be a Liouville number, and I guess that the same holds true for $\xi_{a,b}$ at least in the case $a = b$ (but I don't have a clue on how to prove it). So my second question is: - - -Q2. Is it known whether $\xi_{a,b}$ is never a Liouville number? If [yes, no], could you provide a reference to a published work where this is [proved, disproved]? - - -Based on Q2, it is now natural to ask the following: - - -Q3. What is known about the irrationality measure of $\xi_{a,b}$? In particular, is it known whether [each, any] of the $\xi_{a,b}$ has an irrationality measure $> 2$? - - -Thanks in advance for any help. - -REPLY [8 votes]: This result is due originally to K. Mahler, and holds true more generally with any algebraic $a$ having $|a| > 1$ (so that the series converges absolutely). I can recommend Masser's lecture in the CIME 2000 school on diophantine approximations (LNM 1819), where the main idea of Mahler's proof is outlined as an illustration of the typical transcendence proof. The complete argument can be found in the opening chapter (Theorem 1.1.2) of K. Nishioka's book "Mahler Funtions and Transcendence" (LNM 1631), where you may also find various related results and generalizations. -Mahler's proof of the transcendence of $f(a^{-1})$ is based on the functional equation $f(z^b) = f(z) - z$ of the series $f(z) := \sum_{n \geq 1} z^{b^n}$. A different approach, which is based on Schmidt's Subspace theorem and allows for much more general transcendence statements, was discovered by P. Corvaja and U. Zannier in their article "Some new applications of the subspace theorem" (Compositio math, 2002).<|endoftext|> -TITLE: When are orbits of semisimple group representations closed? -QUESTION [5 upvotes]: Let $G$ be a connected, simply-connected complex semisimple linear algebraic group, and let $V$ be a finite-dimensional complex $G$-module. Is there a nice description of those $v\in V$ for which the $G$-orbit of $v$ is Zariski-closed in $V$? Another question: Which non-zero $v\in V$ have the property that the $G$-orbit is closed in $V\setminus\{0\}$? - -REPLY [7 votes]: There's some terminology here that might be helpful for a literature search: $v$ is said to be semisimple if $Gv$ is closed in $V$ and said to be nilpotent if $v\neq0$ and $Gv$ is not closed in $V\setminus\{0\}$. The set of all nilpotent $v\in V$ is called nullcone of $V$. If $V$ is the adjoint representation, then "semisimple" and "nilpotent" have their usual meaning. -The nullcone of $V$ is of course simply the set $\{v \in V \colon 0 \in \overline{Gv}\}$, or equivalently, it's the zero set of $\mathbb C[V]_+^G$, the nonconstant homogeneous $G$-invariant regular functions on $V$. A perhaps more useful characterization is given to us by the Hilbert–Mumford criterion: a nonzero $v\in V$ is in the nullcone of $V$ if and only if there is a 1-parameter subgroup $\lambda \colon \mathbb C^\times \to G$ such that $\lim_{t\to 0} \lambda(t)v = 0$. There's a lot of research that's been done in this direction: you can start by looking up "nullcone" on MathSciNet. -On the other hand, I'm not aware of any nice, complete characterizations of the semisimple elements in an arbitrary $V$. But there is a nice sufficient condition due to Dadoc and Kac, Polar representations, J. Algebra 92 (1985), 504���524, which goes as follows. Fix a maximal torus $T\subset G$, let $V = \bigoplus_\lambda V_\lambda$ be the corresponding weight-space decomposition of $V$, and let $\Phi$ denote the set of roots. Now, given $v\in V$, write $v = \sum_{i=1}^k v_{\lambda_i}$ with $v_{\lambda_i} \in V_{\lambda_i}\setminus\{0\}$. If - -$0$ is in the convex hull of $\lambda_1,\ldots, \lambda_k$, and -$\lambda_i - \lambda_j \not\in\Phi$ for all $i\neq j$, - -then $v$ is semisimple. (The first condition guarantees that $Tv$ is closed, by the Hilbert–Mumford criterion; the second condition is there to allow us to conclude that $Gv$ is then closed.) For example, every $v\in V_0$ is semisimple.<|endoftext|> -TITLE: Is it possible to construct a formal group law from a Lie group without choosing coordinates? -QUESTION [11 upvotes]: There is a three-way correspondence between: - -Real (connected and simply connected) Lie groups of dimension $n$; -$\mathbb R$-Lie algebras of dimension $n$; -Formal group laws in $n$ variables over the reals. - -To get from the first to the second, one only really needs the fact that manifolds have tangent spaces which are vector bundles, and that the tangent space functor is strong (so that one can talk about left-invariant vector fields). -To get from the second to the third, one uses the Baker--Campbell--Hausdorff formula, which in the realm of formal power series is pure abstract algebra. -However, the only way I know of getting from the first to the third directly involves choosing analytic coordinates around the identity and Taylor expanding the Lie group multiplication. -My question is: can one give a direct geometric construction of a formal group law out of a Lie group without choosing coordinates? The sort of thing I had in mind might, for example, involve using jet bundles. - -REPLY [9 votes]: Sure: if $G$ is the given analytic group then consider the functor $F$ on finite local $\mathbf{R}$-algebras with residue field $\mathbf{R}$ given by defining $F(A)$ to be the set of maps of real-analytic spaces ${\rm{Sp}}(A) \rightarrow G$ "based" at the identity. This is a group-valued functor, and it is visibly pro-represented by the completion $O_{G,e}^{\wedge}$ of the noetherian analytic local ring $O_{G,e}$ of $G$ at the identity $e$. The group law on the functor thereby defines a group-object structure on the formal scheme Spf($O_{G,e}^{\wedge}$), or more specifically defines a formal Hopf algebra structure on $O_{G,e}^{\wedge}$ (i.e., a map $m^{\ast}:O_{G,e}^{\wedge} \rightarrow O_{G,e}^{\wedge} \widehat{\otimes} O_{G,e}^{\wedge}$ satisfying the usual conditions). -Voila, that's it. The same procedure works for analytic groups over non-archimedean fields in char. $p > 0$ (such groups might not be smooth, so "local coordinates" might not even exist, yet the method still works), as well as for group schemes of finite type over any field, and so on. No need for Lie algebras or a BCH formula. The notion of "formal group" makes good sense in terms of complete local noetherian rings even without any notion of formal smoothness (so no "local coordinates").<|endoftext|> -TITLE: Can one cover the plane with less than continuum of lines? -QUESTION [15 upvotes]: I will be working in ZFC, but I am not assuming the Continuum Hypothesis (or Martin's Axiom). I know that it is consistent with ZFC that one can cover the real line with less than continuum of meager sets. My questions are about coverings by sets which are "nicer" than just meager (for instance, their proper intersections are finite). By "plane" below I mean the Euclidean plane and by "line" I mean an affine line. Of course, by Baire's theorem, Questions 1 and 2 both have negative answer if the sets $A$ are countable. -Question 1. Let $A$ be a set so that ${\aleph }_0=|{\mathbb N}|<|A| < 2^{{\aleph }_0} =|{\mathbb R}|$. Can one cover the plane by a family of lines $L_\alpha$, $\alpha\in A$? -In case I am missing some simple geometric consideration, -Question 2. For $A$ as above, can one cover the plane by a family of (real, irreducible) algebraic curves $X_\alpha$, $\alpha\in A$? -It feels as if these questions are related to the list compiled -in this post by Joel David Hamkins, but I cannot quite figure out how. - -REPLY [35 votes]: For your question 1, the answer is no. If $\{X_{\alpha}\}_{\alpha\in A}$ cover a circle, then $|A|=2^{\aleph_0}$ (just because each line can cover at most two points of the circle). -For the question 2, as Todd pointed out, the answer is also no.<|endoftext|> -TITLE: Line bundle ample iff induced morphism finite, looking for reference -QUESTION [5 upvotes]: Let $X$ be a proper scheme over a field $\Bbbk$. Let $\mathscr L\in\mathrm{Pic}(X)$ be a globally generated line bundle. If for some choice of global sections $V\subseteq\mathscr L(X)$, the induced morphism $\phi_V:X\to\mathbb P(V)$ is finite, then $\mathscr L$ is ample. The converse also holds. However, I have searched through Görtz-Wedhorn, Hartshorne, Liu, Mumford's red book, the stacks project and also EGA II without finding a reference for this statement. Does anyone know any? - -REPLY [8 votes]: Robert Lazarsfeld, Positivity in Algebraic Geometry I, Corollary 1.2.15 page 28.<|endoftext|> -TITLE: Dynamical properties of injective continuous functions on $\mathbb{R}^d$ -QUESTION [28 upvotes]: Let $\varphi:\mathbb{R}^d\to\mathbb{R}^d$ be an injective continuous function. -Denote by $\varphi_n$ the $n$-th iterate of $\varphi$, i.e. -$\varphi_n(x)=\varphi_{n-1}(\varphi(x))$ for all $x\in\mathbb{R}^d$. -Consider two properties of $\varphi$: -1) For every compact set $K\subset\mathbb{R^d}$ there exists $n\in\mathbb{N}$ such that -$\varphi_n(K)\cap K=\emptyset$. -2) For every compact set $K\subset\mathbb{R^d}$ there exists $n_0\in\mathbb{N}$ such that -for every $n\geq n_0$ we have -$\varphi_n(K)\cap K=\emptyset$. -The question is the following: is 1 equivalent to 2? -Remark 1. It is obvious that 2 implies 1. -Remark 2. If $d=1$, then the answer is YES and it is a simple consequence of the fact -that an injective function must be monotonic. - -REPLY [2 votes]: Martin's negative answer leaves open the case of $n=2$, and it seems this is for good reason. -Indeed, I think that, for a homeomorphism $\phi$ of $\mathbb{R}^2$, the two properties are indeed equivalent. -Firstly, note that $\phi$ can be assumed to be orientation-preserving (otherwise, replace $\phi$ by its second iterate). -Let $\phi$ be an orientation-preserving homeomorphism of $\mathbb{R}^2$ without fixed points. The Brouwer plane translation theorem states that every point is contained in a domain of translation (which is disjoint from both its image and preimage). See -http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2143652 -So you can cover your compact set $K$ with finitely many such translation domains. It should follow (though I have not checked the details) that $K$ is disjoint from all of its sufficiently large iterates. -(I noticed that you did not say that $\phi$ should be a homeomorphism, but only injective and continuous. It appears plausible that any counterexample in this setting would also give rise to a homeomorphic one, but one should always be careful with such topological questions - as indeed Brouwer himself discovered in the course of the proof of his theorem, I believe.)<|endoftext|> -TITLE: Why is there a connection between enumerative geometry and nonlinear waves? -QUESTION [36 upvotes]: Recently I encountered in a class the fact that there is a generating function of Gromov–Witten invariants that satisfies the Korteweg–de Vries hierarchy. Let me state the fact more precisely. Define $$\langle \tau_{k_1}\cdots \tau_{k_n}\rangle := \int_{\overline{\mathcal{M}}_{g,n}} \psi_1^{k_1}\cdots \psi_n^{k_n},$$ where $\overline{\mathcal{M}}_{g,n}$ is Deligne–Mumford space and $\psi_i$ is the first Chern class of the line bundle over $\mathcal{M}_{g,n}$ whose fiber at a given curve is the cotangent line at the $i$-th marked point of that curve. Next, define $$F(t,\lambda) := \sum_{g=0}^\infty \lambda^{2g-2} \sum_{n = (n_1,\ldots,n_k)} \frac {t^n} {n!} \langle \tau^n\rangle_g,$$ where $t = (t_0, t_1, \ldots)$ and $\lambda$ are formal variables. Then for all $n \geq 1$, $F$ satisfies the following PDE: $$(2n+1)\lambda^{-2}\partial_{t_n}\partial_{t_0}^2F = \partial_{t_{n-1}}\partial_{t_0}F\partial_{t_0}^3F + 2\partial_{t_{n-1}}\partial_{t_0}^2F\partial_{t_0}^2F + \frac 1 4\partial_{t_{n-1}}\partial_{t_0}^4 F.$$ Note that when $n=1$, it follows that (up to coefficients and $\lambda$) $\partial_{t_0}^2F$ satisfies the KdV equation: $$q_t = qq_x + \frac 1 2q_{xxx}.$$ -I was very surprised that a generating function whose coefficients come from the geometry of Deligne–Mumford space should satisfy a nonlinear PDE for waves in shallow water. My question is: - -Is there any "moral" reason for why a water wave PDE should have any connection with Gromov–Witten invariants? - -My understanding (correct me if I'm wrong) is that Kontsevich's proof of this (which went by forming a cell decomposition of $\overline{M}_{g,n}$ using ribbon graphs) doesn't shed light on my question. - -REPLY [26 votes]: Maybe not so surprising. -The rather unobtrusive infinigens $z^{n+1}\frac{d}{dz}$ weave a web of connections among hydrodynamical equations, moduli spaces of Riemann surfaces, and the combinatorics of associahedra. -$$Summary$$ -The most salient link between hydrodynamical equations and moduli spaces seems to be the infinite dimensional Witt Lie algebra/group and its central extension, the Virasoro-Bott Lie algebra/group. Hydrodynamical Euler equations give the geodesics of these groups, which govern the topology of the moduli space of the punctured Riemann surfaces of string theory. Somewhere in between lurk the Stasheff polytopes, the associahedra, whose combinatorics can be related to flow fields and the collisions of particles on a line that are related to the topology of punctured Riemann surfaces. -Example 1) Flows, the geometry of associahedra, and moduli spaces for marked Riemann sufaces of genus 0 -A) Flows, streamlines, integral curves, and compositional inversion: -Let the inverse of the formal power series $\omega=h(z)=a_1\:z+ a_2 \: z^2+ \cdots$ be $z=h^{-1}(\omega)=b_1 \: \omega + b_2 {\omega}^2 + \cdots$ ; then, with $g(z)=1/[dh(z)/dz]$, a flow field is generated by -$$\exp \left[ {t \cdot g(z)\frac{d}{{dz}}} \right]z = \exp \left[ {t\frac{d}{{d\omega }}} \right]{h^{ - 1}}(\omega ) = {h^{ - 1}}[t + \omega] = {h^{ - 1}}[t + h(z)]=W(t,z),$$ -and it is easy to show that the flow map has the following features; -$$\:\:\: W(0,z)= z $$ -$$\:\:\: W(t,0)= h^{(-1)}(t)$$ -$$$$ -$$\:\:\: \frac{dW(0,z)}{dt} = g(z) = [h^{(-1)}]^{'}(h(z))$$ -$$\:\:\: g(h^{(-1)}(\omega)) = [h^{(-1)}]^{'}(\omega)$$ -$$\:\:\: W[s,W(t,z)] = W(s+t,z)$$ -$$\:\:\left [\frac{d}{dt}-g(z)\frac{d}{dz} \right ]\:W(t,z) = 0,$$ -so $(1,-g(z))$ are the components of a vector orthogonal to the gradient of $W$ and, therefore, tangent to the contour of $W$ at $(t,z)$. -B) Compositional (Lagrange) inversion and associahedra (cf. Loday): -The iterated derivatives acting on $z$ and evaluated a $z=0$ generate the coefficients of the inverse power series. E.g., -$$b_5=\frac{1}{5!}[g(z)\frac{d}{{dz}}]^{5}z|_{z=0} = \frac{1}{a_1^{9}} [14\: a_2^{4} - 21\: a_1 a_2^2 a_3 + a_1^2[6 \:a_2 a_4+ 3\: a_3^2] - 1\: a_1^3 a_5].$$ -This is related to a refined f-vector (face-vector) for the 3-D Stasheff polytope, or 3-D associahedron, with 14 vertices (0-D faces), 21 edges (1-D faces), 6 pentagons (2-D faces), 3 rectangles (2-D faces), 1 3-D polytope (3-D faces). Subtracting two from the index of $a_n$, and ignoring the resulting indeterminates with indices with values less than one, allows one to read off the geometry of the associahedron from cartesian products of the lower dimensional associahedra (Loday), e.g., $3\: a^2_3$ becomes $3\: a^2_1$, the cartesian product of the 1-D associahedron with itself, which is a tetragon, or square in some reps. -This correspondence between the refined f-vectors of the $n$-D associahedron and $b_{n+2}$ holds in general, (see OEIS-A133437). -C) Associahedra and marked Riemann surfaces of genus 0: -Brown and Bergstrom in "Inversion of series and the cohomology of the moduli spaces of $M_{0,n}^\delta$" state: -For $n \geq 3$, let $M_{0,n}$ denote the moduli space of genus $0$ curves with $n$ marked points, and $\overline{M}_{0,n}$ its smooth compactification. ... In this paper, we prove that the inverse of the ordinary generating series for the Poincare polynomial of $H^\bullet(M_{0,n})$ is given by the corresponding series for $H^\bullet(M^{\delta}_{0,n})$, where $M_{0,n}\subset M^{\delta}_{0,n} \subset \overline{M}_{0,n}$ is a certain smooth affine scheme. -And on page 3, they give the abbreviated formula -$$M^{\delta}_{0,6}=14\; M_{0,3} \cup 21\: M_{0,4} \cup [6\: M_{0,5} \cup 3\: M^2_{0,4}] \cup M_{0,6}.$$ -So, we have a connection between flows determined by the combinatorics of the associahedra and moduli spaces. -Example II) The inviscid Burgers-Hopf equation and associahedra -Define $$U(x,t)=\frac{x-A(x,t)}{t}$$ and $$A^{-1}(x,t)=x+t\;F(x).$$ Then it is easy to show that with $A(0,t)=0$ that $U$ satisfies the inviscid Burgers equation -$$U_t(x,t)+U(x,t)U_x(x,t)=0 , \:\:\:\: U(x,0)=F(x).$$ -For details, see my sketch "Compositional inverse pairs, the Burgers-Hopf equation, and associahedra" at my mini-arxiv. -With $F(x)=c_2\:x^2+c_3\:x^3+ \cdots\;$, we have as asserted in Example I that -$$A(x,t)=x+(-c_2t)x^2+(-c_3t+2c_2^2t^2)x^3+(-c_4t+5c_2c_3t^2-5c_2^3t^3)x^4+(-c_5t+(6c_2c_4+3c_3^2)t^2+21c_2^2c_3t^3+14c_2^4t^4)x^5+\cdots\:,$$ -the associahedra again. For $F(x)=x^n$, with $n>2$, $A(x,t)$ is the o.g.f. for the Fuss-Catalan numbers, which are related to dissections of polygons (cf. OEIS-A001764, particularly the Schuetz/Whieldon link). For $n=2$, we obtain the celebrated Catalan numbers and relations to Brownian motion, Lax pairs, random matrix theory, and Wigner's semicircle law/distribution, as discussed by Govind Meno in "Burgers turbulence: kinetic theory and complete integrability" and a similarly titled paper by Ravi Srinivasan. Victor Buchstaber in "Toric Topology of Stasheff Polytopes" even derives the Catalan numbers from an infinite set of conservation laws reminiscent of those for the KdV equation. -$$General\:\:\: Discussion$$ -The Lie algebra of the diffeomorphism group of a manifold, Diff(M), consists of all vector fields on M, i.e., the infinitesimal generators $g(z)\frac{d}{dz}$ in Example I above (1-D or 2-D case, real or complex $z$), which induce an infinitesimal change in the coordinates $z \rightarrow z+t\:g(z)$. A basis for this algebra is the infinite dimensional Witt Lie algebra with elements $l_n=-z^{n+1}\frac{d}{dz}$. The geodesics for this group are given by the particular Euler eqn. the inviscid Burgers-Hopf equation (Example II). Already, with the subgroup $(l_{-1},l_0,l_1)$, related to linear fractional transformations, we can see connections to the moduli space of the Riemann sphere through the Riemann-Roch theorem as discussed by Gleb Arutyunov on page 87 of "Lectures on String Theory". -Making a central extension of the Witt algebra (on a circle) leads to the Virasoro algebra and group, whose geodesics are related to the KdV equation -$$\partial_t U+U\:\partial_xU=-c\partial^3_xU,$$ -which is essentially a perturbed inviscid Burgers-Hopf with the constant parameter $c$ being the "depth" of the fluid. For more on this, see "Hydrodynamics and infinite dimensional Riemannian geometry" by Jonathan Evans (a review of The Geometry of Infinite Dimensional Groups by Boris Khesin and Robert Wendt), "Groups and topology in the Euler hydrodynamics and KdV" by Khesin, or " Euler equations on homogeneous spaces and Virasoro orbits" by Khesin and Gerard Misiolek. -The Virasoro algebra in conformal field theory governs the topology of the string world-sheet interactions generating the moduli spaces of Riemann surfaces with punctures corresponding to particles interacting on a line segment (Zwiebach, A First Course in String Theory, pg. 310). The Stasheff associahedra make another cameo appearance being intimately related to the moduli spaces of colliding particles (Devadoss, Devadoss/Heath/Vipismakul, Devadoss/Fehrman/Heath/Vashist, and the beautifully illustrated book Discrete and Computational Geometry by Satayan Devadoss and Joseph O'Rourke). -Alexander Givental in "Gromov–Witten invariants and quantization of quadratic Hamiltonians" relates a Virasoro algebra to the Witten–Kontsevich tau-function/potential and Euler fields. (The corresponding Witt algebra rep is rife with enumerative combinatorics. See my sketch of the algebra in "Infinitesimal generators, the Pascal Triangle, and the Witt and Virasoro algebras".) -So, the connecting element that these hydrodynamical and topological characters seem to share are the simple infinigens--the ghosts of Lie. -Another example (update Oct 12, 2015) -A Ricatti equation related to quadratic infinigens in sl(2) is linked to a soliton solution $1-tanh^2(x-ct)=d[tanh(x-ct)]/dx$ of a Kdv equation in The Elliptic Lie Triad (following up on my comment below on Rzadowski's paper). The hyperbolic tangent can be regarded as an exponential generating function for the number of connected components in the space of M-polynomials in hyperbolic functions (ref. in OEIS A000111) or for a proportionality factor in the Kervaire-Milnor formula in homotopy theory for hyper-spheres involving normalized Bernoulli numbers. -More generally, the bivariate e.g.f. for the Eulerian numbers (A008292/A123125) with its associated quadratic (sl2) infinigen provides a soliton solution of the 1-D KdV equation, and the Eulerians are rife with ($A_n$ and $B_n$) connnections to enumerative algebraic geometry, as discussed by Hirzebruch, Losev and Manin, Batryev and Blume, Cohen, et al.<|endoftext|> -TITLE: Can one characterize the category of finite-dimensional vector spaces? -QUESTION [5 upvotes]: Let $K$ be a field. Does the category of finitely generated $K$-modules have a nice characterization, for example as the unique abelian category satisfying a certain simple condition? For example, we know that: - -Every short exact sequence is split. -The Euler characteristic of every bounded exact sequence is zero. - -Are either of those enough to characterize the category? - -REPLY [4 votes]: The answer is no, and an easy counterexample is provided by the category of finite-dimensional modules over a division algebra such as the quaternions. Of course this is not a very good example because you can easily add small modifications to your question to get rid of it. This category is for instance not symmetric monoidal, unlike vector spaces over a field. As Oskar points out, there is an already answered question in MO which gives a positive answer to your question under somewhat different conditions. You'll like to look at it. I warn you that your Euler characteristic condition may be complicated to state in an abstract setting.<|endoftext|> -TITLE: Is the consistency of $\mathcal{L}_{\infty\omega}$-sentences absolute? -QUESTION [7 upvotes]: The question is exactly that of the title. Suppose $\varphi\in V$ is an $\mathcal{L}_{\infty\omega}$-sentence, and $W$ is an inner model of $V$ such that $\varphi\in W$. Is the statement - -$\varphi$ has a model - -absolute between $V$ and $W$? It is clearly upwards absolute, by induction on rank, so I'm really asking about the downwards direction. -The main obstacle that I see right now is the lack of a Lowenheim-Skolem theorem, which prevents us from using Shoenfield absoluteness. As a specific example, let $A$ be $\omega_1^W$ as a linear order, and suppose $V$ is a forcing extension of $W$ in which $\omega_1$ is collapsed. In $V$, there is a sentence $\varphi\in\mathcal{L}_{\omega_1\omega}\subset\mathcal{L}_{\infty\omega}$ whose only countable model up to isomoprhism is $A$: the Scott sentence of $A$. Now there is a formula $\psi\in \mathcal{L}_{\infty\omega}\cap W$ such that, in $V$, $\psi$ and $\varphi$ are equivalent; this is a roundabout way of saying "$\varphi\in W$," which is not exactly true. But then $\psi$ has no countable models in $W$, since such a model would have to be isomorphic to $A$ in $V$. -I assume this is well-known, but I haven't been able to find an answer myself. - -REPLY [7 votes]: Here is a somewhat easier counterexample. -Let $I$ be a countable set, which is uncountable in $W$. Let $c_n$ and $d_\alpha$ be constant symbols, for $n\in\mathbb{N}$ and $\alpha\in I$. Consider the formula $\varphi$ that asserts that all the $d_\alpha$'s are different, but that every $d_\alpha$ is equal to some $c_n$. That is, -$$\varphi=\bigl(\bigwedge_{\alpha\neq\beta\in I}d_\alpha\neq d_\beta\bigr)\wedge\bigl(\bigwedge_\alpha\bigvee_n d_\alpha=c_n\bigr).$$ A model of this sentence is essentially providing an injective function from $I$ to $\mathbb{N}$, mapping $\alpha\mapsto n$ when $n$ is least for which $d_\alpha=c_n$. Thus, the sentence is not satisfiable in $W$, since $I$ is uncountable there, but it is satisfiable in $V$, since $I$ is countable in $V$. -Using the same idea, one can make a counterexample in a finite signature language as follows. Suppose that $P(\mathbb{N})^W$ is countable in $V$. First, we can make a sentence that asserts that we have a copy of $\langle\mathbb{N},S\rangle$, by asserting that everything in that sort is a finite successor of $0$. Next, we can have another sort that will be a copy of $P(\mathbb{N})^W$, each subset represented by a vertex pointing exactly at its elements. In a single formula (of size continuum in $W$), we can say that every subset of $\mathbb{N}$ that it is represented by some vertex. Finally, with one more binary relation, we can say that every subset is associated with a distinct natural number. The point now is that the overall assertion is not satisfiable in $W$, since $P(\mathbb{N})^W$ is uncountable in $W$, but it will be satisfiable in $V$, since the set is countable there. -Update. Meanwhile, the satisfiability of sentences in $\cal{L}_{\omega_1,\omega}$ logic, that is, using only countable meets and joins, is a $\Sigma^1_1$ assertion and hence absolute between models with the same countable ordinals.<|endoftext|> -TITLE: When can number rings be spanned (as $\mathbb{Z}$-modules) by units? -QUESTION [9 upvotes]: Let $\mathcal{O}$ be the ring of integers in an algebraic number field. Define $R \subset \mathcal{O}$ to be the set of all $\mathbb{Z}$-linear combinations of units. Since the product of two units is a unit, the set $R$ is a ring. -Question : Under what circumstances do we have $R = \mathcal{O}$? -Of course, this holds for $\mathcal{O} = \mathbb{Z}$. It is also clearly false for imaginary quadratic number rings aside from the Gaussian integers. But it is true for all the real quadratic number rings I have played with. Maybe it holds whenever $\mathcal{O}$ has infinitely many units? This might be too optimistic... -In case the above has a wild or overly complicated answer, the following question might be easier. -Question : Under what circumstances is $R$ finite-index in $\mathcal{O}$ as an abelian group? - -REPLY [12 votes]: As was already pointed out, the only imaginary quadratic fields with $R=\mathcal O$ are $\mathbb Q(i)$ and $\mathbb Q(\zeta_3)$. This follows pretty easily from the structure of $\mathcal O^\times$. On the other hand, for real quadratic fields $\mathbb Q(\sqrt{d})$, with $d \in \mathbb Z$ square-free, one has $R=\mathcal O$ in the following cases only: - -$d\not\equiv 1 \pmod{4}$ and either $d+1$ or $d-1$ is a perfect square. -$d\equiv 1 \pmod{4}$ and either $d+4$ or $d-4$ is a perfect square. - -This is was first proved by Belcher, Integers expressible as sums of distinct units, Bull. Lond. Math. Soc. 6 (1974), 66–68. -There are similar results for certain cubic and quartic fields but the problem for general number fields seems to be wide open. Nonetheless, an interesting positive result in this direction was obtained by Frei, On rings of integers generated by their units, Bull. London Math. Soc. 44 (2012), 167–182: - -For every number field $K$ there exists a number field $L$ containing $K$ such that $\mathcal O_L = R_L$.<|endoftext|> -TITLE: On the pathwise uniqueness of solutions of SDEs(Stochastic Differential Equations) -QUESTION [8 upvotes]: Suppose that $(\Omega,\mathscr{F},P)$ is a complete probability space equipped a filtration $\{\mathscr{F}_t\}$ satisfying the usual conditions. $B_t$ is a 1-dimentional Brownian motion with respect to the filtration $\{\mathscr{F}_t\}$. Consider the 1-dimentional SDE -\begin{equation}dx(t)=f(x(t),t)dt+g(x(t),t)dB(t),\ (*)\end{equation} -with the initial value $x(t_0)=\xi\in L^2(\Omega;\mathbb{R})$, which statisfies the Lipschitz condition and the Linear growth condition. -Let -\begin{equation}\Omega_0=\{\omega,\ there\ exist\ at\ least\ two\ different\ sample\ paths\ x_{\alpha}(t,\omega), x_{\beta}(t,\omega)\}\end{equation} -Where $x_{\alpha}(t)$, $x_{\beta}(t)$ are both solutions of $(*)$. -By the existence and uniqueness theorem, $(*)$ has the pathwise unique solution, which means that if there are two solutions $x(t)$ and $\overline{x}(t)$, then $P\{x(t)=\overline{x}(t),\forall t\ge t_0\}=1$. However, from the proof(...$E\big(\sup_{t_0\le t\le T}|x(t)-\overline{x}(t)|^2\big)=0$) of existence and uniqueness theorem(cf. Stochastic Differential Equations and Applications(second Edition) by Xuerong Mao, page 53), we know that for any two solutions $x(t)$, $\overline{x}(t)$, there is a $P$-null set $\Omega_0(x,\overline{x})$ such that $\forall\omega\in\Omega\setminus\Omega_0(x,\overline{x})$, $x(t)=\overline{x}(t), \forall t\ge t_0$, if there are another two solutions $y(t)$, $\overline{y}(t)$, the $P$-null set $\Omega_0(y,\overline{y})$ may not equal to $\Omega_0(x,\overline{x})$. -It is straightforward to see that -\begin{equation}\Omega_0=\bigcup_{x,y\ are\ solutions\ of\ (*)} \Omega_0(x,y)\end{equation} -My question is: whether $P(\Omega_0)=0$. In other words, whether we have a unique solution to the following equation for almost surely fixed $\omega$? -\begin{equation}x(t,\omega)=x(t_0,\omega)+\int_{t_0}^tf(x(s,\omega),s)ds+\int_{t_0}^tg(x(s),s)dB(s)(\omega)\end{equation} -Of course, if $P(\Omega_0)\ne0$, we can construct two new processes destroying the pathwise uniqueness, but are these prosesses $\mathscr{F}_t$-adapted? -Thank you. - -REPLY [3 votes]: The answer really depends on what you mean exactly by "a solution" to your SDE. If it is a solution in the classical (Itô calculus) sense then, since these can always be modified on a null set, you can never hope to have a property of the type you want, as correctly pointed out by thomas. Maybe a more interesting question then is whether there does exist a notion of solution that is sufficiently strong so that it completely determines the solution on some fixed set of full measure. -If $f$ and $g$ are sufficiently smooth ($\mathcal{C}^{2,\delta}$ for some $\delta > 0$ is enough), then this is indeed the case. In the case you consider (i.e. when $B$ is one-dimensional) this was pointed out by Doss and Sussmann in the seventies. (See Doss, Ann. IHP 13 1977 and Sussmann, AoP 6 1978.) In the case when $B$ is multidimensional, this can still be done, but it is a little bit more complicated (see Lyons, Rev. Mat. Iberoamericana 14 1998 or my recent book with Peter Friz).<|endoftext|> -TITLE: Do constructible sets have Krull dimension? -QUESTION [5 upvotes]: Let $(I,\leq)$ be a poset. Recall that the Krull dimension of $I$ is defined as follows: --- $K.dim(I)=-1$ if and only if $I=\{0\}$; --- if $\alpha$ is an ordinal and we already defined what it means to be a poset with Krull dimension $\beta$ for any ordinal $\beta<\alpha$, we say that $K.dim(I)=\alpha$ if and only if $K.dim(I)\neq \beta$ for all $\beta<\alpha$ and, for every descending chain -$$x_1\geq x_2\geq x_3 \geq \ldots \geq x_n\geq \dots$$ -of elements of $I$, there exists $\bar n\in \mathbb N_+$ such that $K.dim([x_n,x_{n+1}])=\beta_n$ for all $n\geq \bar n$ and $\beta_n$ an ordinal $<\alpha$; -(here the notation $[a,b]$ is used for the segment between $a$ and $b$, that is, the subset of $I$ of all elements $\geq a$ and $\leq b$) -The above definition is mostly applied to posets which are indeed lattices. Notice that the lattices with $0$ Krull dimension are precisely the Artinian lattices, it is known (and not difficult to prove) that Noetherian lattices always have Krull dimension (this in fact generalizes to the following statement: "a lattice has Krull dimension if and only if its dual lattice has Krull dimension"). -Let $A\subseteq \mathbb K^n$ be an affine algebraic variety over an algebraically closed field $\mathbb K$. It is well-known that, when endowed with its Zariski topology, $A$ is a Noetherian topological space, that is, the lattice of open subsets is Artinian, equivalently, the lattice of closed subsets is Noetherian. In particular, both these lattices have Krull dimension. There is a third natural lattice to consider, that is, the lattice $\mathcal C(A)$ of constructible subsets of $A$, which is the smallest lattice containing open and closed subsets of $A$ (that is, $\mathcal C(A)$ is obtained taking finite unions of intersections of an open and a closed subset). -Now, here is my question: -Can we say that $\mathcal C(A)$ has Krull dimension? Is this dimension equal to the dimension of the lattice of closed subsets? - -REPLY [8 votes]: For an arbitrary noetherian topological space $X$, I think the lattice $\mathcal{U}(X)$ of open subsets and the lattice $\mathcal{C}(X)$ of constructible subsets (the boolean algebra generated by open subsets) have the same Krull dimension (I use the standard ordering, to avoid confusion (see Ramiro's comment), so the lattice of closed subsets has Krull dimension zero by definition). I'll show they're both equal to a third number, based on the (finer) notion of ordinal length. -Let $\alpha\ge 1$ be an ordinal. Define $\log_\omega(\alpha)$ as the unique ordinal $\kappa$ such that $\omega^\kappa\le\alpha<\omega^{\kappa+1}$ (we agree that $\log_\omega(0)=-1)$. -Let $X$ be a noetherian topological space. Define inductively its length $\ell(X)=\sup(\ell(Y)+1)$ where $Y$ ranges over closed subsets of $X$ distinct of $X$ ($\sup\emptyset=\emptyset$). This is a standard inductive definition (take $Y$ minimal such that $\ell(Y)$ is not defined to get a contradiction). Note that $\ell(X)=0$ iff $X$ is empty. - -Proposition: for every noetherian topological space $X$, the Krull - dimension of both $\mathcal{U}(X)$ and $\mathcal{C}(X)$ is equal to - $\log_\omega(\ell(X))$. - -Let's proceed to the proof. -Notation: We say that an ordinal $\alpha\ge 1$ is irreducible if $\alpha=\omega^\kappa$ for some $\kappa$ (necessarily equal to $\log_\omega(\alpha)$), or equivalently if $\beta,\gamma<\alpha$ implies $\beta+\gamma<\alpha$. -Also $\mathcal{C}(X;A,B)$, for $A,B\subset X$, is the interval between $A$ and $B$, i.e. those $Y\in\mathcal{C}(X)$ such that $A\subset Y\subset B$. Similarly $\mathcal{U}(X;A,B)$ is defined. -The proposition follows from the two following claims (1) (2). - -(1) The Krull dimension of $\mathcal{C}(X)$ is at most - $\kappa=\log_\omega(\ell(X))$. - -We need to know the following facts on the length function (obtained by an easy induction): if $Y$ is a subset of $X$ with the induced topology, then $\ell(Y)\le\ell(X)$. Also if $F$ is a closed subset of $X$, then $\ell(X)\ge\ell(F)+\ell(X\smallsetminus F)$. -We prove (1) this by induction on $\ell(X)$. -Let $(C_n)$ be a descending chain of constructible subsets of $X$ and let us prove that $\mathcal{C}(X;C_{n+1},C_n)$ has Krull dimension $<\kappa$. If $C_n$ is eventually empty, $\mathcal{C}(X;C_{n+1},C_n)$ is eventually a singleton, thus of Krull dimension $-1<\kappa$. Now assume the contrary. -If for some $n$, $C_n$ is not dense, we can work inside its closure and argue by induction to get the desired conclusion. So we assume that each $C_n$ is dense (and not empty). It follows that the interior $U_n$ of $C_n$ is dense as well (in any noetherian topological space, a dense constructible subset has dense interior). In particular, $U_n$ is nonempty as well. -Then $(\ell(U_n))$ is a descending chain of ordinals, hence stabilizes, say for $n\ge n_0$. Set $V=U_{n_0}$. Then for $n\ge n_0$ we have $\ell(V)\ge\ell(V\smallsetminus U_n)+\ell(U_n)=\ell(V\smallsetminus U_n)+\ell(V)$. An easy consequence is that $\ell(V\smallsetminus U_n)<\omega^\kappa$ (using that every ordinal $<\omega^{\kappa+1}$ can be written as $\omega^\kappa\cdot n+\beta$ for a unique integer $n\ge 0$ and ordinal $\beta<\omega^\kappa$, particular case of the "Cantor form"). By induction, it follows that $V\smallsetminus U_n$ has Krull dimension $<\kappa$ for all $n\ge n_0$. Hence $\mathcal{C}(X,U_{n},V)$ has Krull dimension $<\kappa$ for all $n\ge n_0$. -Let $F$ be the complement of $V$. Then $\ell(F)<\ell(X)$, because $V\neq\emptyset$, so $F$ has Krull dimension at most $\kappa$. Hence there exists $n_1\ge n_0$ such that $\mathcal{C}(F;C_{n+1}\cap F,C_n\cap F)$ has Krull dimension $<\kappa$ for all $n\ge n_1$. The lattice $\mathcal{C}(X;C_{n+1},C_n)$ naturally embeds into the product of the lattices $\mathcal{C}(F;C_{n+1}\cap F,C_n\cap F)$ and $\mathcal{C}(V;C_{n+1}\cap V,C_n\cap V)$, and the latter is contained in $\mathcal{C}(V;U_{n+1},V)$. Since a product of lattices of Krull dimension $<\kappa$ also has Krull dimension $\kappa$, we deduce that $\mathcal{C}(X;C_{n+1},C_n)$ has Krull dimension $<\kappa$ for all $n\ge n_1$. This proves that $\mathcal{C}(X)$ has Krull dimension $\le\kappa$. - -(2) The Krull dimension of the lattice $\mathcal{U}(X)$ is $\ge\kappa$. - -We'll use the immediate fact that if $X$ is a noetherian space, then for every ordinal $\gamma\le\ell(X)$ there exists a closed subset $Y$ of $X$ such that $\ell(Y)=\gamma$. -We also use that if $\kappa\ge 1$ and $Y$ is a closed subset of $X$ then $\ell(X)\le \ell(Y)\oplus\ell(X\smallsetminus Y)$, where $\oplus$ is the commutative sum of ordinals: $\alpha\oplus\beta=\max\{\sup_{\gamma<\alpha}((\gamma\oplus\beta)+1),\sup_{\delta<\alpha}((\alpha\oplus\delta)+1))\}$. In particular if $\ell(Y)$ and $\ell(X\smallsetminus Y)$ are both $<\omega^\kappa$ (for $\kappa\ge 1$) then so is $\ell(X)$. -To prove the claim (2), if $\kappa=0$ there is nothing to prove, so suppose $\kappa\ge 1$. -Suppose by contradiction the Krull dimension of $\mathcal{U}(X)$ is an ordinal $\beta<\kappa$. To get a contradiction, we have to construct a descending sequence $(U_n)$ of open subsets such that for each $n$, the Krull dimension of $\mathcal{U}(X;U_{n+1},U_n)$ is $\ge\beta$. Let $Y\neq X$ be a closed subset of $X$ with $\ell(Y)=\omega^\beta$. Define $U_1$ as the complement of $Y$. Since $\ell(Y)=\omega^\beta<\omega^\kappa$, it follows that $\ell(U_1)\ge\omega^\kappa$. -So we can continue by induction (on the integers) to obtain a descending sequence $(U_n)$ of open subsets, so that for each $n$ we have $\ell(U_n\smallsetminus U_{n+1})=\omega^\beta$. By induction (on ordinals), the Krull dimension $\mathcal{U}(U_n\smallsetminus U_{n+1})$ is $\ge\beta$. The latter lattice can be identified with the lattice $\mathcal{U}(X,U_{n+1},U_n)$. Hence we have a contradiction and it follows that the Krull dimension of $\mathcal{U}(X)$ is $\ge\kappa$.<|endoftext|> -TITLE: Isotopy extension theorems -QUESTION [22 upvotes]: I'm looking for the origins of the isotopy extension theorem in categories other than the smooth category. -Precisely, in the smooth category, the isotopy extension theorem says that if $f : [0,1] \times M \to N$ is a smooth 1-parameter family of embeddings (with $N$ boundaryless, and $M$ compact) then there exists $F : [0,1] \times N \to N$ a smooth $1$-parameter family of diffeomorphisms so that $F(0, \cdot) = Id_N$ and $F(t,f(0,x)) = f(t,x)$ for all $(t,x) \in [0,1] \times M$. This was seen to be a "very natural" theorem by Palais, with the generalization stating that the restriction map $Diff(N) \to Emb(M,N)$ was not only a Serre fibration but a locally trivial fibre bundle. In this ideal case, the references are: - -Palais, Richard S. Local triviality of the restriction map for embeddings. Comment. Math. Helv. 34 1960 305–312. -E. L. Lima, On the local triviality of the restriction map for embeddings, Commentarii Mathematici Helvetici Volume 38, Number 1, pp 163-164. - -Let $Aut(N)$ be the automorphisms of the manifold $N$ in whichever category of manifolds it lives in (topological, $PL$ or smooth). My understanding is its known that the restriction map -$$Aut(N) \to Emb(M,N)$$ -is known to be a Serre fibration provided $N$ is a co-dimension $0$ submanifold of $N$, in any of the above three manifold categories. -My questions: -Q1: Where were these results first proven in the PL and TOP cases? Are they known for all dimensions? -Q2: If one allows $M$ to have co-dimension $> 0$, what is known about this map being or not being a fibration? -I'm in the process of trying to both learn the basics and get an overview of smoothing theory. Any help is appreciated. - -REPLY [8 votes]: It looks like the Palais-style variant of the PL isotopy extension theorem is proven in: - -Hudson, John F. P. Extending piecewise-linear isotopies. -Proc. London Math. Soc. (3) 16 1966 651–668. - -In this article Hudson says the result is first proven in M.C. Irwin's dissertation at Cambridge. -By this I mean he's showing $Aut(N) \to Emb(M,N)$ is a Serre fibration. He also assumes co-dimension $\geq 3$.<|endoftext|> -TITLE: Best record toward Selberg's eigenvalue conjecture? -QUESTION [11 upvotes]: What's the best record toward Selberg's eigenvalue conjecture: -a Maass form on $\Gamma_0(N)$ has eigenvalue greater than or equal to 1/4? - -REPLY [16 votes]: \[\frac{1}{4} - \left(\frac{7}{64}\right)^2 = \frac{975}{4096} \approx 0.238037\ldots\] -as far as I know. Established in an appendix by Kim and Sarnak to Kim's "Functoriality for the exterior square of $GL_{4}$ and the symmetric fourth of $GL_{2}$", Journal of the American Math. Society, 16 (2003). -The above is "over the rationals" (which was/is my reading of the question), in comments work on the problem for general number fields got mentioned by Kim and Shaidi, Cuspidality of symmetric powers with applications. Duke Math. J. 112 (2002) and the more recent work of Blomer and Brumley, On the Ramanujan conjecture over number fields. Annals of Math. 174 (2011) which among others establishes the above for general number fields. -For generalizations in a different direction let me also mention Bourgain, Gamburd, Sarnak, Generalization of Selberg's 3/16 Theorem and Affine Sieve, Acta Math. 207 (2011) establishing 'a generalization of Selberg's theorem for infinite index "congruence" subgroups of SL(2,Z).'<|endoftext|> -TITLE: chern classes of push-pulled vector bundles -QUESTION [5 upvotes]: Let $f:X\to Y$ be a finite cover of smooth algebraic varieties, branched along a divisor $R\subset Y$. Let $E$ be a vector bundle on $Y$. What is the relation between the chern classes of $E$ and the chern classes of $F:=f_*(f^*E)$? - -REPLY [7 votes]: You need to know what is the splitting of $f_* \mathscr{O}_X$, which depends on the cover. -Let us consider the double cover case, which is the easiest one. Then $$f_* \mathscr{O}_X = \mathscr{O}_Y \oplus L^{-1},$$ where $L$ is a line bundle on $Y$ such that $R \in |2L|$. Then, given any vector bundle $E$ on $Y$, by using projection formula one obtains -$$F=f_*(f^*E)=f_*(f^*E \otimes \mathscr{O}_X) = E \otimes f_* \mathscr{O}_X = E \oplus (E \otimes L^{-1}),$$ -so the computation of the Chern classes of $F$ is now a straightforward application of the splitting principle. -For general finite covers of degree $d$, one can only say that $$f_* \mathscr{O}_X = \mathscr{O}_Y \oplus V,$$ -where $V$ is a vector bundle on $Y$ of rank $d-1$. As before, one obtains $$F=f_*(f^*E)=f_*(f^*E \otimes \mathscr{O}_X) = E \otimes f_* \mathscr{O}_X = E \oplus (E \otimes V),$$ -but now the computations of Chern classes will be more difficult since in general $V$ is indecomposable. -In some lucky cases, for instance when the cover $f \colon X \to Y$ is Galois with abelian Galois group, $V$ splits completely into line bundles and then, if one is able to identify $V$, the computation can be made again by using the splitting principle. - -REPLY [4 votes]: By the projection formula $f_*(F^*E)\cong E\otimes f_*\mathcal{O}_X$, so $\mathrm{ch}(f_*(f^*E))=\mathrm{ch}(E).\mathrm{ch}(f_*\mathcal{O}_X)$. By Grothendieck-Riemann-Roch $\mathrm{ch}(f_*\mathcal{O}_X)$ is the pushforward of the relative Todd class -$\mathrm{Todd}_{X/Y}$, so in principle you can compute the Chern character, hence the Chern classes, of $f_*(F^*E)$.<|endoftext|> -TITLE: How many combinations of intersections of n hyperplanes there are in which a hyperplane is not crossed more than k times? -QUESTION [5 upvotes]: Consider you have $n$ hyperplanes with dimension $k$ that are not coplanar. Each hyperplane intersects the others $n-1$ times. Any intersection of $k$ such hyperplanes produces a vector. The number of such intersections is therefore -$$d=\left(\begin{array}{c} -n\\ -k -\end{array}\right)$$ -From these $d$ intersections, I am interested in choosing $n$ intersections, but only those combinations that will have any hyperplane intersected in exactly $k$ positions. -Here is an example with $n=4$ and $k=2$: - -The lines represent the planes and the dots represent the intersections of the planes. The top row shows valid combinations, while the bottom row shows invalid ones. -I would like to calculate how many valid combinations exist, for any $k$ and $n>k$. - -REPLY [5 votes]: I added the tag combinatorics and am doubtful about the tag arithmetic-geometry because the particular hyperplane arrangement is irrelevant. The question is simply this: given a set of $n$ symbols (which may as well be $\lbrace 1,2,\cdots,n\rbrace$) find $n$ (different) subsets of size $k$ so that every symbol is in exactly $k$ of the subsets. This could be called a symmetric incomplete block design $(v,b,r,k)=(n,n,k,k)$ but that term is usually reserved for symmetric balanced incomplete block design where we require each pair of symbols to be in $\frac{k(k-1)}{n-1}$ common blocks. -later computing very small counts and using the OEIS finds information for the case $k=2$ and some for the case $k=3$. -In the special case that $k=2$ it is not hard to see that the number of ways for $n=3,4,5,6,7$ are $1,{\LARGE3},12,70,465.$ That is enough to locate this sequence in the OEIS. There you will find references, a recurrence relation, asymptotics and the exponential generating function. Also the concept of a frame: - -[N]umber of 'frames' on n lines: given n lines in general position (none parallel and no three concurrent), a frame is a subset of n of the C(n,2) points of intersection such that no three points are on the same line. - -The answer can be determined fairly quickly for any particular $n$ as a sum over various partitions (all parts at least $3$) of a product of multinomial coefficients (with some extra factorials in the quotient). evidently, all this can be encoded into an exponential generating function. -details: When $k=2$ one could view the chosen blocks to be $n$ of the $\binom{n}{2}$ edges of the complete graph $K_n$ chosen so that each node is on two edges. This is called a $2$-factor and consists of one or more disjoint cycles with combined length $n$. With only one cycle there are $\frac{(n-1)!}{2}$ solutions. This is all for $n=3,4,5.$ When $n=6$ there are $\frac{5!}{2}=60$ one cycle solutions along with $\binom{6}{3}\frac{1}{2!}\cdot 1\cdot 1=10$ solutions with two $3$-cycles. For $n=7$ we have $360$ one cycle solutions along with $\binom{7}{4} \frac{(3-1)!}{2} \frac{(4-1)!}{2}=105$ ways to separate the symbols into groups of size $3$ and $4$ and pick a cycle for each group. -For $n=11,k=2$ the "cycle-type" could be $11$ or $8,3$ or $7,4$ or $6,5$ along with $3,3,5$ and $3,4,4$. In general (for $k=2$) one can give an expression as a summation over the "cycle-types" of product/quotients of multinomial coefficients and factorials. -LATER For the case $k=3$ the number of ways for $n=4,5,6$ are $1,12,330$ That is enough to locate the sequence number of 3-regular 3-hypergraphs on n labeled vertices in the OEIS. A generating function (or at least a huge differential equation satisfied by one) and (complicated) recurrence are given. The usefulness seems limited as the numbers are found only out to $n=15$. In retrospect that would have been a reasonable name to search for. In general your question could be phrased as: - - - -How many $k$-uniform $k$-regular hypergraphs are there on $n$ labelled vertices?<|endoftext|> -TITLE: Are all (possibly infinite dimensional) irreducible representations of a commutative algebra one-dimensional? -QUESTION [5 upvotes]: If $A$ is a commutative algebra over an algebraically closed field $k$, and $\rho:A \rightarrow End(V)$ is an irreducible representation of $A$ (where, a priori, $V$ may be infinite dimensional), can we conclude that $V$ must be one-dimensional? This is easy to show if we assume $V$ is finite dimensional. It is also true if $A$ is a $C^*$-algebra, $V$ is a (complex) Hilbert space, $End(V)$ denotes bounded operators, $\rho$ preserves conjugation (but is not required to be continuous), and irreducible means no closed subrepresentations, using some spectral theory. But is it true in the algebraic sense with no further restrictions on $A$ or $V$? -Edit: Given Dag Oskar Madsen's comment, I will need to place some restrictions on $A$... what if $A$ is finitely-generated? -Double Edit: Faisal's comment takes care of $A$ finitely-generated (and countably generated) over $\mathbb{C}$. - -REPLY [4 votes]: Here are the details on Faisal's suggestion in the comments. -Lemma (Dixmier): Let $k$ be an algebraically closed field, let $A$ be a $k$-algebra with $\dim A < |k|$, and let $V$ be a simple left $A$-module. Then $D = \text{End}_A(V) \cong k$. -Proof. By Schur's lemma, $D$ is a division algebra over $k$. Then $V$, as a $D$-module, is a sum of copies of $D$, and in particular $\dim D \le \dim V \le \dim A < |k|$. Since $k$ is algebraically closed, if $D$ is strictly larger than $k$ then it must contain a transcendental $t$, hence $k(t) \subseteq D$. But the elements $\frac{1}{t - a}, a \in k$ in $k(t)$ are linearly independent, hence $\dim k(t) \ge |k|$; contradiction. $\Box$ -If $A$ is commutative, the image of $A$ in $\text{End}(V)$ is contained in $D$, so $A$ acts by scalars and the conclusion follows. (Incidentally, this gives a short proof of the weak Nullstellensatz over uncountable (algebraically closed) fields.)<|endoftext|> -TITLE: Why does closed string theory have only one dilaton field instead of $22$? -QUESTION [6 upvotes]: Looking at $5D$ Kaluza-Klein theory, the Kaluza-Klein metric is given by -$$ -g_{mn} = \left( -\begin{array}{cc} - g_{\mu\nu} & g_{\mu 5} \\ - g_{5\nu} & g_{55} \\ -\end{array} -\right) -$$ -where $g_{\mu\nu}$ corresponds to the ordinary four dimensional metric and $g_{\mu 5}$ is the ordinary four dimensional Maxwell gauge field, $g_{55}$ is the dilaton field. -As there is one dilaton for one extra dimension, I naively would expect that the zero mass states of closed string theory, which can be written as -$$ -\sum\limits_{I.J} R_{I.J} a_1^{I\dagger} \bar{a}_1^{I\dagger} ¦p^{+},\vec{p}_T \rangle -$$ -and the square matrix $R_{I.J}$ can be separated into a symmetric traceless part corresponding to the graviton field, an antisymmetric part corresponding to a generalized Maxwell gauge field, and the trace which corresponds to the dilaton field. -Why is there only one dilaton field given by the trace of $R_{I.J}$, instead of $22$ dilaton fields corresponding to $22$ extra dimensions of closed string theory which has critical dimension $D = 26$? For example, why are there not $22$ dilaton fields needed to parameterize the shape of the 22 extra dimensions if they are compactified? - -REPLY [4 votes]: You would actually not expect 22 dilatons. Let me try to explain. -As you have pointed out, a putative field theory limit of the closed bosonic string would consist of a metric, a 2-form (which is the potential for a 3-form) and a dilaton. -Let us assume that such a theory exists and let us dimensionally reduce to four dimensions à la Kaluza-Klein. -The 26-dimensional dilaton $\Phi$ gives a scalar field in four dimensions. The 26-dimensional metric $g_{MN}$ gives the following four-dimesional fields: a metric $g_{\mu\nu}$, 22 gauge fields $g_{\mu n}$ and $\binom{23}2 = 253$ scalars $g_{mn}$. Finally, the 26-dimensional 2-form $B_{MN}$ gives the following: $\binom{22}{2} = 231$ scalars $B_{mn}$, $22$ gauge fields $B_{\mu n}$ and $\binom{4}{2} = 6$ scalars dual to the 2-forms $B_{\mu\nu}$. (In four dimensions, 2-forms can be dualised to scalars: via $\star dB = d\varphi$.) -So from a 4-dimensional perspective the role of the “dilaton” is now the $22 \times 22$ symmetric matrix $g_{mn}$. -This does not imply that you should expect any number of scalars in the massless sector of the closed string. The massless sector of the string theory corresponds to a field theory in 26 dimensions and in order to interpret it in 4-dimensional terms, requires dimensional reduction.<|endoftext|> -TITLE: How to project a vector onto a very large, non-orthogonal subspace -QUESTION [12 upvotes]: I have a difficult problem. -I have a very large, non-orthogonal matrix $A$ and need to project the vector $y$ onto the subspace spanning the columns of $A$. If this were a small matrix, I would use Gram-Schmidt or just compute $A(A^TA)^{-1}A^Ty$. Unfortunately, $A$ is just too big to do this in a feasible amount of time. Is there a good trick to do this projection? -EDIT2: Some background info: I work in signal processing and work with Fusion Frames quite a bit. For those unfamiliar, it is the problem of representing a signal with a union of subspaces from an overcomplete dictionary, typically using a block-sparse solver (in my case, a program called SPGL1). The problem is, I need to project my signal vector onto this dictionary before performing the inversion. The dictionary is coherent so under normal circumstances I need to use the formula above or a method like Gram-Schmidt to get an orthonormal basis. -Right now, my working solution is to approximate this projection by using the truncated SVD and using the $U$ matrix as an approximate orthonormal basis (a la PCA). But, it would be nice if anyone knew any iterative projection algorithms. I know this isn't a scientific computing site, but I figured there may be someone here who knows some linear algebra tricks. - -REPLY [4 votes]: It makes a considerable difference if you need to project a vector just once or repeatedly inside some loop of another algorithm. If you would be in the latter case than it would be indeed a good idea to follow suv...rits suggestion and use an iterative algorithm to solve the normal equations (conjugate gradients, probably with preconditioning, suggest themselves). In this case you can often just use a small and fixed number of CG iterations and restart the next CG iteration with what you've got from the previous run. If the point you want to project does not change very much in the outer loop, CG should be converging eventually. -However, your EDIT2 suggests that your only project you data once. Without knowing your precise approximation problem, it could be possible to make a reformulation of the problem: If $y$ is your signal but you need to put $z = A(A^T A)^{-1}A^Ty$ in your algorithm, you could just put in $z$ and add the constraints $z = Aw$ and $A^T Aw = A^T y$, i.e. you introduce two new variables and two new linear constraints. Then use some algorithm for your new inversion which can handle linear constraints. Without knowing more details I can't be more concrete here…<|endoftext|> -TITLE: Categorical Semantics for Second-Order Logics -QUESTION [9 upvotes]: I am currently doing some work using a categorical semantics of first-order logic. The specific semantics I am using is due to Andrew Pitts, as described in: - -Categorical Logic, Andrew M. Pitts, chapter in the Handbook of Logic in Computer Science, Volume VI, 1995. - -In this treatment, the syntax uses terms-in-context and formulae-in-context, instead of just terms and formulae. A context of a term is, loosely, the set of free variables in the term and their types. A formula in classical first-order logic is interpreted using a category $\mathcal{C}$ with finite products, and a contravariant functor $\mathit{Prop}_\mathcal{C}$ from $\mathcal{C}$ to the category of posets and monotone functions. Propositional connectives are characterized in terms of meets and joins in the category and (generalized) quantifiers arise as adjoints of projections. -This semantics has the property that over the category of sets, satisfaction agrees with satisfaction in the Tarskian sense. This property is important for the work I am doing. -Question. Is there a categorical treatment of (Monadic) Second-Order Logic with the same property? Ideally, a categorical semantics that is transparent in the sense that it is almost obvious that it agrees with the Tarskian semantics on the category of sets (or a Boolean topos, which I expect is what I need). -My impression is that the term higher-order in categorical logic typically applies to higher-order functions rather than uninterpreted predicates. Please let me know if this impression is incorrect. -For your information,I am aware that there are slightly different treatments of these concepts starting from Lawvere's Adjointness in Foundations paper and I've consulted the work of Makkai and Reyes, Lambek and Scott, Crole, course notes by Awodey and Bauer and the work of Pitts on Tripos theory. - -REPLY [2 votes]: I think the main references for this are Jacobs's book Categorical Logic and Type Theory (Chapter 8), and also Crole's Categories for Types. -The rough idea is to do categorical logic over the `algebraic theory of types'. -That is, you consider a category whose objects are finite products of a certain object $\mathit{Type}$, and whose morphisms $A \to \mathit{Type}$ are thought of as types with free type variables in $A$. Then, models are fibrations over such categories, of which you ask for more structure to be able to quantify over predicates. By carefully tuning what you demand, you may obtain quantification over all kinds of higher-order predicates, but there are ways of sticking to second-order logic. -Regarding your terminological question about the term `higher-order', I'd say `higher-order' refers both to higher-order functions and to quantification over higher-order predicates, as in, e.g., System $F\omega$. This is of course annoying.<|endoftext|> -TITLE: Are torus knot groups linear? -QUESTION [8 upvotes]: The fundamental group $T(p,q)$ of the complement of a $(p,q)$-torus knot (in $S^3$) admits the presentation $\langle a, b \mid a^p=b^q \rangle $. Is $T(p,q)$ linear, i.e., is there a faithful homomorphism of $T(p,q)$ into a matrix group over some field? -For example, $T(2,3)$ and $T(3,3)$ are linear, since both can be embedded into the 3-strand braid group $B_3$. (Indeed, $T(2,3)=B_3$.) - -REPLY [7 votes]: By the work of Agol, Liu, Przytycki-Wise and Wise we now know that fundamental groups of irreducible 3-manifolds which are not closed graph manifolds are `virtually special', in particular they are linear over $\Bbb{Z}$. -Oddly enough, the only open case for linearity are fundamental groups of certain closed graph manifolds.<|endoftext|> -TITLE: Simplicial replacements in smoothing theory -QUESTION [10 upvotes]: As far as I can tell, ever since Milnor's Microbundles and differentiable structures (1961) paper, whenever people talk about $Diff(\mathbb R^n)$ or $PL(\mathbb R^n)$ or $Homeo(\mathbb R^n)$, they always bring in an auxiliary simplicial space, given various names like the associated "c.s.s. complex" (Milnor), or the associated singular complex $Sing_\cdot(Diff(\mathbb R^n))$. It seems to me like there are as many names for essentially the same thing as there are authors in this subject. -My question is, why do people bother with this complex? i.e. why not just use the original space $Aut^{CAT}(\mathbb R^n)$? -In Milnor's paper he does not address this question. It seems to be a trend in the literature to avoid this discussion. -So to be more precise, I suppose I want to ask the question, - -Q: Is there any point in the smoothing literature where the associated simplicial complex to $Aut^{CAT}(\mathbb R^n)$ is absolutely necessary to a proof, and if so, why? - -If there is an affirmative answer to that question, I would like to know what it would take to remove the usage of the associated simplicial complex. What information are we missing about $Aut^{CAT}(\mathbb R^n)$ that we can avoid when we use the associated simplicial complex? -The only reference I've found that makes any attempt to address something like this question is Lurie's notes on smoothing theory (lecture 6, bottom of page 1). But this isn't quite what I'm looking for. - -REPLY [3 votes]: I present here a reference for Peter May's comment to Tom Goodwillie's answer in this thread. It also corroborates the comment by John Klein below the question stating that there is no obvious topology on the set of PL homeomorphisms which recovers the correct homotopy type. -Corollary 2 in the article -Ross Geoghegan and William Haver, On the space of piecewise linear homeomorphisms of a manifold, Proceedings of the A.M.S., volume 55, number 1, 1976, pages 145-151 -contains the following result: - -Let $M$ be a compact PL manifold (without boundary) of dimension different from 4. Let $PLH(M)$ denote the space of PL homeomorphisms of $M$ with the compact-open topology. Further, let $H^\ast(M)$ denote the space of homeomorphisms of $M$ which are isotopic to PL homeomorphisms, equipped with the compact-open topology. In other words, $H^\ast(M)$ is the union of all the path components of the space of homeomorphisms of $M$ which contain a PL homeomorphism. Then the inclusion of $PLH(M)$ into $H^\ast(M)$ is a weak equivalence. - -Unfortunately, I am not in a position to say anything about the proof of the above result. Perhaps someone else can provide more information.<|endoftext|> -TITLE: Powers of singular matrices and pairs of identical rows -QUESTION [7 upvotes]: Let $A$ be a square real or complex matrix. We’ll call $A$ special if among its rows (or among its columns) there are two identical ones, different from the zero vector, (Added:) and if it has no nilpotent Jordan block. -Obviously, if $A$ is special, then so are the powers $A^2,A^ 3,\dots$. I am wondering if the converse is true, more precisely: - -Is there a $k\in\mathbb N$ and a singular matrix $A$ such that $A^k$ is special, but not $A$ itself ? - -REPLY [5 votes]: Here is a non nilpotent example. -Take $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 1 & -1 \\ 3 & 2 & 0 \end{pmatrix}$. Then $A$ is not special, but $A^2 = \begin{pmatrix} 8 & 5 & 1 \\ 0 & 0 & 0 \\ 8 & 5 & 1 \end{pmatrix}$ is. Moreover, $A^{r+2} = 3^r\cdot A^2$ for any $r\geq 1$, so $A$ is not nilpotent. - -Note that for $2\times 2$ matrices, if a power of a matrix is special, then the matrix itself is. Indeed, assume that the two rows of $A^k$ are equal and non-zero. Since any special matrix is singular, then there is a scalar $\lambda$ such that the second row of $A$ is $\lambda$ times the first. This is still true for any power of $A$; thus the second row of $A^k$ is $\lambda$ times the first. Since they are equal and non-zero, $\lambda = 1$, so $A$ is special. - -Added to fit the comments to the question: -If you assume that only one of the eigenvalues of the $n\times n$ matrix $A$ is $0$, then the answer becomes negative. Indeed, in that case, $A^r$ has rank $n-1$ for any $r\geq 1$. Without loss of generality, we can assume that the first $n-1$ rows of $A$, say $R_1, \ldots, R_{n-1}$, are linearly independent. The last row is then some linear combination $R_n = L(R_1, \ldots, R_{n-1})$ of the others. -But then, if $R_1^{(r)}, \ldots, R_n^{(r)}$ are the rows of $A^r$, we still have that $R_n^{(r)}=L(R_1^{(r)}, \ldots, R_{n-1}^{(r)})$, so the first $n-1$ rows of $A^r$ are still linearly independent (since $A^r$ has rank $n-1$). If $A^r$ is special, then $R^{(r)}_n$ has to be equal to some $R_i^{(r)}$; this means that the coefficients in $L$ are all zero, except the one in front of $R_i^{(r)}$, which is $1$. Thus $R_n = R_i$, and $A$ is special.<|endoftext|> -TITLE: An attractive introduction to the history of modular forms and its applications -QUESTION [9 upvotes]: I am a graduate student majoring in number theory. Recently I have to give a report to graduate students studying mathematics. I am interested in this field, but I know little about it. Can you give me some advice? Welcome to providing some references! Thanks! - -REPLY [2 votes]: Two other references which you may like and which in particular emphasize the historical aspects: - -"Elliptic Curves, Function Theory, Geometry , Arithmetic" by Henry McKean and Victor Moll. -"Linear Differential Equations and Group Theory, from Riemann to Poincaré" by Jeremy Grey.<|endoftext|> -TITLE: Probability of coprime polynomials -QUESTION [14 upvotes]: Given positive integer $N$, we choose $m_1, m_2, n_1, n_2$ independently and with equal probabilities from $\{0,1,\ldots,N\}$, and let -$f_1 = x^{m_1} + (1+x)^{n_1}$ and $f_2 = x^{m_2} + (1+x)^{n_2}$ as polynomials in the indeterminate $x$ over the field ${\mathbb F}_2$ of two elements. Let $P(N)$ be the probability that $f_1$ and $f_2$ are coprime. What can be said about $P(N)$, in particular its asymptotics as $N \to \infty$? -By explicit enumeration in Maple, the first few values are -$$\eqalign{P \left( 1 \right) &={\frac {9}{16}},P \left( 2 \right) ={\frac {56}{ -81}},P \left( 3 \right) ={\frac {45}{64}},P \left( 4 \right) ={\frac { -489}{625}},P \left( 5 \right) ={\frac {1019}{1296}},\cr P \left( 6 - \right) &={\frac {1895}{2401}},P \left( 7 \right) ={\frac {3299}{4096} -},P \left( 8 \right) ={\frac {5308}{6561}},P \left( 9 \right) ={\frac -{2023}{2500}},P(10) = \frac{11954}{14641}\cr}$$ -Random sampling seems to indicate $P(100) \approx 0.83$. The sequence $(N+1)^4 P(N)$ -does not appear to be in the OEIS. -EDIT: That sequence is now in the OEIS as A245488. -$P(100) = 86648767/101^4 \approx 0.8326776196$. - -REPLY [5 votes]: This seems to be a hard and interesting problem. Here's a heuristic on the correct answer, but I have little hope that it can be made into a proof. Below let $D$ denote a polynomial in ${\Bbb F}_2[x]$ and let $\mu(D)$ denote the Mobius function. We will only be interested in $D$ that are coprime to $x(1+x)$. For such $D$ consider the group generated by $x$ and $1+x$ in $({\Bbb F}_2[x]/D)^*$; denote this by $\langle x,1+x\rangle_D$ and its order by -$|\langle x,1+x\rangle_D|$. The conjectured probability is -$$ -\sum_{D, (D,x(1+x))=1} \frac{\mu(D)}{|\langle x, 1+x \rangle_D|^2}. -$$ -To see why this is, note that we can identify whether $(f,g)=1$ by summing $\sum_{D|f, D|g} \mu(D)$. Thus the problem asks for -$$ -\sum_{D, (D,x(1+x))=1} \mu(D) \Big( \frac{1}{(N+1)^2} \sum_{0 \le a, b\le N; D|x^a+(1+x)^b} 1\Big)^2. -$$ -The probability that $D$ divides $x^a+(1+x)^b$ is the same as the probability that -$D$ divides $x^m (1+x)^n +1 = x^m (1+x)^n -1$. Since $x^m (1+x)^n$ ranges uniformly over the elements of $\langle x,1+x \rangle_D$, this probability is clearly $1/|\langle x,1+x\rangle_D|$. This justifies heuristically the conjecture. The argument could be made precise by splitting $a$ and $b$ into intervals of size the order of $x$ in $({\Bbb F}_2[x]/D)^*$ and the order of $1+x$ there. The trouble is that there will be an error of -size $O(1/N)$ in doing so, and this cannot be controlled as the sum over $D$ includes exponentially many terms. -It seems plausible that the sum over $D$ in the conjecture converges, but I don't see any way to prove this. It would be interesting to compute it numerically. One can get something rigorous by doing the above analysis with only those $D$ whose irreducible factors have degree below $\log N$ say. In this way one obtains an upper bound for the desired probability.<|endoftext|> -TITLE: Possible troubles in Shelah's book "Cardinal Arithmetic" -QUESTION [22 upvotes]: I found some possible troubles in Observation 5.3(7) in the Chapter II of the Shelah's book "Cardinal Arithmetic" (page 86). For convenience, I quote the result and the proof in the book here (together with Observation 5.3(4) that is used in the proof of 5.3(7)): - -Observation 5.3 -(4) If $\lambda > \kappa \left( \geq \theta > \sigma \right)$, $\sigma$ regular then - $$ -\operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) = -\sum_{\mu \in \left[ \kappa , \lambda \right]} -\operatorname{cov} \left( \mu , \mu , \theta , \sigma \right) . -$$ -(7) If $\lambda \geq \kappa \geq \theta > \sigma = \operatorname{cf} (\sigma)$, - $\operatorname{cf} (\kappa) \geq \theta$, $\lambda_{0} = \lambda$, - $$ -\lambda_{n+1} = \sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , \tau^{+} , \tau \right) : -\kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \tau \in \left[ \sigma, \theta \right) -\right\rbrace -$$ - then - $$ -\operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) \leq -\bigcup_{n < \omega} \lambda_{n} . -$$ -Proof: - 7) Let $\chi$ be regular large enough, by induction on $n$ choose $N_{n} \prec \left( H(\chi) , \in \right)$ - of cardinality $\lambda_{n}$ such that - $$ -\left\lbrace N_{0}, \ldots , N_{n-1}, \lambda , \kappa , \theta , \sigma \right\rbrace \cup -\left( \lambda_{n} + 1 \right) \subseteq N_{n} , -$$ - and - $$ -\mathcal{P}_{n} = -\left\lbrace A \in N_{n} : \left| A \right| < \kappa , A \subseteq \lambda \right\rbrace -$$ - and $\mathcal{P}_{\omega} = \bigcup_{n < \omega} \mathcal{P}_{n}$. - Suppose $X \subseteq \lambda$, $\left| X \right| < \theta$ and for no - $\mathcal{P} \subseteq \mathcal{P}_{\omega}$, $\left| \mathcal{P} \right| < \sigma$ is - $X \subseteq \bigcup_{A \in \mathcal{P}} A$; let $I$ be the $\sigma$-complete ideal on $X$ - generated by $\left\lbrace X \cap A : A \in \mathcal{P}_{\omega} \right\rbrace$, - so $X \notin I$. Let - $$ -\theta_{n} = \min -\left\lbrace -\left| \mathcal{P} \right| : \mathcal{P} \subseteq \mathcal{P}_{n} , -\bigcup_{A \in \mathcal{P}} A \cap X \notin I -\right\rbrace ; -$$ - now $\theta_{n} \leq \left| X \right| < \theta$ and - $\operatorname{cf} \left( \theta_{n} \right) \geq \sigma$ and $\theta_{n+1} < \theta_{n}$ - (use 5.3(4) applied to - $\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right)$), contradiction. - $\square$ - -First, it is easy to show that $\theta_{n} \leq \left| X \right| < \theta$, -$\operatorname{cf} \left( \theta_{n} \right) \geq \sigma$ and -$\theta_{n+1} \leq \theta_{n}$. To show that $\theta_{n+1} < \theta_{n}$, -we must apply Observation 5.3(4) to -$\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right)$ -($\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right)$ is an "uninteresting" covering number: $\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right) \leq \lambda_{n}$ - consider the family ${[\lambda_{n}]}^1$). -Now, the major difficulty is that we need -$\operatorname{cf} (\theta_{n}) = \theta_{n}$ -to use 5.3(4), but I don't see how to prove this. -Supposing $\operatorname{cf} (\theta_{n}) = \theta_{n}$, I wrote a detailed proof for 5.3(7): applying 5.3(4), we can show that -$$ -\operatorname{cov} -\left( \left| \mathcal{P}_{n} \right| , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq -\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1} = \left| N_{n+1} \right| , -$$ -and then use this to "cover" the set in -${[\mathcal{P}_{n}]}^{\theta_{n}}$ that testifies the definition of -$\theta_{n}$, with a set in ${[\mathcal{P}_{n+1}]}^{< \theta_{n}}$. -My questions are: (answers specific to the case $\sigma = \aleph_0$ are welcome too) -1) Is possible to prove that $\operatorname{cf} (\theta_{n}) = \theta_{n}$? -2) Is possible to prove that -$\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1}$ -when $\operatorname{cf} (\theta_{n}) < \theta_{n}$? -3) Is possible to prove that the sequence -$(\theta_{n})$ -is not eventually constant? -Some observations: -i) This is my proof that -$\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1}$, -if $\operatorname{cf} (\theta_{n}) = \theta_{n}$: -$$ -\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) = -\sum_{\mu \in \left[ \kappa , \lambda_{n} \right]} -\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq -$$ -$$ -\lambda_{n} \cdot \sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) : -\kappa \leq \mu \leq \lambda_{n} -\right\rbrace . -$$ -Now, it is easy to show that -$\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) = \operatorname{cf} (\mu)$ -when $\operatorname{cf} (\mu) \neq \theta_{n}$. Thus, -$$ -\sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) : -\kappa \leq \mu \leq \lambda_{n} -\right\rbrace \leq -$$ -$$ -\lambda_{n} + \sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) : -\kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \theta_{n} -\right\rbrace = -$$ -$$ -\lambda_{n} + \sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : -\kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \theta_{n} -\right\rbrace \leq -$$ -$$ -\lambda_{n} + \sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : -\kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta -\right\rbrace = -$$ -$$ -\lambda_{n} + \lambda_{n+1} = \lambda_{n+1}, -$$ -hence -$$ -\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq -\lambda_{n} \cdot \lambda_{n+1} = \lambda_{n+1} . -$$ -$\square$ -ii) It is more convenient to define -$$ -\lambda_{n+1} = \lambda_{n} + \sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : -\kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta -\right\rbrace . -$$ -iii) If $\eta$ is any cardinal with -$\sigma \leq \operatorname{cf} (\eta) = \eta < \theta$, -then we can show that -$$ -\operatorname{cov} \left( \lambda_{n} , \kappa , \eta^+ , \eta \right) \leq -\lambda_{n+1} -$$ -(same argument of my observation (i)). -iv) The proof works when -$\sigma < \theta \leq \aleph_{\sigma}$, -since $\sigma \leq \operatorname{cf} (\xi) \leq \xi < \aleph_{\sigma}$ implies -$\operatorname{cf} (\xi) = \xi$. -v) When $\sigma = \aleph_0$ (the case that interests me): if we define -$$ -\theta_{\omega} = \min -\left\lbrace -\left| \mathcal{P} \right| : \mathcal{P} \subseteq \mathcal{P}_{\omega} , -\bigcup_{A \in \mathcal{P}} A \cap X \notin I -\right\rbrace , -$$ -then we can show that $\theta_{\omega}$ is regular. If the sequence $(\theta_{n})$ -is eventually constant, then there is $k \in \omega$ -such that $\theta_{n} = \theta_{k}$ for any $n \geq k$, and -$$ -\aleph_0 = \sigma \leq \operatorname{cf} (\theta_{\omega}) = \theta_{\omega} -\leq \operatorname{cf} (\theta_{k}) < \theta_{k} < \theta . -$$ -A more elaborated argument shows that -$\operatorname{cf} (\theta_{k}) = \aleph_0$. -Hence, -$$ -\aleph_0 = \sigma = \operatorname{cf} (\theta_{\omega}) = \theta_{\omega} -= \operatorname{cf} (\theta_{k}) < \theta_{k} < \theta . -$$ -vi) Considering observation (iv), everything works for -$$ -\operatorname{cov} -\left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega} , \aleph_0 \right) . -$$ -Does the same occur with -$$ -\operatorname{cov} -\left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega + 1} , \aleph_0 \right) ? -$$ - -REPLY [4 votes]: I found a way to prove in detail a slightly weaker version of the Observation 5.3(7) cited above: - -Definition - Given an ordinal $\alpha$, and any cardinals - $\mu$, $\eta$, $\theta$ and $\sigma$, - with $\eta \geq \theta$, define - $$ -\operatorname{s}_{\mu , \eta , \theta , \sigma} = -\sup -\left\lbrace -\operatorname{cov} \left( \nu , \nu , {\left( \operatorname{cf} (\nu) \right)}^{+} , \operatorname{cf} (\nu) \right) : -\eta \leq \nu \leq \mu , \sigma \leq \operatorname{cf} (\nu) < \theta -\right\rbrace -$$ - and - $$ -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \alpha \right) = -\left\lbrace -\begin{array}{ll} -\mu , & \textrm{if } \alpha = 0 ; \\ -\sup \left\lbrace -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) : \beta < \alpha -\right\rbrace , & \textrm{if } \alpha \textrm{ is a limit ordinal;} \\ -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) + -\operatorname{s}_{\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) , \eta , \theta , \sigma} , -& \textrm{if } \alpha = \beta + 1 . -\end{array} -\right. -$$ -Proposition - Let $\mu$, $\eta$, $\theta$ and $\sigma$ be cardinals such that - $$ -\mu \geq \eta = \operatorname{cf} (\eta) \geq \theta > \sigma = \operatorname{cf} (\sigma) \geq \aleph_{0} . -$$ - Then, - $$ -\operatorname{cov} \left( \mu , \eta , \theta , \sigma \right) \leq -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega^{2} \right) . -$$ - -I needed this proposition in the following paper, where I included, for the reader's convenience, this proposition as Proposition 3.17, with a detailed proof. - -A.M.E. Levi. Reflection for lindelöf degree and inaccessible cardinals, - Acta Mathematica Hungarica 144(1) (2014), 182-195. - doi:10.1007/s10474-014-0442-0 - -A manuscript version can be viewed here. -When $\sigma = \aleph_{0}$, we show that -$$ -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega \right) < -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega + 1 \right) \leq -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega^{2} \right) ; -$$ -but when $\sigma > \aleph_{0}$, we prove that -$$ -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega \right) = -\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \alpha \right) -$$ -for every ordinal $\alpha > \omega$. Hence, we can prove the following result, which is essentially the book's one, with the additional hypotheses -$\kappa = \operatorname{cf} (\kappa)$ and -$\sigma > \aleph_{0}$: -Proposition -If -$\lambda \geq \kappa = \operatorname{cf} (\kappa) \geq \theta > \sigma = -\operatorname{cf} (\sigma) > \aleph_{0}$, -$\lambda_{0} = \lambda$, -$$ -\lambda_{n+1} = \lambda_{n} + -\sup -\left\lbrace -\operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : -\kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta -\right\rbrace , -$$ -then -$$ -\operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) \leq -\bigcup_{n < \omega} \lambda_{n} . -$$<|endoftext|> -TITLE: How do you define (infinity,1) categories in Homotopy Type Theory? -QUESTION [35 upvotes]: One of the major motivations of Homotopy Type Theory is that it naturally builds in higher coherences from the beginning. One important setting where higher coherence requirements get annoying is higher category theory. It's easy to talk about $\infty$-groupoids in HoTT, they're just types and you build them as higher inductive types. What about the next step? How do you talk about $(\infty,1)$-categories? I looked around at the nlab and the relevant blogs, but didn't find anything. -The natural setup is that you have a type of objects and a (dependent) type of morphisms. But composition seems to run into all the usual difficulties of coherence in higher category theory. Does the HoTT point of view simplify things at all here? -Feel free to assume that I'm familiar with the discussion of ordinary categories in the HoTT book and the background in the HoTT book. On the other hand, also assume that I find all definitions of higher categories beyond dimension 2 at least somewhat confusing. My motivation is that I'm trying to understand what you would need to do in order to give a formal proof of the cobordism hypothesis in dimension 1. - -REPLY [9 votes]: Emily Riehl and Mike Shulman now have a preprint which builds a version of type theory which includes $(\infty,1)$-categories as certain kinds of types.<|endoftext|> -TITLE: Infinite-dimensional hex -QUESTION [9 upvotes]: Suppose $n$ players take turns selecting vertices of the grid $[k]^n = \left\{0, 1, 2, \ldots, k-1\right\}^n$. Each player is assigned a pair of opposite faces of the grid, and wins the game if they connect their assigned faces by a path (potentially using diagonal moves). Once a vertex is chosen, it cannot be chosen by any other player, and the game ends once every vertex has been chosen. -This game must always be won by at least one player. David Gale gives a simple proof here. If we restrict the notion of "adjacent" we can obtain a game more like Hex, which we can prove has a unique winner using a slick induction on $k$. -But what happens if we take $n = \omega$? -More formally, suppose we partition $[k]^\omega$ into $A_1, A_2, \ldots$. We say that two points $x,y$ are adjacent if $|x(i) - y(i)| \leq 1$ for each $i$. Does there necessarily exist an $i$ and a path $x_1, x_2, \ldots, x_n \in A_i$ such that $x_m$ is adjacent to $x_{m+1}$ for each $m$, $x_1(i) = 0$, and $x_n(i) = k-1$? - -One reason to think this might hold up in the infinite-dimensional case is that it works when $k = 2$, which suggests that the infinite dimension isn't fundamentally a deal-breaker. If $k = 2$, then every pair of points is adjacent, so in order for us to fail each $A_i$ must either be a subset of $\left\{x : x(i) = 0\right\}$ or $\left\{x : x(i) = 1\right\}$. Without loss of generality we can assume the former. But then the constant function $x(i) := 1$ is not in any of the $A_i$. -Edit: the case $k = 3$ also works, although I found it a lot harder. (I doubt this proof is the best one. It is strikingly different in character from more typical fixed point arguments, which seems related to the fact that it produces paths of length exactly $3$. I doubt the idea will generalize, though I think this still provides more evidence that the claim is true.) -A partial assignment is an $a \in \left([k] \cup \left\{ *\right\}\right)^{\omega}$, and we say that $x$ satisfies $a$ if $\forall i : a(i) \in \left\{*, x(i)\right\}$. The partial assignments form a poset in the natural way ($a > a'$ iff satisfying $a'$ is easier than satisfying $a'$). -Say that a partial assignment is good if it never takes on the value $1$, and if whenever coordinate $i$ is unequal to $*$, there are no points satisfying the assignment which receive label $i$. It is easy to check that the intersection of any chain of good partial assignments is itself good, so there is some maximal good assignment $a$. -Define $x$ by $x(i) = 1$ if $a(i) = *$, $x(i) = a(i)$ otherwise. Let $i$ be the label of $x$. Because $a$ was good, coordinate $i$ cannot be set in $a$. Because $a$ was maximal, setting coordinate $i$ to $0$ must yield a bad assignment. This means that some point consistent with $a$ which has label $i$ and which has $i^{\text{th}}$ coordinate $0$. Similarly, there is some point consistent with $a$ which has label $i$ and which has $i^{\text{th}}$ coordinate $2$. But together with $x$, these three points form a path which has all of the desired properties. - -REPLY [3 votes]: It seems the claim is false, though I don't have a good sense for what the actual counterexample looks like. With luck, I've made an error somewhere. -Consider the normed space $\ell^{\infty}$. One can show that there is a Lipschitz retraction from the unit ball $[-1, 1]^{\omega}$ to the unit sphere. -Let $f$ be this retraction, let $k$ be some integer much greater than the Lipschitz constant of $f$, and consider my claim for $[k]^{\omega}$. We'll define an outcome of the game as follows: -For a point $x \in [k]^{\omega}$, consider the point $b(x) = \frac {2x}{k} - 1$ in the unit ball. By definition $f(b(x))$ has some coordinate in the interval $[-1, 1/k-1] \cup [1-1/k, 1]$. Let $i$ be one such coordinate, and assign the point $x$ the color $i$. I claim the resulting grid has no winning player. -Suppose for a contradiction that player $i$ won. That implies there is some sequence of adjacent points $x_1, x_2, \ldots, x_m$ all of color $i$ for which the $i^{\text{th}}$ coordinate -goes from $0$ to $k-1$. The $i^{\text{th}}$ coordinate of $x_1$ is $0$, so the $i^{\text{th}}$ coordinate of $b(x_1)$ is $-1$ and hence the $i^{\text{th}}$ coordinate of $f(b(x_1))$ is also $-1$. Similarly, the $i^{\text{th}}$ coordinate of $f(b(x_m))$ is $1$. But since each $x_j$ is assigned color $i$, the $i^{\text{th}}$ coordinate of each $f(b(x_j))$ is in $[-1, 1/k-1] \cup [1 - 1/k, 1]$. -So there must be some $j$ such that $f(b(x_j)) \in [-1, 1/k-1]$ and $f(b(x_{j+1})) \in [1/k-1, 1]$. But since $x_j$ and $x_{j+1}$ are adjacent, $||b(x_j) - b(x_{j+1})||_{\infty} < 2/k$. Since $||f(b(x_j)) - f(b(x_{j+1}))||_{\infty} > 1 - 2/k$, this contradicts the assumption that $f$ has Lipschitz constant much less than $k$.<|endoftext|> -TITLE: The Image of the Mod 2 Homology of BSp in the Homology of BSO -QUESTION [8 upvotes]: I'm essentially trying to figure out exactly what the title asks for. I've been scouring old Seminaires Henri Cartan and books by Stong to try to see exactly how to do this, but the combination of French and older terminology has made it rough going. -In particular, I've been able to work out (I think!) that, modulo 2, the map $$H_\ast(BSp;\mathbb{Z}/2)\to H_\ast(BU;\mathbb{Z}/2) $$ takes $y_{4k}$ to $(x_{2k})^2$, and that the map -$$H_\ast(U;\mathbb{Z}/2)\to H_\ast(SO;\mathbb{Z}/2)$$ -takes $a_{2k-1}$ to certain polynomials $t_{2k-1}$ (the primitives) defined recursively as $$t_1=c_1,t_3=c_3+c_1c_2,\ldots,t_{2k-1}=\sum_{\substack{i+j=2k+1\\0\leq i -TITLE: Probability of having many unique elements -QUESTION [6 upvotes]: If you sample $n$ integers from the range $1$ to $n$ inclusive it seems intuitive that you are likely to get a lot of numbers exactly once. Call $X_n$ the number of integers you get that occur exactly once in your sample. Is there a nice simple way of showing the following for all $n\geq 2$. -$$\exists a,b >0\text{ such that } P(X_n \leq an) \leq 2^{-bn} $$ - -REPLY [9 votes]: $X_n$ is a random variable on the product space $\Omega = \{1,\ldots,n\}^n$ such that changing the value of any one coordinate of $\omega \in \Omega$ changes $X_n(\omega)$ by at most 2. This is a classic case in which martingale concentration inequalities can be applied to show that $$\mathbb{P}(|X_n - \mathbb{E}(X_n)| \geq t) \leq 2e^{-t^2/8n}.$$ So in fact $X_n$ almost always takes a value in some interval of length about $\sqrt n$. $\mathbb{E}(X_n) = n(1-1/n)^{n-1} \sim n/e$, so you can find a $b$ for any $a < 1/e$.<|endoftext|> -TITLE: Is $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ a commutative algebra? -QUESTION [11 upvotes]: Consider $l^\infty(\mathbb{Z})$ the Banach space of bounded complex valued functions on the abelian group $\mathbb{Z}$ with the supremum norm. It has a natural action by $\mathbb{Z}$ given by $(zf)(g):=f(z^{-1}g)$. Then let $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ be the algebra (multiplication by composition) consisting of all bounded $\mathbb{Z}$-equivariant operators $l^\infty(\mathbb{Z})\rightarrow l^\infty(\mathbb{Z})$. -Question: Is this a commutative algebra? -If $l^\infty$ is replaced by $l^2$ the algebra is the von Neumann algebra $L(\mathbb{Z})$ and in this cases $\mathbb{C}(\mathbb{Z})$ lies densely (wrt WOT or SOT) in $L(\mathbb{Z})$ as a subalgebra and therefore $L(\mathbb{Z})$ is commutative. But I think $\mathbb{C}(\mathbb{Z})$ does not lie densely in $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$, so the situation is a bit different. -Question: What happens if $\mathbb{Z}$ is replaced by an arbitrary abelian group? - -REPLY [5 votes]: The following is an abstract (Banach) algebraic take on Werner's construction. Let $A=\ell^1(\mathbb Z)$ with convolution (but in general $A$ is any Banach algebra). We turn the dual space $A^*$ into an $A$-bimodule (though in our example, $A$ is commutative) by dualising the actions: -$$ (a\cdot f)(b) = f(ba), \quad (f\cdot a)(b) = f(ab) \qquad -(f\in A^*, a,b\in A). $$ -Then you can check that $T:A^*\rightarrow A^*$ is $\mathbb Z$-equivariant if and only if $T$ is a module homomorphism: I'll write $T\in\hom_A(A^*)$ to mean $T(f\cdot a) = T(f)\cdot a$ (the right action is more appropriate in the non-commutative group setting). If $e\in A$ is the unit then -$$ T(f)(a) = T(f)(ae) = (T(f)\cdot a)(e) = T(f\cdot a)(e) -\qquad (T\in \hom_A(A^*), f\in A^*, a\in A). $$ -So again $T$ is determined uniquely by $m\in A^{**}, m(f) = T(f)(e)$. Conversely, given $m\in A^{**}$ we can define $T\in\hom_A(A^*)$ by this relation. So $\hom_A(A^*)\cong A^{**}$. A similar result holds if $A$ only has a bounded approximate identity. -So $A^{**}$ becomes an algebra for the product induced from $\hom_A(A^*)$. Define $$ (m\cdot f)(a) = m(f\cdot a) \qquad (m\in A^{**}, f\in A^*,a\in A).$$ -Then if $m_i$ is associated to $T_i$ for $i=1,2$, -$$ (m_1m_2)(f) = (T_1\circ T_2)(f)(e) = T_1(T_2(f))(e) -= m_1(T_2(f)). $$ -However, for $a\in A$, -$$ T_2(f)(a) = m_2(f\cdot a) = (m_2\cdot f)(a) $$ -and so -$$ (m_1m_2)(f) = m_1(m_2\cdot f) $$ -This is precisely the "1st Arens product", say $\Box$, as studied in Banach algebra theory. By making the symmetric choices we get the "2nd Arens product", say $\diamond$, which might differ. If $A$ is commutative then $m_1\Box m_2 = m_2\diamond m_1$ and so $\hom_A(A^*)$ is commutative precisely when the Arens products agree. -However, Young showed in "The irregularity of multiplication in group algebras" -Quart J. Math. Oxford Ser. (2) 24 (1973), 59–62, MathSciNet that for any locally compact group $G$, we have that $L^1(G)$ is not Arens regular. So Werner's question has a negative answer for all Abelian groups (with the axiom of choice!)<|endoftext|> -TITLE: Obstruction to extension of non-abelian groups - finite example? -QUESTION [13 upvotes]: Let $G$ be a non-abelian group, let $\Pi$ be a group, and let $\eta: \Pi\rightarrow Out(G)$ be a homomorphism, where $Out(G)$ is the group of automorphisms of G modulo the normal subgroup of inner automorphisms. The obstruction to the existence of an exact sequence -$1\rightarrow G \rightarrow B \rightarrow \Pi \rightarrow 1$ -that induces $\eta$ is a certain cohomology class in $H^3(\Pi,Z(G))$, where $Z(G)$ is the center of $G$ (c.f. Homology, Mac Lane, IV.8.). -Does anyone know an example of $(G,\Pi, \eta)$ where this obstruction is nonzero, with $G$ finite? What is the smallest group $G$ for which there exist $\Pi,\eta$ with a non-trivial obstruction? - -REPLY [13 votes]: The smallest example when $\eta: \Pi\to \text{Out}(G)$ does not give rise to an extension -of $\Pi$ by $G$ is when $G=D_{16}$, -the dihedral group has order 16. (Apologies to those who write it as $D_8$.) -Pick $c,h\in D_{16}$ -a rotation of order 8 and a reflection. -The elements of $\text{Out}(D_{16})\cong C_2\times C_2$ -correspond to four types of automorphisms -$$ - \begin{array}{cccccccc} - I &:&& c\mapsto c^{\pm 1} && h\mapsto h\cdot c^{\text{even}} && \text{(inner)}\cr - II &:&& c\mapsto c^{\pm 3} && h\mapsto h\cdot c^{\text{even}} \cr - III &:&& c\mapsto c^{\pm 1} && h\mapsto h\cdot c^{\text{odd}} \cr - IV &:&& c\mapsto c^{\pm 3} && h\mapsto h\cdot c^{\text{odd}}, \cr - \end{array} -$$ -and there is no extension of $\Pi=C_2$ by $G=D_{16}$ -with $\Pi$ acting on $G$ as a type $IV$ automorphism. -One way to see this is just to go through the 51 groups of order 32; -only five of them contain $D_{16}$, number 18,19,39,42,43 in GAP -or Magma. The first two $D_{32}, SD_{32}$ give a type III action, next two -$C_2\times D_{16}, (C_2\times C_8):C_2$ a type I action, and the last one $D_8:C_2^2$ a type -II action. -Another way is to use the fact that dihedral, semi-dihedral and generalized quaternion -groups are the only non-abelian 2-groups whose commutator subgroup has index 4 (and index 1 or 2 is impossible). The -commutator subgroup $G'$ of $G=D_{16}$ is generated by $c^2$, and an extension $B$ of -type $III$ or $IV$ acts non-trivially on $G/G'=\{1,c,h,ch\}$, so $B' -TITLE: Crystal structure, lattice, Graph and coloring -QUESTION [5 upvotes]: I am working across mathematics, physics and engineering. And I am looking for whether there exists already formally established knowledge in the field. -Given a periodic graph (actually a physical lattice or crystal structure), we want to examine some periodic coloring ( or ordered but not periodic coloring) of the graph. This coloring in physics may be considered as filling the lattice site with atoms, ions and so forth. Is there some already formally established mathematical area that deals with the this? -Further, we could construct functions from these kinds of coloring to a real number. ( in physics we associate average energy with the specific structure) and we want to minimize it by searching the periodic or ordered graph space. Is there such a branch in mathematics dealing with this also? -Furthermore, I wish to construct a program to analyze any arbitrary periodic colored graphs (that is to analyze different crystal structure. ) By analyze I mean search the coloring that would give the minimum energy. Is there existing well established knowledge for doing this? -An reduced problem would be is there some established mathematical theorems tackling the coloring of an arbitrary periodic graph, in any possible way? -Thank you very much. - -REPLY [5 votes]: There is a theory of color symmetry in mathematical crystallography and in particular, it is applied to lattices. The basic idea is that each coset of a sublattice is assigned a unique color, and what one usually aims for is a symmetrically colored lattice. -Some introductory papers on the topics are: 1.Color groups associated with square and hexagonal lattices 2. Bravais colourings of planar modules with N-fold symmetry<|endoftext|> -TITLE: What is the rational rank of the elliptic curve x^3 + y^3 = 2? -QUESTION [6 upvotes]: In Unsolved Problems in Number Theory, 3rd edition, section D1, 2004, R. K. Guy says, "Andrew Bremner has computed the rational rank of the elliptic curve x^3 + y^3 = Taxicab(n) as equal to 2, 4, 5, 4 for n = 2, 3, 4, 5, respectively." Since Taxicab(1) = 2, my question can be restated: -What is the rational rank of the elliptic curve x^3 + y^3 = Taxicab(1)? - -REPLY [15 votes]: To add to Abhinav's answer: the fact that $x^3+y^3=2$ has no solutions -other then $x=y=1$ is attributed by Dickson to Euler himself: -see Dickson's History of the Theory of Numbers (1920) Vol.II, Chapter XXI -"Numbers the Sum of Two Rational Cubes", page 572. The reference -(footnote 182) is -"Algebra, 2, 170, Art. 247; French transl., 2, 1774, pp. 355–60; -Opera Omnia, (1), I, 491". In the next page Dickson also refers to -work of Legendre that includes this result (footnote 184: "Théorie -des nombres, Paris, 1798, 409; ..."). -This result is actually easier than the $n=3$ case of Fermat, -because the curve has a $2$-torsion point, so only a $2$-descent -is required, and here one soon finds that there are only two -rational points (the second being the point at infinity -$(x:y:1) = (1:-1:0)$). [It also happens that this curve has -conductor $36$, while it's known that the smallest conductor -of an elliptic curve of positive rank is $37$; but that's -using a huge modern cannon to dispatch a classical fly.] -An easy way to bring the curve $x^3+y^3=2$ into Weierstrass form -is to factor $x^3+y^3 = (x+y)(x^2-xy+y^2)$ and substitute -$(x,y)=(r+s,r-s)$ to get $r^3+3rs^2=1$. Thus $s^2=(1-r^3)/(3r)$, -so $3r(1-r^3)$ is to be a square. Now substitute $r=3/X$ to get -$s^2 = (X^3-27)/(3X)^2$, or equivalently $Y^2=X^3-27$ where $y=3sX$. -The solution $(x,y)=(1,1)$ corresponds to $(r,s)=(1,0)$ and then -$(X,Y)=(3,0)$, which we recognize as a 2-torsion point because $Y=0$.<|endoftext|> -TITLE: On Generalizations of Fermat's Conjecture -QUESTION [12 upvotes]: We know the following facts: -(1) For all $1\leq n\leq 2$ the equation $x_{1}^{n}+x_{2}^{n}=x_{3}^{n}$ has a solution in $\mathbb{N}$. -(2) For all $3\leq n$ the equation $x_{1}^{n}+x_{2}^{n}=x_{3}^{n}$ has no solution in $\mathbb{N}$. -Question: Is the following generalization true? -For all $2\leq m$ both of the following statements are true: -(1) For all $1\leq n\leq m$ the equation $x_{1}^{n}+...+x_{m}^{n}=x_{m+1}^{n}$ has a solution in $\mathbb{N}$. -(2) For all $m+1\leq n$ the equation $x_{1}^{n}+...+x_{m}^{n}=x_{m+1}^{n}$ has no solution in $\mathbb{N}$. - -REPLY [25 votes]: No this is not true. For $n=4$ one has -$$2682440^4 + 15365639^4 + 18796760^4 = 20615673^4$$ -found by Elkies (as part of an infinite family of solutions). -Also earlier it was known, for example, for $n=5$, -$$27^5 + 84^5 + 110^5 + 133^5 = 144^5$$ -by Lander and Parkin. -But part 2 was a conjecture of Euler so you are in good company, and $n \ge 6$ is still open. See that page for further details. - -REPLY [10 votes]: The smallest counterexample for $n = 4$ is -$95800^4 + 217519^4 + 414560^4 = 422481^4$. -This has been found out by Roger Frye in 1988, cf. -http://euler.free.fr/docs/euler88.ps.<|endoftext|> -TITLE: Name for Kneser/Johnson-like graphs? -QUESTION [6 upvotes]: I wonder if the following simple generalization of Johnson and Kneser -graphs has a name? Let the vertex set of the graph $G(n,k,t)$ be the -set of $k$-element subsets of an $n$-set, with two $k$-sets adjacent if their -intersection has cardinality $t$. -So for $t=0$ we have the Kneser graphs, and for $t=k-1$ we have the Johnson graphs. -Does this family of intersection graphs have a name, or a standard notation? - -REPLY [2 votes]: Since $G(n,k,0)$ are called Kneser graphs and $G(n,k,k-1)$ are called Johnson graphs, it makes sense to call $G(n,k,t)$ the generalized Kneser graphs or the generalized Johnson graphs. I prefer to refer to $G(n,k,t)$ as the generalized Johnson graphs because then the generalized Kneser graphs can be reserved to refer to the family of graphs $G(n,k, \le s)$. This latter graph is defined to be the graph whose vertices are the $k$-subsets of an $n$-set, with two vertices joined iff the cardinality of their intersection is at most $s$. The special case $s=0$ gives the Kneser graphs, so the terminology generalized Kneser graphs is justified for $G(n,k, \le s$). These graphs have also been studied - for eg, in the papers [Chen and Wang, Discrete Math., 2008] and [Denley, Eur. J. Comb, 1997]. -In the literature, the graphs $G(n,k,t)$ have been called the generalized Johnson graphs (see the paper at arxiv.org/pdf/1202.3455.pdf, or the SAGE code on Godsil's website) or uniform subset graphs (see the papers of [Chen and Lih, JCTB, 1987] and [Chen and Wang, Discrete Math., 2008]). I prefer to use the phrase "uniform" to just refer to the fact that the edges of a hypergraph all have the same cardinality (for eg, independent sets in $G(n, k, \le s)$ refer to $k$-uniform $(s+1)$-intersecting families in the set of subsets of an $n$-set). So I now call $G(n,k,t)$ the generalized Johnson graphs.<|endoftext|> -TITLE: How hard (P, NP, NP-hard) is it to compute Schur norms of matrices (as multipliers)? -QUESTION [13 upvotes]: Given a matrix $A\in M_n(\mathbb{C})$, I will denote by $||A||_\infty$ the operator norm of $A$, as seen acting on the Hilbert space $\mathbb{C}^n$. This makes $M_n(\mathbb{C})$ into a Banach space (actually a Banach algebra, actually a $C^*$ algebra). -We will define a new norm on $M_n(\mathbb{C})$ in the following way: A matrix $T$ acts on $M_n(\mathbb{C})$ by pointwise multiplication (i.e. Schur product), giving rise to an operator $M_T:M_n(\mathbb{C})\longrightarrow M_n(\mathbb{C})$. For example for -$$T=\left[\begin{array}{cc} -1 & 2 \\ -3 & 4 -\end{array}\right]$$ we have -$$M_T\left(\left[\begin{array}{cc} -a & b \\ -c & d -\end{array}\right]\right)= -\left[\begin{array}{cc} -a & 2b \\ -3c & 4d -\end{array}\right]$$ -We will denote by $||T||_{Schur}$ the norm of the operator $M_T$, as an operator of the Banach space $M_n(\mathbb{C})$. How hard is it to compute/estimate this norm? Are there polynomial algorithms (in $n$ and the approximation error) that can do it? Is it actually NP? - -REPLY [14 votes]: It turns out that this norm can be computed efficiently (i.e., it is in $P$). This wasn't known at the time that the Davidson paper (originally linked in a comment above) was written, which is why it suggests that the computation is hard. -To compute the norm, first use the equivalence of points (1) and (2) in Theorem 1.1 of the Davidson paper. This tells us that $\|T\|_{schur} = \|T\|_{cb}$, where $\|\cdot\|_{cb}$ refers to the "completely bounded" norm of $T$. -It was shown in "J. Watrous. Semidefinite programs for completely bounded norms. Theory of Computing, 5:217-238, 2009" that the completely bounded norm can be computed efficiently via semidefinite programming. -For what it's worth, using this method tells me that the norm $\|\cdot\|_{schur}$ of the map stated in the question is $4$. However, this norm is not always equal to $\max_{i,j}\{t_{i,j}\}$. The simplest example I have been able to find to demonstrate this fact is -$$ -T = \begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix}, -$$ -which has $\|T\|_{schur} = 2/\sqrt{3} > 1$. -Edit: Here is some (surprisingly simple) MATLAB code that computes this norm. You must first install CVX before running this code. Just modify the first line of code with whatever operator $T$ you want to compute the norm of. -T = [1 2;3 4]; - -n = length(T); -cvx_begin sdp quiet - cvx_precision high; - variable Y0(n,n) hermitian - variable Y1(n,n) hermitian - minimize max(diag(Y0))/2 + max(diag(Y1))/2 - subject to - [Y0, T; T', Y1] >= 0; - Y0 >= 0; - Y1 >= 0; -cvx_end -cvx_optval<|endoftext|> -TITLE: Failure of Palais-Smale Condition C and the Mini-Max Principle -QUESTION [5 upvotes]: To get a thorough analysis of the critical point structure of a smooth function $f:M\to\mathbb{R}$ on a smooth Hilbert manifold $M$, a compactness assumption gets us far. That assumption is Condition C of Palais and Smale (note that this is automatically satisfied for $M$ compact). It ultimately implies (for instance) the Deformation Lemma and the Mini-Max Principle. -But what if Condition C fails? I'm interested in to what extent I lose control over the function. For instance, I am aware of Uhlenbeck's "perturbation method", where if $f$ doesn't satisfy Condition C then we can look at $f_\varepsilon=f+\varepsilon\cdot g$ (fix function $g$) that satisfies Condition C and try to get critical points of $f$ as limits of those of $f_\varepsilon$. -When Condition C fails, is there a work-around to save the Mini-Max Principle? - -REPLY [2 votes]: There are some works on this, but maybe THE expert on this topic is Abbas Bahri. Take a look at his works or google "critical point at infinity". -See, for example: this paper or his book ""Critical Points at Infinity in Some Variational Problem".<|endoftext|> -TITLE: Local smallness and (higher) topoi -QUESTION [9 upvotes]: The $2$-category of topoi and geometric morphisms is not locally small. For example, if $A$ is the classifying topos for abelian groups, the category of geometric morphisms from $Set$ to $A$ is equivalent to the category of all abelian groups, which is not small. This problem of course persists for higher topoi, since ordinary topoi embedd into them. -Question: Suppose that $\mathcal{E}$ and $\mathcal{F}$ are two topoi (or higher topoi) such that I know that $Hom\left(\mathcal{E},\mathcal{F}\right)$ is small. It is easy to show that for any $F$ in $\mathcal{F},$ $Hom\left(\mathcal{E},\mathcal{F}/F\right)$ is also small. Can it also be shown that for all $E$ in $\mathcal{E},$ $Hom\left(\mathcal{E}/E,\mathcal{F}\right)$ is small? - -REPLY [2 votes]: Consider a theory which has no models in $\mathrm{Set}$, but has a model in $\mathrm{Sh}(L)$ for some locale $L$. For example, the theory $\mathcal{CLF}$ of complete linearly ordered fields with more than $\sharp \mathbb{R}$ number of elements will do. For $ L $ we can take Barr or Diaconescu covering of $\mathbb {B}\mathcal {CLF} $. Take some non-small theory, like abelian groups $\mathcal{A}b$, and consider a classifying topos $\mathrm {B} T$ for pairs $T = (K: \mathcal{CLF}, A: \mathcal{A}b)$. For the topos $\mathcal E $ consider $$\mathrm {Sh} (L + pt) = \mathrm {Sh}(L) + \mathrm {Set} $$ -An etale map $ L \to L + pt $ corresponds to some sheaf $ E \in \mathcal E $, and $\mathcal {E}/E = \mathrm {Sh}(L) $. -$ T$ will have no models in $\mathcal{E}$, but a non-small category of models in $\mathcal{E}/E$. -Added later: -Instead of $\mathcal{CLF}$ we could take any theory without $\mathrm{Set}$-models. For example we could take a geometric theory, corresponding to some locale without points (just in case if there happen to be some obscure smallness problems with the following statements). -For any topos $\mathcal{F}$ we have $$Mod_T (\mathcal{F}) = Mod_{\mathcal{CLF}}(\mathcal{F}) \times Mod_{\mathcal{A}b} (\mathcal{F})$$ -since a model of $T$ is just a pair of models ($Mod_T(\mathcal{F})$ is the category of T-models in $\mathcal{F}$). Since $L$ is the Barr covering for $\mathcal{CLF}$, $Mod_{\mathcal{CLF}}(\mathrm{Sh}(L))$ is a non-empty category. Thus to prove that $Mod_T(\mathcal{F})$ is non-small, we only need to prove it for $Mod_{\mathcal{A}b}(\mathcal{F})$. The category $Mod_{\mathcal{A}b}(\mathrm{Set})$ is clearly non-small. I claim that the category of constant $\mathcal{A}b$-valued sheaves on $\mathrm{Sh}(L)$ is a non-small subcategory of $Mod_{\mathcal{A}b}(\mathcal{F})$. -The terminal morphism $p\colon L \to pt$ gives an adjoint pair of functors $p_* \colon \mathrm{Sh}(L) \leftrightharpoons \mathrm{Set} \colon p^*$, $p^* \vdash p_*$. Sheaves on locale $L$ correspond to etale spaces over $L$, and the pullback functor $p^*$ is simply the pullback of corresponding etale spaces, i.e. it maps $S\in \mathrm{Set}$ to $\left( L\otimes S \to L \right) \in Et(L)$, the morphism to $L$ being the obvious "collapsing fibers" projection $L\otimes S \to L \otimes pt = L$. Here $L \otimes S$ stands for a coproduct of $S$ copies of $L$ in category of locales. Now the statement that $p^*: \mathcal{A}b(\mathrm{Set}) \to \mathcal{A}b(\mathrm{Sh}(L))$ is injective looks like something that should be obviously true (for $A\in \mathcal{A}b(\mathrm{Set})$ the sheaf of abelian groups $p^*(A)$ should have global sections roughly like $A^{\pi_0(L)}$), but at the moment I can only formalize the proof in the following roundabout way. -An isomorphism $p^*(A) \simeq p^*(B)$, $A,B \in \mathcal{A}b(\mathrm{Set})$ would give an isomorphism of etale spaces $L\otimes A \simeq L\otimes B$. Since $\mathcal{L}oc = \mathcal{F}rm^{op}$, this would give an isomorphism $\mathcal{O}(L)^A \simeq \mathcal{O}(L)^B$, $\mathcal{O}\colon \mathcal{L}oc^{op} \simeq \mathcal{F}rm$. The product in $\mathcal{F}rm$ is induced from the product in $\mathrm{Set}$, like in any algebraic theory. Thus if $|A| > |\mathcal{O}(L)|$ and $|B| > |\mathcal{O}(L)|$, then $\mathcal{O}(L)^A$ and $\mathcal{O}(L)^B$ are non-isomorphic even as sets, moreso as frames. Here $|A|$ stands for cardinality. Since we have a non-small set of groups with cardinality greater than $|\mathcal{O}(L)|$ (e.g. all big enough free groups), the statement is proved.<|endoftext|> -TITLE: Continuity on a measure one set versus measure one set of points of continuity -QUESTION [16 upvotes]: In short: If $f$ is continuous on a measure one set, is there a function $g=f$ a.e. such that a.e. point is a point of continuity of $g$? -Now more carefully, with some notation: Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces, with Borel sets $\mathcal B_X$ and $\mathcal B_Y$, respectively. Let $f:(X, \mathcal B_X) \rightarrow (Y, \mathcal B_Y)$ be a measurable function and let $\mu$ be a probability measure on $(X, \mathcal B_X)$. -Say that $f$ is almost continuous if there is a $\mu$-measure-one Borel set $D \subseteq X$ such that $f$ is continuous on $D$, i.e., the restriction $f|_D:D \rightarrow Y$ is a continuous function where $D$ is given the subspace topology. -As usual, we say that $x \in X$ is a point of continuity (for $f$) if, for every Cauchy sequence $(x_n)_{n\in\mathbb{N}}$ that converges to $x$, we have that $(f(x_n))_{n \in \mathbb{N}}$ is a Cauchy sequence that converges to $f(x)$. -Let $C$ be the set of all points of continuity of $f$. (A classic result [Kechris, I.3.B Prop 3.6] shows that $C$ is a $G_\delta$ set.) Say that $f$ is almost everywhere (a.e.) continuous if $C$ is a $\mu$ measure-one set., i.e. $\mu$-a.e. point is a point of continuity. -Finally, say that a measurable function $g$ is a version of $f$ if $f=g$ $\mu$-a.e. -Clearly, if $f$ is a.e. continuous, then it is almost continuous on the set $C$. The converse does not hold in general: Consider for $f$ the indicator function for the rationals in $[0,1]$. Then $f=0$ on the irrationals, a Lebesgue-measure-one set, but $f$ is discontinuous everywhere. -However, $g=0$ is a version of $f$ and $g$ is a.e. continuous. Which raises the question: -If $f$ is almost continuous, is there a version $g$ of $f$ such that $g$ is a.e. continuous? -If it is helpful, one may assume that both $X$ and $Y$ are also complete and separable, i.e., Polish. -Note that by a result of Kuratowski [Kechris, I.3.B Thm. 3.8], if $Y$ is complete, we can, possibly changing versions, assume, without loss of generality, that $f$ is continuous on a $\mu$-measure-one $G_\delta$-set $D$. (In the example above, $f$ is almost continuous on the irrationals, which are of course $G_\delta$.) - -[Kechris: "Classical Descriptive Set Theory" 1995] - -REPLY [9 votes]: As Jason and Gerald predicted, the answer is yes for Polish $X, Y$. (Indeed, it is sufficient for $X$ to be merely separable and metrizable and for $Y$ to be merely complete and metrizable.) -As Nate observed, we may assume that $D$ is $G_\delta$. Here is the key fact: -Lemma. If $D \subseteq X$ is a nonempty $G_\delta$-set then there is a Borel map $h:X \to D$ such that $\lim_{x\to x_0} h(x) = x_0$ for every $x_0 \in D$. -Suppose $D = \bigcap_{n\lt\omega} U_n$, where $(U_n)_{n\lt\omega}$ is a descending sequence of open sets such that $U_n \subseteq \bigcup_{x_0 \in D} B(x_0,1/(n+1))$. Any Borel retraction $h:X \to D$ with the property that if $x \in U_n \setminus U_{n+1}$ then $d(h(x),x) \lt 1/(n+1)$ will be as required. By definition, it is always possible to find a suitable $h(x) \in D$ for each $x \in U_0 \setminus D$. To ensure that $h$ is Borel, fix an enumeration $(d_i)_{i \lt \omega}$ of a countable dense subset of $D$ and, if $x \in U_0 \setminus D$, define $h(x)$ to be the first element in this list that matches all the necessary requirements. (We must have $h(x) = x$ for $x \in D$ and it doesn't matter how $h(x)$ is defined when $x \notin U_0$ so long as the end result is Borel.) -Now, if $f:X \to Y$ is Borel and continuous on $D$, then $g = f\circ h$ is a Borel function that agrees with $f$ on $D$ and $$\lim_{x \to x_0} g(x) = f(\lim_{x \to x_0} h(x)) = f(x_0) = g(x_0)$$ for all $x_0 \in D$.<|endoftext|> -TITLE: Minimal polynomial of sums of roots of unity with constant term $\pm1$ -QUESTION [5 upvotes]: Let prime $p$ and given $\zeta_p = e^{2\pi i/p}$. It is well-known that the minimal polynomial of $x = \zeta_p + \zeta_p^{p-1}$ has a constant term either $\pm 1$ and, for certain $p$, the sum of four terms $x = \zeta_p +\zeta_p^{a}+\zeta_p^{p-a}+\zeta_p^{p-1}$ have as well. However, we also have, -$$F(x) = x^5 + x^4 - 28x^3 + 37x^2 + 25x + 1=0,\;\; x = \sum_{k=1}^{14} (\zeta_{71})^{23^k}$$ -$$F(x) = x^6 + x^5 - 15x^4 - 28x^3 + 15x^2 + 38x - 1=0,\;\; x = \sum_{k=1}^{6} (\zeta_{37})^{11^k}$$ -$$F(x) = x^7 + x^6 - 48x^5 + 37x^4 + 312x^3 - 12x^2 - 49x - 1=0,\;\; x = \sum_{k=1}^{16} (\zeta_{113})^{35^k}$$ -$$F(x) = x^{11} + x^{10} - 40x^9 - 19x^8 + \dots - 1=0,\;\; x = \sum_{k=1}^{8} (\zeta_{89})^{12^k}$$ -Question (edited): -What is the constraint on $p$ such that there is a minimal polynomial $F(x)$ with, - -root $x = \sum_{k=1}^{h} (\zeta_{p})^{a^k}$ -$4 -TITLE: Is there a reasonable definition of TQFTs for n-cobordisms with connected inputs/outputs? -QUESTION [12 upvotes]: A question that's been on my mind for a while is whether any precise statement to the effect of "Heegaard Floer homology is a TQFT," for some reasonable definition of TQFT, can be made. -Of course, a lot of effort is being spent right now on HF as an extended TQFT (e.g. the bordered theory of Lipshitz/Ozsvath/Thurston and, more recently, the cornered theory of Douglas/Lipshitz/Manolescu). But right now I'm just wondering about the 3+1 structure. -The issue in 3+1 dimensions (leaving aside the mixed invariants and how to derive them from a TQFT framework) is that only cobordisms between connected 3-manifolds induce maps on HF. This is, in some sense, a fundamental feature of the theory, since the induced maps for closed 4-manifolds are zero. This was discussed in the MO question Seiberg-Witten theory on 4-manifolds with boundary. -So, what if one were to make this feature into a definition? "Some variant of TQFT" := a functor which only allows these cobordisms with connected inputs and outputs? Does this correspond to some definition that's already out there? Is it a reasonable thing to consider in the framework of, e.g., Lurie's classification of fully extended TQFTs? Or is there some other definition which could be used instead, more amenable to this framework? -I'm putting a "reference-request" tag on this question, because answering it as stated probably would consist of pointing out a relevant paper or two, but I'd be interested more generally in anything that continues the discussion from the MO question I linked above. - -REPLY [3 votes]: Katrin Wehrheim has this issue too for 2+1 dimensions; she's referred to it as "connected TFT" (at least in private communication). She and Chris Woodward are currently working on it (using Lagrangian correspondences and Cerf theory), and she has posted on her website the preprint: Floer Field Theory. Look at Definition 2.2.1. It is related to her other notes on the Symplectic 2-Category. In particular, because the surfaces are required to be connected, you don't have the product axiom. -Morphisms are assigned to the cobordisms through the Cerf relations. This is spelled out in their other note (available on her webiste): Connected Cerf Theory.<|endoftext|> -TITLE: Natural $\Pi^1_2$ (or worse) classes of structures? -QUESTION [8 upvotes]: (To clarify, my interest is mainly lightface, that is, $\Pi^1_2$ instead of $\bf \Pi^1_2$, although it doesn't particularly matter.) -This is just an idle curiosity. In logic, I find myself frequently working with the countable ordinals; this is a $\Pi^1_1$ set of structures, in the sense that the set of reals coding well-orderings with domain $X\subseteq\omega$ is $\Pi^1_1$. This is also a fairly natural class of structures, of interest outside logic as well; and other interesting classes of structures - e.g., torsion groups, Artinian rings, infinite graphs into which a given infinite graph $G$ does not embed - are also $\Pi^1_1$ (and usually are vastly simpler than $\Pi^1_1$). -Obviously, there are more complicated classes of structures; however, all the examples I know are pretty artificial. So my question is: what are some natural classes of countable structures, which are worse than $\Pi^1_1$? -In particular: - -Is there a natural class of countable structures which is $\Pi^1_2$ complete? - -(Of course, what "natural" means is subjective. What I mean by natural here is "of roughly equal interest to non-logicians as well-orderings," but examples from within logic would be okay if there is an interesting argument to be made that they will someday be of interest outside logic.) - -REPLY [8 votes]: An example directly generalizing well-orderings is that the better quasi-orderings are $\Pi^1_2$-complete, shown in Marcone's aptly named The Set of Better Quasi-Orders is $\Pi^1_2$.<|endoftext|> -TITLE: What is the maximum of the ratio $\vartheta(G)/\alpha(G)$? -QUESTION [6 upvotes]: A maximum independent set is a largest independent set for a given graph $G$ and its size is denoted $\alpha(G)$. And the Lovász number of $G$ is denoted $\vartheta(G)$. $\vartheta(G)\geq \alpha(G)$ by definition, then the question is what will the -$$\max\limits_G \frac{\vartheta(G)}{\alpha(G)}$$ -be? Any known results about this topic will be welcome! One can give examples to show how big could the ratio $\vartheta(G)/\alpha(G)$ be or counter-examples to show the ratio will go to infinity. Any help or suggestions will be appreciated! -P.S. here is a related question:Cliques, Paley graphs and quadratic residues. - -REPLY [6 votes]: It is infinite, in fact much stronger versions are also true, see e.g., Theorem 1 here: -http://arxiv.org/abs/cs/0608021 -(Shannon capacity is between $\alpha$ and $\vartheta$.)<|endoftext|> -TITLE: structure of norm one group for quadratic extension of p-adic fields -QUESTION [6 upvotes]: Let $F$ be a p-adic field (finite extensions of $\mathbb{Q}_p$ for some prime $p$), and $E/F$ be a quadratic extension. Use $\sigma$ to denote the nontrivial element in the Galois group $Gal(E/F)$. For any $x\in E$, $N(x)=x\sigma(x)$ denotes the norm of $x$. -Now consider $U(1)=\{x\in E; N(x)=1 \}$, the group of elements of norm one in $E$.We know that $U(1)$ is a subgroup of $U_{E}$, the group of units of ring of integers in $E$, thus it is a compact subgroup (correct me if I'm wrong). -I'm wondering if there are more structural results about $U(1)$? I don't have any specific meaning about 'structure', so any answer reasonable in certain sense is welcome. -Sorry for the question is not so clear, any reference is much appreciated. - -REPLY [4 votes]: Aside from arithmetic point of view provided by Dalawat and Edoardo, -let me give a group theoretical point of view about this group. -If $E/F$ is a separable quadratic field extension with the Galois group $\{id, \sigma\}$ then the map -$$\phi:E^{\times}\rightarrow U(1), \ \ x\mapsto \frac{x}{\sigma(x)}$$ is surjective (as pointed out by Edoardo). We also have $\ker\phi=F^{\times}$. In other words we have a group isomorphism $$U(1)\simeq \frac{E^{\times}}{F^{\times}}.$$<|endoftext|> -TITLE: The homeomorphism problem for hyperbolic 3-manifolds and the virtual Haken theorem -QUESTION [11 upvotes]: If $N$ and $N'$ are two closed hyperbolic 3-manifolds, then one would like to have an algorithm which determines whether or not $N$ and $N'$ are homeomorphic. -If $N$ and $N'$ are Haken, then such an algorithm is provided by Haken. The homeomorphism problem is solved in general by Sela's work on hyperbolic groups and Mostow rigidity. -I was wondering though whether in light of Agol's work one can now give a "purely $3$-dimensional" solution to the homeomorphism problem. More precisely, Agol showed that given any hyperbolic 3-manifold there exists a finite cover with $b_1>0$. So given $N$ and $N'$ I can find a finite cover $\tilde{N}$ with $b_1(\tilde{N})>0$. I can take the subgroup $\pi_1(\tilde{N})$ to be of the type "intersection of all kernels of homomorphisms to a fixed finite group". I can then use the `same' finite index subgroup to get a cover $\tilde{N}'$ with $b_1(\tilde{N}')>0$. So now one can apply Haken to check whether $\tilde{N}$ and $\tilde{N}'$ are homeomorphic. To push this result down to $N$ and $N'$ it seems to me that one needs to calculate the mapping class group of $\tilde{N}$. This group is finite by Mostow rigidity, but are there algorithms for determining it? Are there any bounds on the size of the group? - -REPLY [10 votes]: There is a bound on the size of the mapping class group of a hyperbolic manifold, namely the volume divided by the minimal volume of a hyperbolic 3-orbifold (0.03905...). -There is a (non-rigorous) algorithm to compute the isometry group by Hodgson-Weeks, which is used in the program Snappea. -Addendum: Another possible approach to you question would be to compute the Thurston norm of the cover (which in principle is possible using normal surface algorithms developed by Tollefson). Then one has a short exact sequence of the mapping class group action on the Thurston norm unit ball, and the kernel of this action. If there are fibered faces, then in principle one can compute both by solutions to the conjugacy problem in the mapping class group, which would avoid some of the difficulties involved with hierarchies and Waldhausen/Hemion solution to the homeomorphism problem. I should say though that this entire approach is likely to be computationally infeasible, especially compared to a program like snappea which works well in practice, or using Sela's algorithm as referred to in Henry's answer. So although understanding isometry groups of manifolds is an interesting question, I think your whole approach to the homeomorphism problem is quixotic, except possibly as a theoretical exercise.<|endoftext|> -TITLE: Donsker Theorem Billingsley -QUESTION [5 upvotes]: Theorems $16.1$ and $16.3$ in Billingsley Convergence of measures. -$16.1$ reads : Random variables $u_1,\ldots$ on $(\Omega,\mathcal{B},\mathbb P)$ -and are i.i.d. with $0$ mean and finite variance $\sigma^{2}$, define $X_n(t,\omega) = \frac{1}{\sigma\sqrt{n}}S_{[nt]}(\omega)$ (here $t\in [0,1]$). Then $X_n \to W$ (converges in law to Wiener process). -$16.3$ Theorem $16.1$ remains valid if $\mathbb P$ is replaced by an arbitrary probability measure $\mathbb P_0$ such that $\mathbb P_0 \ll\mathbb P$. -I have a problem understanding why is Theorem $16.3$ needed at all?$16.1$ makes no assumption about the measure $\mathbb P$. If someone has a copy of the great book by Billingsley, can you please look into it? This has been troubling me for quite sometime and I have to admit I am on my own with no advanced probability course being offered in my school. don't delete it quickly because it dont meet your standards....let someone into probability theory give a short reply then delete it - -REPLY [5 votes]: I think the statement of Theorem 16.3 Billingsley meant is - -Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, $(\xi_i,i\geqslant 1)$ be i.i.d. zero mean random variables (for $\mathbb P$), and $X_n(t,\omega):=\frac 1{\sigma\sqrt n}\sum_{i=1}^{\lfloor nt\rfloor}\xi_i(\omega)$. Assume that $\mathbb P_0\ll\mathbb P$. Then - $$X_n\to W\mbox{ in distribution with respect to }\mathbb P_0,$$ - where $W$ is a standard Brownian motion. - -The result is non trivial because the random variables $(\xi_i,i\geqslant 1)$ are not independent with respect to $\mathbb P_0$ in general. So it's not only Theorem 16.1 used with $\mathbb P_0$ instead of $\mathbb P$. -We thus have that if $g$ is a non-negative integrable function with respect to $\mathbb P$, and $F\colon D[0,1]\to\mathbb R$ is continuous and bounded, then -$$\int_\Omega F(X_n(\cdot,\omega))g(\omega)\mathrm d\mathbb P\to \int_\Omega F(W(\cdot,\omega))g(\omega)\mathrm d\mathbb P(\omega).$$ -Theorem 16.1 gave it for $g=1$ while Theorem 16.3 extends to each $g\in L^1_+(\mathbb P)$.<|endoftext|> -TITLE: Can the equation of motion with friction be written as Euler-Lagrange equation, and does it have a quantum version? -QUESTION [20 upvotes]: My (non-expert) impression is that many physically important equations of motion can be obtained as Euler-Lagrange equations. For example in quantum fields theories and in quantum mechanics quantum equations of motion are obtained from the classical ones only if a Lagrangian (or Hamiltonian) is known for the classical case. Is my understanding too oversimplified? -Are there examples of physically important equations which are not Euler-Lagrange for any Lagrangian? -More specifically, let us consider the classical motion of a particle in $\mathbb{R}^3$ with friction: -$$\overset{\cdot\cdot}{\vec x}=-\alpha \overset{\cdot}{\vec x},\, \, \alpha>0,$$ -namely acceleration is proportional to velocity with negative coefficient. -Is this equation Euler-Lagrange for an appropriate Lagrangian? Is there a quantum mechanical version of it? -Added later: As I mentioned in one of the comments below, I do not really know how to make formal what is "quantum mechanical version". As a first guess one could try to write a Schroedinger equation with general (time dependent?) Hamiltonian such that some version of the Ehrenfest theorem would be compatible with the classical equation of motion with friction. - -REPLY [2 votes]: The answer appears to be yes. See Chad R. Galley, "The classical mechanics of non-conservative systems," Phys. Rev. Lett. 110, 174301 (2013) [Editors' Highlight]<|endoftext|> -TITLE: generators of free group -QUESTION [9 upvotes]: Give a rank $n$ free group $G=\langle a_1,a_2,\dots,a_n\rangle$, let $g_1,g_2,\dots,g_n \in G$, -$b_j=g_j^{-1}a_jg_j$ . If $b_1,b_2,\dots,b_n$ can generates the whole $G$, what can we say about $g_1,g_2,\dots,g_n$ ? -I have tried to take $G$ as a fundamental group,and give some topology properties of $g_1,g_2,\dots,g_n$ but failed . -I am not a professional mathematician,may be this question is a trivial, but I still need your help. I will be grateful if any of you can give me some useful advices. -Thanks! - -REPLY [4 votes]: Letting $a_i$ and $b_i$ be as in your question, observe that the homomorphism $F_n \rightarrow F_n$ that takes $a_i$ to $b_i$ for all $1 \leq i \leq n$ is an isomorphism if and only if the $b_i$ generate $F_n$ (this uses the classical fact that a set of $n$ elements of $F_n$ that generate $F_n$ must be a free generating set). This suggests examining the resulting subgroup of $\text{Aut}(F_n)$. -Define $\Gamma_n$ to be the subgroup of $\text{Aut}(F_n)$ consisting of automorphisms $f$ such that $f(a_i)$ is conjugate to $a_i$ for all $1 \leq i \leq n$; this group is known as the pure symmetric automorphism group. Your question basically boils down to asking about the structure of $\Gamma_n$. Here there is quite a bit known. For instance, it is a classical theorem of Nielsen that $\Gamma_2$ is exactly the set of inner automorphisms of $F_2$, which is equivalent to what Lee did in his answer. Things are more complicated for higher $n$ (I don't think there is an easy description of the possible $g_i$ in your question for $n \geq 3$ ), but a large amount is known about $\Gamma_n$. For instance, the paper -Humphries, Stephen P., -On weakly distinguished bases and free generating sets of free groups. -Quart. J. Math. Oxford Ser. (2) 36 (1985), no. 142, 215–219. -proves that it is generated by the elements $c_{ij}$ for distinct $1 \leq i,j \leq n$ defined by -$$c_{ij}(a_k) = \begin{cases} a_k & \text{if $k \neq i$},\\ a_j a_i a_j^{-1} & \text{if $k = i$} \end{cases}.$$ -Also, McCool found a relatively simple presentation for it in terms of these generators in -McCool, J., -On basis-conjugating automorphisms of free groups. -Canad. J. Math. 38 (1986), no. 6, 1525–1529. -There is also a large literature that comes from the theorem of Dahm that says that $\Gamma_n$ is isomorphic to the pure braid group of a set of $n$ unlinked circles in $\mathbb{R}^3$; see -Goldsmith, Deborah L. -The theory of motion groups. -Michigan Math. J. 28 (1981), no. 1, 3–17. -for a proof of this (Dahm never published his result) and also see -Brendle, Tara E.; Hatcher, Allen, -Configuration spaces of rings and wickets. -Comment. Math. Helv. 88 (2013), no. 1, 131–162. -for many interesting generalizations of this. One important result that comes out of this fact is a computation of the integral cohomology ring of $\Gamma_n$ in -Jensen, Craig; McCammond, Jon; Meier, John, -The integral cohomology of the group of loops. -Geom. Topol. 10 (2006), 759–784.<|endoftext|> -TITLE: Joins of, and limits in, $(\infty,1)$-categories via profunctors -QUESTION [7 upvotes]: I'm trying to interpret the join of $(\infty,1)$-category in a more conceptual way. Let me try to explain what I have in mind. -In the classical setting it is almost a triviality to express the join of two categories as a collage along the terminal profunctor: given $\cal C,D$ and the profunctor $\varphi\colon \cal C\to D$ sending all $(C,D)$ to the terminal object in $Set$, the join ${\cal C}\star{\cal D}$ can be written as $ {\cal C}\uplus_\varphi{\cal D}$, where I denote $\uplus_\varphi$ the collage operation. -Now I wonder if, choosing simplicially enriched categories as a model instead of the quasicategorical one, I can describe the join of two $(\infty,1)$-categories $\cal C,D\in{\bf sSet}\text{-}{\bf Cat}$ to be the collage ${\cal C}\uplus_\varphi{\cal D}$ along the terminal $\bf sSet$-profunctor sending any two objects $(C,D)$ to the terminal simplicial set. -Joins of $\infty$-categories are particularly useful to define limits and colimits of diagrams $p\colon K\to\cal C$ to an $(\infty,1)$-category: following T.1.2.13.4 one can define the limit and colimit of $p$ respectively to be the terminal and initial objects in the $(\infty,1)$-categories $\hom_p(\Delta^\bullet\star K,\cal C)$ and $\hom_p(K\star\Delta^\bullet,\cal C)$. -I wonder if this point of view can be generalized to include weighted limits/colimits in $(\infty,1)$-categorical language: here is the question. - -Consider the simplicial category $\mathfrak C[\Delta^n]$, for any $n$; given a simplicial functor $p\colon\cal K\to C$ between simplicial categories, how can I figure out the join (in the sense described before) $\mathfrak C[\Delta^n]\star \cal K$, and the (simplicial) set of simplicial functors $\hom_p(\mathfrak C[\Delta^n]\star \cal K,\cal C)$ which equal $p$ when restricted to $\cal K$? What is its terminal object, when it exists? - -I also wonder if it is possible to encode weighted limit theory using collages along arbitrary profunctors, and then translate also this theory in the $(\infty,1)$-categorical setting. Given a generic $\bf sSet$-profunctor $W\colon \mathfrak C[\Delta^n]\to \cal K$, what should the collage $\mathfrak C[\Delta^n]\uplus_W\cal K$ be? What should the terminal object in $\hom_p(\mathfrak C[\Delta^n]\uplus_W\cal K, C)$ be? -Last but not least: - -Does the "classical" definition of join between simplicial sets $K\star S$ coincide with the coherent nerve of the simplicial category $\bf K\star S$ obtained as the collage along the terminal $\bf sSet$-profunctor? This would automatically entail that the coherent nerve in $\star$-monoidal, as it is known to be. - -Edit: It is maybe useful to give an idea about where I began wondering about this: all stemmed from reading this page, which I was trying to complete. - -REPLY [3 votes]: I am not sure I can say anything useful about the weighted limit question; I want to remark about the other two however. -The join of simplicial (properly speaking, simplicially enriched, which I still denote as $sCat$) categories is indeed preserved by the coherent nerve construction. However, the functor $\mathfrak C$ is not monoidal. For instance, we observe that $\mathfrak C[\Delta^1]=\Delta^1$ joined with itself in $sCat$ is $\Delta^3$, not $\mathfrak C[\Delta^3]$. The right statement is that for two simplicial sets, the natural map $\mathfrak C[S \star T] \to \mathfrak C[S] \star \mathfrak C[T]$ is an equivalence of simplicial categories (but not in general an isomorphism). This all is mentioned in HTT, after Remark 1.2.8.2. -In relation to the discussion about (homotopy) limits, it is important to observe that the join $\star: sCat \times sCat \to sCat$ is not monoidal for the Bergner model structure (my example above with $\mathfrak C [\Delta^1]$ is an instance of that: $\Delta^3$ is not cofibrant). In particular, looking at simplicial sets of the form $sCat(\mathcal{D} \star \mathfrak C[ \Delta^\bullet ], \cal C)$ is not a good idea as (besides the fact we don't see it is a quasicategory) the domain is not necessarily cofibrant and hence the thing we get might have incorrect homotopical behaviour. -We might replace $\mathcal{D} \star \mathfrak C[ \Delta^\bullet ]$ by $\mathfrak C[N\mathcal{D} \star \Delta^\bullet ]$ and then $$sCat(\mathfrak C[N\mathcal{D} \star \Delta^\bullet ], \mathcal C) \cong sSet (N\mathcal D \star \Delta^\bullet, N \mathcal C)$$ which (after taking fibres over the map $p: \mathcal D \to \mathcal C$ of interest) is the familiar cocone quasicategory which is used to define a colimit of $Np$ (and T.4.2.4.1 tells us the relation between quasicategorical and homotopy (co)limits). -To recap, the problem I see with the simplicial categorical join (the collage along the terminal profunctor) is that it does not interact well with the model structure. As the join of simplicial sets has correct homotopical behaviour (T.4.2.1.3), the category $Ho(QCat)$ and hence $Ho(sCat)$ do have a derived monoidal join operation, but it is not clear if there is a way to properly represent it on $sCat$. One might have more luck doing this in a different cartesian model for $(\infty,1)$-categories (like Segal categories for injective model sturcture).<|endoftext|> -TITLE: pseudo-Anosovs with given action on homology -QUESTION [6 upvotes]: It is well-known that the Mapping Class Group of a closed surface of genus $g$ surjects onto $Sp(2g, \mathbb{Z})$ (see, for example the Farb-Margalit book). However, I was wondering if there is a simple proof that the set of pseudo-Anosov elements in MCG surjects onto $Sp(2g, \mathbb{Z})$ (and also a reference for where this might have been stated first) -- I can construct a somewhat sophisticated proof, but this should be easier. - -REPLY [3 votes]: Suppose that $f$ is a pseudo-Anosov and let $\lambda^\pm$ be its stable and unstable laminations. Suppose that $g$ is any mapping class with the following property: $g(\lambda^+) \neq \lambda^-$. Then, for all sufficiently large $n$, the composition $f^n g$ is also pseudo-Anosov. This is proved using the north-south dynamics of $f$. -Now, as in Mosher's answer, choose $f$ in Torelli and let $g$ be a lift of the desired element of $Sp$. If needed, compose $g$ with a bounding pair map to arrange the side-condition. We are done. -Added later: Here are some of the details of the dynamical argument. I've edited this several times to try and make it correct. Full disclosure - I first heard this from Yair Minsky. -Pick $U$, a small neighborhood of $\lambda^+$, chosen so that $\lambda^-$ is not contained in $g(U)$. I'll also want $f|U$ to be strictly contracting. Using north-south dynamics, there is an $m^+ > 0$ so that for all $n > m^+$ we have $f^{m^+} g(U) \subset U$. The contracting property implies that the map $f^n g$ has a unique fixed point in $U$. -On the other hand, pick a small neighborhood $V$ of $g^{-1}(\lambda^-)$, so that $\lambda^+$ is not in $V$. Also, we require $f^{-1}|g(V)$ to be strictly contracting There is a power $m^- > 0$ so that for all $n > m^-$ we have $V \subset f^n g(V)$. Deduce that the map $f^n g$ has a unique fixed point in $V$. -We place one more restriction on $n$. We need $f^n g(V^c)$ to be contained in $U$. -I now claim that $h = f^n g$ has no other fixed points, and so is pseudo-Anosov.<|endoftext|> -TITLE: Maximizing quadratic form on the hypercube -QUESTION [5 upvotes]: I want to maximize a quadratic form $\mathbf x^T\mathbf Q\mathbf x$ and also want to find out which vector $\mathbf x$ maximizes the quadratic form when - -$\mathbf Q$ is an $n\times n$ positive definite matrix, and -$\mathbf x$ is a vertex of an $n$-dimensional hypercube; that is, the convex hull of all sign permutations of the coordinates $\mathbf x = (x_1,\cdots,x_n) \in (\pm 1, \cdots, \pm 1)$. - -In my setting, $\mathbf Q$ has only two types of eigenvalues: -$$ -\text{eigenvalues of $\mathbf Q$} = -\{ -\underbrace{1,\cdots,1}_{k\quad\text{copies}}, -\underbrace{\frac{1}{n+1}\cdots,\frac{1}{n+1}}_{n-k \quad\text{copies}} -\} -$$ -where $n$ is the size of the matrix $\mathbf Q$. -As $\mathbf Q$ is positive definite, the functional $\mathbf x^T\mathbf Q\mathbf x$ is strictly convex and hence its maximum on any polytope $\mathcal P$ is attained at $\mathcal P$'s vertices. -So I can formulate my optimization problem as a quadratic programming: -$$ -\begin{array}{rl} -\text{maximize} & \mathbf x^T\mathbf Q\mathbf x\\ -\text{subject to} & -1 \leq x_i \leq 1 \quad\text{for $i\in\{1,\cdots,n\}$} -\end{array} -$$ -My Goal - -I want to prove that the maximum of $\mathbf x^T\mathbf Q\mathbf x$ is $n$, and -I want to find out a vector $\mathbf x\in (\pm 1, \cdots, \pm 1)$ that maximizes the quadratic form. (There can be many maximizers but any one will do.) - -Unfortunately, I don't have much background on quadratic programming. -It seems there are a handful of solution methods out there such as interior point, active set, gradient projection, just to name a few. But I don't know which one is the best fit for my overly simple setting of quadratic programming. -Question: Which solution method will be the best approach for my problem? - -Intuition behind my guessing -(In case you wonder why I'm guessing that the maximum of the quadratic form is $n$.) -As $\mathbf x$ is a vertex of a hypercube in my setting, I can regard the quadratic form as a scaled version of Rayleigh quotient -$R(\mathbf Q, \mathbf x) =\frac{\mathbf x^T \mathbf Q \mathbf x}{\mathbf x^T \mathbf x}$. -\begin{equation*} -\mathbf x^T \mathbf Q \mathbf x -= -n \frac{\mathbf x^T \mathbf Q \mathbf x}{\mathbf x^T \mathbf x} -= -nR(\mathbf Q, \mathbf x) -\end{equation*} -as $\mathbf x\in(\pm 1,\cdots, \pm 1)$ ensures that $\mathbf x^T\mathbf x = n$. -In this formulation, I know the maximum of the Rayleigh quotient because - -$R(\mathbf Q, \mathbf x) \leq \lambda_{max}(\mathbf Q) = 1$, and -$R(\mathbf Q, \mathbf v) = \lambda_{max}(\mathbf Q) = 1$ when $\mathbf v$ is the dominant eigenvector of $\mathbf Q$; that is, $\mathbf v$ is an eigenvector of $\mathbf Q$ corresponding to the largest eigenvalue $\lambda_{max}(\mathbf Q)=1$. - -Therefore, the max of the scaled Rayleigh quotient $\mathbf x^T \mathbf Q\mathbf x = n R(\mathbf Q, \mathbf x)$ is $n$. -But this approach does not guarantee that the dominant eigenvector $\mathbf v$ is a scaled version of $(\pm 1,\cdots, \pm 1)$ and it cannot be an answer of my quadratic programming. - -REPLY [7 votes]: This maxQP problem is hard, it includes MaxCUT as a special case---see this paper by M. Charikar on MAXQP. Having $Q$ be positive definite does not really help (take the MAXQP problem in Charikar's paper, and since $x_i \in \{\pm1\}$, you can add a suitable multiple of the identity to turn the problem with a symmetric matrix to one with a symmetric positive definite matrix. -That said, chasing the cited paper and the papers that cite it, you should be able to find a wealth of literature to help you approximately solve your problem (including semidefinite programming relaxations)<|endoftext|> -TITLE: A Hausdorff abelian group with no character? -QUESTION [7 upvotes]: Pontryagin Duality for locally compact abliean groups gives plenty of continuous (unitary) characters $\chi : A \to \mathbb{R} / \mathbb{Z}$, but if we do not assume local compactness, can anything be said? In particular, is the following true? - -Every abelian Hausdorff topological group has a nontrivial continuous (unitary) character - -I believe this to be false, but cannot find a concrete example. - -REPLY [8 votes]: If a topological vector space $X$ is not locally convex, then it usually has not non-zero linear continuous functionals, and this means that there are no non-trivial continuous characters on $X$. For example, you can take the space of all measurable functions on $[0,1]$ with the metrics -$$ -d(x,y)=\int_0^1\frac{|x(t)-y(t)|}{1+|x(t)-y(t)|}d t. -$$<|endoftext|> -TITLE: On a conjecture of Lions for the wave equation -QUESTION [6 upvotes]: In Control of Distributed Singular Systems p 236, JL Lions makes the conjecture : -Let $\Omega$ be a domain in $\mathbb{R}^n$, $Q = \Omega \times ]0,T[$ and consider -$\phi'' - \triangle \phi = F$ -$\phi(x,0) = \phi^0(x), \phi'(x,0) = \phi^{1}(x), \phi^0 \in H^1(\Omega), \phi^1 \in L^2(\Omega)$ -$\frac{\partial \phi}{\partial \nu} = 0 $ on $\Sigma$ -where $F \in L^1(0,T;L^2(\Omega))$. -Then, the conjecture is that $\phi \in H^1(\Sigma)$. -Since this book was written almost 30 years ago, I was wondering if there are any results about this conjecture. Thanks in advance for any insight. - -REPLY [4 votes]: You will find your answer in this article: -http://www.sciencedirect.com/science/article/pii/0022247X89902059<|endoftext|> -TITLE: moduli spaces are kahler? -QUESTION [15 upvotes]: I often heard from experts that "moduli spaces are Kahler". This sounds as a meta-theorem asserting that every time one defines reasonable moduli spaces, then there is a standard strategy to see (proof) if that space is Kahler or not. -I'm mainly interested in moduli spaces of geometric structures on surfaces: Teichmuller (ok, that's I know is Kahler), Moduli space of complex projective structures (is this space kahler? where can I find a reference?), et cetera... -Is there a standard (or not) reference for this kind of problems? - -REPLY [2 votes]: The moduli space of holomorphic normal projective connections is an affine space, or empty, as it is identified with the collection of all holomorphic 1-cocycles whose coboundary is a suitable ``traceless Atiyah class'' of the tangent bundle. The paper -Robert Molzon and Karen Pinney Mortensen, -The Schwarzian derivative for maps between manifolds with complex projective connections, Trans. Amer. Math. Soc. 348 (1996), no. 8, 3015–3036. MR 1348154 (96j:32028) 27, 55 -is perhaps the best introduction to the theory of complex projective connections. I don't know a good reference for the relation to the Atiyah class, though it appears somewhere in the work of Kobayashi. The moduli space of flat holomorphic projective connections on a compact complex manifold is a complex subvariety, not known to be smooth.<|endoftext|> -TITLE: Is the sequence $a_n=c a_{n-1} - a_{n-2}$ always composite for $n > 5$? -QUESTION [14 upvotes]: Numerical evidence suggests the following. -For $c \in \mathbb{N}, c > 2$ define the sequence $a_n$ by -$a_0=0,a_1=1, \; a_n=c a_{n-1} - a_{n-2}$ -For $ 5 < n < 500, \; 2 < c < 100$ there are no primes in $a_n$ though -semiprimes exist. - -Is it true that $a_n$ is always composite for $n > 5$ -If yes is there explicit partial factorization? - -Searching OEIS solved the case $c=6$ -with a Pell equation. -Counterexamples are welcome. - -REPLY [6 votes]: Here is another approach to show that $a_n$ is not prime when $c \gt 2$ and $n \gt 2$ -We have (proof at end) $$a_{n+m}=a_na_{m+1}-a_{n-1}a_{m} \tag{*}$$ -So, by induction on $j \ge 1$, $$a_{n+jn}=a_{n}a_{jn+1}-a_{n-1}a_{jn}$$ is always divisible by $a_n.$ -Hence the only question is for $p \gt 2$ prime. But for any odd index $2m+1$ we have $$a_{(m+1)+m}=a_{m+1}^2-a_{m}^2=(a_{m+1}+a_m)(a_{m+1}-a_m)$$ -That is about enough. We should check that $a_{m+1}-a_{m} \gt 1$. In fact, in the case $c=2$, the sequence is $0,1,2,3,4,5,\cdots$ but for $c \gt 2$ we have $a_{n+1}-a_{n}$ increasing since $$a_{n+1}-a_n =(c-1)a_n-a_{n-1}=(c-2)a_n+(a_n-a_{n-1}).$$ - -To prove $(*)$, replace $m+n$ by $s$ and write $$a_s=a_{s-m}a_{m+1}-a_{s-m-1}a_{m}$$ where $m \lt s$ so -$\begin{align*} -a_s &= a_{s-1}a_2-a_{s-2}a_1 -\\ &= a_{s-2}a_3-a_{s-3}a_2 - \\ &=a_{s-3}a_4-a_{s-4}a_3 -\\ &=\cdots. -\end{align*}$ -The first line is jut the defining recurrence relation $a_s=a_{s-1}c-a_{s-2}1$ -and then the difference between successive lines is $$\left(a_{s-m}a_{m+1}-a_{s-m-1}a_{m}\right)-\left(a_{s-m-1}a_{m+2}-a_{s-m-2}a_{m+1}\right)$$ $$=\left(a_{s-m}+a_{s-m-2}\right)a_{m+1}-a_{s-m-1}\left(a_{m+2}+a_m \right)$$ $$=\left(ca_{s-m-1}\right)a_{m+1}-a_{s-m-1}\left(ca_{m+1} \right)=0$$ - -Observe that putting $c=i$ yields Fibonacci numbers times powers of $i$: $$0,1,i,-2,-3i,5,8i,-13,\cdots $$ -This suggests that the divisibility result above (considering the $a_n$ as monic polynomials in variable $c$) can be sharpened to $$\gcd(a_n,a_m)=a_{\gcd(n,m)} $$ The usual proof applies mutatis mutandis. -This means that for any chosen integer value of $c$ the same fact holds, if $p$ is prime and $k=m$ is the least positive index with $p \mid a_k$ then $p \mid a_n$ exactly when $m \mid n$.<|endoftext|> -TITLE: Subset of Spec(A) realized as inverse image of some Spec(B) -QUESTION [5 upvotes]: Let $A$ be a (commutative) ring and $U$ be an arbitrary subset of $\mathrm{Spec}(A)$. Do there exist a ring $B$ and a ring homomorphism $\varphi\colon A\to B$ such that $\varphi^{-1}(\mathrm{Spec}(B))=U$ ? If $\varphi\colon A\to B$ and $\varphi'\colon A\to B'$ are two ring homomorpisms such that $\varphi^{-1}(\mathrm{Spec}(B))=\varphi'^{-1}(\mathrm{Spec}(B'))=U$, do there exist a ring homomorpism $f\colon B\to B'$ such that $f\varphi=\varphi'$ ? - -REPLY [3 votes]: There are already two nice counterexamples. Let me just point out for question (1) a characterization of images of $\mathrm{Spec}(B)$ in $X:=\mathrm{Spec}(A)$: - -A subset $U$ of $X$ is the image of some $\mathrm{Spec}(B)\to X$ if and only if $U$ is pro-constructible. - -See EGA IV.1.9.5. In a noetherian space, $U$ is pro-constructible if it is an arbitrary intersection of constructible subsets (i.e. finite unions of locally closed subsets) of $X$. A constructible subset of $\mathrm{Spec}(\mathbb Z)$ (or any irreducible noetherian scheme of dimension $1$) is either finite or contains the generic point, so a pro-constructible subset is either finite or contains the generic point. This "explains" the example of Laurent.<|endoftext|> -TITLE: The pushout product as an operation -QUESTION [6 upvotes]: Motivation: In his utterly famous paper, Rezk (here, (pag. 7)) defines a structure called "Quillen ring". I'm wearing my algebraist's hat today, so I was wondering if this definition is chosen to suggest that somewhere a "true" ring structure is hidden: now I'm in particular interested in dualizing the definition to obtain the "pushout-product" arrow, which I would like to taxonomize here with your kind help. [What I'm kindly asking here is: does the "pullback-exponential" operation give a "ring operation", in some reasonable sense? Does its dual give a co-ring operation?] -Let me start from the beginning. -For the moment I don't even need a model structure, as I am not interested in cofibration-preservation properties (which can be used to define in a compact way the notion of "left Quillen bifunctor", as I learned about an hour ago). -=== -Take your favourite finitely bicomplete category $\cal C$ (in particular I'll ask for pullbacks and pushouts). -Define a binary operation on the set (of isomorphism classes) of arrows in $\cal C$, taking $f\colon A\to B$ and $g\colon C\to D$ and sending them to the arrow -$$\newcommand{\diam}[4]{\left(#1 \times #4\right)\coprod_{#1 \times #3}\left( #2 \times #3 \right)} -f\diamond g\colon \quad \diam{A}{B}{C}{D}\longrightarrow B\times D$$ -I would like to unravel the algebraic properties of this operation $\diamond \colon {\rm Mor}({\cal C})\times {\rm Mor}({\cal C})\to {\rm Mor}({\cal C})$. - -Is it associative? -Is it commutative? -It seems to have a neutral element (the arrow $\varnothing\to 1$). -Is there an operation on which $\diamond$ distributes over? - -Associativity seems painful, as it is linked to simplification properties which I'm not able to deduce from easy arguments. -This is the scary form of the isomorphism which proves the associativity $f(gh) \cong (fg)h$ for $f\colon A\to B,g\colon C\to D, h\colon V\to W$ -$$ -\diam{\diam{A}{B}{C}{D}}{B\times D}{V}{W} \quad\cong \quad \diam{A}{B}{\diam{C}{D}{V}{W}}{D\times W} -$$ -Great honour to whom will be able to simplify it! - -REPLY [12 votes]: Whenever you start with a biclosed monoidal category $(\mathcal C, \otimes, I)$, the push-out product is a (biclosed) monoidal structure on the category of arrows $\operatorname{Mor}(\mathcal C)$. The trick is to think of $\operatorname{Mor}(\mathcal C)$ as the category of functors $\mathbf 2\rightarrow\mathcal C$, where $\mathbf 2$ is the poset $\{0<1\}$. Two morphisms $f$ and $g$ give rise to a functor $f\otimes g\colon \mathbf 2\times \mathbf 2\rightarrow\mathcal C$ -and their push-out product can be expressed as -$$f\diamond g=\operatorname{colim}\limits_{(\mathbf 2\times \mathbf 2)\setminus \{(1,1)\}}f\otimes g\longrightarrow \operatorname{colim}\limits_{\mathbf 2\times \mathbf 2}f\otimes g$$ -The map is induced by the inclusion of posets. As for the target, notice that $\mathbf 2\times \mathbf 2$ has a final object, $(1,1)$, so the colimit is the value at the final object. -A more or less elementary computation shows that if $\otimes$ preserves push-outs in each variable (and this happens if $\mathcal C$ is biclosed), then the iterated push-out product of morphisms $f_1,\dots,f_n$ is simply -$$f_1\diamond\cdots\diamond f_n=\operatorname{colim}\limits_{\mathbf 2^n\setminus \{(1,\dots,1)\}}f_1\otimes\cdots\otimes f_n\longrightarrow \operatorname{colim}\limits_{\mathbf 2^n}f_1\otimes\cdots\otimes f_n.$$ -This is the way of checking associativity. -As you point out, it is easy to see that $\varnothing\rightarrow I$ is a tensor unit. You can use that $X\otimes\varnothing\cong \varnothing\cong\varnothing\otimes X$ for any object $X$ since $\mathcal C$ is biclosed. -If $\mathcal C$ is symmetric, hence so is the category of arrows. This is really easy and does not require the property of being biclosed. -The pentagon and other coherence axioms follow from the corresponding axioms for $\mathcal C$ and the universal property of colimits.<|endoftext|> -TITLE: Are there any known non-trivial functions that takes on squarefree values with the right density? -QUESTION [5 upvotes]: A number $n \in \mathbb{N}$ is squarefree if for every divisor $d | n, d > 1$, we have $d^2 \nmid n$. It is known that the squarefree numbers have a density of $6/{\pi^2}$ over $\mathbb{N}$. It is a question of interest to know whether polynomials take on infinitely many squarefree values. -I am asking for an explicit example of a function (perhaps a polynomial) that is known to take on infinitely many squarefree values with the correct density. This is trivial if $f(x) = x$, and it is known that all polynomials $f(x) \in \mathbb{Z}[x]$ satisfying basic non-degeneracy conditions and of degree 2 or 3 take on infinitely many squarefree values. It is conjectured to hold for all $f(x) \in \mathbb{Z}[x]$ satisfying non-degeneracy conditions (that is, the content of $f$ should be square-free, and that for all primes $p$, there exist $n \in \mathbb{Z}$ such that $f(n)$ is not divisible by $p^2$), and this conjecture was proved to be true by Granville if one assumes the abc conjecture. -However, the question of whether $f$ assumes squarefree values with the 'correct' density is not investigated as often, at least to my knowledge. That is, the density of the set -$$\displaystyle S = \{n \in \mathbb{Z}: f(n) \text{ is squarefree. }\}$$ -over $\mathbb{Z}$. The case of interest would be when the density of $S$ is exactly $6/\pi^2$, which would show that $f$ has no bias towards squarefree values. -Are there any arithmetic function where this statement is known to be true, other than obvious ones like $f(x) = x$? - -REPLY [7 votes]: Indeed the Piatetski-Shapiro numbers are such an example: see Theorem 4 of this paper http://arxiv.org/pdf/1203.5884.pdf by Baker, Banks, Brudern, Shparlinski and Weingartner. -Note: I understood the question to ask for the density to be exactly $\pi^2/6$; which the PS example gives. One can also compute the density for polynomials of small degree (e.g. degrees $2$ and $3$) but the density won't be $\pi^2/6$. Finally a trivial example that also gives $\pi^2/6$ is to take $f(n) = n+ [\sqrt{n}]$ and use the distribution of squarefree numbers in short intervals. In my opinion, the PS example is perhaps the most interesting (at least the most interesting that I know).<|endoftext|> -TITLE: Intuition for holomorphic bisectional curvature -QUESTION [6 upvotes]: I would like to know the intuition behind the holomorphic bisectional curvature of Hermitian manifolds. I already know that the classical sectional curvature of a Riemannian (not necessarily complex) manifold roughly tells us how geodesics spread apart. What is the analogue for holomorphic bisectional curvature? Do you know any reference, where I can read more about this? - -REPLY [2 votes]: Ngaiming Mok, The uniformization theorem for compact Kähler manifolds of nonnegative holomorphic bisectional curvature, J. Differential Geom. Volume 27, Number 2 (1988), 179-214. -Mok improves Mori's and Siu-Yau's proof of Fraenkel's conjecture claiming that a compact Kahler manifold with positive HBC is biholomorphic to ${\Bbb C} P^n$. He shows that any compact Kahler manifold with non-negative HBC is locally isometric to a product of symmetric spaces and ${\Bbb C} P^n$. It is interesting that Mok uses Hamilton's evolution equation (essentially the same as used by Perelman 15 years later).<|endoftext|> -TITLE: Operator Valued Kadison--Singer Problem -QUESTION [17 upvotes]: The Paving Conjecture, which is equivalent to the famous Kadison--Singer Problem, was spectacularly settled in the affirmative by Marcus--Spielman--Srivastava (arxiv:1306.3969). Let $E$ denote the canonical projection from ${\mathcal B}(\ell_2{\mathbb N})$ onto the subalgebra $\mathcal D$ of the diagonal operators (which is isomorphic to $\ell_\infty\mathbb N$). The Paving Conjecture asserts that for every $\epsilon>0$ there is $K=K(\epsilon)\in{\mathbb N}$ which satisfies the following property: For any $x\in{\mathcal B}(\ell_2{\mathbb N})$ such that $E(x)=0$, there is a partition ${\mathbb N}=S_1\sqcup\cdots\sqcup S_K$ such that $\|P_{S_i} x P_{S_i}\|<\epsilon\|x\|$ for every $i$. Here $P_S$ is the orthogonal projection from $\ell_2{\mathbb N}$ onto $\ell_2S$, which belongs to $\mathcal D$. -Given that the Paving Conjecture is settled, I think it is natural to wonder whether it can be generalized to the operator valued setting. Let $M$ be a von Neumann algebra. Is the following true? For every $\epsilon>0$ there is $K=K(M,\epsilon)$ which satisfies the following property: For any $x\in{\mathcal B}(\ell_2{\mathbb N})\bar\otimes M$ such that $(E\otimes\mathrm{id})(x)=0$, there is a prtition $1\otimes 1 = \sum_{i=1}^K P_i$ of the identity by orthogonal projections $P_i$ in ${\mathcal D}\bar\otimes M$ such that $\|P_i x P_i\|<\epsilon\|x\|$ for every $i$. Sorin Popa pointed out to me that even the case where $M=L^\infty[0,1]$ is not clear. Also, what is the growth of $K(M_n({\mathbb C}),\epsilon)$ as $n\to\infty$? -As an operator algebraist, I am frustrated by the fact that the important problem in operator algebra theory was solved by outsiders by linear algebra. My motivation is that, although I don't have an application, the above generalization, if it's true, would require a new proof of the Paving Conjecture (now the Marcus--Spielman--Srivastava theorem) that involves operator algebra theory. - -REPLY [8 votes]: The case $M = L^\infty[0,1]$ seems like a simple measurability question. If every $x \in B(l^2)$ can be $K$-paved to $\|\sum P_ixP_i\| < \epsilon \|x\|$, then every $x \in B(l^2)\otimes L^\infty[0,1] \cong L^\infty([0,1],B(l^2))$ has a pointwise a.e. $K$-paving that does the same thing. So we just need a measurable selection. Surely one of the standard measurable selection theorems suffices here, have you checked this? -I really disagree with your characterization of the problem as having been solved without operator algebra theory. You didn't mention the brilliant reduction of Akemann and Anderson that gets us down to this combinatorial question. That part of the solution is completely operator algebraic and it is quite nontrivial. In his paper "Filters in C*-algebras" Tristan Bice develops some very beautiful machinery for achieving this reduction, among other operator algebraic applications. It sounds as if you are not even aware of this part of the story.<|endoftext|> -TITLE: Quadratic residues and nonresidues of arbitrary patterns -QUESTION [8 upvotes]: Let $p_1, p_2, \dotsc, p_n$ be distinct primes, and let $\epsilon_1, \epsilon_2, \dotsc, \epsilon_n$ be an arbitrary sequence of $1$ and $-1$. -There is an integer $a$ such that $\left( \frac{a}{p_1} \right) = \epsilon_1, \left( \frac{a}{p_2} \right) = \epsilon_2, \dotsc, \left( \frac{a}{p_n} \right) = \epsilon_n$, where $\left( \frac{a}{p_i} \right)$ denotes Legendre's symbol. -What can we say about the number $a$? -- I couldn't find any results for such numbers. -In particular, I am interested in bounds for the smallest possible $|a|$. -Could you recommend any papers or books on the topic? -EDIT -It's better to write "distinct odd primes" for $p_1, p_2, \dotsc, p_n$. -I have checked for several primes $< 1000$ by using computer. The smallest $a > 0$ does not seem to go beyond the product of two largest primes. - -REPLY [6 votes]: The number of integers $a$ in $[0, x]$ with the desired property is -$$ -2^{-n}\sum_{a=1}^x\prod_{i=1}^n\left(1+\epsilon_i\left(\frac{a}{p_i}\right)\right). -$$ -Expand the right hand side to obtain one term $2^{-n} x$, and $2^n-1$ sums of the form $2^{-n}\sum_{a\leq x}\left(\frac{a}{q}\right)$, where $q$ is the product of some of the $p_i$. Since sums over non-trivial characters can be bounded, for $x$ sufficiently large the term $2^{-n} x$ dominates, and we obtain that some $a$ exists. -To get a bood upper bound on $a$ you need some specific bound on character sums. There are several such bounds, which to use depends on the information you have for the $p_i$. In general one can use Burgess bounds, and obtains that for any $\epsilon>0$ there exists some $C_\epsilon$, such that $a<2^{C_\epsilon n}q^{\frac{1}{4}+\epsilon}$, where $q=\prod p_i$. If the $p_i$ are all small, one should obtain better results from zero-free regions for Dirichlet $L$-series. If the $p_i$ are of different magnitude, then most of the occurring charactersums are much smaller then $\prod p_i$, thus one can reduce the influence of $n$.<|endoftext|> -TITLE: Vaught conjecture for uncountable languages -QUESTION [7 upvotes]: Recall Vaught conjecture: the number of countable models of a first-order complete theory in a countable language is finite or $\aleph_0$ or $2^{\aleph_0}.$ -Now let $\lambda$ be an uncountable cardinal. Is the following version of Vaught conjecture true: -Vaught conjecture for uncountable languages. If $T$ is a complete theory in a language of size $\lambda,$ and if $T$ has more than $\lambda-$many non-isomorphic models of size $\lambda$, then $T$ has $2^\lambda$ many non-isomorphic models models of size $\lambda.$ -Also, do we have the following version of Moreley theorem: -Morley theorem for uncountable languages. If $T$ is a complete theory in a language of size $\lambda,$ and if $T$ has more than $\lambda^+-$many non-isomorphic models of size $\lambda$, then $T$ has $2^\lambda$ many non-isomorphic models models of size $\lambda.$ - -REPLY [4 votes]: Suppose that $2^{\aleph_{1}} > 2^{\aleph_{0}} > \aleph_{2}$, and consider the language with unary predicates $P_{n}$ $(n \in \omega)$ and $Q_{\alpha}$ $(\alpha < \omega_{1})$. Consider the theory $T$ which says that (1) each $P_{n}$ is satisfied by infinitely many points (2) no point satisfies more than one $P_{n}$ (3) no point satisfies any $Q_{\alpha}$. -This appears to me to be complete, as for each $n \in \omega$ you can argue by complexity of formulas that for any $k$-ary formula $\phi$ not using any of $P_{m}$ ($m \geq n$), any two models $M$ and $N$ of $T$ and any two tuples $\langle a_{0},\ldots,a_{k-1} \rangle$ from $M$ and $\langle b_{0},\ldots,b_{k-1} \rangle$ from $N$, if -$M \models P_{p}(a_{i})$ if and only if $N \models P_{p}(b_{i})$ for each $p < n$ and each $i < k$ then $M \models \phi(a_{0},\ldots,a_{k-1})$ if and only if $N \models \phi(b_{0},\ldots,b_{k-1})$. Using this fact, one can show that the second player wins the back-and-forth game using $M$ and $N$ of any fixed finite length, restricting to any finite set of the $P_{n}$'s. -Now $T$ appears to have exactly continuum many models of cardinality $\aleph_{1}$ up to isomorphism, determined by how many points satisfy each $P_{n}$ (two options in each case), and how many satisfy none of them (countably many options).<|endoftext|> -TITLE: Securing privacy of "who communicates with whom" under Orwell-like conditions -QUESTION [23 upvotes]: Assume that there is a big and powerful country with an -information-greedy secret service which has backdoors to all internet nodes -throughout the world which permit him to observe all exchanged data and all -computations done inside the nodes. -Is it still possible under these conditions to ensure by mathematical means -that this secret service cannot find out who communicates with whom, -if one designs internet protocols in a suitable way? -My feeling is that the answer is likely "yes", but I am not working in cryptography. --- Probably a cryptographer can tell more. - Clarification (added after the first two answers): -A good answer to this question could either consist of a description of -a method together with substantial heuristic arguments in support of its -suitability for the given purpose, or it could give substantial heuristic -arguments that there is no such method. Mere handwaving arguments in -favor of a positive or a negative answer do not answer the question. -On the other hand, the question is only meant to ask whether there are -mathematical methods which in practice serve the purpose, -just like in practice RSA can be used as a public key cryptosystem. -It is not asking for a proof or disproof (of what precisely??), since this -would not make sense. - Added after the first 5 answers (excluding deleted one(s)): -The answers given so far mostly take the question as a soft question, -which it is not. So far, Goldstern's answer comes closest to answering -the question in that it proposes a concrete method -- but as it stands, -it is still quite a way to go to get to anything practical. -Let me also emphasize that any transmission of data is "communication", -including transmissions of publicly available webpages, downloads etc.. -So firstly everybody has a lot of communication partners, and -secondly efficiency is a very important concern. - -REPLY [4 votes]: Unconditional sender privacy is a well-known problem; -D. Chaum - "The dining cryptographers problem" -could be a reasonable starting point. -Having decided on existance, the question remains what price (in terms of communication complexity) could be considered acceptable to name it efficient enough (that is, practical). -"Dining" communication scheme had a impact and was extended in a number of ways, including an option for lost messages: -M. Waidner, B. Pfitzmann - The dining cryptographers in the disco: Unconditional sender and recipient untraceability with computationally secure serviceability. -For a practical (depending on a computationally hard problem) solution, "remailer" was designed long ago for email. -Papers of Lance Cotrell and Ulf Moeller there could be of interest. -This was mentioned by Henry Cohn as "mix network" already.<|endoftext|> -TITLE: Reference for counting points over finite fields -QUESTION [5 upvotes]: The following fact is extremely well known: - -Fact. Let $Y$ be a geometrically irreducible variety (not necessarily smooth or proper) over a finite field $k$. - Then there is a constant $B$, depending only on cohomological invariants of the base change of $Y$ to an algebraic closure of $k$, such that: if $l$ is any finite extension of $k$ satisfying $|l|>B$, then $Y(l)$ is non-empty. - -Unfortunately, I can't find any simple reference for this well-known fact. If $Y$ is projective (and possibly singular), then it follows from the Lang-Weil estimates. If $Y$ is smooth and proper over $k$, then it follows from Deligne's proof of the Weil conjectures. Given that we don't mind replacing $Y$ with an open subvariety, there's a plausible argument for reducing the general case to one of these cases. -Also, the phrase "depending only on cohomological invariants of..." is rather vague, especially when $Y$ might be non-proper and singular. So maybe it's better to stick to more concrete statements, such as that $B$ should remain constant in certain types of families. -I would like to cite this fact without having to devote half a page to deducing it from something else. So my question is simply: - -What is a good reference for the fact stated above? - -REPLY [3 votes]: If you know that your algebraic variety is given (or can easily be given, or some open subset can) by a bounded number of polynomials in a bounded number of variables of bounded degree, and you don't care about actual constancy of the Betti numbers, you can combine easily the trace formula expression of D. Petersen's answer with bounds for Betti numbers as in Katz ("Sums of Betti Numbers in Arbitrary Characteristic", Finite Fields and Their Applications 7, 29-44 (2001)). -A convenient reference for the trace formula would be SGA 4 1/2, for instance.<|endoftext|> -TITLE: How long does it take to compute a class number? -QUESTION [6 upvotes]: I was wondering if there are any known (upper and lower) bounds for the complexity of computing the class-number of a finite extension of the rationals. (A general bound should be in function of the discriminant, I guess.) -This would also be of interest in special cases like fields with a given degree or signature. The simplest one would be the family of imaginary quadratic fields: in this setting there is Swan's algorithm (which actually computes reduced bases for all ideals of least norms in their class), which takes (I think) at most $m^3$ steps to do $\mathbb{Q}(\sqrt{-m})$ when implemented without more thought. Is there a better algorithm known, or is there a speedy implementation of Swan's algorithm? - -REPLY [2 votes]: Index calculus works, and much faster (although it is randomized); search for "subexponential class group".<|endoftext|> -TITLE: function that is the average of affine transformations of itself -QUESTION [6 upvotes]: Consider the function $f : \mathbb{R} \to [-1,1]$ with -$$ - f(x) = \begin{cases} - -1 & x \le -1 \\ - +1 & x \ge +1 \\ - \frac{f(\frac32 (x-\frac13)) + f(\frac32 (x+\frac13))}{2} & -1 \le x \le +1\,. - \end{cases} -$$ -So $f$ is the average of two affine transformations of itself. -The picture shows the graph of $f$ (in red) and the two affine transformations from the definition (in blue). - -What is known about such functions? Is $f(0.5) \approx 0.694064$ rational? - -REPLY [5 votes]: This function (with a parameter $0 < \lambda < 1$, more generally than $2/3$) arises in Probability as cumulative distribution function $f_\lambda$ of a sum $S:=\sum_{j=0}^\infty \pm \lambda^j$ with random signs, independently and identically chosen with equal probability $(1/2,1/2)$ (the $k$-th iterate considered here by Per Alexandersson corresponds to a partial sum, if I'm not wrong). In this context, a functional equation like the one you wrote is a simple consquence of the independence, partitioning into two cases, according to the sign of the first term. -For $\lambda <1/2$ one finds a function whose variation is concentrated on a Cantor like set -exactly the Cantor ternary set when $\lambda =1/3$. It is also easy to see that $\lambda=1/2$ corresponds to the uniform distribution $f(x) = x$ (due to elementary facts of the binary representation). What happens for $\lambda > 1/2$ is more difficult to understand; the problem is studied since the $30$s starting with Erdős, and found interesting connections in different areas of mathematics. I'm not adding other chats, but the keyword for a search, which is Bernoulli convolutions. -Here is a good survey paper on the subject: Sixty years of Bernoully convolutions, by -Y.Peres W.Schlag, B.Solomyak .<|endoftext|> -TITLE: Does the internal axiom of choice imply Lagrange's theorem? -QUESTION [5 upvotes]: In a topos ${\mathcal{A}}$, given a group object $G$ and a subgroup $H$, the object $H\backslash G$ of right cosets is the coequalizer of two maps $G\times H\rightrightarrows G$, namely the group multiplication and the projection onto $G$. Denote the coequalizing map by $H\backslash -$: -$$G\times H\rightrightarrows G\xrightarrow{H\backslash -}H\backslash G$$ -Say that a map $f:X\to Y$ is a product projection if it is part of a span $Z\leftarrow X\xrightarrow{f} Y$ that writes $X$ as a product of $Y$ and $Z$. One can show that $H\backslash -:G\to H\backslash G$ has a section if and only if it is a product projection. If so, then in fact $G$ is the product of $H$ and $H\backslash G$: -$$G\simeq H\times H\backslash G$$ -In the topos ${\mathcal{FinSet}}$ of finite sets, the external axiom of choice holds, which is to say that every epimorphism has a section. The map $H\backslash -$ is an epimorphism, hence has a section if $G$ is a finite group, hence $G\simeq H\times H\backslash G$. This is (a strong form of) Lagrange's theorem. -Say that a topos ${\mathcal{A}}$ satisfies Lagrange's theorem if, for each group object $G$ and a subgroup $H$, the map $H\backslash -:G\to H\backslash G$ has a section (or equivalently, is a product projection). For example, the topos ${\mathcal{Top}}$ of topological spaces and continuous maps does not satisfy Lagrange's theorem as the Hopf fibration $S^1\to S^3\to S^2$ is a counterexample. -The external axiom of choice implies Lagrange's theorem. Hence the topos ${\mathcal{Set}}$ satisfies Lagrange's theorem. The internal axiom of choice, namely that exponentials $-^X:{\mathcal{A}}\to{\mathcal{A}}$ preserve epimorphisms, is weaker than the external axiom of choice. My question is: - -Does there exist a topos that satisfies the internal axiom of choice but not Lagrange's theorem? - -REPLY [14 votes]: Consider the topos $\mathcal E$ of $\mathbb Z/2$-sets. It satisfies the internal axiom of choice but not the external one. The Klein four-group $(\mathbb Z/2)\times(\mathbb Z/2)$ admits an action of $\mathbb Z/2$, namely interchanging the two factors, which makes it an object of $\mathcal E$. Let $H$ be the diagonal subgroup, $\{(0,0),(1,1)\}$ and note that it is pointwise fixed by the $\mathbb Z/2$-action. The quotient $H\backslash G$ consists of two points, both of which are fixed by the $\mathbb Z/2$-action. Therefore, the product $H\times(H\backslash G)$ consists of four points, each fixed by the $\mathbb Z/2$-action. But $G$ has only two fixed-points, the other two points, namely $(0,1)$ and $(1,0)$, being interchanged by the action.<|endoftext|> -TITLE: Comparison of cycle maps -QUESTION [5 upvotes]: Let $X$ be an algebraic variety over $\bar{\mathbb{Q}}$ of dimension $d$, then there is the l-adic cycle map $\mathrm{cl}_{et}:\mathrm{CH}^i(X)\rightarrow\mathrm{H}^{2i}(X,\mathbb{Q}_\ell(i))$ from the Chow group into the l-adic etale cohomology group, where $\mathbb{Q}_\ell(i)$ refers to the Tate twist. Its definition is as follows: given an irreducible subvariety $Z$ of codimension $i$, there is a pullback map $j^\ast:\mathrm{H}^{2(d-i)}_c(X,\mathrm{Z}/n\mathrm{Z})\rightarrow H^{2(d-i)}_c(Z,\mathrm{Z}/n\mathrm{Z})$; based on the Poincare duality $\mathrm{H}^{2(d-i)}_c\big(X,(\mathrm{Z}/n\mathrm{Z})(d-i)\big)\times \mathrm{H}^{2i}\big(X,(\mathrm{Z}/n\mathrm{Z})(i)\big)\rightarrow \mathrm{H}^{2d}_c(X,(\mathrm{Z}/n\mathrm{Z})(d))\cong \mathrm{Z}/n\mathrm{Z}$, one defines $\mathrm{cl}_{et}(Z)$ to be the class in $\mathrm{H}^{2i}\big(X,\mathrm{Z}/n\mathrm{Z}(i)\big)$ representing the homomorphism $j^\ast(d-i)$ on $\mathrm{H}^{2(d-i)}_c\big(X,\mathrm{Z}/n\mathrm{Z}(d-i)\big)$. By passing to the inverse limit, one arrives at the coefficient sheaf of $\mathbb{Z}_\ell$ and $\mathbb{Q}_\ell$. -When $X$ is nonsingular, there is also the cycle map $\mathrm{cl}_\mathrm{C}:\mathrm{CH}^i(X)\rightarrow\mathrm{H}^{2i}(X(\mathbb{C}),\mathrm{C})$ into the Singular cohomology group. So what is the relation between these two cycle maps? (Is there a reference on this question?) If one embeds $\mathbb{Q}_\ell$ into its topological algebraic closure $\mathbb{C}_\ell\cong \mathbb{C}$, would $\mathrm{cl}_{et}$ and $\mathrm{cl}_\mathrm{C}$ coincide or, if not, how to compare the $\mathrm{Z}$-module structure of the two images? -In Milne's lecture notes on Etale cohomology, he states a comparison theorem (Thm. 21.1) between etale cohomology and the singular cohomology based on the complex topology. According to it, there is a canonical isomorphism $\mathrm{H}^{2i}_{et}(X,\mathrm{Z}/n\mathrm{Z})\cong \mathrm{H}^{2i}(X(\mathbb{C}),\mathrm{Z}/n\mathrm{Z})$. It is with respect to such an isomorphism that one asks about the compatibility of the cycle maps. - -REPLY [6 votes]: Everything you could wish for is true :-). Passing to the inverse limit in Milne's theorem 21.1 one gets an isomorphism -$$ H^i_{et}(X, \mathbf{Z}_\ell) = H^i_{sing}(X(\mathbf{C}), \mathbf{Z}) \otimes \mathbf{Z}_{\ell}$$ -(you don't have to extend scalars to $\mathbf{C}$ or $\mathbf{C}_\ell$ or anything nasty like that). Moreover, the same works if you put $H^i_c$ instead of $H^i$ on both sides, and the isomorphism is compatible with cycle classes. The way to see that is as follows: it's obvious if the cycle is the whole of $X$; and the comparison map is functorial in $X$ and respects Poincare duality, so it follows for an arbitrary cycle. (The same compatibility also works for Chern classes on higher $K$-theory, with the cycle-class case being $K_0$.)<|endoftext|> -TITLE: De Rham isomorphism for noncompact manifolds? -QUESTION [7 upvotes]: Maybe someone has a quick answer. Thanks. -For noncompact manifolds, is the De Rham cohomology isomorphic to the singular cohomology? Is the De Rham cohomology defined with the cochain of compactly supported differential forms isomorphic to the singular cohomology with compact support? -The references I checked on De Rham isomorphism state the theorem only for compact manifolds. - -REPLY [6 votes]: The answer is Yes. You can find a proof (via sheaves theory) for example in Demailly's book - Chapter IV (6.7) for ordinaire cohomology and (7.10) for compactly supported cohomology -http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf<|endoftext|> -TITLE: Is every premodular category the *full* subcategory of a modular category? -QUESTION [5 upvotes]: In Müger's article "Conformal Field Theory and Doplicher-Roberts Reconstruction", he defines the "modular closure" of a braided monoidal category. So every braided monoidal category (and therefore every premodular category, which is a special case, see Bruguières articles) is the subcategory of a modular category, namely of its closure. -Is it known when this inclusion is full? If not, is it possible for an arbitrary premodular category to construct a modular category that contains it as a full subcategory? - -REPLY [3 votes]: I believe the answer to your first question is no: If you take a symmetric tensor category, Müger's construction should give the trivial tensor category. -The answer to your second question is yes. There is always a full inclusion of a braided tensor category $\mathcal{C}$ into its center $\mathcal{Z}(\mathcal{C})$, which is modular if $\mathcal{C}$ is pre-modular.<|endoftext|> -TITLE: Monoids and groups of fractions -QUESTION [6 upvotes]: Let $G$ be a group containing a monoid $M$ that spans $G$ as a group. Is it possible to have a proper quotient $\varphi \colon G \to Q$ of $G$ such that the restriction of $\varphi$ to $M$ is injective? -More specifically I'm interested in the following: if $M$ is an Ore monoid (cancellative and admitting least common right multiples) then it embeds into its group of right fractions $Q$. There is also a universal group $G$ through which any map from $M$ to a group factors (it has presentation $\langle M \mid m \cdot n = (mn) \text{ for }m,n \in M\rangle$). So $Q$ is a quotient of $G$. Can it be a proper quotient? -In other words: $Q$ is by definition universal among the groups $\iota \colon M \to H$ subject to the condition that $H = \iota(M) \cdot \iota(M)^{-1}$. Is it nontheless universal among all groups? - -REPLY [7 votes]: For an Ore monoid the universal group is the group of right fractions. This is proved just as the universal property for localization of commutative rings is proved. It is irrelevant whether $H=\iota(M)\iota(M)^{-1}$, you can simply send a fraction $(m,n)$ to $\iota(m)\iota(n)^{-1}$ and check that this gives a well defined homomorphism. Or you can look in a category theory book for the words "calculus of right fractions" where they will prove a more general universal property for localizing categories. Think of a monoid as a one object category and their condition that the monoid admit a calculus of right fractions is the Ore condition. -In general, the answer is there can be proper quotients. The free group on two generators is generated as a group by the free monoid on two generators. There are lots of groups generated by two elements which generate a free submonoid, like the free metabelian group or the lamplighter group. -Edit: Alternatively you can show that if $M$ satisfies an Ore condition and $\iota\colon M\to H$ is a homomorphism to a group $H$ then $\iota(M)\iota(M)^{-1}$ is a subgroup of $H$ and so the two universal properties are the same. It is basically the same argument as to why you can form the product of 2 fractions.<|endoftext|> -TITLE: The interplay between certain aspects of interpretability, model theory and category theory -QUESTION [14 upvotes]: I have some questions about the interplay of interpretability, model theory and category theory. Since I had difficulties in finding literature or other helpful information about this topic, it would be great if somebody of you can help me. -For a first-order theory $T$, let $Mod(T)$ denote the category of all models of $T$ where the arrows are the homomorphisms (or - if you want so - take the elementary embeddings). Let $I_T$ denote the spectrum of $T$. That is, $I_T$ is a function which takes some cardinal $\kappa$ as argument and which outputs the number of non-isomorphic models of $T$ having cardinality $\kappa$. I will also refer to the syntactical notion of interpretability (i.e. "theory $T_1$ interprets theory $T_2$") and mutual interpretability ("$T_1$ interprets $T_2$ and vice versa"). For what follows, assume that $T_1$ and $T_2$ are arbitrary first-order theories. Here are the three questions for which I would like to know whether there are any answers: - -Are $Mod(T_1)$ and $Mod(T_2)$ equivalent (in the category theoretical sense) whenever $T_1$ and $T_2$ are mutually interpretable? Does the converse holds? -Are $Mod(T_1)$ and $Mod(T_2)$ equivalent (in the category theoretical sense) whenever $I_{T_1}=I_{T_2}$? Does the converse holds? -Does $I_{T_1}=I_{T_2}$ holds whenever $T_1$ and $T_2$ are mutually interpretable? Does the converse holds? - -I will be grateful for any comments, answers and also for references to literature (papers, books,...) in which these questions or similar ones are discussed. Thank you! -Note that I do not expect that these questions are "profound conjectures" or something like that. Maybe that among the experts, their answers are already known and even "trivial". I was just thinking about some model theoretical topics and in this context, the three questions from above started to occupy me. Since I am not able to find any literature concerned with similar questions, I just would like to know whether there are some people in this forum who know more than I do and could share their knowledge with me. -Edit: Following some remarks of some of the commentators of this post, the above questions may also be interesting if we replace "mutual interpretable" with "bi-interpretable". - -REPLY [4 votes]: Regarding the first question: -(a) Theories with equivalent categories of models are not necessarily bi-interpretable. One can come up with simple examples in propositional logic (first-order logic with only zero-place relation symbols). For example, let $T_1$ be the empty theory in the language $\{ p_0,p_1,\dots \}$, and let $T_2$ be the theory with axioms $\{ p_0 \vdash p_i \}$ for all $i$. Both $Mod(T_1)$ and $Mod(T_2)$ are discrete categories with $2^{\aleph _0}$ elements, hence they are equivalent. -(b) Bi-interpetable theories do have equivalent categories of models. This result follows from the fact that an interpretation $F:T_1\to T_2$ gives rise to a functor $F^*:Mod(T_2)\to Mod(T_1)$; and if $T_1$ and $T_2$ are bi-interpretable, then $F$ has a "pseudo-inverse" $G$ which gives rise to a inverse $G^*$ to $F^*$. -The previous paragraph was a bit sketchy. A more elegant argument can be given using categorical logic: the theory $T_i$ gives rise to a syntactic category $C_{T_i}$. If $T_1$ and $T_2$ are bi-interpretable, then the pretopos completion of $C_{T_1}$ is equivalent to the pretopos completion of $C_{T_2}$. It then follows that the category $LF(C_{T_1},Sets)$ of logical functors from $C_{T_1}$ into the category of sets is equivalent to the category $LF(C_{T_2},Sets)$ of logical functors from $C_{T_2}$ into the category of sets. But $LF(C_{T_i},Sets)$ is equivalent to $Mod(T_i)$. [These ideas are all contained in the latter half of Makkai and Reyes, First order categorical logic.] -Note: I am taking "bi-interpretable" as synonymous with "definitional equivalence." Note that mutual interpretability does not imply bi-interpretability. See Andreka et al., "Mutual definability does not imply definitional equivalence, a simple example." http://www.math-inst.hu/pub/algebraic-logic/amn-definote.pdf<|endoftext|> -TITLE: History of Koszul complex -QUESTION [24 upvotes]: This is a question about the history of commutative algebra. I'm curious why the Koszul complex from commutative algebra is called the Koszul complex? All of Koszul's early papers are about Lie algebras and Lie groups, in particular about the Chevalley-Eilenberg complex. He never published papers on commutative algebra. It looks like the Koszul complex (under this name) was first used in early versions of Serre's work on multiplicities around 1950. What was the motivation to associate this complex with Koszul? -It is clear to me that that any linear form $f$ on a module $M$ over a commutative ring $R$ gives a Lie algebra structure on $M$ with bracket $[x,y]=f(x)y-f(y)x$. The Chevalley-Eilenberg complex (with coefficients in the trivial representation) of this Lie algebra is the Koszul complex of $f$. So the Koszul complex is a special case of the Chevalley-Eilenberg complex for one stupid Lie algebra. -Was this observation a bridge between Lie algebra cohomology and commutative algebra that actually gave us the Koszul complex around 1950? -By the way, is there a special name for such Lie algebras associated to a form? - -REPLY [23 votes]: Although the germ of the idea might've appeared in Koszul's earlier work on the cohomology of Lie algebras and homogeneous spaces, it seems that the first full-fledged appearance of the Koszul complex/resolution is in Koszul, Sur un type d'algèbres différentielles en rapport avec la transgression, Colloque de topologie (espaces fibrés), Bruxelles (1950), 73–81. The primary motivation there is topological/geometric (cohomology of fiber bundles), but Koszul does give fairly abstract algebraic results and definitions. Specifically, consider a principal $G$-bundle $p\colon E\to B$, where $G$ is a compact connected Lie group. Write $x_1,\ldots,x_l$ for the primitive generators of $H^\ast(G)$, so that $H^\ast(G)=\bigwedge^l_{i=1} x_i$. Then there are $G$-invariant differential forms $\{\omega_i\}$ on $E$ whose restrictions $\{\xi_i\}$ to a fiber $G$ are bi-invariant forms that represent the classes $\{x_i\}$ and such that $d\xi_i$ is the image under $p^\ast$ of some form $c_i$ on the base $B$. The exterior algebra $\Omega^\ast(B)$ of $B$ may be viewed as a module over the polynomial ring $A=\mathbb R[c_1,\ldots,c_l]$. Koszul is led to the "Koszul complex" -$$ {\textstyle \bigwedge_{i=1}^l} x_i \otimes \Omega^\ast(B) $$ -(with the appropriate differential) through topological considerations: he notes that in certain cases (for nice enough $B$), one can replace $\Omega^\ast(B)$ above with $H^\ast(B)$ and then the resulting complex $\bigwedge x_i \otimes H^\ast(B)= H^\ast(G) \otimes H^\ast(B)$ computes the cohomology of $E$. -Koszul takes a look at the general properties of "Koszul complexes" of the form $E \otimes M$ where $E=\bigwedge_{i=1}^l x_i$ and $M$ is a module over $A=k[x_1,\ldots,x_l]$, and calls the resulting cohomology $H^\ast(M)$ the cohomology of the $A$-module $M$. He proceeds to use this machinery to give a generalization of Hilbert's syzygy theorem. This is, e.g., the context in which the Koszul complex arises in Cartan & Eilenberg's book on homological algebra (see Ch. VIII, sections 4 and 6)---and this is probably (?) the first textbook appearance of the construction. -See also A. Haefliger, Des espaces homogènes à la résolution de Koszul, Ann. Inst. Fourier 37(4) (1987), 5–13, for some interesting historical commentary on Koszul's work.<|endoftext|> -TITLE: traces of sobolev spaces under additional assumptions -QUESTION [5 upvotes]: Let $p\in [1,\infty]$, $\Omega$ an open bounded domain with (smooth, if necessary) boundary $\partial \Omega$. -Is there a subspace $X\subset L^p(\Omega)$ - a simply describable space, ideally a Sobolev or Besov or Triebel-Lisorkin or somebody else's space - such that the trace operator is surjective from $\{f\in X:\Delta f\in L^p(\Omega)\}$ onto $L^p(\partial\Omega)$? -The answer is affirmative if $p=2$: One can take $X=H^\frac{1}{2}(\Omega)$. That the trace operator is surjective from -$$ -\{f\in H^\frac{1}{2}(\Omega):\Delta f\in L^2(\Omega)\}\quad \hbox{onto}\quad L^2(\partial\Omega) -$$ -has been checked in [DOI 10.1007/s00020-002-1163-2], Lemma 3.1: The idea is to use a certain isomorphism that allows to construct a harmonic $H^\frac{1}{2}(\Omega)$-function with given initial data, but the proof is essentially an application of Lions-Magenes theory, which is restricted to the Hilbert case. (For me, the obvious place to look for a Banach space extension is the book of Grisvard, but I could not find anything there, either). -EDIT: My conjecture is of course that, more generally, $X=W^{\frac{1}{p},p}(\Omega)$ can be taken. This would yield the space $W^{\frac{1}{p},p}(\Omega)\cap D(\Delta;L^p(\Omega))$ with the notation of Grisvard. Such spaces do not seem to have been introduced in the books of Adams-Fourier or Demengel-Demengel. -EDIT #2: Perhaps it is convenient to explain my interest in this question. The point is that I would like to define not only a right inverse of the trace operator, as customary; but rather an operator that maps a given function on the boundary into the (unique) solution of an eigenvalue equation for $\Delta$ whose boundary values are prescribed by the given function - something that was considered e.g. in many articles of the Japanese school active on Dirichlet forms in the 1960s (Fukushima, Sato etc.). -Now, I am willing to consider very weak notions of solution, but at the very least I want to make sure that both sides of the the eigenvalue equation are in $L^p$. - -REPLY [3 votes]: Partial answer: according to Triebel (Theory of function spaces, 1983, Remark 2.7.5, p. 139), the trace of the Besov space $B^{1/p, p}_1 (\Omega)$ is $L^p (\partial \Omega)$, but the linear extension operator from $L^p (\partial \Omega)$ to $B^{1/p, p}_1 (\Omega)$ cannot be linear. In particular, the harmonic extension fails to give such an extension. -Edit 1: The proof can be found in the paper V.I. Burenkov, M.L. Gold'man, On extension of $L^p$ functions, Proceeding of the Steklov Institute of Mathematics, 1981, p. 33-54 (translated from a 1979 paper in Russian).<|endoftext|> -TITLE: Infinitely many $N$ such that $\langle p\rangle=\langle q\rangle$ mod $N$ -QUESTION [16 upvotes]: Suppose that $p,q>1$ are two relatively prime integers. Are there infinitely many positive integers $N$ such that - -$N$ is relatively prime to $p$ and $q$; -there exists positive integers $k,l$ such that $p^k\equiv q\mod N$ and $q^l\equiv p \mod N$? - -REPLY [8 votes]: @GH from MO. Yes, we can find infinitely many primes $N$ without Dirichlet as well though the argument is a bit too long to be posted as a comment. -Let $A$ be a large number. Let $U$ be a finite set of primes that can divide $p^k-q$ in principle. Note that if we want to have $p^k\equiv p^\ell\mod u^m$ with $u\in U$, then, by the lifting the exponent lemma, we need to ensure that $v_u(k-\ell)\ge m-C(u,p)$. Let $S$ be the set of primes between $A$ and $2A$. Now take $k\in S$ and construct $N$ as above. Recall that $N\equiv p-q\mod k$. If we are in trouble, we must have $N$ consisting of primes in $U$ only. Since $N\ge k+p-q$, we must have $u^{v_u(N)}\ge A^{\delta(U)}$ for some $u\in U$. Let $\ell$ be the next prime in $S$ corresponding to the same $u\in U$ with this property. Then $v=v_u(\ell-k)$ also satisfies $u^v\ge c(U)A^{\delta(U)}$ and thereby $\ell-k\ge c(U)A^{\delta(U)}$. This results in $|S|\le C(U)A^{1-\delta(U)}$, which (if true for all sufficiently large $A=2^k$) is bad enough to contradict something as simple as Euler's theorem on the divergence of inverse primes.<|endoftext|> -TITLE: Countable Product of Class Forcing Notions -QUESTION [7 upvotes]: Is the following consistent? -There are definable class forcing notions $\lbrace \mathbb{P}_{n}\rbrace_{n\in \omega}$ such that: - -The product of any finitly many of them preserves $\text{ZFC}$ and all cardinals. -The $\prod_{n\in \omega}\mathbb{P}_{n}$ collapses all uncontable cardinals to $\omega$. - -REPLY [10 votes]: Yes, this can happen. -Start with $V=L$ (but $V=HOD+GCH$ suffices), and assume there are no inaccessible cardinals. For each regular cardinal $\delta$, partition the ordinals of cofinality $\delta$ below $\delta^+$ into $\omega$ many disjoint stationary sets $\text{Cof}_\delta\cap\delta^+=\bigsqcup_n S^\delta_n$. Let $\mathbb{P}_n$ be the Easton support product that at stage $\delta$ shoots a club disjoint from $S^\delta_n$, destroying its stationarity, using the club-shooting forcing of $L$, which consists of closed bounded subsets disjoint from that set, ordered by end-extension. At each stage, this is $\lt\delta$-closed and $\leq\delta$-distributive, a result due to Baumgartner, Harrington and Kleinberg (the case of $\delta^+=\omega_1$ is explained in lemma 17 of this paper, but the general argument is similar). -Thus, the forcing is gradually killing off components of the stationary partition. The usual Easton arguments, combined with the analysis of club-shooting forcing, show that each $\mathbb{P}_n$ preserves ZFC and all cardinals and cofinalities, and this is also true of any finite product of the $\mathbb{P}_n$'s. -But the combined forcing of all $\mathbb{P}_n$ will kill off the entire partition, which collapses the cardinal. In particular, the full product will collapse every successor cardinal $\delta^+$, since the intersection of all the clubs at any given stage must be empty. It follows that every cardinal will be collapsed to $\omega$.<|endoftext|> -TITLE: What is the translation in Fourier transform for a function to have exp. decay at $x\to -\infty$ -QUESTION [6 upvotes]: It is known that smooth functions with exponential decay at $\pm\infty$ are functions whose Fourier transform have analytic continuation in some suited complex strip. I was wondering what happens if we ask for exponential decay from only one side. More precisely : -The context : In Fourier Analysis, Self-Adjointness by Reed and Simon (Methods of Modern Mathematical Physics, Vol. 2), Theorem IX.14 tells us that (I'll take dimension 1 for simplicity) : - -if $T$ is a tempered distribution on - $\mathbb R$ such that : - -$\hat T$ has an analytic continuation to $|\Im z| < a$ for - some $a > 0$ -on each slice $\mathbb R + i\eta$ with $|\eta| < a$, $\hat T$ is - integrable -for each $0 < b < a$, the supremum of the integrals on the slices - $\mathbb R + i\eta$ with $|\eta| < b$ - is finite - -then $T$ is a bounded continuous - function and for any $0 < b < a$, there - exists $C_b \geq 0$ s.t. $$|T(x)| \leq C_b e^{-b|x|}$$ - -This result can be "desymmetrised" easily : by assuming analytic continuation in $-a_- < |\Im z| < a_+$ with $a_-,a_+ > 0$ we get decay faster than $e^{-a_+x}$ as $x\to -\infty$ and than $e^{a_-x}$ as $x\to+\infty$. -In http://arxiv.org/abs/math/0007097, authors mention a similar result which is an equivalence in $\mathcal S(\mathbb R)$ morally saying - -Functions with such kind of decay are functions whose Fourier transform are of rapid decrease and admit a holomorphic continuation in the associated strip such that the continuation on each slice of the strip is of rapid decrease. - - -The question : -I was wondering if anyone would knew what happens if we only ask for exp. decay at $-\infty$ : for instance for a $L^\infty$ function, is there some reasonable equivalent condition on the Fourier transform (in the sense of tempered distributions) for it to have $e^{ax}$ decay as $x\to-\infty$ ? -Not assuming exp. decay at $+\infty$ makes that such functions can have "ugly" Fourier transforms (since they're not in $L^1$ or $L^2$ in general), for instance $e^x H(-x) + H(x)$ where $H$ denotes the Heaviside step function yields $\delta_0 + i \text{pv}\frac{1}{\xi} - \frac{i}{\xi-i}$. Nonetheless, the "functional part" (in some kind of bad and undefined sense...) of this transform indeed has holomorphic continuation in the half-plane $\Im z < 1$ (but is not $L^1$ on any slice). This is the kind of observations that decided me to ask this question here :) - -REPLY [7 votes]: Let me summarize what has been already said. If $f$ is a locally integrable function, you can break it into two parts: $f=f_1+f_2$, where $f_1$ is supported on the positive ray, and -$f_2$ on the negative ray. Then you consider two "halves" of the Fourier transform: -$$F^-(z)=\int_0^\infty e^{-izt}f_1(t)dt$$ -and -$$F^+(z)=\int_{-\infty}^0 e^{-itz}f_2(t)dt.$$ -The first function is analytic in the lower half-plane, and the second is analytic in -the upper half-plane. This pair of analytic functions is called "generalized Fourier transform in the sense of Carleman", or "Fourier transform in the sense of hyperfunctions". If "usual" Fourier transform exists in some sense, -like when $f\in L^2$ or $f\in L^1$, or $f$ is a temperate distribution, then it -is recovered as the sum of boundary values of these two analytic functions. Boundary values -must be understood in the appropriate sense, dependng on your conditions on $f$. -Now if $f$ decreases exponentially, say on the negative ray, this means that $F^+$ -has an analytic continuation from the upper half-plane to a larger half-plane $\Im z>-a$. -That is Fourier transform on the real line is a sum of of a function analytic in the -lower half-plane and another one analytic in the larger upper half-plane $\Im z>-a$. -This is the characterization you asked for. -In general, a temperate distribution on the real line has a canonical representation -as sum of the boundary values of two analytic functions, one in the upper half-plane, another in the lower half-plane. So this property (analytic continuation of one of -these functions to a larger half-plane) is an intrinsic property which exactly characterizes the exponential decrease of the original on one of the semiaxes. -References: - -Carleman, L'Intégrale de Fourier et Questions que s'y Rattachent, Publications Scientifiques de l'Institut Mittag-Leffler, 1. Uppsala, 1944. -A. Kaneko, Introduction to hyperfunctions, Kluwer, Dordrecht, Tokyo, 1988.<|endoftext|> -TITLE: Which dense inclusions of sites are ∞-dense? -QUESTION [8 upvotes]: An inclusion of sites f: D→C is dense if it induces an equivalence between the categories of sheaves on C and D. -Likewise, f is ∞-dense is it induces an equivalence between the ∞-categories of ∞-sheaves on C and D. -An ∞-dense inclusion is dense but the opposite is false in general, see Counterexample 6.5.4.2 in Lurie's Higher Topos Theory. -A nontrivial example of an ∞-dense inclusion of sites is given by the inclusion Cart→Man -of cartesian spaces into all smooth manifolds, as explained in -Is the site of (smooth) manifolds hypercomplete? -As it turns out, in the 1-categorical case we have an ample supply of dense inclusions of sites, -as explained by Proposition C2.2.16 of Johnstone's Sketches of an Elephant: -essentially small sites of definition of a given Grothendieck topos E that are subcanonical -can be identified with full subcategories of E whose objects form a separating family, -with the Grothedieck topology induced from the canonical topology on E. -Every inclusion of such sites is dense, but not necessarily ∞-dense. -I wonder if there is a (practical) criterion that allows us to check whether a dense inclusion is also ∞-dense. -We can assume the source site D to be hypercomplete, if necessary. -The example that I have in mind takes D to be the site of smooth manifolds (or cartesian spaces) -and C a subcategory of the category of sheaves of sets on smooth manifolds -that contains some infinite-dimensional smooth manifolds (e.g., smooth mapping spaces Map(M,N) between finite-dimensional manifolds). -What additional conditions are needed to ensure that a dense inclusion of sites is ∞-dense? - -REPLY [2 votes]: If you content yourself with the hypercomplete case, a sufficient practical criterion is given in my paper joint with Tony Yue Yu, that you can find here: http://arxiv.org/abs/1412.5166. The result we are concerned with is lemma 2.35. -It is the analogue of the well known result in algebraic geometry, that can be found here: http://stacks.math.columbia.edu/tag/03A0.<|endoftext|> -TITLE: Beginner's questions on the post-doc application -QUESTION [30 upvotes]: I am going on the (research oriented) post-doc job market (mainly in the US) this winter and would like to ask some questions. I am aware of the site http://academia.stackexchange.com, but I am writing here because want more math-specific answers. -Two beginner's questions that were not previously asked in MO are: - -Can I send more than what employer institutions ask for? For example, if your teaching evaluation of your previous class looks great, can I include it even if they won't require? Does extra reference letters hurt my application? A reason to ask this is that I am afraid of my application to be overlooked among hundreds of applications. I want to make a difference, but shouldn't I take any risk? -What is the difference between "publication list" and "publication list section in CV"? Should the publication list be more than a mere list? Include abstracts? A reason to ask this is that I have only a few preprints and the list looks a bit miserable. - -These days one post-doc position often receives a very large number applications, and it is not likely that the hiring committee read all the applications. What filter is there in the first round? Names of your school, advisor and letter writers? -Thank you very much for your advice. - -REPLY [32 votes]: I've served on the postdoc hiring committee at Rice for several years and am chair of it this year. My answers are geared to research-oriented postdoc applications. -For question (1), it is fine to include extra letters of recommendation, but don't go nuts. We ask for 3 including a teaching letter, but most of our applicants include 3-4 research letters plus a teaching letter. A letter from someone that doesn't know you very well is damaging to your application (so don't sacrifice quality for quantity), and if you get more than 4 research letters and 1 teaching letter then it starts to look weird. -However, don't include things like teaching evaluations we don't ask for. We don't read them and make an effort not to let them influence our decisions, but it does look a little weird. -For question (2), just include a list of publications excerpted from your cv (to emphasize, this list should also be in your cv). To be honest, I'm not really sure why we ask for a separate list of publications and we don't really look at it; probably it is there because of some decision made by a hiring committee long ago, and inertia is a powerful force. -As far as your unnumbered question, a member of our committee does actually read every application. The first read (to form a long list of people who are read more carefully) is quick, but it does happen. The most important things we look at in the first round are research area (to make sure it is at least vaguely related to someone in the department) and the research letters (both who they are from and what they say). - -REPLY [11 votes]: In addition to other useful remarks: -The first filter is your "pedigree", which is PhD school and/or post-doc institution, advisor, and letter-writers. -The length of your research (and teaching) statement(s) can be whatever you want... however, in those statements you are "selling yourself" in the face of competition from many other roughly comparable applicants. The point is that a boring or tedious or too-technical research statement will "put off" hiring-committee people in other fields, so they won't read to the end, or even past the first page. Thus, the research statement has to "sell" you both to specialists and to non-specialists, since, after all, there's probably just one or two people on the hiring committee who could appreciate the nuances of your work. Even though it typically happens only marginally, thinking in terms of "broader appeal" can be very helpful. Thus, in effect, the _first_page_ of your research statement should have optimized visual impact, and be an artful combination of info-for-experts, and of overview-and-broader-significance for non-experts. -On similar principles, there should be an immediately-visible version of publication list that lets both specialists and disinterested-non-specialists see "numbers" and "journals" at a glance. Including abstracts and further information is fine, on subsequent pages, but bear in mind that many people will not look past the first page, so that's your chance to make an impact. -Also, grants, fellowships, honors have an impact on "disinterested non-specialists", since these are intelligible. Get that on the first page of CV. -And, disagreeing with some other remarks, and agreeing with some: avoid lukewarm letters at all costs. Hiring committees do not "average" the letters. Rather, even if you have two glowing letters, a single negative or even ambivalent letter can be fatal. The reason is that, in effect, there are so many applicants that there is some incentive to find a reason to eliminate as many candidates as possible, before looking at the pluses of the remaining ones. Thus, a "blemish" can be a target for hiring committee members in other fields, who might be happier to hire someone with interests closer to theirs anyway. - -REPLY [6 votes]: Let me start with (2) since fewer people have addressed this already. I certainly have wondered about it the few times I've been on the job market. However, I've also seen many applications at my institution. I think most people: -Include the publication list in the CV and in the separate document. -I don't think it is necessary to include abstracts in the publication list (some people do, most people don't). The main part of this data presumably should also be in your research statement. -For (1), having extra letters of reference do not hurt. Especially for tenure track jobs at big schools (like Penn State) having the minimum is actually very unusual. -On the other hand, it seems unusual to include things that are not asked for.<|endoftext|> -TITLE: Can motivic E_∞-ring spectra be strictified to commutative motivic symmetric ring spectra? -QUESTION [5 upvotes]: Theorem 4.5.4.7 (4.4.4.7 in the old version) in Lurie's Higher Algebra (or Theorem 4.3.22 in DAG III) states (roughly speaking) that under certain conditions -the ∞-category of commutative ∞-monoids in a given symmetric monoidal ∞-category C -is equivalent to the underyling ∞-category of the model category of strictly commutative monoids -in a symmetric monoidal model category that presents C. -In other words, E_∞-monoids can be strictified to strictly commutative monoids, provided -that the relevant symmetric monoidal model category satisfies certain (rather strong) conditions, -like being freely powered. -In Example 4.3.25 in DAG III Lurie verifies the conditions of the above theorem -for the symmetric monoidal model category of symmetric simplicial spectra equipped with the S-model structure. -In particular, he obtains that E_∞-ring spectra can be strictified to (strictly) commutative symmetric ring spectra. -I wonder if a similar result is true for motivic symmetric spectra. -Some of the arguments in Lurie's writeup and references therein seem to be specific to simplicial sets, -so it's not completely obvious whether all arguments carry through to the motivic case. -More generally, the same question can be asked for symmetric spectra -in an arbitrary model category M. -Obviously, some additional assumptions must be imposed on M, -so I wonder what the full list of conditions might be. -Can motivic E_∞-ring spectra be strictified to commutative motivic symmetric ring spectra? -References on this matter will be appreciated. - -REPLY [3 votes]: This is very related to some work in progress of mine, which began as part of a larger project of Aaron Mazel-Gee and Markus Spitzweck. Details can be found in the long version of my research statement, section 5.1. If you want more details, please email me separately, as there are still things I want to work out before making a draft public. -I am much more accustomed to model categories so let me phrase my answer in that context. Basically, one way to get a model structure for stable motivic homotopy theory is to use Mark Hovey's stabilization machine from Spectra and Symmetric Spectra in General Model Categories. This gives you a model structure on motivic symmetric spectra which generalizes the usual one on symmetric spectra. Classically, in order to get the strictification you desire (which operad people might call rectification), you would need to pass to the positive stable model structure on symmetric spectra, introduced in Brooke Shipley's A Convenient Model Category for Commutative Ring Spectra. The basic point of this model structure is to force the unit to no longer be cofibrant, removing Gaunce Lewis's obstacle to having a good model category spectra. In this model category, commutative monoids do inherit a model structure (with weak equivalences and fibrations maps which are such as maps of symmetric spectra). Furthermore, a cofibrant commutative monoid must be cofibrant as a symmetric spectrum (this is what "convenient" means). Finally, rectification holds as shown for example in Theorem 1.4 in Elmendorf-Mandell Rings, Modules, and Algebras in Infinite Loop Space Theory. -The theorem in my research statement shows that Hovey's machine can be tweaked to output a positive stable model structure, at least when the input model category is combinatorial. In particular, we checked that this applies to motivic symmetric spectra. To apply the Elmendorf-Mandell result it seems you must also assume the model category is simplicial, i.e. satisfies the SM7 axiom. Thankfully, motivic symmetric spectra does satisfy SM7. Determining in what generality these rectification results hold is still future work for me, so I don't want to say anything too definitively on this yet. For sure both $E_\infty$ ring spectra and strict commutative ring spectra inherit model structures, and it seems highly likely they are Quillen equivalent, so you have a good hope for rectification coming from these considerations. -EDIT: It has been pointed out to me that this answer was slightly lacking in a couple of places. For one thing, the idea of the positive model structure originally goes back to Jeff Smith, so I should have included his name above. Secondly, the real feature of importance in the positive model structure is that for any cofibrant spectrum $X$, the natural map $(E\Sigma_n)_+ \wedge_{\Sigma_n} X^{\wedge n} \to X^{\wedge n}/\Sigma_n$ is a weak equivalence, i.e. the map from the the homotopy colimit (which is the extended powers) to the colimit (symmetric powers). I believe I can prove this for motivic symmetric spectra, so the rest of Shipley's paper should go through (I'm still writing this up, though, so for now you should treat this as hearsay). If you're interested in learning more about this property, I asked a couple of questions about it on MO and got some very informative answers from Peter May. -I should also mention that classically if you instead work with orthogonal spectra rather than symmetric spectra then you also need to pass to a positive variant in order to get the property above, and hence rectification. See Mandell-May-Schwede-Shipley Model categories of diagram spectra. This has also been done for equivariant orthogonal spectra in a paper of Mandell-May. I don't know anything about orthogonal motivic spectra and I don't know how that would be done, because motivic spaces are built on simplicial sets rather than topological spaces, and I don't know how to represent $O(n)$ in that setting. Classically, if you use S-modules then rectification comes for free because commutative spectra and $E_\infty$ spectra are the same thing (see EKMM, the point is that the monoidal product already builds in higher homotopies). Po Hu has developed motivic S-modules but does not seem to mention commutative monoids or $E_\infty$ at all in her paper. I don't know that side of the story well enough to know if it still comes for free or not.<|endoftext|> -TITLE: Does an ultrapower of an Aronszajn tree have an $\omega_{1}$-branch? -QUESTION [14 upvotes]: Throughout this question, I shall let $A^{\mathcal{U}}$ denote the ultrapower of a structure $A$ by an ultrafilter $\mathcal{U}$. Suppose that $T$ is an Aronszajn tree and $\mathcal{U}$ is an ultrafilter on a countable set $I$. Let $T_{\alpha}$ denote the $\alpha$-th level of a tree $T$. Let $T^{(\mathcal{U})}$ denote the tree where the $\alpha$-th level of $T^{(\mathcal{U})}$ is the ultrapower $(T_{\alpha})^{\mathcal{U}}$ and where if $(x_{i})_{i\in I}/\mathcal{U}\in(T_{\alpha})^{\mathcal{U}},(y_{i})_{i\in I}/\mathcal{U}\in (T_{\beta})^{\mathcal{U}}$, then $(x_{i})_{i\in I}/\mathcal{U}<(y_{i})_{i\in I}/\mathcal{U}$ iff $\{i\in I|x_{i} -TITLE: Bi-orderability of Baumslag-Solitar group $\langle a,b \mid a^{-1} b^m a = b^n\rangle$ and of $\langle a,b \mid a^{-1} b a^m = b^n\rangle$ -QUESTION [12 upvotes]: We say that a group $(A, \cdot)$ is bi-orderable if there exists a total order $\preceq$ on $A$ such that $xz \prec yz$ and $zx \prec zy$ for all $x,y,z \in A$ with $x \prec y$. -Let $m,n$ be non-zero integers, and let ${\rm BS}(m,n)$ denote the Baumslag-Solitar group $\langle a, b \mid a^{-1} b^m a = b^n\rangle$. I was happily surprised to learn from Yves Cornulier (see here) that ${\rm BS}(1,n)$ is bi-orderable for $n \ge 1$, which naturally leads to the following: - - -Q1. Is there a ``nice'' characterization of those pairs $(m,n)$ such that ${\rm BS}(m,n)$ is bi-orderable? - - -E.g., ${\rm BS}(m,n)$ is not bi-orderable if $mn < 0$ (wlog, say that $1 \prec b$. Then, $b^{-1} \prec 1$, with the result that $ 1 \prec a^{-1} b^m a$ and $b^n \prec 1$ for $m > 0$, and dually $a^{-1} b^m a \prec 1$ and $1 \prec b^n$ otherwise) or $m = n \ge 2$ (wlog, say that $ab \prec ba$. Then, $a^n b \prec a^{n-1} b a \prec \cdots \prec aba^{n-1} \prec ba^n$). But what about the other cases? -Update on Q1. The question has been completely answered: ${\rm BS}(m,n)$ is bi-orderable if and only if $mn > 0$ and $\min(|m|,|n|) = 1$. For details, see Yves' answer below. -On a related note, let ${\rm D}(m,n)$ be the two-generator one-relator group $\langle a, b\mid a^{-1} b a^m = b^n\rangle$. This is sort of a variant of ${\rm BS}(m,n)$, and, while different in many ways, I'd like to know if both have a similar behavior as for orderability (by the way, does ${\rm D}(m,n)$ have a conventional name, so that I can look for references by myself?). In particular: - - -Q2. Is there a ``nice'' characterization of those pairs $(m,n)$ such that ${\rm D}(m,n)$ is bi-orderable? - - -In fact, ${\rm BS}(1,n) = {\rm D}(1,n)$, so the above (and, especially, Yves' argument) proves that ${\rm D}(1,n)$ is bi-orderable if and only if $n \ge 1$. But what about the other cases? E.g., is ${\rm D}(n,n)$ bi-orderable for $n \ge 2$? -Here is a question similar to Q1 (with total orders replaced by partial ones). - -REPLY [10 votes]: (After the discussion in the comments.) $BS(m,n)$ is bi-orderable iff $mn>0$ and $\min(|m|,|n|)=1$. -Since it was already mentioned that $mn<0$ implies $BS(m,n)$ not bi-orderable and that $BS(1,n)$ is bi-orderable for all $n\ge 2$, all remains is to check that $2\le m\le n$ implies $BS(m,n)=\langle t,x\mid tx^mt^{-1}=x^n\rangle$ is not bi-orderable. -Define $y=txt^{-1}$. Then $y^m=x^n$. So $x$ commutes with $y^m$. So (*) $x$ commutes with $y$. But this is not the case. -(*) in a biorderable group, as mentioned by Salvo, $[X,Y^m]=1$, $m\ge 1$ implies $[X,Y]=1$, because otherwise if $XY -TITLE: Action of automorphisms of a $K3$ surface on its $(-2)$-curves -QUESTION [5 upvotes]: Consider a complex $K3$ surface $X$ and take its group of automorphisms $Aut(X)$. It is a known fact that the action of $Aut(X)$ on the set of rational $-2$ curves of $X$ has only finite number of orbits. -Questions. What kind of ideas one has to use to prove this fact? Is there some nice exposition? Does this fact follows somehow from global Torelli theorem for $K3$'s? - -REPLY [13 votes]: The group of symplectomorphisms $Aut(X)$ of a K3 is the group $O(\Lambda)$ of -automorphisms of its period lattice $\Lambda=H^{1,1}(M,{\Bbb Z})$. For each -(-2)-cohomology class $\eta\in H^{1,1}(M,{\Bbb Z})$, either $\eta$ or $-\eta$ is -represented by a curve (this follows from the Riemann-Roch formula). This curve -is unique, because its self-intersection is negative. Therefore, finiteness of the -number of orbits of $Aut(X)$ on curves is equivalent to the finiteness of the number of orbits of $O(\Lambda)$ on the set of (-2)-classes in $\Lambda$. This result follows from an answer to a question that I asked on Mathoverflow: -orbits of automorphism group for indefinite lattices -(automorphism group $O(\lambda)$ of any lattice $(\Lambda, q)$acts on the set $R_C$ of classes -$\eta$ with $q(\eta, \eta)=C$ with finitely many orbits).<|endoftext|> -TITLE: Computing van Kampen diagrams -QUESTION [15 upvotes]: If G is a finitely presented group (with generating set X) and w is a word over X such that -w=1 in G, then the latter can be witnessed by a so called van Kampen diagram for w, which is -a planar diagram where for each region the boundary cycle is labelled with a group relator, -and the boundary of the whole diagram is labelled with the word w. -Is there a finitely presented group G (with generating set X) such that the following hold? - -G has polynomial Dehn function -G has a polynomial time word problem -There is no polynomial time algorithm for the following problem: -INPUT: A word w over X such that w=1 in G -OUTPUT: A van Kampen diagram for w - -In other words: We can efficiently check whether w=1 in G but we cannot compute efficiently a witness for this fact (but we know that a small witness exists). -If the answer to the above question is positive, one certainly needs some complexity theoretic assumption. A reasonable starting point might be to assume that a nondeterministic -polynomial time Turing machine M exists such that: - -One can check in deterministic polynomial time whether an input w is accepted by M. -There is no polynomial time Turing machine that computes for a given word w (that is -accepted by M) an accepting computation of M for input w. - -REPLY [6 votes]: This is equivalent to $P\ne NP$ by the result of the paper : Birget, J.-C.; Olʹshanskii, A. Yu.; Rips, E.; Sapir, M. V. Isoperimetric functions of groups and computational complexity of the word problem. Ann. of Math. (2) 156 (2002), no. 2, 467–518. -Update See the discussion here.<|endoftext|> -TITLE: A problem in the domino shuffling algorithm -QUESTION [8 upvotes]: The domino shuffling algorithm first appeared in the following paper by Propp and Kuperberg: -Alternating-sign matrices and domino tilings -They used this algorithm to give a fourth proof that the number of domino tilings of the $n$th aztec diamond is $2^{\frac{n(n+1)}{2}}$. -Now the domino shuffling has been a powerful tool in the combinatorics of domino tilings. The algorithm is quite simple, but its correctness is very subtle to illustrate. -Consider the infinite chessboard on the lattice $\mathbb{Z}^2$, color the cells black or white such that adjacent cells get different colors. Then any $1\times2$ domino (if placed on the chessboard) will cover exactly one white cell and one black cell. -Now suppose some dominos (the number of the dominos may be infinite) are placed on the chessboard with out overlapping with each other, they cover the whole chessboard partially, so we call it a partial tiling $T$. -A $2\times2$ square is called odd block with respect to $T$ (following Propp's convention), if it contains exactly two parralle dominos of $T$, and has a black cell in its upper left-hand corner. Any $2\times2$ square with a black cell in its upper left-hand will be called a block. -The partial tiling $T$ is called odd-deficient, if it has no odd blocks, and it's free region (cells not covered by $T$) can be tiled with disjoint blocks. -The domino shuffling algorothm states that, given any odd-deficient partial tiling $T$, one can produce a new partial tiling $S(T)$ which is also odd-deficient, and the mapping $T\to S(T)$ is an involution. ($S(S(T))=T$) -The algorithm goes as follows: for every domino $A$ in $T$, we find the unique block $B$ contains $A$, then we move $A$ to the oppsite position in $B$. -It's easy to see that $S(T)$ would not contain any odd block, because odd blocks are unchanged under the shuffling procedure, and since $T$ contains no odd blocks, $S(T)$ would not either. -But the crucial point is that the free region of $S(T)$ can also be tiled by disjoint blocks. This is a very subtle problem in the algorithm, the original proof in Propp's paper did not explain much in this direction, which puzzled me for quite a long time. -In Aigner's book "A course in Enumeration", he used a 2-coloring method to handle this problem,but I think his wany is still too involved. -My question is: is there any easy way to deduce that $S(T)$ is also odd-deficient? - -REPLY [5 votes]: Propp redoes, clarifies and generalizes domino shuffling in the following lovely paper: -Propp, James. "Generalized domino-shuffling." Theoretical Computer Science 303, no. 2 (2003): 267-301. http://arxiv.org/abs/math/0111034 -The redone explanation is a way of thinking of the domino shuffle map locally, using a graph transformation called urban renewal; these days people also call it the spider move. The move replaces a 4-cycle in a graph (the "square", we might call it) with an 8-vertex graph made of a 4-cycle with a pendant edge on each vertex (the "spider"); the feet of the spider get attached where the vertices of the square were. -Spider moves, when combined with a suitable reweighting of the edges in the graph, preserve the generating function for perfect matchings up to an explicit constant. The argument is much like the one in Elkies-Kuperberg-Larsen-Propp which you're asking about, but it is substantially streamlined because you only have to think about the square and the spider. -In answer to your question: once you understand this point of view, odd-deficiency is obvious. When you reconstruct the domino shuffle using spider moves (which Propp explains how to do in the same paper), the squares and the spiders comprise the odd blocks, and all the action happens within them. The even blocks do not enter into the picture at all. -As further evidence that the spider move is the correct point of view, Kenyon-Goncharov http://arxiv.org/abs/1107.5588 recently used the spider move to define a certain cluster integrable system for the dimer model.<|endoftext|> -TITLE: Are compact simple groups homotopically non-abelian? -QUESTION [9 upvotes]: Take a compact connected simple centreless Lie group $G$. - Can the commutator map $G\times G\to G$ sending $(x,y)$ to $[x,y]$ be homotopic to a constant map? - -I am interested mostly in the case, where $G={\rm PSU}(n)$. - -As far as I understand, the commutator map is homologically trivial (right?). -There is a related question. - -REPLY [7 votes]: The following is the main theorem in -Araki, S.; James, I. M.; Thomas, E., "Homotopy-abelian Lie groups", Bull. Amer. Math. Soc. 1960. - -Theorem: A compact connected Lie group is homotopy-abelian only if it is abelian.<|endoftext|> -TITLE: Heegaard genus of the hyperbolic dodecahedral space (is it 3 or 4?) -QUESTION [9 upvotes]: I have a question on the hyperbolic dodecahedral space, first described by C.Weber and H. Seifert in 1933 [Die beiden Dodekaederr\"aume, Math Z. 37 (1933), 237-253]. Is it known whether it admits a Heegaard splitting of genus 3? -We can see (even with elementary combinatorial tools, like discrete Morse theory) that it does admit a Heegaard splitting of genus 4; but I was wondering if this is best possible. - -REPLY [2 votes]: The Whitehead link fibers with a twice-punctured genus 1 fiber. Since the Seifert-Webber Space is a cyclic 5-fold covering of the Whitehead link, the image of Whitehead link in this cover should again be a fibered link with a twice-punctured genus 1 fiber. So this will give a genus 3 Heegaard splitting. -I reckon this argument should work to show each of the two isometry classes of 5-fold cyclic covers (as mentioned in the comments to this question) have genus 3 Heegaard splittings. -Edit: Of course Agol's comment is right. I wasn't considering the other possible 5-fold cyclic branched covers. -So if I've got my head on straight about it, these other covers should be dual to a thrice-punctured genus 1 fiber of the link exterior. (Such a fiber is obtained from adding the twice-punctured torus Seifert surface with a suitably oriented once punctured torus bounded by one component and disjoint from the other.) Then in the cover, as well as in S^3, the two boundary components of the fiber on the same link component join together to make a non-orientable surface with $\chi = -3$ and one boundary component. The boundary of a neighborhood of this surface then gives a genus 4 Heegaard splitting. Presumably, this splitting is equivalent to the ones Bruno obtained.<|endoftext|> -TITLE: Bruhat decomposition for reductive groups in characteristic zero? -QUESTION [5 upvotes]: Let $G$ be a reductive, linear algebraic group (variety) over an algebraically closed field $\Bbbk$ of characteristic zero. If $G$ is connected, I know from Humphrey's book that for any Borel subgroup $B\subseteq G,$ there is some opposite Borel subgroup $B^-$ such that $T=B\cap B^-$ is a Torus, and denoting by $U$ and $U^-$ the unipotent radicals of $B$ and $B^-$ respectively, there is an open immersion -\begin{align*} -\beta: U\times T\times U^- &\longrightarrow G \\ -(u,t,v) &\longmapsto utv -\end{align*} -I am now in a situation where my group is not connected, and I am wondering if the above still holds (to some degree). Searching MO, I found this post which suggests that the above holds for "split reductive" groups. I am not familliar with the term "split reductive", and searching for a definition has yielded very little so far, but wikipedia seems to suggest that this is some issue of separability which I would not have to worry about over an algebraically closed field of characteristic zero. -So, my question is: If $G$ is not connected, can something similar or even the same be said? A reference would be the icing on the cake, I would love to read more about this. Thanks a lot in advance. - -REPLY [5 votes]: Maybe I can clarify some of the issues here. First, the distinction between "reductive" and "semisimple" is minor, since the difference between these lies in the maximal torus and doesn't really affect the Bruhat decomposition or the big cell $U^- T U$. But most books and survey articles do assume that $G$ is connected, mainly to avoid trivialities which complicate the notation. So it's hard to find a formulation of the Bruhat decomposition or the big open cell without the connecteness assumption. -Fortunately the issue is minor, since a disconnected reductive (or semisimple) group is a finite union of translates of the identity component $G^\circ$ which are geometrically the same. The most common example of a disconnected semisimple group is the full orthogonal group, in which the special orthogonal group has index 2. Here you can choose a diagonalizable matrix $g$ of determinant $-1$ lying outside the identity component as a coset representative for $G/G^\circ$ and get two disjoint copies of the big cell whose union is still open in the Zariski (or euclidean) topology. A similar thing happens in general for disconnected groups. -Concerning the notion of a "split" maximal torus, this is covered in all textbooks when fields of definition are discussed. As in other algebraic theories, you may need to extend an arbitrary field in order to diagonalize a semisimple matrix or such. (Typically this is always possible for an algebraic structure by taking a finite separable extension, which is where "separable" probably comes in here.) Over an algebraically closed field there is no problem. For a torus, as mentioned in comments, it just means being isomorphic over the given field to a direct product of copies of the multiplicative group. The contrast is with non-split situations that occur for compact Lie groups, etc.<|endoftext|> -TITLE: What is the geometric meaning of homology group and cohomology group? -QUESTION [7 upvotes]: I am reading some algebraic topological book, where they said that the p-th homology group tells us how many p-dimensional holes inside the set. How should I understand this? I know that in the planar case, the 1-th order homology group tells us exactly how many 1-holes the set has, but I do not have a good feeling about the p-holes, could anybody explain this by some very simple examples? -Also they said the cohomology groups are dual to homology group, is there also a good geometric meaning of cohomology group? I heard some duality result, like the Alexander duality between homology group and cohomology group, but I do not know the real meanings. - -REPLY [9 votes]: I recommend you to have a look at the first chapter of Roger Fenn's book "Techniques of geometric topology". - -Going back to Poincaré's first approach to homology (where homology is defined as a geometric calculus), you can think to $n$-cycles of a space $X$ as maps from $n$-manifolds (with singularities) to $X$ up to a bordism relation. A nice reference is Mathias Kreck's book: "Differential algebraic topology", in that text manifolds with singularities are treated as stratifolds. -There are several ways to think to cohomology in a geometric way. One goes back to Dan Quillen's treatment of complex cobordism and is described in the case of a closed manifold in Kreck's book (in this setting the duality between homology and cohomology is transparent). -I also want to emphasize on Dennis Sullivan's principle that says that cohomology is represented by geometric cocycles. Geometric cocyles are cycles with "normal geometry". Roughly speaking, if your space is stratified they are cycles transverse to the stratification. In that geometric setting, the cup product is given by the geometric intersection, going back to the roots of algebraic topology. This principle is described in Roger Fenn's book, in Clint MacCrory's PhD thesis (available on his homepage), in Mark Goresky's paper "Whitney stratified chains and cochains" and in Buoncristiano-Rourke-Sanderson's monography "A geometric approach to homology theory". - -Thus you can think of the homology of a space $X$ as equivalences classes of maps $V\rightarrow X$ where $V$ is a "manifold with singularities". And you must play with singular spaces in order to recover singular homology, this is a deep and beautiful result due to René Thom. For "reasonable" spaces $X$ (including polyedra) together with and additional geometric structure: a stratification, these geometric cycles can be taken as sums of subspaces of $X$ (substratified spaces). The second lesson is that cohomology can be build from particular cycles those who are in general position with respect to this stratification. This last fact is deeply rooted into a geometric interpretation of Poincaré duality in terms of the polyedral dual decomposition $C$ to a triangulation $T$ of a manifold $M$. - -Thanks to $T$ you get cycles for the homology of $M$, -Thanks to $C$ you get geometric cocycles for the cohomology of $M$, by definition they are transverse to $T$.<|endoftext|> -TITLE: Avoiding integers in the spectrum of the Laplacian of a Riemann surface -QUESTION [21 upvotes]: Let $\Sigma^2$ be an orientable compact surface of genus $gen(\Sigma)\geq2$, and denote by $\mathcal M(\Sigma)$ the moduli space of hyperbolic metrics on $\Sigma$, i.e., Riemannian metrics of constant curvature $-1$. Recall that, from Teichmüller theory, this is a finite-dimensional subspace of the space of all metrics on $\Sigma$, and actually has dimension $6gen(\Sigma)-6$. For a given metric $g$, denote by $Spec(\Delta_g)=\{0=\lambda_0<\lambda_1\leq\lambda_2\leq\dots\}$ the spectrum of the Hodge-Laplacian of $g$, acting on $C^\infty(\Sigma)$. - -Question 1: Given a natural number $n\in\mathbb N$, does there exist a hyperbolic metric $g\in\mathcal M(\Sigma)$ such that $n\notin Spec(\Delta_g)$? - -I have attempted various ``deformation'' arguments to try to prove that if some $g\in\mathcal M(\Sigma)$ has $n$ as an eigenvalue, then a small perturbation $g'\in\mathcal M(\Sigma)$ would no longer have that eigenvalue, but without success... Instead of avoiding any given natural number, one can try the (seemingly) easier task: - -Question 2: Does there exist a hyperbolic metric $g\in\mathcal M(\Sigma)$ such that $n\in Spec(\Delta_g)$ only for finitely many numbers $n\in\mathbb N$? - -Or even more, - -Question 3: Does there exist a hyperbolic metric $g\in\mathcal M(\Sigma)$ such that $\mathbb N\not\subset Spec(\Delta_g)$? - -Although the answers to Q2 and Q3 seem to be "obviously yes", I have not been able to find a rigorous argument to prove that. I have tried to argue by contradiction, to show that if $Spec(\Delta_g)$ contains infinitely many natural numbers (or all of them), then we would somehow violate Weyl's formula (see this post) which says that $\lambda_k\sim \frac{k}{gen(\Sigma)-1}$ as $k\to+\infty$, but again without success (even in the case $gen(\Sigma)=2$). - -REPLY [16 votes]: I was informed by Sugata Mondal at the MPI that Scott Wolpert proved the following result in his 1994 Annals paper Disappearance of cusp forms in special families: - -Theorem 5.14. The eigenvalues of the Laplacian above $\tfrac14$ on a closed hyperbolic surface vary nontrivially under analytic deformations. - -That is, if $g_t$ is a real-analytic path of hyperbolic metrics on $\Sigma$, then no real-analytic branch $\lambda_k(t)$ of an eigenvalue of the Laplacian $\Delta_{g_t}$ can be constant. Recall that by the Kato Selection Theorem, up to relabeling the indices $k$'s, the functions $\lambda_k(t)$ are real-analytic. -I also contacted Scott Wolpert, who confirmed that this indeed answers all my above questions. In particular, he had called Q1 the "middle C embarrassment question", since prior to the above result it could have been that middle C on the piano is a frequency for every hyperbolic Laplacian. Finally, note that his result also clearly allows to prescribe the genus of the closed hyperbolic surface $\Sigma$ at the same time as avoiding any given real number $\lambda>\tfrac14$ in its spectrum.<|endoftext|> -TITLE: Looking for a reference (on GW invariants of quintic) -QUESTION [5 upvotes]: 1) Apparently, physicist can calculate the GW invariants of quintic CY 3-fold up to genus 51. -I am looking for a reference that has a table of these number for some low degrees (say up to degree 5) and low genera (at least until g=3). -2) For each genus g, there is a lower bound $d(g)$ such that for every $d -TITLE: Two definitions of modules in monoidal category -QUESTION [5 upvotes]: The standard definition of a (left) simplicial module $V$ over some simplicial algebra $A$ is the map of simplicial vector spaces $A\otimes V\to V$ that gives the usual modules component-wise. Here $\otimes$ denotes the component-wise tensor product of simplicial vector spaces (it gives to the category $sVect$ of simplicial vector spaces a symmetric monoidal structure). -But there is a natural structure of simplicial algebra on $sHom(V,V)$ (the internal $Hom$ induced by the product $\otimes$). So we can also define structure of a simplicial module over $A$ on $V$ by fixing a morphism of simplicial algebras $A\to sHom(V,V)$. -I think we can do it in more generality, for any symmetric monoidal category $C$ with internal $Hom$. If we have a monoid $A$ in $C$ and some object $V\in C$, then we can define the structure of $A$-module on $V$ in two different ways: via $A\otimes V\to V$ and $A\to \underline{Hom}(V,V)$, where $\underline{Hom}$ is the internal $Hom$. - -The question is: are these two definitions equivalent for any symmetric monoidal category with internal $\underline{Hom}$? If not, is it true for simplicial vector spaces? - -Thank you very much! - -REPLY [9 votes]: I will write $[B, C]$ instead of $\underline{\mathrm{Hom}}(B, C)$. Recall the tensor–hom adjunction: -$$\mathrm{Hom}(A \otimes B, C) \cong \mathrm{Hom}(A, [B, C])$$ -Thus there is a canonical bijection between morphisms $\alpha : A \otimes V \to V$ and $\tilde{\alpha} : A \to [V, V]$. Let us show that $\alpha$ is an $A$-action on $V$ if and only if $\tilde{\alpha}$ is a monoid homomorphism. For simplicity I will work in a strict monoidal category. - -The unit axiom for $\alpha$ says, $\alpha \circ (e \otimes \mathrm{id}_V) = \mathrm{id}_V$; and the the unit axiom for $\tilde{\alpha}$ says, $\tilde{\alpha} \circ e = \eta_V$, where $\eta_V : I \to [V, V]$ is the right adjoint transpose of $\mathrm{id}_V : V \to V$. The naturality of the tensor–hom adjunction in the first variable implies that these two conditions are equivalent. -The compatibility axiom for $\alpha$ says, $\alpha \circ (m \otimes \mathrm{id}_V) = \alpha \circ (\mathrm{id}_A \otimes \alpha)$; and the compatibility axiom for $\tilde{\alpha}$ says, $\tilde{\alpha} \circ m = \mu_V \circ (\tilde{\alpha} \otimes \tilde{\alpha})$, where $\mu_V : [V, V] \otimes [V, V] \to [V, V]$ is the right adjoint transpose of the composite -$$[V, V] \otimes [V, V] \otimes V \xrightarrow{\mathrm{id}_{[V, V]} \otimes \epsilon_{V,V}} [V, V] \otimes V \xrightarrow{\epsilon_{V,V}} V$$ -where $\epsilon_{B,C} : [B, C] \otimes B \to C$ is the left adjoint transpose of $\mathrm{id} : [B, C] \to [B, C]$. Now, naturality in the first variable implies the right adjoint transpose of $\alpha \circ (m \otimes \mathrm{id}_V)$ is $\tilde{\alpha} \circ m$, and naturality in the first variable implies the left adjoint transpose of -$\mu_V \circ (\tilde{\alpha} \otimes \tilde{\alpha})$ is the following composite, -$$A \otimes A \otimes V \xrightarrow{\tilde{\alpha} \otimes \tilde{\alpha} \otimes \mathrm{id}_V} [V, V] \otimes [V, V] \otimes V \xrightarrow{\mathrm{id}_{[V, V]} \otimes \epsilon_{V,V}} [V, V] \otimes V \xrightarrow{\epsilon_{V,V}} V$$ -but $\alpha = \epsilon_{V,V} \circ (\tilde{\alpha} \otimes \mathrm{id}_V)$, so the above reduces to $\alpha \circ (\mathrm{id}_A \otimes \alpha)$. - -Thus the claim is proved.<|endoftext|> -TITLE: extending elementary embeddings from initial segments of V to all of V -QUESTION [12 upvotes]: Suppose $j:V_\lambda \rightarrow V_\eta$ is (elementary and) cofinal. Can $j$ be extended to all of $V$? -(Subsidiary question: What conditions are there on an ultrafilter/extender/whatever so that the induced $j:V \rightarrow M$ has $V_\eta$ as an initial segment of $M$? Note that $j(\kappa)$ might well be less than $\eta$, where $\kappa$ is the critical point.) - -REPLY [12 votes]: At successor ordinals, the answer is no, not necessarily, assuming the consistency of a nontrivial instance of your hypothesis. -For a counterexample, let $\kappa$ be the least $1$-extendible cardinal. So there is $j:V_{\kappa+1}\to V_{\eta+1}$, and this map is elementary and cofinal, mapping $\kappa$ to $j(\kappa)=\eta$. -I claim that this map cannot extend to $j:V\to M$ for a transitive set $M$, with $V_{\eta+1}\subset M$. The reason is that the size of $j\upharpoonright V_{\kappa+1}$ is $|V_{\kappa+1}|=2^\kappa$, which is therefore coded by a size $2^\kappa$ subset of $V_{\eta+1}$, which is therefore an element of $V_{\eta+1}$. So $M$ will see that $\kappa$ is $1$-extendible, and so by elementarity there must be a $1$-extendible cardinal below $\kappa$, contradicting minimality. -Similarly, if $\kappa$ is the least $(\theta+1)$-extendible cardinal, then there is $j:V_{\kappa+\theta+1}\to V_{\eta+1}$. But since $j\upharpoonright V_{\kappa+\theta+1}$ is a size $|V_{\kappa+\theta+1}|$ subset of $V_{\eta+1}$, it is coded by an element of $V_{\eta+1}$, and so if $j:V\to M$ extends $j$, then $M$ would see that $\kappa$ is $(\theta+1)$-extendible, violating minimality. -Update. But at limit ordinals $\lambda$, the answer is yes, all such elementary cofinal embeddings $j:V_\lambda\to V_\eta$ lift to $j:V\to M$ for some transitive class with $V_\eta\subset M$. To see this, for any $a\in V_\eta$, let $\mu_a$ be the measure generated via $j$ by $a$, so that $X\in \mu_a\iff a\in j(X)$. This measure will concentrate on any set $D_a\in V_\lambda$ for which $a\in j(D_a)$, and such a set exists since the map was cofinal. Let $j_a$ be the ultrapower of $V$ by $\mu_a$. Elementary seed theory shows that $j_a\upharpoonright V_\lambda$ is a factor embedding of $j$, and furthermore, the range of $j_a\upharpoonright V_\lambda$ is precisely the seed hull $X_a=\{\ j(f)(a)\mid f:D_a\to V_\lambda,\ f\in V_\lambda\ \}\prec V_\eta$. These elementary substructures form a directed system, and the map $j$ is the direct limit of the maps $j_a\upharpoonright V_\lambda$. But now the point is that the maps $j_a$ are defined on all of $V$, since they are simply ultrapower maps by the measures $\mu_a$. And so one may take the corresponding direct limit of the full maps $j_a:V\to M_a=\text{Ult}(V,\mu_a)$. The result is an elementary embedding $j:V\to M$ into a transitive class (after the collapse), which extends the given map $j:V_\lambda\to V_\eta$. -Basically, what is happening here is that we use the original map $j:V_\lambda\to V_\eta$ to define an extender, which is then applied to the whole of $V$. This strategy didn't work fully at succcessor ordinals $j:V_{\theta+1}\to V_{\eta+1}$, since there was no way to ensure that all of $V_{\eta+1}$ was picked up in the range of the induced extender, as those seeds $a$ are on the top level of $V_{\eta+1}$ and not covered by any element of $V_{\theta+1}$ (and perhaps this could be considered a violation of cofinality of $j$, even though $j(\theta)=\eta$). But even in the successor ordinal case, we can still define the induced extender, and the resulting $j:V\to M$ will have $V_\eta\subset M$ and it will agree with the original $j:V_{\theta+1}\to V_{\eta+1}$, and in particular include the image of $V_{\theta+1}$ under $j$, although it may not have all of $V_{\eta+1}$ contained in $M$.<|endoftext|> -TITLE: Solovay's Theorem on Partitions of Stationary Sets and Weak Choice Principles -QUESTION [5 upvotes]: There is a weak choice principle called $DC_\lambda$ which holds in $L(V_{\lambda+1})$ under the assumption of a non-trivial elementary embedding $$j:L(V_{\lambda+1})\prec L(V_{\lambda+1})$$ and it is known that this choice principle is not sufficient to split the ordinals below $\lambda^+$ (a regular cardinal) which have cofinality $\omega$ into disjoint stationary sets. - -Is there a choice principle $\Phi$, which, when augmented with $DC_\lambda$ and strictly weaker than full AC that suffices to prove Solovay's Theorem on Partition of Stationary Sets? A little more specifically, what is the minimal $\Phi$ such that $T=ZF + DC_\lambda + \Phi$ where $$T\vdash \text{All stationary subsets of }\lambda^+\text{ have a disjoint partition into stationary sets}?$$ -One such candidate could be $Unif(V_{\lambda+1}\times V_{\lambda+1})$: given any $R\subseteq V_{\lambda +1}\times V_{\lambda +1}$ there exists some function $f\subset R$ with the same domain as $R$. -(This question is related to both A proposed axiom of Laver (updated) and Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? .) -Are there other weak choice principles $\Phi$ that could be considered? -Are there (perhaps) some partition properties with infinite exponents that would prohibit Solovay's Theorem? - - -EDIT: While I am interested in the more general question (which I believe Asaf Karagila has addressed in the comments and chat), I am really specifically interested in the context that Woodin's Axiom $I_0$ holds. Specifically, any assertion $\Phi$ I'm looking for can't imply that $[\lambda]^\omega$ is well-ordered (in conjunction with $DC_\lambda$). - -REPLY [3 votes]: Although it may not be what you are looking for, this paper (called Splitting stationary sets from weak forms of Choice) contains some results along the lines of your questions. Theorem 3.1 of the paper shows (in the simplest case to state) that, assuming DC, if $\lambda$ is a regular cardinal and there is a function that chooses a countable cofinal subset of each ordinal below $\lambda$ of countable cofinality, then the set of ordinals of countable cofinality below $\lambda$ can be split into $\lambda$ many stationary sets. Corollary 3.3 shows that there is no elementary embedding from $V$ into $V$ given a stronger assumption : DC plus the statement that for every ordinal $\lambda$ there is a wellorderable $\mathcal{A} \subseteq [\lambda]^{\aleph_{0}}$ such that every element of $[\lambda]^{\aleph_{0}}$ intersects a member of $\mathcal{A}$ infinitely. The stronger hypothesis is needed to get that $\kappa^{+}_{\omega}$ (the ordinal where the stationary set splitting occurs in the standard argument) is regular. I believe that there are weaker principles that suffice to get the regularity of $\kappa_{\omega}^{+}$ in some of Shelah's papers on Choice-less PCF theory (number 835, for instance). I hope this helps.<|endoftext|> -TITLE: Infinite desert with waterpoints -QUESTION [20 upvotes]: UPDATE: I created a simple web-application, that allows the user to move waterpoints around, then automatically calculates a maximum set of interior-disjoint squares between the points (the algorithm is exponential so it works only up to about $n=11$ points). - -There are $n$ waterpoints in an infinite desert. Assume each waterpoint is a point in $\Bbb R^2$, and the x and y values of the points are all different. -Define a paradise as an axis-aligned square which touches at least two waterpoints. -Define $p(n)$ as the maximum number of interior-disjoint paradises -in the worst possible arrangement of $n$ waterpoints. What is $p(n)$? -SOME SIMPLE CASES: -If the waterpoints are on a diagonal (e.g. $\{(i,i)|1\leq i\leq n\}$), then there can be only a single paradise between each pair of adjacent waterpoints, for a total of $n-1$. Hence: $p(n) \leq n-1$: - -For $n \leq 5$, it can be checked that $p(n) = n-1$: - -For $n=1, 2$ this is obvious. -For $n=3$, put 2 paradises in the following way. Order the waterpoints in an increasing order of their x value ($x_1 < x_2 < x_3$). Cut the plane to two half-planes using a vertical line through $x_2$. The western half-plane contains two waterpoints $x_1$ and $x_2$, so you can put a paradise there. The same is true for the eastern half-plane. Hence: $p(3)=2$. - - - -For $n=4$, Cut the plane to two half-planes using a vertical line through $x_2$. Put a single paradise in the western half-plane through $x_1$ and $x_2$. In the eastern half-plane, order the 3 paradises vertically in an increasing order of their y values and cut to two quarter-planes using a horiznotal line through $y_2$. Each quarter-plane contains 2 waterpoints so you can have a single square in each. Hence: $p(4)=3$. -For $n=5$, Cut the plane to two half-planes using a vertical line through $x_3$. Cut the two half-planes to two quarter-planes each using horizontal lines. Each quarter-plane contains 2 waterpoints (some of them on the boundary) so you can put a single paradise in each quarter-plane. Hence: $p(5)=4$. - -This cutting scheme doesn't work anymore when $n>5$, because if we end up with 3 waterpoints in a quarterplane, it might be impossible to have 2 paradises. -On the other hand, in all cases that I checked, I always managed to find $n-1$ paradises. -So the question is: is it always possible to have $n-1$ paradises, for every $n$? If so, how? If no, what is the minimum? - -PREVIOUS WORK: -Several months ago I published a similar question in math.SE, with a major difference: instead of an infinite desert, there was a finite (square) cake (i.e. all points and squares should be within a large predefined square). -With the help of an answer from prof. Boris Bukh, I managed to prove an upper bound of: $$p(n) \leq \lceil{n \over 2}\rceil - 1$$ The proof was by constructing a set of $n=2k$ points, that are organized such that it is not possible to put more than $p=k-1$ squares. The following figures illustrate the construction for $k=6$: - -Later I also proved a lower bound of: -$$ p(n) \geq {(n+2) \over 6} \ \ \ \ \ [n \geq 2]$$ -The proof was by describing a recursive algorithm that, given a square with $n$ points, divides the square to 4 smaller squares, and puts paradises separately in each of them. -Now I am trying to adapt these results to the infinite desert case. -The upper bound construction crucially relies on the assumption that the cake is bounded in at least two adjacent directions. It can work in a quarter-plane, but not in an unbounded desert ($\Bbb R^2$). -(In a half-plane, a similar construction leads to an upper bound of $p(n) \leq \lceil{2(n-1) \over 3}\rceil$). -The lower bound algorithm will obviously work in an infinite desert, however, after only a single iteration, we will have 4 quarter-planes, and thus the lower bound will not be much better than in the bounded-cake case. - -REPLY [9 votes]: After a lot of playing around, I found this counter-example with $n=6$ points and only 4 squares. This link alone is not a sufficient proof because there might be a bug in my function that calculates the maximum disjoint set. So here is a formal description of the arrangement, and an attempt to prove that indeed no more than 4 squares are possible: - -First, consider only the 4 points near the x axis (blue). We can consider only squares between adjacent points, because squares between non-adjacent points can be shrinked to touch only adjacent points. Between each of the 3 pairs of adjacent points, there are 2 options to draw a square: either above the x axis, or below it. There is a small overlap between these 2 squares, so only one square per pair is possible: - -Second, consdier the point at the top of the y axis (red). Any square that uses this point must have a side-length of at least 6 in order to approach the x axis. Any such square necessarily blocks at least two of the cyan squares above the x axis. Additionally, the leftmost square is slightly larger than 6 and thus also blocks the middle cyan square below the x axis. The two extreme options are plotted below in magenta: - -The case for the point at the bottom of the y axis (red) is the same. -All in all, a maximum set of disjoint squares can be drawn in only the following ways (ordered in an increasing order fo the number of magenta squares): - -One magenta square at the top-left(bottom-right), and 3 cyan squares below(above) the x-axis. -Two magenta squares at the top(bottom), and 2 cyan squares below(above) the x-axis. Or, one magenta square at the top-left(right), one magenta square at the bottom-right(left), one cyan square at the left(right) below the x axis, and one cyan square at the right(left) above the x axis. -Two magenta squares at the top(bottom), one magenta square at the bottom-left(top-right), and one cyan square at the right(left) below(above) the x axis. -Two magenta squares at the top and two at the bottom. - -In all ways, there are only 4 squares. Hence $p(6)\leq 4$. - -This construction can be duplicated and translated a long distance to the left and bottom (for example). For every duplication, we get more 6 points and more 4 squares, but an additional square can be added between two adjacent duplicates. This gives a bound of: $p(6k)\leq 5k-1$ for every integer $k$, i.e. $p(n)/n \lesssim 5/6$. -A further improvement can probably be achieved by creating 6 copies of the construction, arranged in just the same way as the original construction. This can be continued recursively like a fractal, and gives $p(6^k)\leq 4\cdot(6^{k-1}+...+1) = 4\cdot\frac{6^k}{5}$ for every integer $k$, i.e., $p(n)/n \lesssim 4/5$. - -This upper bound is still far away from the lower bound of $p(n)/n \gtrsim 1/6$...<|endoftext|> -TITLE: The cone on a manifold -QUESTION [22 upvotes]: I believe that I have run across the statement that if $X$ is a compact smooth manifold and $CX$ is the cone on $X$, i.e. $[0,1] \times X$ modulo $(0,x)\sim(0,y)$ for all $x,y \in X$, then $CX$ admits the structure of a smooth manifold with boundary, with $\left\{1\right\} \times X$ smoothly immersed in it as the boundary, just when $X$ is diffeomorphic to a sphere in Euclidean space with its standard smooth structure. I would like a reference, or a reference to a similar statement. - -REPLY [23 votes]: There is no need for smoothness here : if $X$ is a compact topological $n$-manifold and $CX$ is the cone on $X$, then $CX$ is a topological manifold if and only if $X$ is homeomorphic to a sphere. This is trivial for $n=1$, so assume that $n \geq 2$. The backward implication is trivial, so assume that $CX$ is a topological manifold. Let $p_0 \in CX$ be the cone point. The local homology groups $H_{k}(CX,CX-p_0)$ are then $\mathbb{Z}$ for $k=0,n+1$ and $0$ otherwise. Looking at the long exact sequence for the pair $(CX,CX-p_0)$, we then get that $H_{k}(X)$ is $\mathbb{Z}$ for $k=0,n$ and $0$ otherwise. In other words, $X$ is a homology $n$-sphere. Next, since $n+1 \geq 3$ and $CX$ is an $(n+1)$-manifold, the space $CX$ must satisfy the following condition : for all point $q \in CX$ and all neighborhoods $U$ of $q$, there must be a neighborhood $V$ of $q$ such that $V \subset U$ and $V-q$ is simply connected. Around the cone point $p_0$, it is easy to see that this condition implies that $X$ must be simply-connected. The Poincare conjecture thus implies that $X$ is homeomorphic to an $n$-sphere.<|endoftext|> -TITLE: $Pic$ of the stack of elliptic curves vs. $Pic$ of the coarse space -QUESTION [9 upvotes]: There's a natural map $f:\overline{\mathcal{M}}_{1,1}\to \overline{M}_{1,1}\cong \mathbb{P}^1$ from the stack of elliptic curves to the coarse space. Both spaces have $Pic=\mathbb{Z}$ hence $f^*:\mathbb{Z}\to\mathbb{Z}$ is an homomorphism. What homomorphism? My guess is: $x \mapsto 24 x$ since the generator of the stack is the Hodge class, which has degree 1/24. Do you agree? - -REPLY [7 votes]: I believe the number is 12. -I will assume the characteristic of the base field is not 2 or 3 so that I can use $\overline{\mathcal{M}}_{1,1} \simeq \mathbf{P}(4,6)$. Recall that $\mathbf{P}(4,6)$ is constructed by dividing $V = \mathbf{A}^2 \smallsetminus \{ (0,0) \}$ by the weight $(4,6)$-action of $\mathbf{G}_m$. -The line bundle $\mathcal{O}(1)$ (which coincides with the Hodge bundle and generates the Picard group) can be constructed as the equivariant line bundle $V \times \mathbf{A}^1$ on $V$ with $\mathbf{G}_m$ acting by weights $(4,6,1)$. Then $\mathcal{O}(4)$ has a canonical section $g_4$ and $\mathcal{O}(6)$ has the section $g_6$. In $\mathcal{O}(12)$ we have the two sections -$1728 g_4^3$ -$\Delta = g_4^3 - 27 g_6^2$. -This linear series defines $j : \mathbf{P}(4,6) \rightarrow \mathbf{P}^1$. -Note that $j$ has degree $1/2$ because of the generic automorphism on $\mathbf{P}(4,6)$. That is, $j_\ast j^\ast$ is multiplication by $1/2$. Thus the Hodge class $\lambda$ satisfies -$\int_{\overline{\mathcal{M}}_{1,1}} \lambda = \int_{\mathbf{P}(4,6)} c_1(\mathcal{O}(1)) -= \frac{1}{12} \int_{\mathbf{P}(4,6)} c_1(j^\ast \mathcal{O}(1)) = \frac{1}{12} \int_{\mathbf{P}^1} j_\ast j^\ast c_1(\mathcal{O}(1)) = \frac{1}{24}$. -Another thing that may be confusing here is that $j$ is generically unramified, so that a local equation for a point in $\mathbf{P}^1$ pulls back under $j$ to a local equation in $\overline{\mathcal{M}}_{1,1}$. Thus $j^\ast \Delta = \delta$ (if $\Delta$ denotes the boundary in the coarse moduli space and $\delta$ the boundary in the stack).<|endoftext|> -TITLE: Yoneda on a not so small category -QUESTION [7 upvotes]: I am working with "usual" category theory, maybe over ZFC, and I have a functor $F : Set \to Set$. I'd like to apply Yoneda lemma to $F$, i.e. obtain: -$$ [Set, Set](h_A, F) \cong F A $$ -However, most of texts assume that the domain of $F$ should be a small category in order to apply it. I know this has to do with the size of $[Set, Set]$ and the existence of $h_A$. How could I fix my foundations so I can use Yoneda lemma like this? Or I'll never be able to do such thing? - -REPLY [10 votes]: Zhen Lin states one direction of the Yoneda Lemma: given $x\in F A$, the natural transformation $\theta_x:H_A\to F$ on $f:A\to B$ is $\theta(f)=F f\cdot x$. The other direction is that every $\theta:H_A\to F$ is of this form, where $x=\theta(\mathsf{id}_A)$. I leave it as an exercise to show that $\theta_x$ is natural and that this correspondence is bijective. -What I have just stated is the natural mathematical argument for the central idea of the Yoneda Lemma, albeit without the calculations required for the proofs. None of it has anything to do with Set Theory, whether that be ZFC, NBG or even NF. -This argument makes sense in a world in which objects and morphisms are things of a different sort from one another. Functors and natural transformations, on the other hand, are schemes or recipes that transform objects and morphisms into objects and morphisms. Functors and natural transformations are not themselves things. -We are quite used to this many-sorted language in algebra or first order logic. Even when the algebra itself has only one sort (if you don't know what I mean by that, I mean exactly the situation that you know), elements of the algebra, homomorphisms, predicates, ideals etc. belong to different sorts of the language in which we discuss them. -It was the innovation of Set Theory to reify schemes, i.e. turn them into things. In its unrestricted form, this innovation led to the famous antinomies or paradoxes, such as Russell's. The outcome of this was to develop restricted reification, such as Zermelo Set Theory. -Unfortunately, the mathematical community became addicted to Cantor's "paradise". So, even in disciplines such as Category Theory where we should have known better, we continue to find attempts at unrestricted reification such as Grothendieck Universes. -The disciplined alternative is the type-theoretic style, in which we nominate certain forms of reification, for example that of functions ($\lambda$-abstraction), and stick to just those (in the context of a particular discourse). -Now to reply to Michal Przybylek. I have just said and he agrees that "If people think that they need any kind of Set Theory to speak about Yoneda lemma, then they are wrong." However, far from being "philosophical" or "waving my hands", I have explained quite precisely what is the methodological set-up ("foundations" if you wish) behind the argument. -At least, I discussed the ordinary form of the argument. Of course, as mathematicians have always done extremely profitably with pre-existing arguments, one may re-interpret it in more exotic settings. Enriched, internal and locally cartesian closed categories are examples of such settings. Moreover they are also examples of desciplined restricted reification operations. - -REPLY [4 votes]: The problem is not whether you can use Yoneda lemma or not, but to associate a precise meaning to your sentences. As a rule of thumb, Yoneda lemma works whenever you can formally say what it actually means. -I do not agree with Zhen's answer which says that it is not really important that the objects are not inside the universe --- the truth is exactly the opposite --- it is the only issue here. In the usual ZFC formalization, objects $FA$ are real sets, whereas objects $\mathit{nat}(h_A, F)$ are not. Of course you can "lift" $FA$ to the meta-universe, but then you will compare apples with apes.<|endoftext|> -TITLE: Expected symmetry in the diophantine approximations of an irrational number -QUESTION [7 upvotes]: Given $x \in \mathbb{R}$ we will write $\{x\}$ for the fractional part of $x$ and $\|x\|$ for the distance of $x$ from the nearest integer, in such a way that $\{x\} = x - \lfloor x \rfloor$ and $\|x\| = \min(\{x\}, 1 - \{x\})$, where $\lfloor x \rfloor$ is, as usual, the greatest integer $\le x$. -Let $x$ be a fixed irrational number and $\mu$ its irrationality measure, namely the infimum of the set of all positive exponents $s$ such that -$$0 < \left|x - \frac{m}{n}\right| < \frac{1}{n^s}$$ -for finitely many pairs $(m,n) \in \mathbb{Z} \times \mathbb{N}^+$. Assume that $\mu$ is finite (added later: see N. Elkies' answer and the comments below). Recall that $\mu \ge 2$ by (a corollary of) Dirichlet's (approximation) theorem. Then, given $s \in \{2\} \cup [2,\mu[$, define $G_s$ as the set -$$\{n \in \mathbb{N}^+: \|nx\| < n^{1-s}\},$$ -and let $G_s^{-} := \big\{n \in G_s: \|nx\| = \{nx\}\big\}$ and $G_s^+ := G_s \setminus G_s^{-}$. -We have that $G_s$ is infinite (by Dirichlet's theorem and the definition of $\mu$), so at least one of $G_s^-$ or $G_s^+$ is infinite too, and I'm tempted to claim that each of them must be infinite. I don't have a serious argument in support of this: My only point is that it would be weird, I believe, to observe a similar "asymmetry" in the diophantine approximations of $x$ (yet, I would be happy to hear that my expectation is completely wrong). This leads to the following: - - -Q1. Is it true that $G_s^-$ is infinite for each $s \in \{2\} \cup [2,\mu[$? Q2. If Q1 is well-established, would you kindly provide me with a reference? - - -Some remarks: (i) If the answer to Q1 is yes, then it is as well true that $G_s^+$ is infinite for each $s \in \{2\} \cup [2,\mu[$. (ii) I can prove that the answer to Q1 is positive if $s = 2$ (this follows from Dirichlet's theorem and some elementary properties of the simple continued fraction expansion of $x$). (iii) By (ii) and Khintchine's theorem (which yields that the irrationality measure of most real numbers is equal to $2$), the answer to Q1 is yes for almost all $x$. - -REPLY [5 votes]: This can't be right in general. For example -$$ -x = 0.101000001000000000000000001\ldots - = \sum_{k=0}^\infty 10^{-3^k} -$$ -has plenty of lower approximants with exponent just under 3 -(the partial sums) but no upper ones.<|endoftext|> -TITLE: Average rank of elliptic curves, excluding those of low rank -QUESTION [15 upvotes]: It's conjectured that, asymptotically, half of elliptic curves have rank 0, half have rank 1, and elliptic curves of rank $\geq 2$ have density 0. But what if we disregard elliptic curves of rank 0 or 1: Are there any conjectures about the average rank of elliptic curves of rank $\geq 2$? -More generally, for any integer $n \geq 2$, what value should one expect for the average rank of elliptic curves of rank $\geq n$? - -REPLY [8 votes]: Regarding the work of Watkins as mentioned by Matt Young, he has a different paper where the question of rank distribution in quadratic twist families is considered. There the RMT prediction for rank 2 is well shown by the Rubinstein data to demonstrate about $D^{3/4}$ twists of rank 2 (or more) in the even parity subclass, while Watkins suggests that 3/4 is too high for rank 3. He does not posit anything exactly, but the data for the congruent number curve, divided into 2 natural classes (Section 3.3) gives best-fit exponents of 0.44 and 0.55, so about $D^{1/2}$ is probably a best current guess for rank 3, though in initial regions of data collection the logarithmic factors can be difficult. -http://archive.numdam.org/article/JTNB_2008__20_3_829_0.pdf‎ -The natural linear extrapolation for these types of problems might be semi-valid at least initially, so $D^{1/4}$ for rank 4 and then some power of logarithm for rank 5 are as knowledgeable as guesses as any. But the 2-torsion plays a role here, and it is not clear whether it affects the exponent on the $D$-power. Watkins has a recent preprint (joint with 5 others) where data for the congruent number curve is given, finding "lots" of rank 6 examples, but "few" of rank 7 (and none of rank 8). Granville has a heuristic (see Section 4 loc. cit.) that might suggest ranks 5 and 7 are the correct cutoffs in the generic and 2-torsion cases. -http://magma.maths.usyd.edu.au/~watkins/papers/RANK7.pdf -But really no one has any factual idea, and a number of caveats can be listed, concerning specially parametrised (sparse) families. Indeed, the rank 28 example of Elkies starts from a rank 17 special family and he gains 11 from searching on specializations, and the same was approximately true for the NSA curve, they had rank 24 starting from I think a rank 13 family of Mestre or Nagao. So again 11 more than the family rank. However, generically it seems one should not expect more than 10 "small" points (polynomial height) on an elliptic curve except in such special families, and one actually reachieves the bound of 21 when appending 11 as above. -Edit: Elkies says that the NSA searched in a rank 11 family, so they beat the family rank by 13 in fact, see page 5 of his arxiv.org/pdf/0709.2908v1<|endoftext|> -TITLE: Finitely generated solvable groups all of whose abelian normal subgroups are finite -QUESTION [5 upvotes]: Is there a classification for infinite finitely generated solvable groups all of whose abelian normal subgroups are finite? -I mean by classification something like presentation. -Edited: Is there an infinite finitely generated solvable group $G$ all of whose abelian normal subgroups are finite and $G$ is not residually finite? -Thanks in advance for any help. - -REPLY [7 votes]: To your edited question: - -There is an infinite finitely generated solvable group with no infinite normal abelian subgroup. - -The example is an extension of $\mathbf{Z}/p\mathbf{Z}$ by the lamplighter $(\mathbf{Z}/p\mathbf{Z})^2\wr\mathbf{Z}$. -Fix a prime $p$ congruent to -1 modulo 4 (so that -1 has no square root mod $p$). Consider a group $H_p$ made up of formal sums $\sum a_ne_n\oplus\sum b_nf_n\oplus c$, where $a_n,b_n,c\in\mathbf{F}_p$ (all but finitely being 0), and $e_n,f_n$ is a fixed "basis". Define the "Heisenberg-like product" $(\sum a_ne_n\oplus\sum b_nf_n\oplus c)(\sum a'_ne_n\oplus\sum b'_nf_n\oplus c')=\sum a''_ne_n\oplus\sum b''_nf_n\oplus c''$, where $a''_n=a_n+a'_n$, $b''_n=b_n+b'_n$, $c''=c+c'+\sum a_nb'_n-a'_nb_n$. This is a group (it is the quotient of an infinite direct sum of copies of the Heisenberg by a central subgroup, at least if $p\neq 2$). -We denote by $z$ the element $0\oplus 0\oplus 1$ of $H_p$. -There is an obvious automorphism $\alpha$ of $H_p$ shifting the indices, namely mapping $e_n\mapsto e_{n+1}$, $f_n\mapsto f_{n+1}$, $z\mapsto z$. Let us also consider the involution $s: e_n\mapsto f_n\mapsto e_n$, $z\mapsto -z$, which commutes with $\alpha$. -Consider the semidirect product $G_p=H_p\rtimes(\mathbf{Z}\times\mathbf{Z}/2)$, where $\mathbf{Z}\times\mathbf{Z}/2$ is identified to $\langle\alpha,s\rangle$. Then $G$ is 3-step solvable, and finitely generated (by 3 generators, say $\alpha$, $s$, and $e_0$). -I claim that $G_p$ has no infinite abelian normal subgroup, and more precisely that every abelian normal subgroup $N$ of $G_p$ is contained in the normal cyclic subgroup $\langle z\rangle$ of order $p$. Otherwise, the projection $N'$ of $N$ on $G_p$ is a nontrivial abelian normal subgroup. Since $H'_p=H_p/Z$ is its own centralizer in $G'_p=G_p/Z$, we have $N'\cap H'_p$ nontrivial. So we can suppose that $N\subset H_p$. Thus $N'$ is an $\langle\alpha,s\rangle$-invariant subgroup of $H'_p$, i.e. a $\mathbf{F}_{p^2}[\alpha^{\pm 1}]$-submodule of $H'_p$. Here we defined this module structure by letting a given square root of -1 (which is in $\mathbf{F}_{p^2}\smallsetminus\mathbf{F}_{p}$) act as $s$. Since $N'$ is nonzero, it contains some element of the form $w=\lambda_0e_0+\lambda_1e_1+\dots \lambda_ke_k$, where $k\ge 0$, $\lambda_i\in\mathbf{F}_{p^2}$, and $\lambda_0\lambda_k\neq 0$. So it also contains the element $w'=\lambda_0e_{-k}+\dots \lambda_ke_0$. But any two lifts in $N$ of $w$ and $w'$ do not commute and we get a contradiction. - -Addendum to give details of my comments to the question: - -Every infinite solvable group has an infinite 2-step nilpotent characteristic subgroup with finite derived subgroup. In particular, every infinite solvable group $G$ that is residually finite has an infinite abelian normal subgroup (which can be chosen characteristic if $G$ is finitely generated). - -(A group $G$ is by definition 2-step nilpotent if its derived subgroup $G'$ is central; this includes abelian groups.) -The second assertion follows from the first: choose $N$ as in the first assertion; by residual finiteness, there is a finite index normal subgroup $M$ such that $N'\cap M=\{1\}$; if $G$ is residually finite then $M$ can be chosen to be characteristic. Hence $N\cap M$ is an infinite abelian normal subgroup, and is characteristic if $M$ is characteristic. -For the first assertion, let $N$ be the last infinite term in the derived series. So $N'$ is finite. Hence, the centralizer of $N'$ in $N$ has finite index in $N$, is an infinite characteristic subgroup with central finite derived subgroup, that is, 2-step nilpotent.<|endoftext|> -TITLE: Converse to Stokes' Theorem -QUESTION [32 upvotes]: Does satisfying Stokes' Theorem imply that a form is linear? -Let $M$ be an $n$-manifold. A differential $k$-form $\omega \in \Omega^k M$ assigns to each point $x \in M$ a function $\omega_x : \Lambda^k T_x M \to \mathbb{R}$ which is linear. -My suspicion is that this pointwise statement of linearity is essentially equivalent to the global statement given by Stokes' theorem. To make the question precise requires a notion of "nonlinear differential form," which is provided by the notion of $k$-density, in the sense of Gelfand. For completeness I include the definition given in the link: -Definition: A $k$-density ("nonlinear* $k$-*form") on an $n$-manifold $M$ is a function assigning to each point $x \in M$ a function $\omega_x : S^k T_x M \to \mathbb{R}$ which is homogeneous (of degree 1). Here $S^k T_x M$ is the cone of simple (a.k.a. decomposable) elements of $\Lambda^k T_x M$. -A $k$-density can be pulled back to a $k$-submanifold $N \subset M$ and integrated once $N$ is given an orientation, in exactly the same way that a linear $k$-form is pulled back and integrated. The homogeneity condition ensures that this integration is independent of how $N$ is parameterized, and that the integral changes sign if the orientation of $N$ is reversed. I'm happy to discuss the fact that $\omega$ need only be defined on simple forms if it seems puzzling. -With these preliminaries, I can precisely state my question. -Question: Let $\omega$ be a $k$-density on an $n$-manifold $M$. Suppose there exists a $k+1$-density $\eta$ such that for every compact oriented $k+1$-submanifold with boundary $N \subseteq M$, we have $\int_{\partial N} \omega = \int_N \eta$ (where $\partial N$ is given the induced orientation). Does it follow that $\omega$ is a differential $k$-form, i.e. that $\omega_x$ is linear at each point $x$? Does it follow that $\eta$ is linear? -I'm pretty sure that if $\omega$ is linear, then $\eta$ can only be $\mathrm{d} \omega$, and so is also linear. -Throughout here, I haven't mentioned smoothness assumptions: I assume everything is smooth. But if the answer depends on the degree of regularity, I would find that very interesting. Also, the notion of homogeneity used could be weakened to positive homogeneity. If this affects the answer, again I would find this very interesting. - -Motivation: (edit 11/06): -I'm realizing I never really gave the motivation for this question. It more-or-less comes down to something that Ilya Grigoriev said a few years ago in this MO response. Basically, when asked for intuition about differential forms, he said "They're things you can integrate, plus they're linear." He admitted that the linearity seemed like a weird sacrifice of geometric intuition in favor of algebraic simplicity, since it rules out arc length / surface area and similar measures. -My question here grew out of the sense that differential forms ought to seem natural from some purely geometric perpspective -- otherwise, for one thing, it's hard to explain why they are so perfect for describing electromagnetism. In a comment to Ilya's answer, Haralde Hanche-Olsen pointed to Stokes' theorem as motivation for the linearity of differential forms: my question asked to make this precise. For me, Stokes' theorem is satisfyingly "geometric", even though it's starkly divorced from concepts like arc length and surface area. So with Anton's proof below, I for one would be happy to claim that differential forms are geometrically natural after all: Stokes' theorem carves them out as a separate "regime" of geometry, independent from arc length / surface area and related densities. - -REPLY [10 votes]: Let me just record a few notes on Anton's very nice answer since my comments are already overflowing. This is mostly for my own benefit, and I'm making it community wiki. -Last things first, the argument proceeds by first showing that the "exact density" $\eta$ is linear, and then bootstraps to show that $\omega$ is linear too. In showing that $\eta$ is linear, the only fact used (besides the fact that $\eta$ is a density) is that $\eta$ integrates to 0 over any closed submanifold by the usual Stokes' Theorem-type argument: $\int_S \eta = \int_{\partial S} \omega = \int_{\emptyset} \omega = 0$. This implication is used twice in the bootstrap argument, applied first to $\eta$ and then to $\omega - \omega'$: very elegant. -To show that $\eta$ is linear, Anton formulates a very nice criterion of linearity, that $\sum_{i=1}^n s_i = 0$ should imply that $\sum_{i=1}^n \eta(s_i) = 0$. I believe that we can take $n=3$ here (in particular, let's forget that I used $n$ to denote the dimension of the ambient space way back in the problem statement). -The heart of the proof is to apply the Burago-Ivanov theorem. The point is that the integral in the theorem statement becomes just the sum $s_1 + \dots + s_n = 0$. So the theorem gives us a closed submanifold $S$ whose tangents are concentrated along the $s_i$, with appropriate weighting, except for a set of Euclidean measure $< \epsilon$. Following Anton's comment (with updated notation), -$\int_S \eta = \int_{S_1} \eta + \int_{S_2} \eta = \sum \eta(s_i) + O_{\eta}(\epsilon)$ -where $S_1 \subset S$ is the set of points where $S$ is tangent to some $s_i$ and $S_2 \subset S$ is the remainder. Now, as advertized, we use the fact that $\eta$ integrates to 0 over the closed submanifold $S$: $\int_S \eta = 0$, so $\sum \eta(s_i) = O_{\eta}(\epsilon)$. Since $\epsilon>0$ was arbitrary, we have $ \sum \eta(s_i) = 0$. -The Burago-Ivanov theorem was a little intimidating for me. There is a measure on the Grassmanian and a weird $\Lambda^k(\mathbb{R}^n)$ - valued integral over the Grassmanian (note that "$x$" in the integral denotes the identity function when the Grassmanian is regarded as a subspace of $\Lambda^k(\mathbb{R}^n)$, or alternatively it denotes the inclusion map $G(k,n) \to \Lambda^k(\mathbb{R}^n)$). But it's not so bad. The point of the theorem is that it's actually easy to construct measures on the Grassmanian. The "geometric" way is to take some surface $S$ and pushforward its Euclidean surface area along the Gauss map: the resulting measure tells you how much of the surface is parallel to a given plane represented by a point in the Grassmanian. Burago and Ivanov call this the "weighted Gaussian image" of the surface $S$. The "algebraic" way is to simply pick a finite number of unit $k$-vectors $u_1, \dots, u_N$, and assign weights $\lambda_i$ to them to form a discrete measure $\mu$ with $\mu(u_i) = \lambda_i$. For example, Anton constructs a measure this way in his argument. -We can ask whether an algebraically-constructed measure also arises from the geometric construction, as a weighted Gaussian image. Any finitely-supported measure arises geometrically by taking a bunch of disconnected shapes of appropriate areas which lie in the appropriate planes. But according to Burago and Ivanov, we can do better than this (if we're willing to allow an $\epsilon$ of room): we can can construct such a surface whose boundary is controlled by the integral of the measure (which, again, for a discrete measure is a weighted sum of its supporting $k$-vectors). In particular, if the integral / sum comes out to zero, we can take the surface to be closed: this correspondence between an algebraic condition (related to linearity) and a geometric condition amenable to the Stokes' criterion is exactly what we need. -The $\epsilon$ of room corresponds to parts of the surface whose tangent planes weren't supported by our measure. This is needed because for a general set of planes we can't quite construct a closed polytope with sides parallel to exactly those planes. For example, the planes might be linearly independent, so there's no lower-dimensional subspace for them to intersect along. Given that there are such geometric obstructions, it's pretty remarkable that the Burago-Ivanov theorem actually holds! What does such surfaces look like? Burago and Ivanov provide one example: consider a triangular prism: by stretching the prism on one axis and shrinking on the other, you can shrink the end caps to be arbitrarily small while maintaining constant area of the other sides. This is the sort of thing that has to happen.<|endoftext|> -TITLE: Optimization problem on trace of rotated positive definite matrices -QUESTION [6 upvotes]: Given two $n \times n$ symmetric positive definite matrices $A$ and $B$, I am interested in solving the following optimization problem over $n \times n$ unitary matrices $R$: -$$ -\mathrm{arg}\max_R \,\mathrm{trace}(RAR^TB)~~~\text{s.t.}~~~RR^T = I_n~. -$$ -More generally, given two sets of $m$ positive definite matrices $\{A_i\}_{i=1}^m$ and $\{B_i\}_{i=1}^m$ I would like to solve: -$$ -\mathrm{arg}\max_R \,\sum_i\mathrm{trace}(RA_iR^TB_i)~~~\text{s.t.}~~~RR^T = I_n~. -$$ -If I recall correctly, I have seen the following inequality -$$ -\mathrm{trace}(R\,\mathrm{diag}({\bf c})\,R^T\,\mathrm{diag}({\bf d})) \le {\bf c}^T{\bf d} -$$ -for positive vectors ${\bf c}$ and ${\bf d}$. If this inequality is correct, then $R=I_n$ provides the optimal solution for diagonal matrices and using spectral decomposition of non-diagonal $A$ and $B$ we can solve the problem in the case of $m = 1$. Can somebody please show me how to prove this inequality? More importantly, Is there a way to solve the problem in the more general case of $m > 1$? - -REPLY [3 votes]: If your $B_i$ commute among themselves, and the $A_i$ commute among themselves, -then the $m=1$ solution extends somewhat. Indeed, in that case write $A_i=V C_i V^T$, -$B_i=UD_iU^T$ where $U,V$ are unitary and $C_i,D_i$ diagonal. Now, optimizing over -$R$ becomes the same (after the right and left multiplication by $V^T$ and $U$) -as optimizing over $W$ orthogonal in the expression -$$tr \sum_i WD_iW^TC_i=\sum_{j} \sum_{i,k} w_{jk}^2D_i(k)C_i(j)= -\sum_{j,k} s_{jk} \sum_i D_i(k)C_i(j)=: \sum_{j,k} s_{jk} T(k,j),$$ -where $T(k,j)=\sum_i D_i(k)C_i(j)\geq 0 $. -Here, the $s_{jk}$ form a doubly stochastic matrix. By Birkhoff's theorem, the extreme points -of the set of doubly stochastic matrices are permutations, so the optimal solution for -the above convex problem is a permutation matrix $\{s_{j,k}\}$ ; unlike the case where $\sum_i$ is trivial, -I am not sure there is a simple description of the optimizing permutation. -Remark: when $m=1$, the argument above shows the inequality you wanted: the optimal permutation in that case is the one that orders the eigenvalues of $B$ in the same order as those of $A$.<|endoftext|> -TITLE: Hidden points in polygons -QUESTION [20 upvotes]: Let $h(n)$ be the largest number of mutually invisible points that can be located in a -polygon $P$ of $n$ vertices. Two points $x$ and $y$ are mutually invisible if the segment -$xy$ contains a point strictly exterior to the closed set $P$. -    -I have a vague sense that this question—What is $h(n)$?—may have been studied before, and perhaps even entirely resolved, but I am not recalling references. Any help would be appreciated—Thanks! - -Answered by Wlodek Kuperberg: -  -The same construction shows that $h(n)=n-3$ for hiding points in a -polyhedron of $n$ vertices in $\mathbb{R}^3$. - -REPLY [14 votes]: OK - following the suggestion of Mariano Suárez-Alvarez, a moderator - I post my comment as an answer: -If k points are mutually invisible in an n-sided poygon, then no two of them lie in the same triangle of a triangulation. Therefore $h(n)≤n−2$, and there is a simple example, generalizing your construction for $n=4$ and $5$, of an $(n+2)$-sided polygon with $n$ mutually invisible points in it, for every $n$: Draw a concave-up arc of a circle tangent to your "V" at its upper ends and place $n$ vertices on it. Place your red points very close to the midpoints of the chords of the circular arc. Unless I misinterpret your question, this shows that $h(n)=n−2\ \ $ for every $n≥3$.<|endoftext|> -TITLE: Embedding expanders in CAT(0) spaces -QUESTION [15 upvotes]: It is well-known that expanders are hard to embed into Hilbert (or $\ell^p$) spaces - any embedding of an expander with $n$ vertices has distortion $\Omega(\log n)$. -Can anyone provide a reference (or a quick argument) that the same holds for any "nice" CAT(0) space (like the hyperbolic space) instead of Hilbert space? I'm not giving a precise definition of "nice" here, but this is discussed e.g. in Gromov's "Random Walk in Random Groups". - -REPLY [11 votes]: Here is a class of Hadamard spaces in which, I think, no expander embeds: -Consider a Hadamard space $X$, -which is also "FDSCBB" in the sense -Burago, Gromov, and Perel'man (see http://seven.ihes.fr/~gromov/PDF/3[86].pdf section 7). -By Corollary 7.10 in [BGP] The tangent cone has non-negative curvature in Alexandrov sense. -By Theorem 3.19 from [Bridson & Haeflinger] the tangent cone is also CAT(0), which together means it is Euclidean. -So given a finite subset $S\subset X$, let $x_0\in X$ be its center of mass. -Since the tangent cone around $x_0$ is Euclidean we can have a map into Hilbert space $f:S\cup\{x_0\} \to H$ such that $f(x_0)=0$ for every $s\in S$, -$\|f(s)\|_2= d(x_0,s)$, and the angles in the image are preserved, .i.e. $\angle_0 (f(s),f(s')) = \angle_{x_0}(s,s')$ for every $s,s'\in S$. -The CAT(0) property implies that the mapping $f$ is 1-Lipschitz on $S$. -$f$ also have the "average distortion property", i.e., -$$ \sum_{s,s'\in S} \|f(s)-f(s')\|_2^2 \ge \frac14 \sum_{s,s'\in S} d(s,s')^2 .$$ -Together this implies the ``Poincar\'e inequality" for any graph with respect to Hilbert space transfers to $X$ with a loss of factor of 4, -see for example http://www.cims.nyu.edu/~naor/homepage%20files/spectral-compare.pdf page 5. -We are left prove the average distortion property of $f$. -This is probably standard but since I'm not aware of it, I'll sketch the argument. -For $\lambda\in[0,1]$, -denote by $\lambda S \subset X$ the subset where each $s\in S$ is replaced with the (unique) point $\lambda s$ which is of distance -$\lambda d(x_0,s)$ from $x_0$ along the geodesic connecting $x_0$ and $s$. -Observe that $x_0$ is also the center of mass of $\lambda S$. -Define $f_\lambda:\lambda S\to H$ by $f_\lambda(\lambda s)=\lambda f(s)$. -When $\lambda $ is small $f_\lambda$ preserves distances upto arbitrary precision times $\lambda$, and so, denoting $y_0$ the barycenter of $f_\lambda(\lambda S)$ -$$ \sum_{s,s'\in S} \|f(s)-f(s')\|_2^2 =\lambda^{-2} \sum_{s,s'\in S} \|f_\lambda(\lambda s)-f(\lambda s')\|_2^2 = -\lambda^{-2} 2|S| \sum_{s\in S} \|f_\lambda(\lambda s)-y_0\|_2^2 -\ge \lambda^{-2} |S| \sum_{s\in S} \|f_\lambda(\lambda s)\|_2^2=|S| \sum_{s\in S} d(x_0,s)^2 -\ge \frac14 \sum_{s,s'\in S} d(s,s')^2 .$$ -Update: -Upon reading the literature more carefully, similar arguments were done before. For example, Naor & Silberman In arXiv:1005.4084 have basically the same argument. See -Proposition 4.9 and Corollary 4.10 there.<|endoftext|> -TITLE: Geometric structure of flag manifolds, Borel -Weil-Bott theorem -QUESTION [6 upvotes]: I want to know if there is proof of Borel Weil Bott theorem, that is as geometric as it can be. -Let $G$ be a semisimple compact Lie group and $T$ be a maximal torus. We know that $G/T$ is a projective manifold. A way to show this is to consider the complexification of $G$ and identify $G/T$ as $G_{\mathbb C}/B$. -It seems to me that the following are true: -(1) The root space of $G$ corresponds to $H^{1, 1}\big( G/T\big)\cap H^2(G/T, \mathbb R)$: as $G/T$ is simply connected, $H^2(G/T, \mathbb Z)\cong H_2(G/T, \mathbb Z) \cong \pi_2(G/T)$ and $\pi_2(G/T)$ are generated by those "$\mathbb P^1$" corresponds to each $\alpha \in \Delta^+$ (the $su(2)$ subalgebra $S_\alpha$, to be precise). -(2) The positive Weyl Chamber $C$ corresponds to the Kahler cone of $G/T$, for similar reason as in (1). -So my question is, is (1) and (2) actually holds? -For the following two, I am not so sure. It is related to the proof of Borel Weil Bott theorem. -(3) Is $\rho = \frac{1}{2}\sum_{\alpha\in \Delta^+} \alpha$ corresponds to the First Chern class of $G/T$? Because we know that if a integral weight $\lambda$ satisfies $\lambda + \rho \in C$, then all higher cohomology $H^p(G/T, L_\lambda)$ ($p>0$) vanishes. If $C$ is the Kahler cone, then it seems that the statement is a consequence of Kodaira vanishing theorem, if $\rho$ actually corresponds to the first Chern class. -(4) When $\lambda + \rho$ is not in $C$, there is some correspondance between $H^p(G/T, L_\lambda)$ and $H^0(G/T, L_\mu)$ for some $\mu$ such that $\mu + \rho \in C$. Is there an intuitive geometric reason behind this? -To sum up a bit, can anyone suggest a reference related to these "geometrical aspect" of $G/T$? Thanks in advance. - -REPLY [3 votes]: Correct. You can be fairly explicit here. For each root $\alpha$, let $\omega_\alpha \in \mathfrak g^\ast$ be a left-invariant form on $G$ that is dual to $\mathfrak g_\alpha$. Then for $\lambda \in \mathfrak t^\ast$, we have $d\lambda = \sum_{\alpha\in\Phi^+} \langle \lambda,\alpha\rangle \omega_\alpha \wedge \omega_{-\alpha}$ (up to some scalar). This defines a form on $G/T$ that is of type (1,1) and is positive iff $\lambda$ is in the interior of $C$. Another way of thinking about this whole thing is to note that we have, via transgression, an isomorphism $H^2(G/T;\mathbb R) \cong H^1(T;\mathbb R)$ and this latter space may be identified with $\mathfrak t^\ast$ in the usual manner. -Yes, this is correct as well. -Not quite. It's $2\rho$ that corresponds to $c_1(G/T)$. I'm not sure how you're applying Kodaira, but note that the first Chern class of the canonical bundle of $G/T$ is $-2\rho$. -You have to be a bit careful about what you mean here. In any case, what you're noticing is essentially a manifestation of Serre duality (keep in mind that $L_{-2\rho}$ is the canonical bundle of $G/T$). - -As for references, a good place to start is the pair of articles by Borel and Hirzebruch from the 1950s. It is in these articles that the Borel–Weil–Bott theorem was first conjectured. Bott's original paper (Annals, 1957) is definitely worth a read too, even though it is a bit "heavy-handed" (to quote Bott himself). The proof he gives is very geometric indeed—all the Lie algebra cohomology business can more or less be ignored, if you want. Some papers of Griffiths and Schmid from that era are also fairly geometric and contain a wealth of information. Schmid also has more recent survey articles on geometric representation theory that are worth a look. Finally, let me recommend Dennis Snow'swell-written survey article on homogeneous vector bundles. (Caveat: All these references are a bit short on symplectic geometry and the coadjoint picture. I don't know what to recommend for that.)<|endoftext|> -TITLE: Relation between graphs and groupoid $C^*$-algebras -QUESTION [8 upvotes]: In the paper "Graphs, groupoids and Cuntz-Krieger algebras" by Kumijan, Pask, Raeburn, Renault it was shown (if I understand it correctly) that whenever $G$ is a row-finite directed graph -with no sinks then the corresponding graph $C^*$-algebra is isomorphic to the groupoid $C^*$-algebra -w.r. to a suitable groupoid having the space of infinite paths as its unit space. -My question is whether it is possible to remove the limitation of having no sinks; I know that a typical way of "desingularizing" this sort of graphs is to add an infinite tail to any sink. Does it remain true, then, that the graph $C^*$-algebra is isomorphic to the groupoid $C^*$-algebra associated to the path groupoid? -On a more general note: what is known about functoriality of these three procedures? -$$\begin{aligned} -\textrm{graph}&\longrightarrow C^*\textrm{-algebra}\\ -\textrm{groupoid}&\longrightarrow C^*\textrm{-algebra}\\ -\textrm{graph}&\longrightarrow\textrm{groupoid}\end{aligned}$$ - -REPLY [6 votes]: The process of adding tails to the graph does not preserve the isomorphism class of the graph $C^*$-algebra. One simple example is to allow $E$ the be the graph consisting of one vertex and no edge. Then working from the rules for graph algebras (for graphs with sinks), you obtain that $C^*(E) \cong \mathbb{C}$. But $\tilde{E}$ will be an infinite chain of edges, and you can check that $C^*(\tilde{E})$ will be isomorphic to the compact operators on $\ell^2$, $C^*(\tilde{E}) \cong \mathcal{K}(\ell^2)$. -However, you don't lose everything by desingularizing. Desingularizing is useful because we will have $C^*(E) \sim_{\text{M. E.}} C^*(\tilde{E})$, where $\sim_{\text{M. E.}}$ denotes Morita equivalence. This is an equivalence relation between $C^*$-algebras which is weaker than isomorphism but still strong enough to ensure that many important $C^*$-algebraic properties of $C^*(E)$ are shared by $C^*(\tilde{E})$. For example, the lattice of closed two-sided ideals of $C^*(E)$ will be isomorphic to that of $C^*(\tilde{E})$ (so in particular they either are both simple as $C^*$-algebras or neither is). Many other key properties are preserved under Morita equivalence, such as being AF or purely infinite, having real rank zero, and so on. -A good place to look at this is the paper "The $C^*$-algebras of arbitrary graphs" by Drinen and Tomforde (https://projecteuclid.org/euclid.rmjm/1181069770). They develop the machinery of desingularization in some detail. In fact, their construction goes beyond removing sinks, so that infinite emitters can be removed in similar fashion from a graph without disturbing the Morita equivalence class of the affiliated $C^*$-algebra. Another route is via inverse semigroups, which Paterson carried out in this paper: http://www.theta.ro/jot/archive/2002-048-003/2002-048-003-012.pdf. Note that this other paper only addresses the infinite emitter problem, all graphs are assumed sinkless. These desingularization and inverse semigroup constructions have various generalizations to more general families of $C^*$-algebras: Cuntz-Pimsner algebras, $C^*$-algebras of higher-rank graphs, and so on. -To address your more general note: the graph $C^*$-algebra construction is not usually studied from a functorial approach. This is because in order to obtain a meaningful functor from Graphs to $C^*$-algebras, you have to be very strict as to the sort of morphisms you allow in the Graphs category. For example, suupose that $m: E \to F$ is a morphism of graphs (not saying what that means right now, but certainly it should carry vertices to vertices and edges to edges in such a way that it respects the range and source maps). Then (assuming we want the graph algebra construction to be covariant) we should get some $C^*$-algebra homomorphism $\phi=\phi_m: C^*(E) \to C^*(F)$. You might start by saying that $\phi(s_e) = t_{m(e)}$ for an edge $e \in E^1$ and $\phi(p_v) = q_{m(v)}$ for a vertex $v \in E^0$. Here $C^*(E)$ is generated by $\{s_e,p_v\}_{e \in E^1,v \in E^0}$ and $C^*(F)$ is generated by $\{t_f, q_w \}_{f \in F^1, w \in F^0}$, i.e. the universal Cuntz-Krieger families. This can cause issues if the morphism $m$ is not injective on edges. For if $e,e'$ are two distinct edges of $E$, then $s_e^* s_{e'}^* = 0$ by the definition of a graph algebra. If $m(e)=m(e')$, however, our ingenuous homomorphism will cause a problem -$$ 0 = \phi(0) = \phi(s_e^* s_{e'}) = \phi(s_e)^* \phi(s_{e'})= t_{m(e)}^* t_{m(e')} = t_{m(e)}^* t_{m(e)} = q_{s(m(e))}$$ -Since the universal generators for a graph algebra are non-zero, this causes a problem. You might start working through what the requirements are that a graph morphism must satisfy in order to define a $C^*$-algebra homomorphism in this naive fashion. -The second construction is better behaved. Of course there is a notion of isomorphism for groupoids, and the $C^*$-algebra respects it. Moreover, there is a more general notion of equivalence for groupoids. Two (sufficiently nice) topological groupoids are equivalent if they each act on a common space $Z$ in a compatible fashion. For details, see this useful paper of Muhly, "Equivalence and isomorphism for groupoid $C^*$-algebras" (http://www.theta.ro/jot/archive/1987-017-001/1987-017-001-001.pdf). The main theorem of this paper states that if two groupoids equivalent in the above sense, then their affiliated groupoid $C^*$-algebras are Morita equivalent. I am not sure if people have spent a lot of time looking at the category of topological groupoids in and of itself, or functorial properties of the groupoid $C^*$-algebra construction. One reason for this is that, most of the time, you don't work very often with morphisms between groupoids. They arise, but they are not central in many theorems. It might be that there is a category of groupoids and "one-sided equivalences" and that the equivalence relation defined above is the isomorphism relation in this category. This would line up better with the categorical construction for Cuntz-Pimsner algebras. -For the last direction you asked about, the path groupoid construction is simple enough that you can read most interesting properties directly from the parent graph. That is, the structure of the path groupoid as a topological groupoid (is it Cartan, proper, transitive, etc.) is generally straightforward to determine from the structure of the graph. If you are interested in functoriality you'd again have to decide what kind of graph morphisms you want to use and what kind of "groupoid morphisms" you'd want to use. If you have a graph morphism $m$ (of the vague kind considered in the $C^*$-algebra question), then it will certainly define a map on the infinite path spaces respecting the shift maps, so that it should it will define an algebraic groupoid homomorphism (a map $\phi: G_E \to G_F$ mapping composable pairs to composable pairs and acting like a group homomorphism, $\phi(gg')=\phi(g)\phi(g')$). The issue that could arise is that while I believe it is continuous, $\phi$ might fail to be proper (I think, this is speculative). You might look also at this paper of Putnam (http://www.math.uvic.ca/faculty/putnam/r/9901main.pdf) on Smale spaces for more ideas about functoriality for a related groupoid construction. There the desired morphism of groupoids is an injective groupoid homomorphism which is a homeomorphism onto an open subgroupoid of the codomain. But I'm far away from understanding this stuff.<|endoftext|> -TITLE: Concatenation of strings -QUESTION [5 upvotes]: We have two strings (i. e., finite tuples) $A$ and $B$. -We have to find if for some positive integers $n$ and $m$, the string $A$ concatenated $n$ times equals the string $B$ concatenated $m$ times or not. -I have made an interesting observation but am unable to prove it. It appears that this can happen if and only if $AB=BA$. - -REPLY [4 votes]: Your question seems to be answered by Theorem 1.5.3 of Allouche and Shallit's Automatic Sequences: Theory, Applications, Generalizations (Cambridge University Press, 2003). - -Let $x,y\in\Sigma^+$. Then the following three conditions are equivalent: - (1) $xy=yx$. - (2) There exists integers $i,j>0$ such that $x^i=y^j$. - (3) There exist $z\in\Sigma^+$ and integers $k,l>0$ such that $x=z^k$ and $y=z^l$. - -The book has a proof.<|endoftext|> -TITLE: The First Failure of GCH in Large Cardinals Smaller than Measurables -QUESTION [6 upvotes]: A well known theorem by Scott says: -If $\kappa$ is a measurable cardinal and $\mu$ a normal measure on it and $\mu (\lbrace\lambda\in\kappa~|~2^{\lambda}=\lambda^{+}\rbrace)=1$ then $2^{\kappa}=\kappa^{+}$. -So: -If $\text{GCH}$ is false in a measurable cardinal then there is a smaller cardinal which $\text{GCH}$ is false in it too. -Equivalently: -If $\text{GCH}$ is false then its first failure point is not a measurable cardinal. -The role of existence of a normal measure on the cardinal $\kappa$ seems very essential in the proof of Scott's theorem. So perhaps one can prove the following statement: -For any large cardinal type $\text{A}$ smaller than measurables, it is consistent with $\text{ZFC}$ that $\text{GCH}$ be false and its first failure point be a large cardinal of type $\text{A}$. -Question is: -Question: Is the above statement true? For what type of large cardinals less than measurables is there a known result like above statement? - -REPLY [9 votes]: The statement is not true. -Theorem. If $\kappa$ is strongly unfoldable and the GCH holds below $\kappa$, then it holds at $\kappa$ also. -This is just because the strongly unfoldable cardinals are $\Sigma_2$-reflecting, and the result also holds for any $\Sigma_2$-reflecting cardinal. The strongly unfoldable cardinals are less than measurable in consistency strength, being equiconsistent with an unfoldable cardinal, and in particular, they are consistent with $V=L$. -Another small instance of the GCH overspill phenomenon, by essentially the same argument, occurs when $\kappa$ is $\Sigma_2$-extendible, which means that $V_\kappa\prec_{\Sigma_2}V_\theta$ for some ordinal $\theta$. If the GCH holds up to $\kappa$, then it will also hold up to $\theta$, and so at $\kappa$ itself. Meanwhile, these $\Sigma_2$-extendible cardinals have no strength at all---they provably exist in ZFC, unless you insist that they are also inaccessible, in which case the strength is strictly weaker than the existence of a Mahlo cardinal. -Meanwhile, your statement is indeed true for many of the large cardinals below measurable. Many of the large cardinal properties below measurable have to do with the existence of size $\kappa$ objects, such as embeddings between $\kappa$-models or homogeneous sets for colorings of size $\kappa$. For any such kind of large cardinal property, if $\kappa$ can be made to have this property indestrucitibly after $\text{Add}(\kappa,1)$, then we can make the GCH fail at $\kappa$ simply by forcing with $\text{Add}(\kappa,\kappa^{++})$. The large cardinal property will be preserved since all the size $\kappa$ objects are added by a size $\kappa$-subforcing, which amounts to $\text{Add}(\kappa,1)$. This is how one can show that it is consistent that the GCH fails first at $\kappa$, when $\kappa$ is weakly compact, indescribable, unfoldable, Ramsey, strongly Ramsey and so on. One should simply perform the preparatory forcing to make $\kappa$ indestructible by $\text{Add}(\kappa,1)$, and then appeal to the argument I just gave. -Update. I should not have included the indescribable cardinals on the list, because every $\Pi^2_1$-indescribable cardinal is $(\kappa+1)$-strongly unfoldable and so the failure of the GCH at $\kappa$ will reflect down. - -REPLY [6 votes]: As far as I know the best results in this direction are due to Levinski (see his paper "Filters and large cardinals"). -The following results are taken from the above paper. First a few definitions: -A premeasure on $\kappa$ is a normal filter $F$ on $\kappa$ such that $P(\kappa)/I$ is $\kappa^+-$distributive, $I$ the dual of $F$. $\kappa$ is premeasurable if it carries a premeasure. -A quasi measure on $\kappa$ is a normal filter $F$ on $\kappa$ such that $P(\kappa)/I$ ha a $\leq \kappa-$closed dense subset. $\kappa$ is quasi measurable if it carries a quasi measure. -Theorem A. Assume GCH+ there exists a measurable cardinal $\kappa$ and let $\lambda \geq \kappa^{++}.$ then there is a generic extension in which $\kappa$ is the first cardinal violating GCH, $2^\kappa \geq \lambda$ and $\kappa$ is premeasurable. -Theorem B. Assume there exists a measurable cardinal $\kappa$ of Mitchell order $\kappa^{++}.$ then there is a generic extension in which $\kappa$ is the first cardinal violating GCH, $2^\kappa = \kappa^{++}$ and $\kappa$ is quasi measurable.<|endoftext|> -TITLE: Sequences with integral means -QUESTION [22 upvotes]: Let $S(n)$ be the sequence whose first element is $n$, and from then onward, -the next element is the smallest natural number ${\ge}1$ that ensures that the -mean of all the numbers in the sequence is an integer. -For example, the second element of $S(4)$ cannot be $1$ (mean $\frac{5}{2}$), -but $2$ works: $S(4)=4,2,\ldots$. Then the third element cannot be -$1$ (mean $\frac{7}{3}$) -or $2$ (mean $\frac{8}{3}$), but $3$ works: $S(4)=4,2,3,\ldots$. -And from then on, the elements are all $3$'s, which I'll write as -$S(4)=4,2,\overline{3}$. -Here are a few more examples: -$$S(1)=1,\overline{1}$$ -$$S(2)=2,2,\overline{2}$$ -$$S(3)=3,1,\overline{2}$$ -$$S(4)=4,2,\overline{3}$$ -$$S(5)=5,1,\overline{3}$$ -$$S(11)=11,1,3,1,\overline{4}$$ -$$S(32)=32,2,2,4,5,3,1,\overline{7}$$ -$$S(111)=111,1,2,2,4,6,7,3,8,6,4,2,\overline{13}$$ -$$S(112)=112,2,3,3,5,1,7,3,8,6,4,2,\overline{13}$$ -$$S(200)=200,2,2,4,2,6,1,7,1,5,1,9,7,5,3,1,\overline{16}$$ -Has anyone studied these sequences? -Is there a simple proof that each sequence ends with a repeated number -$\overline{m}$? -Is there a way to predict the value of $m$ from the start $n$ without computing the -entire sequence up to $\overline{m}$? -Might it be that the repeat value $m=r(n)$ satisfies $r(n+1) \ge r(n)$? -This question occurred to me when thinking about streaming computation of means, -a not-infrequent calculation (e.g., computing mean temperatures). - -Added, supporting the discoveries of several commenters: $r(n)$ red, -and now fit with $1.135 \sqrt{n}$ blue for $n\le 10000$: -  - -As per Eckhard's request, with Will's function $\sqrt{4n/3} -1/2$: - -REPLY [16 votes]: Will Sawin's guess that the mean stabilizes around $2\sqrt{n/3}$ is very close to the correct answer, but not quite! Asymptotically the right answer is $\sim 2\sqrt{n/\pi}$. Note that $2/\sqrt{\pi}$ is about $1.12837\ldots$. -To see why this is, suppose that $n$ is large, and let $\ell$ also be large, but -small compared with $n$. Consider the number of steps $k$ needed until the mean first dips below $k\ell$. It is easy to see that $k\sim \sqrt{n/\ell}$. Now if we continue from step $k$, the mean must first drop by $\ell-1$ at each step, until we get to a point $k_1$ such that the mean at this point $k_1$ dips below $k_1(\ell-1)$. It is easy to see that $k_1$ is about $k(2\ell-1)/(2\ell-2)$. From here on at each step the mean drops by $\ell -2$ at each step, until we arrive at a point $k_2$ where the mean has dipped below $k_2(\ell-2)$. Now $k_2$ is about -$$ -k\Big(\frac{2\ell-1}{2\ell-2}\Big) \Big(\frac{2\ell-3}{2\ell-4}\Big). -$$ -And so on. (In following the above calculations, the comment of Sergei Ivanov is useful.) -In this manner we find that the number of steps taken to terminate (which is also the size of the final mean) is about -$$ -\sqrt{n/\ell} \Big(\frac{2\ell-1}{2\ell-2}\Big) \cdots \Big(\frac{5}{4}\Big)\Big(\frac{3}{2}\Big). -$$ -Computing this using Stirling's formula finishes the proof. -Note: Everyone writes, no one reads! (Myself included, of course.) I just looked at the OEIS reference linked by Elkies, and the $\sim 2\sqrt{n/\pi}$ result essentially appears there. Apparently this is first due to Viggo Brun himself, and refined by M.E. Andersson in a paper in Acta Arithmetica (see http://matwbn.icm.edu.pl/ksiazki/aa/aa85/aa8542.pdf ). But the paper is in German and six pages long, so the above short proof may still be useful.<|endoftext|> -TITLE: Ways to define "definability" -QUESTION [12 upvotes]: The notion of a definable set is not expressible in the language of set theory: there is no formula $\delta(x)$ that is equivalent with there being a formula $\phi(y)$ such that $x = \lbrace y : \phi(y)\rbrace$, i.e. for which -$$\delta(x) \leftrightarrow (\exists \phi \in \mathsf{Form})\ x = \lbrace y : \phi(y)\rbrace$$ -is provable. (Please forgive my blunt use of the set-builder notation $\lbrace y : \phi(y)\rbrace$. I know there's the rub.) -The same holds for (unrestricted) definitions with parameters: there is no formula $\delta(x)$ that is equivalent with there being a formula $\phi(y,z_1,\dots,z_n)$ such that $x = \lbrace y : \phi(y,z_1,\dots,z_n)\rbrace$ for some $z_1,\dots,z_n$, i.e. for which -$$\delta(x) \leftrightarrow (\exists \phi \in \mathsf{Form})(\exists z_1,\dots,z_n)\ x = \lbrace y : \phi(y,z_1,\dots,z_n)\rbrace$$ -is provable. -But the notion of ordinal-definable sets is expressible in the language of set theory (see Jech, p. 194), i.e. there is a formula $\delta(x)$ such that - bluntly said - -$$\delta(x) \leftrightarrow (\exists \phi \in \mathsf{Form})(\exists z_1,\dots,z_n)\ x = \lbrace y : \phi(y,z_1,\dots,z_n)\rbrace \wedge \mathsf{ON}(z_i) $$ -is provable. -My questions are: - - -Are there other restrictions on the parameters like $\mathsf{ON}(z_i)$ (the formula that states that $z_i$ is an ordinal) - that give rise to a definable notion of definability, i.e. a class of - accordingly definable sets? -Can these restrictions been characterized? What do they have in common, eventually? What is essential? -Are there restrictions $\Psi(\phi)$ on the formulas that comparably give rise to a definable notion of definability, i.e. there would be a formula - $\delta(x)$ such that - $$\delta(x) \leftrightarrow (\exists \phi \in \mathsf{Form})(\exists z_1,\dots,z_n)\ x = \lbrace y : \phi(y,z_1,\dots,z_n)\rbrace \wedge \Psi(\phi) $$ - is provable. -Or maybe some mixture $\Psi(\phi,z_1,\dots,z_n)$? - - -Question 3 includes: Can we - in a Goedelian way - define for a formula (as a set), that some of its arguments must fulfill some other formula? - -REPLY [6 votes]: With respect to question 3: one way to do this is to restrict the quantifier complexity of the formulas. That is, for every constant $k$, one can define a formula $\delta_k(x)$ expressing “$x$ is definable by a $\Sigma_k$-formula with parameters”. Here, a formula $\phi(\vec x)$ is $\Sigma_k$ (in the Lévy hierarchy) if it can be written in the form $\exists y_1\,\forall y_2\,\exists y_3\,\dots Qy_k\,\theta(\vec x,\vec y)$, where all quantifiers in $\theta$ are bounded. -$\def\Tr{\mathrm{Tr}}$EDIT: Let me clarify how exactly $\Sigma_k$-formulas can be used to answer question 3. What I wrote above is true, but useless by itself, because definability of sets by formulas allowing arbitrary parameters is not an interesting notion (every set is trivially definable from itself as a parameter). But the Lévy hierarchy provides more than that: - -For every $k>0$, there is a truth definition for $\Sigma_k$-formulas, which is itself $\Sigma_k$. That is, there is a $\Sigma_k$-formula $\Tr_k(n,x)$ such that ZF proves -$$\phi(x)\leftrightarrow\Tr_k(\ulcorner\phi\urcorner,x)$$ -for every $\Sigma_k$-formula $\phi(x)$, where $\ulcorner\phi\urcorner$ is the Gödel number of $\phi$. This is essentially optimal: if a class of formulas has a truth definition, it is up to equivalence included in some $\Sigma_k$ (because the truth definition itself is in $\Sigma_k$ for some $k$). -If $k>0$, and $X$ is a class of possible parameters which includes $\omega$ and is closed under pairing (that is, $X\times X\subseteq X$), there is a formula $\delta(x)$ which expresses “$x$ is $\Sigma_k$-definable using parameters from $X$”, namely -$$\delta(x)=\exists n\in\omega\,\exists z\in X\,\forall y\,(y\in x\eq\Tr_k(n,\langle y,z\rangle)).$$ -(Here, the pairing operation on $X$ does not have to be the standard Kuratowski pair. For example, one can define an injection $\mathrm{Ord}\times\mathrm{Ord}\to\mathrm{Ord}$, which allows us to take $X=\mathrm{Ord}$, so this generalizes ordinal definability. Also, I believe the assumption $\omega\subseteq X$ is redundant as long as $|X|\ge2$, because one can reconstruct a suitable copy of $\omega$ inside $X$ using the pairing operation.)<|endoftext|> -TITLE: Examples of optimal ultracontractivity estimates for a Markovian semigroup $T_t$ that do not depend polynomialy on $t$ -QUESTION [6 upvotes]: Let $(X,\mu)$ be a measure space and $T_t : L_2(\mu) \to L_2(\mu)$ for $t \geq 0$ a symmetric Markovian semigroup. Local ultracontractivity estimates of the form: -$$ -\| T_t : L_p(\mu) \to L_q(\mu)\| \leq \frac{C_{p,q}}{t^{\frac{d}{2} ( \frac{1}{p} - \frac{1}{q} )}} \text{, } t \leq 1. -$$ -had been studied in connection with local Sobolev inequalities, see for example theorem II.4.2 of [1]. In particular they are all equivalent for different $p < q$. Abusing slightly the terminology we will say that the bounds on the norm $\| T_t : L_p \to L_q \|$ are polynomial if they are of the form $t^{- \alpha}$. -Question 1 Are there concrete examples of symmetric Markovian semigroups for which the optimal bound of $\| T_t : L_p(\mu) \to L_q(\mu)\|$ with $p < q$ is not polynomial. More generally, given a decreasing function $\Phi : (0,1] \to [0,\infty)$ are there any known techniques to construct $(X,\mu,(T_t)_t)$ such that: -$$ -C_1 \Phi(t) \leq \| T_t : L_2(\mu) \to L_\infty(\mu)\| \leq C_2 \Phi(t) \text{, } t \leq 1. -$$ -Question 2 If $(G,\mu)$ is a LCH unimodular group with its Haar measure and $(T_t)_t$ is a right (resp. left) invariant symmetric Markovian semigroup, can local ultracontractivity estimates be non polynomial? For which pairs $(G, (T_t)_t )$ we are forced to have polynomial bounds? -For the second question there seems to be some partial results known in the literature. For instance, if $G$ is an unimodular Lie group and $T_t = e^{-t L}$ where the infinitesimal generator $L$ can be expressed as: -$$ -L = \sum_{i=1}^{k}{X^{\ast}_i X_i} = - \sum_{i=1}^{k}{X_i X_i} -$$ -for some $\{X_i\}_i$ generating the whole Lie algebra. Then, using the argument of 5.6.1 in [2], one can prove an scale invariant Poincaré inequality. As small balls grow polynomially for the subriemannian metric $d_L$ associated to $L$ we have that $(G,d_L,\mu)$ is a doubling metric measure space for small balls. Following 5.5.1 in [2] a scale invariant Poincaré inequality together with a doubling condition for small balls implies a Parabolic Harnack principle for small regions. Using the parabolic Harnack principle bilateral Gaussian bounds can be found for the heat kernel, obtaining: -$$ -\frac{C_1}{\mu(B_e(\sqrt{t}))} e^{\beta_1 \frac{d(x,e)^2}{t}} \leq h_t(x) \leq \frac{C_2}{\mu(B_e(\sqrt{t}))} e^{\beta_2 \frac{d(x,e)^2}{t}} \text{, } t \leq 1. -$$ -Since the volume of balls for small $t$ is comparable to $t^{d_0}$, where $d_0$ is the homogeneous dimension, the optimal ultracontractivity estimates are polynomial. -Can this construction be extended to more general infinitesimal generators in Lie groups or more general LCH groups? -My intuition is that, since measure spaces with a Markovian semigroup are a very big class, the first question will have a positive answer. For the second question the extra rigidity of working with invariant semigroups will reduce the possible ways of decay and maybe the answer is negative. -[1] Varopoulos, Saloff-Coste & Coulhon: The Analysis and Geometry of Groups. -[2] Saloff-Coste , "Aspect of Sobolev Type Inequalities". - -REPLY [4 votes]: Here are some partial answers to your question -Q1) There are several interesting examples from random walks on groups. The examples that you are looking for are random walks on non-amenable groups, polycyclic groups with exponential volume growth, Lamplighter groups, etc. See for example Theorem 2.3 in www.math.u-psud.fr/~breuilla/SaloffNotes.pdf and references there. -Q2) Yes, the ultracontractivity estimate need not be polynomial. Consider Brownian motion on $G= \mathbb{R} \times S^1$, a cylinder. It has both Volume doubling property and satisfies Poincaré inequality. Hence it satisfies a Parabolic Harnack inequality and as you mentioned a two-sided gaussian estimate. However the volume growth of the balls in not polynomial (small balls have quadratic volume growth and larger balls have linear growth). Hence the ultracontractivity estimate is not polynomial.<|endoftext|> -TITLE: Can the sphere be partitioned into small congruent cells? -QUESTION [36 upvotes]: On the unit $2$-sphere ${\mathbb S}^2$ furnished with the geodesic distance, a subset homeomorphic to a planar disk is called a cell. A finite family of cells is a tiling if their interiors are mutually disjoint and their union is the whole sphere. My main question is: -(1) Can the sphere be tiled by congruent cells of an arbitrarily small diameter? If not, how small can the diameters of the cells be? -An obvious example of a tiling with arbitrarily many congruent cells is obtained by cutting the sphere into $n$ sectors by $n$ uniformly spaced great semicircles, each connecting the North and the South poles. Since the cells' diameter is $\pi$ - the same as the diameter of the whole sphere, they cannot be called small by any means. -A somewhat less obvious example is constructed as follows. Consider the $4k$-faceted polyhedron inscribed in the sphere, consisting of a $2k$-faceted antiprism ($k\ge3$) capped off by two pyramids, as shown below for $k=18$. With the properly chosen altitude of the antiprism, all $4k$ (isosceles-triangular) facets become congruent by design. -${\qquad\qquad\qquad}$ -The central projection of the facets to the sphere produces a tiling of the sphere with $4k$ congruent, isosceles-triangular cells of diameter considerably smaller than $\pi$, but greater than $\pi/3$ and converging to $\pi/3$ as $k\to\infty$. -In the special case of $k=5$, the inscribed polyhedron is the regular icosahedron. In this case, if each of its $20$ equilateral triangular facets is barycentrically partitioned into 6 triangles, the central projection to the sphere yields a tiling with $120$ congruent, triangular cells of diameter well below $\pi/3$. No better examples are known to me, which raises the following, specific two questions: -(2) Is there a tiling of the sphere with an arbitrarily large number of congruent tiles, each of diameter $d\le\pi/3$? -(3) Is there a tiling of the sphere with congruent cells of diameter smaller than that in the subdivided-dodecahedral $120$-cell tiling described above? - -REPLY [22 votes]: Not an answer. But permit me to draw attention to Robert J. MacG. Dawson's website on congruent sphere tilings, -including this beautiful tiling by triangles:<|endoftext|> -TITLE: Is there an irreducible but solvable septic trinomial $x^7+ax^n+b = 0$? -QUESTION [19 upvotes]: The following irreducible trinomials are solvable: -$$x^5-5x^2-3 = 0$$ -$$x^6+3x+3 = 0$$ -$$x^8-5x-5=0$$ -Their Galois groups are isomorphic to ${\rm D}_5$, ${\rm S}_3 \wr {\rm C}_2$ and -$({\rm S}_4 \times {\rm S}_4) \rtimes {\rm C}_2$, respectively. -Question: Is there an irreducible septic trinomial $x^7+ax^n+b=0$ with -solvable Galois group, where $n \in \{1, \dots, 6\}$, $a, b \in \mathbb{Z} \setminus \{0\}$? -P.S. For $n=1$, I did a search only for those with 1 real root and, if I did it correctly, there are none with integer $|a,b|<50$. - -REPLY [5 votes]: The good news is that in this book: -Generic Polynomials: Constructive Aspects of the Inverse Galois Problem - By Christian U. Jensen, Arne Ledet, Noriko Yui (pp. 52 and up). -There is a complete criterion, as follows: -$x^7 + a x + b$ is solvable (we assume it's irreducible) if and only if the polynomial (which I am intentionally leaving in Mathematica input form, so you can type it in yourself). -P35[a_, b_, x_] := - x^35 + 40 a x^29 + 302 b x^28 - 1614 a^2 x^23 + 2706 a b x^22 + - 3828 b^2 x^21 - 5072 a^3 x^17 + 2778 a^2 b x^16 - - 18084 a b^2 x^15 + 36250 b^3 x^14 - 5147 a^4 x^11 - - 1354 a^3 b x^10 - 21192 a^2 b^2 x^9 - 26326 a b^3 x^8 - - 7309 b^4 x^7 - 1728 a^5 x^5 - 1728 a^4 b x^4 + 720 a^3 b^2 x^3 + - 928 a^2 b^3 x^2 - 64 a b^4 x - 128 b^5 -has factorization pattern [degrees of irreducible factors] one of (14, 21), (7, 7, 7, 21), (7, 7, 7, 14), (7, 7, 7, 7, 7) -The bad news is that I have run this for all pairs (a, b) where both coordinates are between -1000 and 1000, and there is not a single success.<|endoftext|> -TITLE: Linearisation of a group -QUESTION [5 upvotes]: If $G$ is a connected Lie group acting on a vector $\mathbb{C}$-space $V$ then it is well known that the algebra of invariants $\mathbb{C}[V]^G$ coincides with the algebra of invariants $\mathbb{C}[V]^L$ of corresponding Lie algebra $L.$ -Question. Let now $G$ be a finite group, $V$ be its representation. Is there exists a Lie algebra $L$ and its representation on $V$ such that $\mathbb{C}[V]^G=\mathbb{C}[V]^L$? -It is easy to see that it impossible for symmetric group $S_n$. But maybe there are classes of finite groups for which can be found the positive answer? -Edit. Let $V=< v_1,v_2,\ldots,v_n >$ be standard representation of the symmetric group $S_n.$ Suppose that there exist a derivation $D=f_1 \frac{\partial}{\partial x_1}+\cdots+ f_n \frac{\partial}{\partial x_n}, f_i \in S(V)$ of the symmetric algebra $S(V)$ such that $D(I)=0$ for all $I \in \mathbb{C}[V]^{S_n}$. -Let $I_1,I_2,\ldots, I_n \in S(V)$ be a minimal generating set for $\mathbb{C}[V]^{S_n}.$ Since $D(I_k)=0, \forall k,$ we get the system of polynomial equations on $f_1,f_2,\ldots,f_n:$ -$$ -f_1 \frac{\partial I_1}{\partial x_1}+\cdots+ f_n \frac{\partial I_1}{\partial x_n}=0,\\ -f_1 \frac{\partial I_2}{\partial x_1}+\cdots+ f_n \frac{\partial I_2}{\partial x_n}=0,\\ -\ldots \\ -f_1 \frac{\partial I_n}{\partial x_1}+\cdots+ f_n \frac{\partial I_n}{\partial x_n}=0. -$$ -The Jacobian of system of invariants $I_1,I_2,\ldots, I_n$ is not zero. It follows the system has only trivial solution $f_i=0.$ - -REPLY [3 votes]: You asked for it! Here's the Hilbert series argument. I need to assume that the derivation $D$ is homogeneous, or equivalently comes from extending a linear map $V \to V$. This covers the case of a Lie algebra action on $V$. -First, recall that $\dim S^n(V^{\ast}) = {\dim V + n - 1 \choose \dim V - 1}$ has asymptotic growth rate $\Theta(n^{\dim V - 1})$ for $\dim V$ fixed and $n \to \infty$. -Let $g_n = \dim S^n(V^{\ast})^G$ denote the dimension of the $n^{th}$ graded component of $\mathbb{C}[V]^G$. An explicit expression for the Hilbert series $\sum g_n t^n$ can be obtained from Molien's theorem which shows in particular that the leading term in the asymptotic expansion of $g_n$ is at least $\frac{1}{|G|} n^{\dim V - 1}$ (hence, as above, that $\mathbb{C}[V]^G$ has Krull dimension $\dim V$). There should be a more elementary way to obtain this estimate. -On the other hand, let $D$ be a derivation induced from a linear map $V \to V$. Let $e_1, ..., e_n$ be a Jordan basis of $D$ with generalized eigenvalues $\lambda_1, ..., \lambda_n$. Then the monomial $e_1^{k_1} ... e_n^{k_n}$ in $\mathbb{C}[V]$ is a generalized eigenvector of $D$ with generalized eigenvalue $\sum k_i \lambda_i$. In particular, the $n^{th}$ graded component of $\mathbb{C}[V]^D$ has dimension at most the number of monomials of degree $n$ of generalized eigenvalue $0$. The exponents of these monomials are determined by the kernel of the map -$$\mathbb{Z}^n \ni (k_1, ..., k_n) \mapsto \sum_{i=1}^n k_i \lambda_i \in \mathbb{C}.$$ -If at least one of the $\lambda_i$ is nonzero, the kernel of the above map is a submodule of rank at most $n-1$, from which it follows that the dimension $d_n$ of the $n^{th}$ graded component of $\mathbb{C}[V]^D$ has asymptotic growth rate at most as fast as $O(n^{\dim V - 2})$. -If all of the $\lambda_i$ are equal to zero, then $D$ is nilpotent. In this case we need to inspect the action of $D$ on monomials more carefully. Permute the Jordan basis so that $D(e_i)$ is either $e_{i-1}$ or $0$ and so that $D(e_2) = e_1$, and consider the lexicographic order on $\mathbb{C}[V]$ with respect to the $e_i$. Then $D(e_1^{m_1} ... e_k^{m_k})$ has leading term $m_2 e_1^{m_1+1} e_2^{m_2-1} ... e_k^{m_k}$, from which it follows that the image of $D$ acting on $S^n(V^{\ast})$ has dimension at least the number of monomials with $m_1 \ge 1$, or the dimension of $S^{n-1}(V^{\ast})$. The two dimensions are polynomials in $n$ with the same leading term and $d_n$ is their difference, so we again find that the asymptotic growth rate of $d_n$ is at most $O(n^{\dim V - 2})$.<|endoftext|> -TITLE: Is the list of "known" 3D compact manifolds complete? -QUESTION [8 upvotes]: "it is an open question if the known compact manifolds in 3-D are complete." - -This is a quote from Eric Weisstein's -CRC Concise Encyclopedia of Mathematics, Second Edition. 2010, p.480. -(Google Books link) -Is this still the case, post-Perelman? What are the known compact manifolds in $\mathbb{R}^3$? -I ask this as someone (obviously!) naive in these areas. -Thanks for educating me! - -REPLY [8 votes]: I like to think that compact 3-manifolds will not be known until we know "the list" of compact, oriented, hyperbolic 3-manifolds, in the way that we know "the list" of compact, oriented, hyperbolic surfaces: the genus 2 surface, the genus 3 surface, the genus 4 surface, ... -We don't know "the list" yet, as the other answers indicate. But the potential for ordering hyperbolic 3-manifolds by volume was established by Jorgensen who proved that the set of volumes of finite volume hyperbolic 3-manifolds is a well-ordered set with order type $\omega^\omega$ (the same order type as polynomials with natural number coefficients), and each volume occurs for only finitely many manifolds. And with that in mind, we know the first entry: the Weeks manifold, which is the unique lowest volume closed hyperbolic 3-manifold, as proved by Gabai, Meyerhoff, and Milley. I'm not up on the very latest developments of this technology, but I think that the next few lowest volumes have also been completely listed, using the MOM technology of the same three authors.<|endoftext|> -TITLE: Are the reduced group Von Neumann algebra/ Group $C^{\ast}$ algebra functorial in the case of LCH groups -QUESTION [9 upvotes]: Let $G$ be a LCH group and $\mu$ be its left Haar measure. Call $\lambda_G : G \to U(L_2(G,\mu))$ the left regular representation. We can define the reduced $C^{\ast}$ algebra and reduced Von Neumann algebra, $C_{\lambda}^{\ast} G $ and $W_{\lambda}^{\ast} G $ respectively, as the smallest $C^{\ast}$, resp. Von Neumann, subalgebras of $B(L_2(G,\mu))$ containing $\lambda_G[C_c(G)]$. It is easy to see that -$$ -C_{\lambda}^{\ast} G \subset M(C_{\lambda}^{\ast} G) \subset W_{\lambda}^{\ast} G -$$ -and that for every $\mu$ in the space of finite Borel measures over $G$, $\lambda(\mu) \in M(C_{\lambda}^{\ast}G )$. -Question 1 Let $j : H \to G$ be a proper continuous injective homomorphism whose image $j H$ we will denote also by $H$. Let $\nu_{H} \in M_{\text{loc}}(G)$ be the locally finite measure described by its action on the compactly supported continuous functions $C_c(G)$ as: -$$ -\nu_H (f) = \int_{H} f |_H (j(h)) d \mu_H (h), -$$ -where $\mu_H (h)$ is the Haar measure on $H$. Or equivalently $j^{\ast} \mu_H = \nu_H$. For every $\varphi \in C_c(H)$ we can define $\varphi \nu_H$ as $j^{\ast} (\varphi \mu_H)$. Since $\varphi \nu_H \in M(G)$ we have that $\lambda_G (\varphi \nu_H) \in M(C_\lambda^{\ast} G)$. Do the map ${J}$ given by: -$$ -J : \lambda_H(\varphi) \mapsto \lambda_G(\varphi \nu_H) -$$ -extends to a normal $\ast$-homomorphism $J: W_{\lambda}^{\ast} H \to W_{\lambda}^{\ast} G$ ?. Does $J$ extends to a non degenerate $\ast$ homomorphism $J: C_{\lambda}^{\ast} H \to M( C_{\lambda}^{\ast} G )$? -Question 2 If $\alpha : G \to G$ is a continuous automorphism then, if $\varphi \in C_c(G)$, does the map $\Phi$ defined by: -$$ -\Phi : \lambda_G (\varphi) \mapsto \lambda_G (\alpha_{\ast} \varphi), -$$ -where $\alpha_{\ast} \varphi(t) = \varphi(\alpha(t))$, extends to a normal $\ast$-homomorphism $\Phi: W_{\lambda}^{\ast} G \to W_{\lambda}^{\ast} G$?. Does $\Phi$ extends to a non degenerate $\ast$ homomorphism $\Phi: C_{\lambda}^{\ast} G \to M( C_{\lambda}^{\ast} G )$? -Question 3 If $q : G \to K$ is a continuous and open surjective group homomorphism and $q^{-1}[\{e\}] = N$ be its kernel. Then given $\phi \in C_c(G)$ we can define $P_N \phi \in C_c(K)$ as: -$$ -(P_N f)(\kappa N) = \int_{N} f(\kappa \eta) d \mu_N (\eta) -$$ -And it is clear (as long as $\Delta_G |_N = \Delta_N$) that $P_N$ is contractive in the $L_1$ norm. Does the map $Q$ given by extension of: -$$ -Q : \lambda_G(f) \mapsto \lambda_K(P_N f) -$$ -extends to a normal $\ast$-homomorphism $Q: W_{\lambda}^{\ast} G \to W_{\lambda}^{\ast} K$, or to a non degenerate $\ast$-homomorphism $Q: C_{\lambda}^{\ast} G \to M( C_{\lambda}^{\ast} K )$ ?. -The first two questions seem to be easy when $\Delta_G |_{j H} = \Delta_H$ and when $\alpha$ is measure preserving, respectively. While the third seems much difficult to me. Indeed, assuming $\Delta_G |_{j H} = \Delta_H$, the $G$-space $X = G/H$ has a $G$ invariant measure $\rho$ such that $(G,\mu_G) = (H \times X,\mu_H \otimes \rho)$, where the equivalence is understood as measurable spaces. That equivalence induces a unitary isometry $u: L_2(G) \to L_2(H) \otimes_2 L_2(X,\rho)$. For every $j(h) \in jH$: -$$ -u \lambda_G(j(h)) = \lambda_H (h) \otimes \text{Id}_{L_2(X,\rho)} u. -$$ -Since $\lambda_G(j(h)) = J(\lambda_H(h))$, we have that $J$ is unitary equivalent to the tensor amplification. A very similar argument works for the second question. Just by constructing a unitary $u: L_2(G) \to L_2(G)$ given by $u(f)(t) = f(\phi(t))$ and see that $u$ intertwines $\Phi$ and the identity. Can this type of arguments be extended to the general case? Or are $\Delta_G |_{j H} = \Delta_H$ and $\alpha_{\ast}\mu_G = \mu_G$ necessary conditions for question 1 and question 2 respectively?. -I am aware that similar questions have been posted on this site, see [1] and [2] but in the context of discrete groups. In that setting the first and second questions have always a positive answer while the third has a positive answer for $C^{\ast}$ algebras if and only if $N$ is amenable and for Von Neumann algebras if and only if $N$ is finite. The converse in the $C^{\ast}$ algebra case seems to be related to the problem of characterizing amenable groups as those groups for which the co-unit $\mathcal{E} : C_\lambda^{\ast} G \to \mathbb{C}$ given by -$$ -\mathcal{E}(\lambda_G(f)) = \int_G f d \mu_G -$$ -is bounded. Are there references for this characterization in the case of non discrete groups? -[1] The functoriality of group C* algebra structure -[2] Is the group von Neumann algebra construction functorial? - -REPLY [6 votes]: The answer for Question 1 is "yes". I believe this to be a little subtle. Firstly, as $j:H\rightarrow G$ is proper and injective, $j(H)$ is closed in $G$, and $j:H\rightarrow j(H)$ is a homeomorphism. So wlog $H$ can be identified with a closed subgroup of $G$, with $j$ the inclusion. You sort of hint at this in the statement of your question. -Then, what is your map $J$? Well, if $\varphi\in C_c(H)$ then $f=\varphi\mu_H$ is a member of $L^1(H)$ (and such elements are dense). So $J\lambda_H(f) = \lambda_G(j^*(f))$ where $j^*$ is the pushforward $M(H)\rightarrow M(G)$. A more functional analytic way to think of this is to note that as $j$ is proper, it defines a map $\theta:C_0(G)\rightarrow C_0(H); g\mapsto g\circ j$ (as $H$ is closed, actually this map is a surjection). Then the Banach space adjoint is $j^*:M(H)=C_0(H)^* \rightarrow C_0(G)^*=M(G)$. -Now, when does $J$ extend to a normal $*$-homomorphism $W^*_\lambda(H) \rightarrow W^*_\lambda(G)$? I like to think of this in an abstract harmonic analysis framework-- the predual of $W^*_\lambda(G)$, denoted $A(G)$, can be given the structure of a commutative Banach algebra: the "Fourier Algebra" as defined by Eymard. If we regard $\lambda_G$ as a map $L^1(G)\rightarrow W^*_\lambda(G)$ then we can regard $\lambda_G^*$ as a map $A(G)\rightarrow L^1(G)^*=L^\infty(G)$, and then this actually maps into $C_0(G)$ densely (this "is" the Gelfand map of the commutative Banach algebra $A(G)$, suitably interpreted-- if $G$ is abelian, it is the Fourier transform, hence the name). -As $J$ is normal, it has a preadjoint $A(G)\rightarrow A(H)$, and if you regard these as non-closed subalgerbas of $C_0(G)$ and $C_0(H)$ respectively, we just get the map $\theta$ described above. So the question becomes equivalent to: does the map $a\mapsto a\circ j$ map $A(G)$ to $A(H)$ boundedly (hence automatically contractively). -It turns out that the answer is: "yes". It's even a quotient map-- every $A(H)$ function arises as the restriction of an $A(G)$ function (identifying $H$ as a closed subgroup of $G$). This theorem is known as "Herz restriction", and the nicest writeup I know is an MSc thesis: Cameron Zwarich's thesis, see Section 4.2. -Once we know $J$ exists, we know that $J\lambda_H(f)=\lambda_G(j^*(f))$ for all $f\in L^1(H)$, and as $j^*(f)\in M(G)$, it follows that $J\lambda_H(f)\in MC^*_\lambda(G)$, so indeed $J$ does restrict to a map $C^*_\lambda(H) \rightarrow MC^*_\lambda(G)$. The example of $\mathbb Z\subseteq\mathbb R$ shows we can't hope to get into $C^*_\lambda(G)$. -Surely the same techniques give a positive answer of Q2. -I think $\mathbb R \rightarrow \mathbb R/\mathbb Z=\mathbb T$ shows that Q3 does not have a positive answer in general; I think the Fourier transform shows that $W^*_\lambda(\mathbb R)=L^\infty(\mathbb R), W^*_\lambda(\mathbb T)=\ell^\infty(\mathbb Z)$ and the map $J$ should send $(e^{2\pi ixt})_{x\in\mathbb R}\in -L^\infty(\mathbb R)$ to $(e^{2\pi int})_{n\in\mathbb Z}\in\ell^\infty(\mathbb Z)$. This doesn't exist at the $L^\infty$ level. You always have a map at the level of full $C^*$-algebras, $C^*(G)\rightarrow C^*(K)$ (surjective even, and no multiplier algebra) but amenability issues might becomes a problem trying to drop to $C^*_\lambda(K)$.<|endoftext|> -TITLE: Ackermann's function over the reals -QUESTION [15 upvotes]: Ackermann's function is defined over integers $x$, $y$, $A(x,y)$, -with conditions for when $x=0$ or $y=0$, and otherwise uses recursive -definitions involving arguments $x-1$ and $y-1$. - -Is there a natural generalizations of $A(x,y)$ for $x,y \in \mathbb{R}$? - -Perhaps this is well-known to the cognoscenti. -I'd appreciate a pointer. - -REPLY [2 votes]: In order to obtain a valid and consistent extension all we need is to define how we calculate one fractional operation for example $H_{\frac{3}{2}}(x,y)$ which is something between addition and multiplication. So let us try to extend one work that says that we can use arithmetic-geometric mean and obtain such consistent extension. First thing first. -Notice that Ackermann function is directly related to hyperoperations. -$$A(m,n) = \begin{cases} - 0[m]n & m=0 \\ - 2[m](n+3)-3 & m>0 \\ -\end{cases}$$ -So let us repeat a more or less standard notation -$$H_n(a,b)=a[n]b$$ -$$H_0(a,b)=b+1$$ -$$H_1(a,b)=a+b$$ -$$H_2(a,b)=ab$$ -$$H_3(a,b)=a^b$$ -$$H_n(a,b)=a[n]b=a[n-1](a[n](b-1))$$ -According to https://www.hindawi.com/journals/mpe/2016/4356371/ we can define $H_{\frac{3}{2}}$ using arithmetic-geometric mean, $M(x,y)$, and its inverse and one constant that we can calculate since for all hyperfunctions we know that -$$H_n(2,2)=4$$ -because $2+2=4$ and $H_n(2,2)=2[n]2=2[n-1](2[n]1)=2[n-1]2$ -(The author is arguing that the resulting $H_{\frac{3}{2}}$ coming from AG mean corresponds to physical process, where the analysis very likely came from, and polynomial interpolation as well.) -So it is $$H_{\frac{3}{2}}(x,y)=M^{(-1)}(M(x,y),\epsilon_{\frac{3}{2}})$$ -$$\epsilon_{\frac{3}{2}}=M^{(-1)}(2,4)$$ -Now that we have $H_{\frac{3}{2}}(x,y)$ we can obtain integer evaluations for $H_{\frac{5}{2}}(x,n)$ simply -$$H_{\frac{5}{2}}(x,n)=H_{\frac{3}{2}}(x,H_{\frac{3}{2}}(x,...\text{n times}))$$ -The same as what we have with other operations $x \cdot n = x+x+...\text{n times}$ -Now that we have integer $H_{\frac{5}{2}}(x,n)$ and $H_{\frac{5}{2}}(x,n+1)$ we can find the middle point. Notice for multiplication and exponentiation this is how we find value in the middle -$$\frac{xn+x(n+1)}{2}$$ -$$\sqrt[2]{x^n x^{n+1}}$$ -So $H_{\frac{5}{2}}(x,n+\frac{1}{2})$ is the solution of -$$H_{\frac{3}{2}}(x,2)=H_{\frac{3}{2}}(H_{\frac{5}{2}}(x,n),H_{\frac{5}{2}}(x,n+1))$$ -And the same for whatever else middle integer we want. -Once we have $H_{n+\frac1{2}}$ we use the same division procedure to obtain $H_{n+\frac{1}{4}}$ and $H_{n+\frac{3}{4}}$ and so on and we are done. Once we have all rational fractional parts, we can extend it to real numbers even though we cannot express it in some closed form. -We can easily extend this down to $H_{\frac{1}{2}}$ using $x[1+r]y=x[r](x[1+r](y-1)), 00 \\ -\end{cases}$$ -where $x[r]y=H_{r}(x,y)$ is an extension over reals obtained by the above procedure using the n/n+1 hyper-mean. -Essentially this procedure is the same if we decide to use some another average operation between addition and multiplication. All comes to define just that and the rest of the construction is all the same.<|endoftext|> -TITLE: Why is the Category of Correspondences not pseudo abelian? -QUESTION [5 upvotes]: i've just gotten into the theory of motives.I understand the construction of the Karoubian envelope (pseudo-abelian completion) to ensure that morphisms have kernels and images in order to get certain decompositions of the diagonal of Motives in $\mathcal{M}_k$.But i don't see why the morphisms in $Corr_k$ which are elements of $CH(X \times Y)$ for the objects $X,Y$ don't have kernels and/or images in general. -A second one: In the category of effective Chow-Motives, do you know an example of non isomorphic varieties,that have the same motive,except for a totally split quadric and projective spaces? - -REPLY [2 votes]: Answer 1. Take the projective line $\mathbb{P}^{1}_{k}$, and let $p$ be the correspondence projecting to a point. Then this has an image, namely $\mathrm{Spec}(k)$. However, it does not have a kernel. One can see this by looking at cohomological realisations. If the kernel existed, it would only have an $\mathrm{H}^{2}$; since the $\mathrm{H}^{0}$ is accounted for by $\textrm{Spec}(k)$. -[Edit] -I realise I am made this more difficult then necessary. Suppose that all idempotents have kernels. Then the category of correspondences is equivalent to its Karoubian envelope. In the Karoubian envelope we have a decomposition $\mathbb{P}^{1}_{k} = \textrm{ker}(p) \oplus \textrm{ker}(1 - p)$. The direct sum in this category is given by the disjoint union of the underlying schemes and correspondences. Under our assumption of equivalence with the category of correspondences, this exhibits the connected scheme $\mathbb{P}^{1}_{k}$ as the disjoint union of $\textrm{Spec}(k)$ and some other scheme. Contradiction. -[/Edit] -Answer 2. Since the Hom-sets have $\mathbb{Q}$-coefficients, every isogeny becomes an isomorphism. But maybe you find this answer a bit cheating.<|endoftext|> -TITLE: Finite spectrum annihilated by multiplication by two -QUESTION [15 upvotes]: Let $X$ be a finite spectrum. Say that $X$ has characteristic two if multiplication by two on $X$ is nullhomotopic. -Does there exist a noncontractible finite spectrum of characteristic two? -(This is mentioned as an open problem in Barratt's 1959 paper "Spaces of finite characteristic.") - -REPLY [15 votes]: This doesn't seem very hard. Am I missing something? -Let $X$ be a non-trivial finite spectrum of characteristic $2$. Then let $R=\mathrm{Hom}(X,X)$, the function spectrum of maps $X$ to $X$. This $R$ is an associative $S$-algebra, with $0=2$ in $\pi_0R$; furthermore, $R$ is finite. -If $X\neq0$, then $H_*(X;F)\neq0$, where $F=\mathbb{F}_2$, and hence $H_*(R;F)\neq0$. Note that in this case the unit map $S\to R$ has non-trivial image in homology (which is the unit of the ring $H_*(R;F)$), and thus $H^0(R;F)\to F$ is surjective. -Consider the forgetful functor $\mathrm{Ass}_S\to \mathrm{Unital}_S$ from associative $S$-algebras to unital spectra. This has a (homotopical) left adjoint $T_*$: given a unital spectrum $M=(M,f\colon S\to M)$, let $T_*(M)$ be the homotopy pushout in $\mathrm{Ass}_S$ of -$$ -T(M) \xleftarrow{T(f)} T(S)\to S, -$$ -where $T$ is the free associative algebra functor. -You can also build $T_*(M) = \mathrm{colim}_n T_*^n(M)$ iteratively a la the James construction: to get $T_*^n(M)$, glue the $n$-fold smash product $M^{\wedge n}$ to $T_*^{n-1}(M)$ along a suitable map $F^n(M)\to M^{\wedge n}$. [I.e., consider the $n$-cubical diagram obtained from smashing $S\to M$ with itself $n$-times; $F^n(M)\to M^{\wedge n}$ is the map from the hocolim of the deleted $n$-cube to the object at the terminal position of the $n$-cube.] -Let $M=S^0\cup_2 e^1\to R$ be the map of unital spectra which exists by our hypothesis on $R$. The map extends to an associative algebra map -$$ -T_*(M) \to R. -$$ -In homology, $H_*(T_*(M);F) \approx F[x]$ with $|x|=1$, since $M^{\wedge n}/F_n(M) \approx S^n$. -In mod 2 cohomology, the generator $u\in H^0(T_*(M);F_2)$ must satisfy $Sq^1(u)\neq0$, and an easy argument shows that therefore $Sq^n(u)\neq0$ for all $n$. But $H^0(R;F) \to H^0(T_*(M);F) \xrightarrow{\sim} H^0(S;F)=F$ is surjective, so there exists $v\in H^0(R;F)$ such that $Sq^n(v)\neq0$ for all $n$, contradicting finiteness.<|endoftext|> -TITLE: Is there an infinite group with exactly two conjugacy classes? -QUESTION [16 upvotes]: Is there an infinite group with exactly two conjugacy classes? - -REPLY [18 votes]: As a psychological curiosity, Per Enflo writes in his Autobiography that the existence of groups with two conjugacy classes was a key insight behind his many solutions to outstanding problems in Functional Analysis. -"I made important progress in mathematics in 1966, but it was more on the level of new insights, than actual results. When thinking about topological groups in the spirit of Hilbert's fifth problem (I had gradually modified the Hibert problem to some very general program: To decide whether different classes of topological groups shared properties with Lie groups) I was wondering whether there exist "very non-commutative" groups i.e. groups, where all elements except e, are conjugate to each other. I constructed such groups, by finding the right finite phenomenon and then make an induction. I understood, that this is a very general construction scheme (or "philosophy"), that probably could be applied to various infinite or infinite-dimensional problems. And actually – this philosophy is behind several of my best papers - the solution of the basis and approximation problem, the solution of the invariant subspace problem for Banach spaces, the solution of Smirnov's problem on uniform embeddings into Hilbert space and more." -( http://perenflo.com/sida9.html )<|endoftext|> -TITLE: Minimal number of generators of subgroups of Noetherian groups -QUESTION [8 upvotes]: A group $G$ is Noetherian (or slender) if all its subgroups are finitely generated. Does this imply that the minimal number of generators of subgroups of $G$ is bounded above? -For example, if $G$ is a polycyclic group that admits a polycyclic series of length $n$ then every subgroup of $G$ can be generated by $n$ (or fewer) elements. This idea also applies for virtually-polycyclic groups. -It is unknown whether all finitely presented Noetherian groups are virtually-polycyclic. On the other hand, there are finitely generated Noetherian groups that are not virtually-polycyclic, for example the Tarski monster. However, all proper subgroups of the Tarski monster are cyclic and hence there is a bound on the minimal number of generators of its subgroups. -(See this post for a related question.) - -Edit: What if we restrict ourselves to finitely presented Noetherian groups? - -REPLY [14 votes]: No. - -For any countable family of countable involution-free groups $G_1,G_2,\dots$, there is a 2-generated group $H$ containing all $G_i$ as proper subgroups such that each proper subgroup of $H$ is either cyclic or a conjugate of a subgroup of some $G_i$. - -This is Obraztsov's embedding theorem. Clearly, it gives a desired example if we put, e.g, -$$ -\{G_i\}=\{\hbox{all finite groups of odd orders}\}. -$$<|endoftext|> -TITLE: Is sigma-additivity of Lebesgue measure deducible from ZF? -QUESTION [11 upvotes]: Is sigma-additivity (countable additivity) of Lebesgue measure (say on measurable subsets of the real line) deducible from the Zermelo-Fraenkel set theory (without the axiom of choice)? -Note 1. Follow-up question: Jech's 1973 book on the axiom of choice seems to be cited as the source for the Feferman-Levy model. Can this be sourced in the work of Feferman and levy themselves? Are these S. Feferman and A. Levy? - -REPLY [11 votes]: This depends on exactly how you define Lebesgue measure since some definitions incorporate countable additivity. However, there is a model of ZF, the Feferman-Levy model, where $\mathbb{R}$ is a countable union of countable sets which ensures that any countably additive measure on $\mathbb{R}$ has to be trivial. - -REPLY [9 votes]: No, you can't have that. It is consistent that the real numbers are a countable union of countable sets, in which case you immediately have that there is no nontrivial measure which is countably additive on the real numbers. -There are other models, however, in which $\aleph_1$ is singular, the countable union of countable sets of real numbers is countable; but every set is Borel. In such models, I believe, you can't have a countably additive Lebesgue measure as well. - -To your question, yes. These are Solomon Feferman and Azriel Levy. The result appears as an abstract in Notices of the AMS from 1964 (give or take a year, this is from memory).<|endoftext|> -TITLE: how to obtain a generalized Morse function out of a fiber bundle? -QUESTION [7 upvotes]: I already asked this question in MSE but did not get any answer/comment yet. -Let $M\to E\to B$ be a smooth fiber bundle. In "Parametrized Morse Theory and Its Applications,(Proceedings of the ICM, 1990)", K. Igusa says that if dim $B$$<$dim $M$, then, there exists a smooth function $f:E\to\mathbb{R}$ such that, when restricted to the fibers, is a generalized Morse function (i.e. a smooth function with only non degenerate critical points and birth-death singularities). For the proof of this fact he refers to "On the homotopy type of the space of generalized Morse functions (Topology, Vol 23, No2., 1984)". There he proves that if $N$ is an $n$-dimensional smooth manifold, then there is an $n$-connected map between the space of such functions on $N$ and certain infinite loop space $\Omega^{\infty}S^{\infty}(BO\wedge N_+)$. -My question is, how come does the connectivity of the latter map imply the existence of the generalized Morse function $f$?. Perhaps more generally, is there a canonical way to assign such a function to the total space of a fiber bundle? - -REPLY [6 votes]: It's basically this: -Associated with a manifold $N$ are two spaces, call them $F_1$ and $F_2$. The first is the space of generalized Morse functions on $N$ and the second is basically $\Omega^\infty S^\infty(BO\wedge N_+)$. These are sufficiently canonical (functorial w.r.t. diffeomorphisms) that, if you have a fiber bundle $E\to B$ with fiber $N$, you get two more fiber bundles with base $B$, having fibers $F_1$ and $F_2$ respectively. Also, there is a canonical map $F_1\to F_2$, yielding a map of bundles. -The $F_2$-bundle has a canonical section. -Igusa shows that the map $F_1\to F_2$ is $n$-connected, where $n=dim(N)$. If $dim(B)\le n$ then it follows that the section of the $F_2$-bundle is homotopic to one which comes from a section of the $F_1$-bundle. A section of the $F_1$-bundle is what you want: a function on $E$ that is fiberwise generalized Morse.<|endoftext|> -TITLE: Sherman-Morrison type formula for Moore-Penrose pseudoinverse -QUESTION [14 upvotes]: Given an $n\times n$ invertible matrix $\mathbf A$ and two column vectors $\mathbf u$, $\mathbf v\in\mathbb R^n$, suppose that $1 + {\mathbf v}^T {\mathbf A}^{-1}\mathbf u \neq 0$. -Then the Sherman-Morrison formula states that -\begin{equation*} -(\mathbf A + \mathbf u \mathbf v^T)^{-1} = -\mathbf A^{-1} - -{\mathbf A^{-1}\mathbf u\mathbf v^T \mathbf A^{-1} \over 1 + \mathbf v^T \mathbf A^{-1}\mathbf u}. -\end{equation*} -Question: I'm wondering whether we have a similar formula when the inverse in the Sherman-Morrison formula is replaced by the Moore-Penrose pseudoinverse in case that $\mathbf A$ is singular matrix. - -REPLY [12 votes]: It's all in Meyer's paper Generalized Inversion of Modified Matrices published 1973 in SIAM Journal on Applied Mathematics. -The material is also available at around p.51 in the Meyer & Campbell book.<|endoftext|> -TITLE: Sufficient condition for coverings between non-orientable surfaces -QUESTION [5 upvotes]: Let $X_k$ be the connected sum of $k$ projective planes. I am interested in necessary and sufficient conditions for the existence of a covering $\pi: X_{k'} \to X_k$, where $k$ -and $k'$ are integers. -A necessary condition is that the Euler characteristic of $X_{k'}$ is a multiple -of the Euler characteristic of $X_k$. Though obtaining a sufficient condition -seems more difficult. -The case of orientable surfaces is easy, but this is more difficult than I think, -and I can't find it anywhere. -I would appreciate any hint about this. - -REPLY [8 votes]: Your necessary condition, rephrased slightly, is sufficient. It should say that the Euler characteristic of $X_{k'}$ is a positive multiple of $\chi(X_k)$. This modification is to take care of the possibility that $X_k$ is $RP^2$, which is not covered by any other non-orientable surface. If $\chi(X_k)= \chi(X_{k'}) = 0$, then both are Klein bottles, and certainly there is a covering. Finally if $\chi(X_k)<0$, then it is a connected sum of a torus with some number of projective planes. So it has non-orientable covers of all degrees; these are determined by their Euler characteristics. So, given $X_{k'}$ with $\chi(X_{k'}) = n \chi(X_k)<0$, choose a degree n cover of $X_{k}$; it will be homeomorphic to $X_{k'}$. -The case of orientable surfaces follows by the same argument.<|endoftext|> -TITLE: Cohen-Lenstra Heuristics reference -QUESTION [9 upvotes]: I am looking for good references (preferably, books) on Cohen-Lenstra Heuristics (on Real Quadratic fields) which explain in detail the reasons behind its fundamental assumption (higher the cardinality of the automorphism group less likely it is cyclic) and its connection with Dedekind Zeta function. -I have looked at the book 'Computational Algebraic Number Theory' by Cohen but it only states the conjecture and there is not much explanation given. -I would also like to know if there has been any work towards making this probabilistic conjecture more concrete so that we can get an answer for the Gauss' class number conjecture for real quadratic fields. - -REPLY [11 votes]: I don't have a book reference, but here are some rambling words about why one might, in general, expect objects $x$ to appear with probability proportional to $\frac{1}{|\text{Aut}(x)|}$. The short version is that this is a very natural number to associate to $x$. -To start with, let $X$ be a finite set. If you wanted to pick a random element of $X$, probably you would do it uniformly, and you expect that in the absence of further structure this is what would happen "in Nature." -Now suppose $X$ comes equipped with the action of a finite group $G$, and you wanted to pick a random orbit of $X$. One way to do this is to pick a random element of $X$ and consider its orbit. The induced probability measure on orbits, rather than assigning each orbit equal weight, assigns each orbit a weight inversely proportional to the size of its stabilizer. -This is one way to motivate the following definition. Let $X$ be a groupoid all of whose objects have a finite automorphism group which is tame in the sense that the sum -$$\sum_{x \in \pi_0(X)} \frac{1}{|\text{Aut}(x)|}$$ -converges (where $\pi_0(X)$ is the set of isomorphism classes of objects of $X$). The above sum is called the groupoid cardinality of $X$, and it induces a natural probability measure on $\pi_0(X)$ where an isomorphism class $x$ occurs with probability inversely proportional to $\text{Aut}(x)$. -One way to think about groupoid cardinality is that it is analogous to the Euler characteristic. The basic intuition is that we expect $\chi(X/G) = \frac{\chi(X)}{|G|}$ for a suitably nice group action of a finite group $G$ on a space $X$, and the one-object groupoid associated to a group $G$ can be thought of as a model of the classifying space $BG = EG/G$, where $EG$ is contractible and so in particular $\chi(EG) = 1$. For further discussion of the naturality of groupoid cardinality see this blog post. -Example. Let $X$ be a finite set on which a finite group $G$ acts. Form the action groupoid, whose objects are the elements of $X$ and which has a morphism $s_1 \to s_2$ labeled by $g \in G$ whenever $gs_1 = s_2$. Then the groupoid cardinality of the action groupoid is -$$\sum_{x \in \pi_0(X)} \frac{1}{|\text{Stab}(x)|} = \frac{|S|}{|G|}$$ -and the induced probability measure on $\pi_0(X)$ is the one we considered above. -Example. Let $G = \mathbb{Z}/2\mathbb{Z}$. Then $BG \cong \mathbb{RP}^{\infty}$ has a cell decomposition with one cell in every dimension, so its Euler characteristic should be -$$1 - 1 + 1 - 1 \pm ...$$ -which has, say, Cesaro sum $\frac{1}{2} = \frac{1}{|G|}$! -Example. Consider the groupoid of finite sets and bijections. Its groupoid cardinality is -$$\sum_{n=0}^{\infty} \frac{1}{n!} = e.$$ -With respect to the corresponding probability measure, a random finite set $S$ occurs with probability $\frac{1}{e |S|!}$. The distribution of cardinalities we get this way is Poisson with mean $1$. -What kind of process produces random finite sets? One candidate is to take the fixed point set of a random permutation $\pi \in S_n$ for $n$ large, and in fact one can show that as $n \to \infty$ the distribution of $|\text{Fix}(\pi)|$ approaches a Poisson distribution with mean $1$. See, for example, this blog post for details. -More generally, the distribution of the number of $r$-cycles of a random permutation in $S_n$ as $n \to \infty$ is Poisson with mean $\frac{1}{r}$. This should have an interpretation in terms of random finite sets of $r$-cycles, and indeed it does: a collection of $n$ $r$-cycles should have an automorphism group of size $r^n n!$ because we can both permute the cycles and cyclically permute the elements in each cycle, and this recovers a Poisson distribution with mean $\frac{1}{r}$. -For a related example of a more number-theoretic flavor, one can say the same thing about irreducible factors of degree $r$ in a random monic polynomial of large degree over $\mathbb{F}_q$, except that now one has to let $q \to \infty$ as well: see this blog post for details. -Example. Here is another example from number theory. Recall that by the Chebotarev density theorem, the Frobenius elements associated to primes $p$ in the Galois group $G = \text{Gal}(K/\mathbb{Q})$ of a number field $K$ are equidistributed in $G$ as $p$ varies. But Frobenius elements are not elements of $G$, they are conjugacy classes, hence objects, well-defined up to isomorphism, in the action groupoid associated to the action of $G$ on itself by conjugacy. So the Chebotarev density theorem can be reinterpreted as saying that a given conjugacy class appears as a Frobenius element with probability inversely proportional to the size of its centralizer.<|endoftext|> -TITLE: Critical points of rank-into-rank embeddings -QUESTION [10 upvotes]: $\DeclareMathOperator{\crit}{\operatorname{crit}}$A rank-into-rank embedding is a non-trivial elementary embedding from a rank initial segment of $V$ into itself: $j:V_\delta\prec V_\delta$. Define the critical sequence of such an embedding by setting $\kappa_0=\crit(j)$ (the first ordinal moved by $j$) and $\kappa_{n+1}=j(\kappa_n)$. Let $\lambda=\crit^\omega(j)=\sup_{n<\omega} \langle \kappa_n\rangle$. It is straightforward to see that $\lambda$ is a strong limit cardinal of countable cofinality. -By a theorem of Kunen, if such an embedding can exist, then $\delta$ must be the ordinal $\lambda$ or $\lambda+1$. -It is not hard to see that $\crit(j)$ must be measurable. In fact, for any $n$, $\crit(j)$ is also $n$-huge as witnessed by the ultrafilter $$U=\{X\subseteq\mathcal{P}(\kappa_n): j"\kappa_n\in j(X)\}.$$ Further, if we let $j^n$ denote $j$ composed with itself $n$ times, then $$V_\lambda\models ``\lambda\text{ is supercompact"}.$$ To see this, suppose $\crit(j)\leq \theta <\kappa_n$, then $$U=\{X\subseteq\mathcal{P}_{\crit(j)}(\theta): j^n"\theta\in j^n(X)\}$$ winesses the $\theta$-compactness of $\crit(j)$ (in $V_\lambda$). -For the last claim, it is enough that $\crit(j)$ is $<\lambda$-supercompact, i.e. not fully supercompact in $V$. In this case, however, $\crit(j)$ could be fully supercompact. -But extendible cardinals are not characterized by the presence of ultrafilters and this motivates my question here. -Question: Can the critical point of a rank-into-rank embedding be extendible? -It may not make sense (I think) to ask for full extendibility of $\crit(j)$: Suppose otherwise that $\crit(j)$ is fully extendible. Let $k$ witness the $\theta$-extendibility of $\crit(j)$ for some $\theta>\crit^\omega(j)$. Then we have $$V_{\crit(j)}\prec V_{\crit^\omega(j)}\prec V_\theta.$$ -This looks suspiciously like Woodin's Enormous Cardinal (though his notion is defined in the context of just ZF). See http://logic.harvard.edu/EFI_Woodin_talk.pdf, slide 20. Thus I'm not sure that $\crit(j)$ can be fully extendible. -Question: Assume $j$ is a rank-into-rank embedding and let $\lambda=\crit^\omega(j)$. Can $\crit(j)$ be $<\lambda$-extendible? -Edit: -I should point out (reminded by Carlo Von Shnitzel's comments below) that there is a sort of local intertwining of supercompact cardinals and extendible cardinals that may be relevant. See Kanamori's book, p.316-318. -Also, there may be some subtlety here concerning $\Sigma_k$ correctness. Suppose $$j:V_\lambda\prec V_\lambda.$$ I think assuming $V_\lambda\prec_3 V$ (or even $V_\lambda\prec_2 V$) is a strictly stronger assumption. If $\crit(j)$ were extendible, then $V_{\crit(j)}\prec_3 V$. But the embedding assumption also gives us that $V_{\crit^\omega(j)}\prec_3 V$, even though $\crit^\omega(j)=\lambda$ is not itself an extendible cardinal. Similarly if we assume $\crit(j)$ is actually supercompact. - -REPLY [3 votes]: Theorem: If $\kappa$ is the critical point of $j\colon V_\lambda\prec V_\lambda$, then $\kappa$ is $\lambda$-weakly extendible. Furthermore, if $\kappa$ is the critical point of $j\colon V_\lambda\prec V_\lambda$, then $\kappa$ is $\lt\lambda$-strongly extendible. -Proof. First off $\kappa+\lambda=\lambda$, as $\kappa_n$ is a cardinal for each $n$ ($\kappa_0=\kappa$), and therefore $\lambda$ is a cardinal $>\kappa$. By definition, there is an elementary embedding $j\colon V_\lambda\prec V_\lambda$ with critical point $\kappa$. Similarly, for each $\alpha\lt\lambda$ such that $\alpha\lt j^n(\kappa)$, $\kappa$ is strongly $\alpha$-extendible as witnseesed by $j^{(n)}\restriction V_\alpha: V_\alpha\prec V_{j^n(\alpha)}$.■ -Theorem: If $\kappa$ is the critical point of $j\colon V_\lambda\prec V_\lambda$ and $\kappa\in C^{(2)}$, then $\kappa\gt$ the least rank-into-rank cardinal. Furthermore, if $\kappa$ is $\lambda$-strongly extendible, then $\lambda>$ the least rank-into-rank cardinal. -Proof. The statements “$\lambda$ is rank-into-rank” and “there exists a rank-into-rank embedding” are both $\Sigma_2$. And so if $V_\kappa\prec_{\Sigma_2} V$, then $V_\kappa\vDash\text{There is a rank-into-rank embedding}$ and if $V_\kappa\vDash\lambda_0\text{ is rank-into-rank}$, then $\lambda_0$ is rank-into-rank. For the second part, let $k: V_\lambda\prec V_{k(\lambda)}$ witness $\lambda$-strong extendibility. Then $k(\kappa)+1\lt k(\lambda)$ and so $V_{k(\kappa)+1}\subseteq V_{k(\lambda)}$ and therefore $k(\kappa)$ is inaccessible. Therefore, $V_{k(\kappa)}\vDash\text{There is a rank-into-rank embedding}$, and so $V_\kappa\vDash\text{There is a rank-into-rank embedding}$ (As $V_\kappa\prec V_{k(\kappa)}$), and if $V_\kappa\vDash\lambda_0\text{ is rank-into-rank}$, then $\lambda_0$ is rank-into-rank, because $\Sigma_2$-formulas are upward absolute in $V_\kappa$ for inaccessible $\kappa$.■ -Note then that the consistency strength of “$\kappa$ is the critical point of $j\colon V_\lambda\prec V_\lambda$, and $V_\kappa\prec_{\Sigma_2} V$” is therefore greater than I3. Furthermore, the consistency strength of “$\kappa$ is the critical point of $j\colon V_\lambda\prec V_\lambda$, and $\kappa$ is $\lambda$-strongly extendible” is therefore greater than I3. -Theorem: If $\kappa$ is I2, then the cardinals which are I3 and extendible in $V_\kappa$, are stationary in $\kappa$. -Proof. Let $X$ be the set of cardinals which are I3 in $V_\lambda$ as witnessed by some $j\subseteq V_\lambda$. Then, if $\alpha\in X$, $\alpha$ is I3, because $V_\alpha^{V_\beta}=V_\alpha$ whenever $\alpha\lt\beta$. Therefore $\kappa\in j(X)$, because $\Sigma_2^1$-properties are prserved, and so the cardinals $Y\in D$, where $Y$ is the set of cardinals which are I3 and $D$ the measure generated by $j$. Similarly, $\kappa$ is extendible in $V_\lambda$ and so $V_{j(\kappa)}$ and so $Z\in D$, where $Z$ is the set of cardinals extendible in $V_\kappa$. Therefore $Y\cap Z\in D$, so that $Y\cap Z$ is stationary, because every club set $C$ has $j(C)\cap\kappa=C$ and so $\kappa\in j(C)$.■ -References: Upward reflection of rank-into-rank cardinals<|endoftext|> -TITLE: A Special Pair of Models for ZFC (New Version) -QUESTION [7 upvotes]: Are there two models $M$ and $N$ for $\text{ZFC}$ such that: -(1) $M\subseteq N$ -(2) $\aleph_{1}^{N}=\aleph_{1}^{M}$ -(3) $\aleph_{2}^{N}=\aleph_{\omega +1}^{M}$ -Update: According to Peter's useful answer it seems this problem is still open. So my question is about "partial" or "similar" results with different (but not too different) $\aleph$s. Please give me some references too. - -REPLY [6 votes]: Regarding your question about the situation at other $\aleph$s: Lemma 3.1 from Cummings' paper "Collapsing successors of singulars," together with the fact (due to Shelah, I believe) that, if $\overrightarrow{f}$ is a scale of length $\aleph_{\omega+1}$, then there is a club $C$ in $\aleph_{\omega+1}$ such that every $\alpha \in C \cap \mathrm{cof}(\geq \aleph_4)$ is good for $\overrightarrow{f}$, yields that, if $5 \leq n < \omega$, then there cannot be inner models $V\subset W$ such that $\aleph_{\omega+1}^V = \aleph_n^W$. -Also, to add to the list of restrictions your desired situation would impose on $V$ and $W$ already given by Andres and Mohammad, it follows from results of Sharon and Viale from their paper "Some consequences of reflection on the approachability ideal" that if $V\subset W$ are inner models and $\aleph_{\omega+1}^V = \aleph_n^W$ for some $n<\omega$, then, in $W$, a form of simultaneous stationary reflection must fail at $\aleph_n^W$. More precisely, in $W$, for all stationary $S\subseteq \aleph_n^W$, there are $\{S_i \mid i<\omega \}$ such that each $S_i$ is a stationary subset of $S$ and there is no $\alpha < \aleph_n^W$ such that each $S_i$ reflects at $\alpha$. (Note that this is trivial if $n=1$, since no stationary subsets of $\aleph_1$ reflect.)<|endoftext|> -TITLE: How many ways can a given permutation be obtained as a product of k 2-cycles? -QUESTION [6 upvotes]: Let $\sigma_1, \ldots, \sigma_b$ be all the 2-cycles in $S_n$. (So, $b = \binom{n}{2}$.) Given $\pi \in S_n$, what is known about how many ways $\pi$ can be obtained as a product of $k$ (not necessarily distinct) elements of $\{\sigma_1,\ldots, \sigma_b\}$? Of course, if $k$ is too small than this number can be zero. It is clear that this depends on $\pi$ only through its conjugacy class. Thus, I was wondering whether there is a nice formula in terms of $n$, $k$ and the sizes of the cycles in the cycle-decomposition of $\pi$. - -REPLY [3 votes]: A single cycle of length $n$ will have $n^{n-2}$ different ways to be decomposed into $n-1$ transpositions (Hurwitz). For a permutation in $S_n$ which is a product of distinct $l$ cycles $\{C_i\}_{i=1}^{l}$ we have a multinomial to interleave the transpositions. -$$ -{n-l \choose n(C_1),\dots,n(C_l)} \prod_{i=1}^{l}\left(n(C_i)+1\right)^{n(C_i)-1} -$$ -Where $n(C_i)=length(C_i) - 1$ is the number of transpositions within the cycle $C_i$. And the binomial coefficient ${n-l \choose n(C_1),\dots,n(C_l)}$ counts the ways to interlace insertions between the cycles. -In this link is a review and generalizations of Hurwitz's result by Strehl. -I know this only deals with the minimal decompositions and you were asking about a general $k$ that could be greater. Hope this helps anyway.<|endoftext|> -TITLE: Are paths in HoTT perhaps just "cost-free" paths? -QUESTION [5 upvotes]: Homotopy type theory (HoTT) doesn't seem to say anything about "mutations" of values in type $T$, an important concept in computer science. Mutations occur when you "change a value" of some variable $x\in T$; e.g. from $x=5$ to $x=6$. Is it possible to extend the notion of homotopy type (in the sense of HoTT) to allow for mutations? Perhaps each mutation is a path in some space, and the HoTT notion of 'identity' is given by the "cost-free" paths? -Let ${\mathbb R}_+=\{x\in{\mathbb R}\;|\;x\geq 0\}\cup\{\infty\}$ be the monoid of extended nonnegative real numbers ('numbers'), with the additive monoid structure (e.g. $0+x=x$, and $x+\infty=\infty$). Suppose $X$ is a topological space equipped with a function $Cost\colon Path(X)\to{\mathbb R}_+$, which I'll call the cost function, from its set of paths $Path(X)=\text{Hom}_{Top}([0,1],X)$, to the monoid of numbers, such that - -constant paths have cost $0$, and -the cost of a concatenation of paths is the sum of their costs (i.e. $Cost(P\star Q)=Cost(P)+Cost(Q)$). - -Note that this notion is not homotopical because we do not require that homotopic paths have equal costs. However, we can construct the cost-free subspace $X_0$ of $X$, given by the union of all paths $P$ with $Cost(P)=Cost(-P)=0$. For a path $P$, let $-P$ denote its composition with $(r\mapsto 1-r)\colon [0,1]\to[0,1]$. Then we construct the space $X_0$ and a continuous map $i_X\colon X_0\to X$ as follows: -$$X_0=\text {colim}_{\{P\colon [0,1]\to X\;|\;Cost(P)=Cost(-P)=0\}}\;(P)\hspace{.5in}i_X(P,r)=P(r),\;\; r\in [0,1].$$ -Note that the subset $X_0\subseteq X$, constructed using the above colimit, will not always have the subspace topology inherited from $X$. However, there is a pretty general subcategory of "path-generated spaces" $X$, i.e. where open sets are detected by paths, for which it will. Regardless, the paths in $X_0$ form a groupoid. -Let's define a mutation space to be a pair $(X,Cost)$ as above, where $X$ a path-generated space. A mutation in $(X,Cost)$ is a path in $X$, and the underlying HoTT homotopy type of $(X,Cost)$ is $X_0$, the subspace generated by cost-free mutations. There are many reasonable notions for morphism of mutation spaces, e.g. continuous maps $f\colon X\to Y$ such that for any path $P\colon[0,1]\to X$, we have $Cost_Y(P\circ f)\leq Cost_X(P)$, under which the mutation spaces would form a category $MuSp$. But any good notion for this category should admit a "cost-free subspace" functor $(-)_0\colon MuSp\to Top$. -My question is especially directed to those with background in HoTT and CS: -Q1: is there anything flawed about the idea "mutations are paths and 'identities' are cost-free paths"? -Q2: is there any work being done on this kind of HoTT generalization, or on anything like it? - -REPLY [4 votes]: This is not exactly an answer to your question, but I think it might gesture towards what you have in mind. -In computer science, one of the best known ways of dealing with mutation is Hoare logic and its derivatives. The basic judgment of Hoare logic is the "Hoare triple" $\{P\}c\{Q\}$, which asserts that a program $c$, modelled as a state transformer $c : S \to S$, will take a state satisfying the predicate $P$ to a state satisfying the predicate $Q$. For instance, here is an example of a valid proof in Hoare logic, adapted from Wikipedia: -$$\frac{\displaystyle\frac{}{\{x=42\}y := x+1\{y = 43\}} \quad \frac{}{\{y=43\}z := y\{z = 43\}}}{\{x=42\}y := x+1; z := y\{z = 43\}}$$ -Now, there is a correspondence, which I believe is folklore, between Hoare logics and fibrations $p : \mathbb{E} \to \mathbb{B}$. In particular, a proof of the triple $\{P\}c\{Q\}$ can be seen as an arrow $\alpha : P \to Q$ in the total category $\mathbb{E}$ lying over the arrow $c : S \to S$ in the base $\mathbb{B}$. The sequential composition rule -$$\frac{\{P\}c_1\{Q\}\quad\{Q\}c_2\{R\}}{\{P\}c_1;c_2\{R\}}$$ -has a natural interpretation in terms of the functoriality of $p$. Finally, the existence of cartesian liftings corresponds to the existence of weakest preconditions. By invoking weakest preconditions, any Hoare triple $\{P\}c\{Q\}$ can be turned into an equivalent "ordinary" logical implication $P \supset c^*Q$ (where I am writing $c^*Q$ for the pullback of $Q$ along $c$, usually written $wp(c)(Q)$ when seen as a weakest precondition). On the other hand, the latter can also be seen as just a special kind of Hoare triple $\{P\}skip\{c^*Q\}$, where "$skip$" is the identity command; in the terminology of fibrations, a proof of $\{P\}skip\{c^*Q\}$ corresponds to a "vertical" arrow in $\mathbb{E}$. -Lastly, there are deep connections between type theory and the theory of fibrations, a lot of which has been worked out in categorical models of dependent type theory. On the other hand, not much of this has been tied yet to the theory of computational side effects. That's in part the motivation for a recent paper I co-wrote with Paul-André Melliès, where we also discuss briefly the Hoare logic example. - -Update: Let me try to make the connection to the original question a little more clear. If you abstract away from the notation of Hoare logic, state transformers can of course be seen as elements of some arbitrary monoid $M$, so that the base category $\mathbb{B}$ is simply its delooping $\mathbb{B} = \mathbf{B}M$. Likewise, the total category $\mathbb{E}$ can be taken as an arbitrary category, interpreted as a category of "states and mutations". Then the functor $p : \mathbb{E} \to \mathbb{B}$ is precisely your "cost assignment". Saying that $p$ is additionally a fibration over $\mathbb{B}$ means that all of the information in $\mathbb{E}$ can be recovered from "cost-free" paths. -As Andrej and Urs have remarked, though, there's a question of whether paths are directed: I've taken $\mathbb{E}$ and $M$ to be categories/monoids, rather than groupoids/groups, because I assumed this was the interpretation you had in mind. In any case, many of the constructions of type theory admit interpretation as operations on categories rather than groupoids, although in the standard formulation of dependent type theory the identity type is symmetric.<|endoftext|> -TITLE: Zariski dense subgroups of linear algebraic groups -QUESTION [5 upvotes]: The theorem of Matthews, Vaserstein and Weisfeiler asserts that if $G$ is a simply connected absolutely almost simple groups over $Q$ and $\Gamma$ a finitely generated subgroup of $G(Q)$ which is Zariski dense in $G$ then for almost all primes $p$, the reduction of $\Gamma$ mod $p$ is $G_p(F_p)$. -I was wondering wether the assumption `absolutely almost simple' is really necessary? -For example, would the conclusion hold for a group like $Res_{F/Q} G$ where $F$ is a number field and $G$ absolutely simple simply connected over $F$? - -REPLY [3 votes]: This isn't an answer, but rather an extended comment to recast the question from another perspective that is given "more directly" over number fields. Perhaps this is something the OP is already well aware of, but hard to tell from the formulation of the question. Hopefully someone else can address the punchline at the end. -In Theorem 8.1 of the paper by those 3 authors, one finds a strong approximation refinement (which immediately implies the "one prime at a time residual surjectivity almost everywhere" property): for a sufficiently large finite set $S$ of places of $\mathbf{Q}$ containing the archimedean place, and $\mathbf{A}^S$ the factor ring of adeles away from $S$, the closure of $\Gamma$ in $G(\mathbf{A}^S)$ is open. -This adelic strengthening seems like the right statement to be trying to generalize over number fields $F$: if $G$ if a connected semisimple $F$-group that is absolutely simple over $F$ and simply connected, and if $\Gamma \subset G(F)$ is a finitely generated subgroup, then under what "algebro-geometric density" conditions on $\Gamma$ can we conclude that the closure of $\Gamma$ in $G(\mathbf{A}_F^S)$ is open for a suitable finite set $S$ of places of $F$ (containing the archimedean places)? -[In general, the connected semisimple $\mathbf{Q}$-groups that are simply connected are precisely the finite products $\prod {\rm{R}}_{F_i/\mathbf{Q}}(G_i)$ for number fields $F_i$ and connected semisimple $F_i$-groups $G_i$ that are absolutely simple and simply connected. So the Weil restrictions in the question are exactly the $\mathbf{Q}$-simple cases; i.e., you're asking to relax "absolutely simple" to "$\mathbf{Q}$-simple".] -So let $H = {\rm{R}}_{F/\mathbf{Q}}(G)$ for such $G$ over a number field $F$, and let $\Gamma$ be a finitely generated subgroup of $H(\mathbf{Q}) = G(F)$. If $\Gamma$ is Zariski-dense in $H$ then it is Zariski-dense in $G$, since $\Gamma$ viewed over $\mathbf{Q}$ visibly factors through the Weil restriction to $\mathbf{Q}$ of its Zariski closure in $G$ over $F$. However, the converse is false, as we see using $\Gamma = {\rm{SL}}_n(\mathbf{Z})$ and $G = {\rm{SL}}_n$ over any number field $F \ne \mathbf{Q}$ (for which $\Gamma$ viewed over $\mathbf{Q}$ has Zariski closure equal to the evident $\mathbf{Q}$-subgroup ${\rm{SL}}_n \hookrightarrow {\rm{R}}_{F/\mathbf{Q}}({\rm{SL}}_n)$). -The preceding counterexample shows that the naive analogue over number fields for the absolutely simple case (using Zariski-density over the number field) is false: the subgroup ${\rm{SL}}_n(\mathbf{Z})$ in $G(F)$ for $G = {\rm{SL}}_n$ and a number field $F \ne \mathbf{Q}$ fails to map onto ${\rm{SL}}_n(O_F/\mathfrak{p})$ for any prime $\mathfrak{p}$ with residual degree $> 1$ over $\mathbf{Q}$. -So in effect, your question is asking whether Zariski-density in the Weil restriction to $\mathbf{Q}$ is the "correct" strengthening of Zariski-density over $F$ to imply openness of the closure in $G(\mathbf{A}_F^S)$ of a finitely generated subgroup $\Gamma \subset G(F)$, for some finite set $S$ of places of $F$ (containing the archimedean places). A quick look on MathSciNet doesn't seem to lead to papers that address this, but perhaps someone more familiar with the relevant literature can point to a suitable reference?<|endoftext|> -TITLE: Is SL_n of an order in a number ring finite-index in SL_n of the number ring? -QUESTION [5 upvotes]: Let $\mathcal{O}$ be the ring of integers in an algebraic number field and let $R \subset \mathcal{O}$ be an order. For instance, we might have $\mathcal{O} = \{\text{$x+i y$ $|$ $x,y \in \mathbb{Z}$}\}$ and $R = \{\text{$x+i y \in \mathcal{O}$ $|$ $y$ even}\}$. Question : Is $\text{SL}_n(R)$ a finite-index subgroup in $\text{SL}_n(\mathcal{O})$? I might be missing something obvious here, but I'm having trouble proving this (maybe because it isn't true?). - -REPLY [11 votes]: Yes. If $f = (\mathcal O : R)$, then $f\mathcal O \subset R$ and consequently $\text{SL}_n(R)$ contains $\ker (\text{SL}_n(\mathcal O) \to \text{SL}_n(\mathcal O/f\mathcal O))$, which has finite index because $\mathcal O/f\mathcal O$ is finite.<|endoftext|> -TITLE: Number of ways to write an integer as a product of irreducibles -QUESTION [8 upvotes]: Is there any way to tell the number of distinct ways to factor $a\in\mathcal{O}_k$ (up to units, of course) when $k$ is not a PID? A simple investigation in $\mathbb{Q}(\sqrt{-5})$ with integer ring $\mathcal{O}=\mathbb{Z}[\alpha]$ and Galois group $G=\langle\sigma\rangle$, with $\alpha = \sqrt{-5}$, gives, for example (and this is incredibly basic, but in case we have algebra people not familiar with the computations of number theorists or outside interested parties to motivate the question). -$6=\begin{cases} 2\cdot 3 \\ (1+\alpha)(1-\alpha)\end{cases}\qquad (*)$ -Since $\mathbb{Z}[\alpha]$ is integrally closed, we can look mod 2 and mod 3 to find factorizations for $(2)$ and $(3)$, which gives $m_\alpha(x)\equiv (x+1)^2\mod 2$ and $m_\alpha(x)\equiv (x-1)(x-2)\mod 3$, and so $(2)$ ramifies as $(2,1+\alpha)^2=\mathfrak{p}_1^2=\mathfrak{p}_2^2$ (the reason for using two labels for the same ideal will be clear later) and since $N_{k/\mathbb{Q}}(1+\alpha)=6$, we know $(2,1+\alpha)$ is not principal, so $2$ is an irreducible integer, and similarly for $(3)=(3,\alpha-1)(3,\alpha -2)=(3,1-\alpha)(3,1+\alpha)=\mathfrak{p}_3\mathfrak{p}_4$ to show 3 is irreducible, and $1\pm\alpha$ are irreducible by norm calculations as well. -So in this case we'd like to see that $(*)$ produces all of the factorizations of $6$ into irreducibles. So we need only look at ideals above 2 and 3. Since $h_k\,(=|\text{Pic}(\mathcal{O}_k)|)$ any pairing of two of the four (we need to count the ramifying prime twice, of course, and this is why we double counted it earlier) ideals we've produced multiply to an irreducible in some factorization of 6 so we see that of the ${4\choose 2}=3$ ways to do this, but clearly matching 1 with 3 is the same as matching 1 with 4 because $\mathfrak{p}_1=\mathfrak{p}_2$. -I'd like to know if there is a general way to tell the total number of factorizations of a given number in an integer ring, $\mathcal{O}_k$ for a number field, $k$. If this is too hard, then I'm also willing to settle on number of factorizations for a rational integer where we are allowed to use integers from $\mathcal{O}_k$ in the factorization. -Obviously the combinatorics of the problem mean that there is not just one number for the entire field, simply looking at the number 12 in the example above illustrates why this is. However some easy reductions: -If we look at the ideal factorization, we can always ignore inert primes, as they clearly have no contribution to making more factorizations, moreover we can ignore principal primes, but these are harder to detect with simple tricks if the norm of the element given to us by a factorization in some reduction with a coprime conductor (this also illustrates the problem when we cannot find a ring with trivial conductor in our integer ring). If the class number (again, I don't know if algebraists use the same terminology, so I'll clarify: this is the number $h_k=|\text{Pic}(\mathcal{O}_k)|$) is 2, then we know we can select non-principal ideals and their product is principal and that the product ideal is generated by an irreducible element without any further information about the exact structure of $\text{Pic}(\mathcal{O}_k)$. -One might argue that one need only pair inverse (non-trivial) classes of primes together to get more information, but as far as I know detecting the exact ideal class of a non-trivial element when $h_k>2$ is a non-trivial feat, but then this is not my prime area of expertise, and I would love to be proven wrong on that front. -EDIT: It seems reasonable that if you have an ideal $\mathfrak{a}=\prod\mathfrak{p}_i^{e_i}$ that if you could detect the ideal classes of the $\mathfrak{p}_i$ you could manage to pair them up based on the abelian group decomposition of the class group into a direct sum of cyclic groups and reducing the question to finding group inverses with minimal exponents for the representing ideal powers, but it's not clear if this would be a prescription for every possibility or not. - -REPLY [10 votes]: It's ironic that you begin your query with an example in a ring with class number 2 because this is precisely the case that's been worked out in detail. -Inspired by a 1960 result of Leonard Carlitz, Robert Valenza published a paper in 1990 entitled ``Elasticity of factorization in number fields'' which works out something akin to what you're asking about. http://www.ams.org/mathscinet-getitem?mr=1072466 He shows that the number of ways one can factor an element into irreducibles grows something like $h_k/2$ - assuming that the ring $\mathcal O_k$ is not a PID. Unfortunately, this is a lot more subtle than it sounds: for example, if $\mathcal O_k = \mathbb Z[\sqrt{-5}]$ is the classic example with class number $h_k = 2$, then the element $6 = 2 \cdot 3 = (1 + \sqrt{-5}) \cdot (1 - \sqrt{-5})$ can only be factored in to 2 irreducibles. In general, if -$ a = \displaystyle \prod_{i=1}^n \varpi_i = \prod_{j=1}^m \pi_j$ -has different factorizations, then the ratio $n/m \leq h_k/2$. Valenza's theorem asserts that we always have $m = n$ if and only if $h_k \leq 2$. As another nice example, Valenza considers the ring $\mathcal O_k = \mathbb Z[(1+\sqrt{-23})/2]$ with class number $h_k = 3$. We observes that -$27 = 3^3 = (2 + \sqrt{-23}) \cdot (2 - \sqrt{-23})$ -so that $n/m = h_k/2 = 3/2$. In general, if $Z_n \hookrightarrow \text{Pic}(\mathcal O_k)$ then Valenza explains how to exhibit a factorization with $m = 2$ and $n = |Z_n|$ . -My graduate student Alex Barrios once wrote up a very nice summary of Valenza's result; it can be found here: http://www.math.purdue.edu/~egoins/seminar/12-10-05.pdf<|endoftext|> -TITLE: The density of numbers of the form $p + 2^k$ -QUESTION [6 upvotes]: In 1934, Romanoff proved that the following set has positive lower density: -$$\displaystyle \mathcal{R}(x)= \{n \in \mathbb{N} : n \leq x, n = p + 2^k \}$$ -where $p$ is a prime and $k \geq 0$ is a non-negative integer. In 2010 Lee proved the analogous result when powers of 2 are replaced by terms in the Fibonacci sequence. Very recently, Ballot and Luca generalized the above to arbitrary non-degenerate linear recurrences (Ballot, Christian, Luca, Florian, On the sumset of the primes and a linear recurrence, Acta Arithmetica 161.1 (2013)). -The basic idea behind the proof is to define $r(n)$ to be the number of ways to write $n$ as the sum of a prime and a term in the linear recurrence considered, and simultaneously obtain a lower bound for $\displaystyle \sum_{n \leq x} r(n) \gg x$ and and upper bound for $\displaystyle \sum_{n \leq x} r(n)^2 \ll x$, and then apply the Cauchy-Schwarz inequality to obtain the desired result. -Thus, the nature of the proof does not yield an explicit constant. So there are two questions that remain unanswered: - -Can one obtain an asymptotic for $\#\mathcal{R}(x)$? Is the natural density expected to exist at all? - -Can one obtain an explicit constant for the lower bound $\#\mathcal{R}(x) \gg x$? - - -Thanks for any insights. It would already be insightful to obtain an answer to the above two inquiries for the case of Romanoff's theorem. - -REPLY [4 votes]: Romani (Calcolo 20 (1983), 319 - 336) gave a heuristic argument leading to the conjecture that the asymptotic density should exist and should be about 0.434... . -The basic idea of the heuristics is that the probability for the event $n=p+2^a$ there are some obvious congruence conditions, but that apart from these conditions these events behave like independent random variables. In this way one can not only obtain a guess for the density, but also a conjectural distribution for the number of integers for which the equation $n=p+2^a$ has exactly $k$ solutions.<|endoftext|> -TITLE: Intuitionistic algebraic topology? -QUESTION [9 upvotes]: Are there results in algebraic topology -- preferably relating to homology or homotopy or phraseable in simplicial sets -- that are not true in an intuitionistic logic? -In other words, are there results that crucially rely on a law of the excluded middle, or on the axiom of choice? - -REPLY [3 votes]: Without AC, it is impossible to prove that every set is equipotent to an ordinal (in ZF !) and it is impossible to prove that a functor is an equivalence of categories if and only if it is full faithful and essentially surjective ; so I guess that a lot of things are wrong without AC, in particular concerning combinatorial model categories because very often we have to use transfinite cardinals. For example, "Implications of large-cardinal principles in homotopical localization" or "Definable orthogonality classes in accessible categories are small" for links between large cardinal axioms and Bousfield localization.<|endoftext|> -TITLE: Do Homotopy Fully Faithful Functors Push-out? -QUESTION [9 upvotes]: A (homotopy) fully faithful functor is a map of $\infty$-categories which induces weak equivalences on mapping spaces. -Are homotopy fully faithful functors preserved under (homotopy) pushout? -More precisely, if $C\to D$ is fully faithful, and $C\to E$ is an arbitrary functor, is the canonical map $E\to E\sqcup_C D$ fully faithful? - -REPLY [11 votes]: The answer is yes, fully-faithful functors are stable under co-base change. -This is a model independent statement and so we can in particular take $\infty$-category to mean Segal categories. Then this follows directly from Cor. 16.6.2 in the arXiv version of Carlos Simpson's book "Homotopy theory of higher categories", in particular see the proof of this corollary. -More precisely, the full-faithfulness condition in terms of hom spaces also makes sense for Segal precategories, and it is clearly preserved for (homotopy) pushouts of these. Simpson's Cor. 16.6.2 shows that this condition is still preserved after you re-complete to get a Segal category again. -Simpson proves this very generally in the context of M-enriched Segal precategories with very mild conditions on the model category M. By using Simpson's result and varying M you get the analogous statement in many other contexts, not only for $\infty$-categories. For example you also get this in the canonical homotopy theory of ordinary categories, and also in the homotopy theory of Cat-enriched Segal categories. Since this later is equivalent to the homotopy theory of bicategories and in this case the fully-faithfulness can be expressed in a homotopically independent way, you can deduce this for bicategories as well.<|endoftext|> -TITLE: The least number of quadratic polynomials needed to cover $[1,N]$ -QUESTION [6 upvotes]: Let $r = r(N)$ denote the least number of quadratic polynomials $f_i \in \mathbb{Z}[x]$, $1 \leq i \leq r$, each with non-negative coefficients, such that for every integer $n \in [1,N]$, there exists $1 \leq i \leq r$ such that $n = f_i(m)$ for some integer $m \in \mathbb{N}$. Clearly, $r(N) \rightarrow \infty$, since for each quadratic polynomial $f$ with non-negative coefficient the set of integral values in $[1,N]$ that can be represented by $f$ is $O(N^{1/2})$, so for any number $c$, the set of values that can be covered by $f_1, \cdots, f_c$ is also $O(N^{1/2})$. This also shows that $r(N) \gg N^{1/2}$. -In fact, $r(N) = O(N^{1/2})$ as well, since the polynomials $f_1(x) = x^2, f_2(x) = x^2 + 1, \cdots, f_r(x) = x^2 + r$ with $r = N - \lfloor N^{1/2} \rfloor^2$ will cover cover $[1,N]$. -My question is what is the smallest positive value $c$ such that $r(N) \geq cN^{1/2}$ for all $N$ sufficiently large? -Thanks for any input. - -REPLY [2 votes]: That's not an answer too, just a result of a quick and naive computation: For $k\ge1$ let $n_k$ be the largest integer such that $r(n_k)=k$. Of course then $r(m)=k$ for all $n_{k-1}+1\le m\le n_k$. We have $n_1, n_2, \ldots = 1, 2, 6, 9, 13, 19, 23, 32, 32, 37, 45, 54, 61, 72, 82, 91, 101, 114, 128, 137\ldots$. And no, OEIS does not know this sequence ...<|endoftext|> -TITLE: Is there an algorithm to compute efficiently the dessin d'enfant from a Belyi pair? -QUESTION [8 upvotes]: Let $(X,f)$ be a Belyi pair, i.e. a Riemann surface $X$ together with a morphism $f: X \to \mathbb{P}^1$, ramified only in $0,1, \infty$. Grothendieck's dessin d'enfant is the pre-image $G$ of the interval $[0,1]$ considered as a graph embedded in $X$ (a dessin d'enfant has more information than that, but let us concentrate on just the graph for the moment). What is the algorithm to compute $G$? -What I mean by an algorithm is some efficient procedure that works in terms of algebraic data used to define $X$ and $f$. We may assume $X$ projective, so let $X$ be given by its homogeneous coordinate ring $R^\bullet$, specified by generators and relations, and $f$ specified by elements of a graded $R^\bullet$-module $M^\bullet$ corresponding to the line bundle $f^*(O(-1))$. -Is there an algorithm that computes the adjacency matrix of $G$ in terms of these data? -update: a dessin d'enfant also contains splitting of the set of vertices into two classes and a cyclic ordering of edges incident to each vertex (which arises from the monodromy action) which can be used to define an action of the free group on two generators on the edges of the dessin. The stabiliser of the action is a subgroup of finite index, and in fact, a conjugacy class of a group of finite index in $F_2$ defines the isomorphism type of a dessin d'enfant. -Therefore my question can also be reformulated accordingly: how to compute efficiently (conjugacy class of) a finite index subgroup of $F_2$ corresponding to a Belyi pair $(X,f)$? - -REPLY [5 votes]: This question is addressed in the recent preprint arXiv:13112529, in section 7. This is a survey of computing Belyi maps from the designs, but section 7 addresses the inverse problem.<|endoftext|> -TITLE: An explicit construction of reals added after some forcing notions -QUESTION [5 upvotes]: Consider the forcing notion(s) introduced by Friedman (or Mitchell or Neeman) for adding a club subset of $\omega_2$ by finite conditions. In the generic extension CH fails, but I can't see the reals added by the forcing. Would you please give an explicit construction of $\aleph_2$-many reals in the generic extension by these forcings. Can we determine if the added reals are Cohen, Random, or .... -Remark. The question applies to many similar forcing constructions, in particular to the forcings introduced in Neeman's paper. -References. -1) Friedman, Forcing with finite conditions. Set theory, 285–295, Trends Math., Birkhäuser, Basel, 2006. -2) Mitchell, Adding closed unbounded subsets of $\omega_2$ with finite forcing. Notre Dame J. Formal Logic 46 (2005), no. 3, 357–371. -3) Neeman, Forcing with sequences of models of two types. - -REPLY [9 votes]: Each of the posets you mention adds $\omega_2$ many Cohen reals. Let $G$ be generic for any of the 3 posets you mentioned. The point is that any collection of $\omega_1$-many reals in $V[G]$ can be captured by an intermediate extension of the form $V[G \cap M]$ where $M$ is a sufficiently elementary substructure of $V$ containing $\omega_1$ as a subset and $G$ contains a strong master condition (in the sense of Mitchell) for the model $M$. Then one can show that the quotient forcing $\mathbb{P}/(G \cap M)$ is strongly proper with respect to all countable models of the form $N[G \cap M]$, where $N$ is a countable model from $V$. This is a general phenomenon, but in the case of these forcings it can be shown rather directly; Neeman's preprint gives the details. The fact that the quotient is strongly proper with respect to stationarily many countable models abstractly implies that the quotient adds lots of Cohen reals; Mitchell discusses this in his paper "On the Hamkins Approximation Property".<|endoftext|> -TITLE: If a subset and its complement are path-connected, an neighborhood of the subset is path-connected -QUESTION [5 upvotes]: I apologize if this is too elementary for this site. -Given a closed subset, $X\subset \mathbb{R}^n$, given $X$, $X^C$ path-connected, show that any path-connected neighborhood of $X$, denoted $M$, has that $M-X$ is path-connected. -Both my professor and I are unable to solve it. It came up in the context of metric geometry (specifically, for continuous maps $S^{k}-\{(0, 1)\}\to \mathbb{R}^{k+1}$) , but it seemed to generalize. -If this is a standard result, where can I find it? -Thank you. - -REPLY [6 votes]: The answer seems to be yes. -Indeed, $M$ and $X^C$ are open subsets which cover $\mathbb{R}^n$ and whose intersection is $M \setminus X$. The Mayer-Vietoris sequence gives -$$H_1(\mathbb{R}^n)\to H_0(M\cap X^C)\to H_0(M)\oplus H_0(X^C) \to H_0(\mathbb{R}^n)\to 0. $$ -This sequence is isomorphic to -$$0 \to H_0(M\setminus X)\to \mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z} \to 0. $$ -Therefore $H_0(M\setminus X)$ is isomorphic to $\mathbb{Z}$, which means that $M\setminus X$ is path-connected. -Note that we did not use the fact that $X$ is path-connected, only that it is closed and that $M$ and $X^C$ are path-connected.<|endoftext|> -TITLE: A question about representations of finite groups -QUESTION [9 upvotes]: Let $G$ be a finite group and let $V$ be an irreducible complex representation of $G$. -Does there exist an element $g \in G$ which acts on $V$ with distinct eigenvalues? -If true, can you provide a proof/reference, and if false, a counterexample? - -REPLY [27 votes]: If $g\in G$ has order $n$, then its eigenvalues are all $n$th roots of unity. So if $V$ is a representation with degree greater than $n$, the eigenvalues of $g$ on $V$ can't all be distinct. -There are plenty of examples of groups with irreducible representations whose degree is greater than the largest order of any element of the group, and so these will certainly give counterexamples. For example, the symmetric group $S_6$ has irreducible representations of degree 9,10 and 16.<|endoftext|> -TITLE: Non-dense uncountable linear orderings -QUESTION [16 upvotes]: While I was working on my paper I came across this question: -$\mathbf{Question}$. Suppose $(X,<)$ is an uncountable linear ordering of cardinality $\kappa$. Is there a subset $Y$ of $X$ such that $|Y|=\kappa$ and for any two elements $y_{1},y_{2}$ of $Y$ with $y_{1} -TITLE: Is there always a maximum anti-rectangle with a corner square? -QUESTION [20 upvotes]: Let $C$ be an axis-parallel orthogonal polygon with a finite number of sides. Define an anti-rectangle in $C$ as a set of small squares in $C$, such that no two of them are covered by a single large rectangle in $C$. Define a maximum anti-rectangle as an anti-rectangle that contains a maximum number of squares. For example, the following L-shape has a maximum anti-rectangle with 2 squares: - -The following C-shape has a maximum anti-rectangle with 3 squares: - -And the following shape has a maximum anti-rectangle with 5 squares: - -When I try to build an anti-rectangle, I usually start with putting squares in the corners of $C$, because intuitively, the corners have the least chance of being covered by a rectangle. This raises the following conjecture: -For every $C$, there is a maximum anti-rectangle, which contains a corner square. -(- a corner square is a square with at least two adjacent sides that are in the boundary of $C$). -I found in Chaikan et al (1981) a proof that the conjecture is true when $C$ is linearly-convex (- contains every vertical or horizontal line that connects two of its points; like the L-shape above). -Their proof is not valid when $C$ is not linearly-convex, but I also cannot find a counter-example. What do you say? - -CONCLUSION: Many thanks to all repliers. The conjecture is false when the polygon may have holes (as shown by dmotorp). It is true when the polygon is hole-free (as proven by Nick Gill and mhum). -UPDATE: I cited this thread in the following working paper. I hope it will be accepted. - -REPLY [7 votes]: This answer is more or less just a formalization of the intuition put forward by Nick Gill in his answer. The beginning parts are mostly just formalities, so you can probably skip down to the diagram. - -Let $C$ be an axis-aligned orthogonal polygon. Following Chaikan et al., we'll consider $C$ as a union of squares. For any square $x \in C$, we define $$R(x) = \{ y \in C \;|\; \exists \text{ a rectangle in $C$ containing $x$ and $y$}\}$$ -with the understanding that $x\in R(x)$. Next, we define a pre-order $\leq$ on $C$ by $$ x \leq y \iff R(x) \subseteq R(y)$$ -Let $x$ and $y$ form an anti-rectangle and let $R(z) \subseteq R(x)$. Then, we can show that $z$ and $y$ also form an anti-rectangle. It follows that given an anti-rectangle $\{x_1, x_2, \ldots, x_k\}$, we can find another anti-rectangle $\{x_1', x_2', \ldots, x_k'\}$ of equal cardinality where each $x_i'$ is chosen to be a minimal element such that $x_i' \leq x_i$. In fact, we can construct a maximum anti-rectangle by selecting one representative from each of the minimal equivalence classes (defined in the usual way by the pre-order). -It now remains to show that if $C$ does not contain any holes, then there exists at least one minimal corner square. We will actually show slightly more: that there is at least one corner square $x$ on a "support edge" (by Chaikan et al's terminology) such that $R(x)$ is a rectangle. Observe that if $R(x)$ is a rectangle, then $x$ is minimal; we leave as an exercise for the reader to construct an example that shows that the converse is false. -Each edge in $C$ can be categorized by the squares at its endpoints. An edge can have either two, one, or zero corners at its endpoints. In the terminology laid out in Nick Gill's answer, these would correspond to $RR$, $RL$ (or $LR$), and $LL$ edges respectively (also, $RR$ edges correspond to "support edges" in Chaikan et al's terminology). -The key lemma we will need is that if $C$ is a simple, orthogonal polygon then it has strictly more $RR$ edges than $LL$ edges. First, we'll take as given that there are strictly more $R$ corners than $L$ corners (again, as defined in Nick Gill's answer) in $C$. Let $rr$, $ll$, and $rl$ be the number of $RR$, $LL$, and $RL$ edges. We will now try to count the number of each type of corner via counting each edge. Each $RR$ edge we'll count as two $R$ corners, each $LL$ edge as two $L$ corners, and each $RL$ edge as one $R$ and one $L$ corner. Counting in this way, we end up counting each corner exactly twice. So, we have $2r = 2rr+rl$ and $2l = 2ll+rl$. Thus, $0 < 2r - 2l = 2rr - 2ll$ and hence there are strictly more $RR$ edges than $LL$ edges. -Consider the following diagram of an $RR$ edge: - -The heavy black lines indicate known boundaries of $C$. The red squares, $x$ and $y$ are the corners of the $RR$ edge. The blue squares $a$ and $b$ are the edge squares directly above $x$ and $y$ respectively (i.e.: all the squares between $x$ and $a$ and between $y$ and $b$ are interior squares). -If there were no edge squares in the shaded region, then $R(x)$ would be a rectangle and we would be done. So, let $c$ be an edge square in the shaded region. Furthermore, we choose $c$ to one of the edge squares closest to the $xy$ edge (i.e.: $c$ is one of the southernmost edge squares in the shaded region). Call the edge corresponding to $c$, $E$. Note that while there may be more than once such edge, for our purposes we will only need to choose one. -We now infer that $E$ must be an $LL$ edge. By construction of the shaded region, the endpoints of $E$ must lie inside the shaded region (otherwise, it would violate one of the clear paths between $x$ and $a$ or $y$ and $b$). Thus, if one of the endpoints of $E$ were a corner, it would contradict the choice of $c$ as one of the southernmost edge squares. Finally, we see that $E$ can be uniquely identified in this way with the $RR$ edge corresponding to $xy$. Otherwise, it would once again contradict the choice of $c$ as closest to $xy$. -So, we conclude that any $RR$ edge where the corners are not minimal can be uniquely identified with an $LL$ edge. Since there are more $RR$ edges than $LL$ edges, one of them must contain a minimal corner.<|endoftext|> -TITLE: Homotopy groups other than $\pi_1$ : what are they good for? -QUESTION [7 upvotes]: Homotopy groups $\pi_k$ were introduced before the homology groups $H_k$ in the 1st topology book I read and the 1st topology course I took. Later on $\pi_1$, $H_k$, and $H^k$ appeared in numerous contexts in a variety of subjects. -On the other hand, I never encountered $\pi_k$, $k\ge 2$ beyond the topology textbooks, with the exception of the $\pi_3(S^2)$ curiosity, which IMHO is just an algebraic sideshow for Hopf fibration. -Therefore the question: are homotopy groups $\pi_k$, $k\ge 2$ useful for something that cannot done with other invariants such as $H_k$? - -REPLY [7 votes]: Higher homotopy groups arise in the study of topological defects in physics. -Many physical systems are described by an "order parameter", a manifold valued field on a region of space, $f:U\subset \mathbb{R}^3\rightarrow M$. For example, in some materials there is a continuous (if you don't look too closely) vector valued magnetization property defined at each point in space. The vector has a constant length fixed by the nature of the material. So we have a field $f:\mathbb{R}^3\rightarrow S^2$. In a certain type of liquid crystal the molecules are shaped roughly like line segments so at each point the orientation of the molecules is given by a line. So we have a field $f:\mathbb{R}^3\rightarrow \mathbb{R}P^2$. Dislocations in crystals give rise to maps $f:\mathbb{R}^3\rightarrow T^3$ where $T^3$ is the 3-torus. -Sometimes these order parameters have singularities, points or curves where $f$ is no longer defined. These can be studied by looking at the region around the "defect". For example, if $f$ has a point singularity at the origin then we can look at $f$ on a sphere, $S^2$, around the defect. This gives rise to the restriction $f':S^2\rightarrow M$ which gives an element of $\pi_2(M)$. If $f$ is continuous everywhere except at the origin, then continuously moving the sphere outwards from the origin won't change which homotopy class $f'$ represents. So if $f'$ represents a non-trivial homotopy class then in some sense the defect can't be hidden or deformed away and can be detected on the surface of a sphere a long way from the defect itself. These phenomena have a variety of physical consequences. -(Defects along a line give rise to elements of $\pi_1(M)$ and elements of $\pi_3(M)$ can give rise to a non-localisable "texture defect".) -There is a substantial amount of literature on these defects and they apply to everything from solid and liquid crystals to ferromagnets and cosmic strings. -There is a wikipedia article on topological defects. Here is a paper with some pictures of point defects in liquid crystals. -Higher homotopy groups also arise in the (hot right now) field of topological insulators.<|endoftext|> -TITLE: Is $\mathbb{H}^n$ quasi-isometric to a leaf of a codimension 1 foliation of a compact manifold? -QUESTION [13 upvotes]: If we extend the action of $\pi_1(\Sigma_g), g\geq 2,$ from $\mathbb{H}^2$ to its boundary $\partial_{\infty}\mathbb{H}^2=S^1$, the surface bundle corresponding to this action of $\pi_1(\Sigma_g)$ on $S^1$ has a leaf that is isometric to $\mathbb{H}^2$. Hence $\mathbb{H}^2$ is even isometric to a leaf in a codimension 1 foliation of a compact manifold. -For $n>2$ the above argument fails, and hence my question in the title: Is it known whether for every $n$ $\mathbb{H}^n$ is quasi-isometric (or even isometric) to a leaf of a foliation of a compact (n+1)-manifold? - -REPLY [2 votes]: Let me add a comment. One knows that a codimension-one foliation without leafwise holonomy must be given by a Z^n action, by a result of Sacksteder -Sacksteder, Richard -Foliations and pseudogroups. -Amer. J. Math. 87 1965 79–102. -So, if the foliation is defined by a group action on the circle, and the orbits are QI to a hyperbolic space, then there must be leaves with holonomy. This means there is some point on the circle for which there is a non-trivial element of the group that fixes the point. Maybe this can be used to prove that no such actions exist. In any case, it proves that not all leaves are QI to H^n.<|endoftext|> -TITLE: Pro-algebraic versus continuous Galois cohomology, and schematic homotopy types -QUESTION [11 upvotes]: I've been thinking about Bertrand Toen's approach to studying the homotopy theory of schemes, and I've come across an inconsistency in my understanding of the subject that I was hoping somebody might be able to iron out for me. -If I take a field $k$ of characteristic $\neq\ell$, then if I have understood this -http://www.aimath.org/WWN/motivesdessins/Toen.pdf -correctly, there should be an $\ell$-adic schematic homotopy type $h(X)$ associated to every smooth and projective variety over $k$, such that the 'absolute' $\ell$-adic cohomology of $X$, which coefficients in a local system, can be calculated as an appropriate hom set inside the derived category of perfect complexes on $h(X)$. Specifically, I understood that the derived category of perfect complexes on $h(X)$ should be equivalent to the full subcategory of the usual $\ell$-adic derived category $D^b_c(X_{\mathrm{et}},\mathbb{Q}_\ell)$ whose cohomology sheaves are lisse. -The reason this is confusing me is because I also thought that the schematic homotopy type associated to the point $\mathrm{Spec}(k)$ should be the classifying stack $BG_k^\mathrm{alg}$ associated to the $\mathbb{Q}_\ell$ pro-algebraic completion $G_k^\mathrm{alg}$ of the absolute Galois group of $k$. -But the derived category of perfect complexes $D_\mathrm{perf}(BG_k^\mathrm{alg})$ on this classifying stack, as far as I can tell, should not be equivalent to $D^b_c(\mathrm{Spec}(k)_\mathrm{et},\mathbb{Q}_\ell)$ - if I take some finite dimensional continuous $G_k$-representation $V$ then the cohomology computed in the former category will be the algebraic group cohomology -$$\mathrm{Hom}_{D_\mathrm{perf}(BG_k^\mathrm{alg})}(\mathbb{Q}_\ell,V[i])=H^i(G_k^\mathrm{alg},V)$$ -whereas the cohomology computed in the latter category will be continuous group cohomology -$$\mathrm{Hom}_{D^b_\mathrm{lisse}(X_\mathrm{et},\mathbb{Q}_\ell)}(\mathbb{Q}_\ell,V[i])=H^i_\mathrm{cts}(G_k,V).$$ -These two won't coincide, and this suggests that there are no suitable subcategories of $D_{\mathrm{perf}}(BG_k^\mathrm{alg})$ and $D^b_c(\mathrm{Spec}(k)_\mathrm{et},\mathbb{Q}_\ell)$ that will coincide. -So what's going on? I've convinced myself that the two expectations -$$ D_\mathrm{perf}(h(X))\cong D^b_\mathrm{lisse}(X_\mathrm{et},\mathbb{Q}_\ell)$$ -and -$$h(\mathrm{Spec}(k))\cong BG_k^\mathrm{alg} $$ -are incompatible - so which one should I hold on to? Or have I missed a trick somewhere? -EDIT: Clarified what I mean by the 'cohomology' calculated in both categories. - -REPLY [2 votes]: This answer is due to Jon Pridham. -While we might not expect $H^i(G_k,V)=H^i(G_k^\mathrm{alg},V)$ for every finite dimensional, continuous $G_k$-representation $V$, there are certain results from the motivic theory that suggest that this might be true if $k$ is a number field (or local field of char $0$), and that the representation 'comes from geometry'. -So although we possible should not expect a schematic homotopy type $h(\mathrm{Spec}(k))$ which is a gerbe, and whose derived category of perfect complexes is equivalent to $D^b_c(\mathrm{Spec}(k)_\mathrm{et},\mathbb{Q}_\ell)$, there may well be a gerbe whose derived category of perfect complexes is equivalent to the subcategory of $D^b_c(\mathrm{Spec}(k)_\mathrm{et},\mathbb{Q}_\ell)$ generated by complexes of geometric origin, at least when $k$ is a number field. -Essentially, the existence of the motivic $t$-structure 'almost' implies that the motivic schematic homotopy type is a gerbe, from which it would follow that the $\ell$-adic schematic homotopy type associated to motivic representations (i.e. representations of geometric origin) is a gerbe.<|endoftext|> -TITLE: Are there quanitative versions of Thurston's geometrization for manifolds which fiber over $S^1$? -QUESTION [12 upvotes]: The geometrization theorem tells us: - -Theorem (Thurston) The mapping torus $M_\phi$ of a pseudo-Anosov diffeomorphism $\phi: S_g \rightarrow S_g$ from a genus $g$ surface to itself admits a complete hyperbolic metric of finite volume. - -Otal has a complete presentation of this result starting from standard graduate material. -My question is: How much can one say about the geometry of the $M_\phi$ as a function of $\phi$ and $g$? Are there upper bounds on volume? Systole? Are there simpler bi-Lipschitz models of $M_g$? Are there any easily computable geometric quanities? -For example -- One has that the Heegaard genus of $M_g$ is $g$. This gives a lower bound on volume in terms of $g$. Is there an upper bound? - -REPLY [6 votes]: To address your last point, for a mapping torus of $S_g$, there is an upper bound on the Heegaard genus of $2g+1$, which is sharp (in fact, $M_{\phi^n}$ will have rank $2g+1$ for $n$ large), but is not always an equality. However, there is no upper bound on the volume of a fiber bundle in terms of Heegaard genus. For example, Jesse Johnson has a description of all fiber bundles of Heegaard genus 2, and it's not hard to see that these examples may have arbitrarily large volume by a Dehn filling argument. -For a fairly explicit and detailed description of the relation between the geometry and the monodromy for punctured torus bundles (in particular, its continued fraction expansion), see the work of Gueritaud. One has that the volume is coarsely related to the continued fraction expansion length (or LR-decomposition, which is the analogue of Brock's pants distance result in this case), with explicit sharp constants. This is preceded by the work of Minsky (who gets coarse estimates on the holonomies of pivot curves) and in fact Jorgensen who originally discovered hyperbolic structures on these manifolds. See also this paper for estimates on the cusp area in terms of the LR-length and other geometric information. -Unfortunately in the higher genus case, there is not such a good comparison of the topological and geometric pictures. In particular, there are not many results which are uniform in the genus (although Brock's upper bound on volume in terms of pants distance has a uniform constant, but the lower bound cannot). For fixed genus, Minsky's model in principle gives a complete coarse comparison of the geometry with the topological description in terms of the action of the monodromy on the curve complex. However, the constants are not explicit, as they are obtained by geometric limit arguments. Also, there is not a strict analogue of continued fraction length; instead, one works with ``tight geodesics" in the curve complex, which are difficult to compute with. All this machinery was developed to resolve the ending lamination conjecture, but of course applies to the fibered case.<|endoftext|> -TITLE: Sums of powers of multinomial coefficients -QUESTION [5 upvotes]: Does anyone know how to estimate the sums of $p$-powers of $s$-multinomial coefficients, with $p,s\in\mathbb N$ arbitrary? Here's some data that I gathered from OEIS, -$$\sum_{a_1+\ldots+a_s=n}\binom{n}{a_1,\ldots,a_s}^p -\simeq s^{pn}\sqrt{\frac{K_{sp}}{(\pi n)^{(s-1)(p-1)}}}$$ -where $K_{1p}=K_{s1}=1$, $K_{2p}=2^{p-1}/p$, $K_{3p}=1,\frac{27}{16},\frac{81}{16},16,\ldots$ and $K_{s2}=1,1,\frac{27}{16},4,\ldots$ -My questions are: (1) is the above formula correct? (2) what's $K_{sp}$, in general? -Many thanks. - -REPLY [2 votes]: $$K_{sp}=\frac{s^{s(p-1)}}{p^{s-1}2^{(p-1)(s-1)}}$$<|endoftext|> -TITLE: The first element in the stable homotopy of a $K(\mathbb{Z}/2, n)$ -QUESTION [5 upvotes]: The first element in the stable homotopy groups of a $K(\mathbb{Z}/2, n)$ (which is outside the range of the Freudenthal suspension theorem) is $\pi_{2n} K(\mathbb{Z}/2, n) \simeq \mathbb{Z}/2$. In particular, there is a unique nontrivial stable map $S^{2n} \to K(\mathbb{Z}/2, n)$. This map is necessarily zero in cohomology. -In what Adams filtration can we detect this map? (Can we detect it using some other cohomology theory?) - -REPLY [8 votes]: The mod 2 cohomology $H^*(K(\mathbb{Z}/2,n))$ is freely generated over the Steenrod algebra by the canonical class $\iota\in H^n$ in degrees $*\leq 2n$, and the only relation introduced in degree $2n+1$ is $Sq^{n+1}(\iota)=0$. Thus the Adams spectral sequence for $K(\mathbb{Z}/2,n)$ has classes in bidegrees $(0,n)$ (from $\iota$) and $(1,2n+1)$ (from the relation), and nothing else in bidegrees $(s,t)$ with $t-s<2n$. Both of these classes must survive the spectral sequence for degree reasons. This shows that there is a nontrivial stable map $S^{2n}\to K(\mathbb{Z}/2,n)$ of Adams filtration 1.<|endoftext|> -TITLE: The j-function and Pell equations -QUESTION [9 upvotes]: Given the j-function, -$$j(\tau)=\frac{1}{q}+744+196884q+21493760q^2+\dots$$ -it is well-known that for $\tau=\tfrac{1+\sqrt{-d}}{2}$, positive integer $d$, then $j(\tau)$ is an algebraic integer of degree = class number $h(-d)$. Thus, -$$j(\tfrac{1+\sqrt{-163}}{2}) = -12^3(231^2-1)^3$$ -$$j(\tfrac{1+\sqrt{-427}}{2}) = -12^3\big((7215+924\sqrt{61})^2-1\big)^3$$ -where the squares are due to a certain Eisenstein series. However, not all d yield a cube. -Conjecture: Is it true that given fundamental discriminant $d = 3m$, $m \neq$ square, $h(-d) = 2^n$, then $j(\tau) = -(U_m)^k\,x^3$ for some integer k, where $U_m$ is a fundamental unit, and $x$ is an algebraic integer of degree $h(-d)$? -For example, given $d = 51, 483, 651$ which have $h(-d) = 2,4,8$, respectively, and fundamental units, -$$U_{17} = 4+\sqrt{17}$$ -$$U_{161} = 11775 + 928\sqrt{161}$$ -$$U_{217} = 3844063+260952\sqrt{217}$$ -then, -$$j(\tfrac{1+\sqrt{-51}}{2}) = -48^3 (U_{17})^2 (5+\sqrt{17})^3$$ -$$j(\tfrac{1+\sqrt{-483}}{2}) = -120^3 (U_{161}) {x_1}^3$$ -$$j(\tfrac{1+\sqrt{-651}}{2}) = -96^3 (U_{217})^2 {x_2}^3$$ -where $x_1,x_2$ are algebraic integers (rather tedious to write down) of degree 4,8, respectively. -Is the conjecture true? -P.S. I tested it with $h(-d) = 6$ and it does not work, so I think it is only for $h(-d) = 2^n$. - -REPLY [10 votes]: So you're really asking when the ideal generated by $j(\tau)$ is the cube of an ideal. There's a famous paper of Gross and Zagier that describes the prime factorization of the ideal generated by the difference $j(\tau_1)-j(\tau_2)$ of two CM $j$-invariants. In your case $j(\tau_2)=0$ is a CM value. This description is likely to be useful in trying to prove (or disprove) your conjecture. The reference is -Gross, Benedict H.; Zagier, Don B. On singular moduli. J. Reine Angew. Math. 355 (1985), 191–220. MR0772491<|endoftext|> -TITLE: Weak threads in $\square (\kappa, <\kappa)$ sequences -QUESTION [9 upvotes]: The following definition is well known ($\kappa$ is regular uncountable cardinal): -Definition: a sequence $\mathcal{C} = \langle \mathcal{C}_\alpha | \alpha < \kappa,\,\alpha \text{ limit ordinal} \rangle$ is a $\square (\kappa, <\lambda )$-sequence when for every $\alpha$, $|\mathcal{C}_\alpha| < \lambda$, every $C \in \mathcal{C}_\alpha$ is a club in $\alpha$, if $\gamma \in \text{acc }C$ then $C\cap \gamma \in \mathcal{C}_\gamma$ and most importantly - there is no thread i.e. there is no club $D\subset \kappa$ such that for every $\alpha \in \text{acc } D$, $D\cap \alpha \in \mathcal{C}_\alpha$. -We say that a club $D$ in $\kappa$ is a weak thread for $\mathcal{C}$ when for every $\alpha \in \text{acc }D$ there is $C \in \mathcal{C}_\alpha$ such that $D\cap \alpha \subset C$. (I don't know if this is the standard terminology) -Clearly every thread is a weak thread. -If $\lambda < \kappa$, there is no weak thread for $\square (\kappa, <\lambda)$ sequence, since it will imply the existence of a thread (we can prove it by showing that the candidates for initial segment of a thread that contain the weak thread, ordered by end extension, forming a tree of height $\kappa$ and width $<\lambda$, so by Kurepa it has a cofinal branch). On the other hand, if we have a $\square(\kappa, <\kappa)$ sequence, we can always build another $\square(\kappa, <\kappa)$ sequence that has a weak thread (but doesn't have a thread). -Question: Assuming the existence of a $\square (\kappa, <\kappa)$ sequence. Can we prove that there is a $\square (\kappa, <\kappa)$ sequence with no weak threads? -Edit: I'll try to explain why this question is interesting. -Todorcevic, in Partitioning pairs of countable ordinals present his method of minimal walks and showed that the existence of $\square(\kappa)$ implies the existence of Aronszajn tree on $\kappa$. His proof easily generalizes to show that $\square(\kappa,<\kappa)$ implies the existence of Aronszajn tree on $\kappa$. On the other hand, from Aronszajn tree it's easy to produce a $\square(\kappa, <\kappa)$ sequence (the elements of the square sequence will be some appropriate closures of the bounded branches of the tree on some club) so is seems to be equivalence. The square sequences derived this way always have weak threads. -Now, if the answer to the above question is positive we conclude that c.c.c. forcing can't destroy the tree property: every $\square(\kappa, <\kappa)$ sequence in $V[G]$ agrees on many elements with coherent sequence of width $<\kappa$ in $V$ (by the c.c.c.), and this sequence has a thread - which will be a weak thread for the original sequence in $V[G]$. So in $V[G]$ every $\square(\kappa,<\kappa)$ sequence has a weak thread and my question translate to "do there are any (thread-less) $\square(\kappa,<\kappa)$ sequences in $V[G]$" (i.e. does the tree property still holds in $V[G]$). - -REPLY [4 votes]: The answer is negative (at least at double successors of regular cardinals): it's possible that every square sequence is weakly threaded and there are Aronszajn trees. -For simplicity let work with $\omega_2$. -We start with supercompact cardinal. Using Unger construction in Fragility and indestructibility of the tree property we collapse it to be $\omega_2$ while forcing the tree property at $\omega_2$ to be indestructible under any $\omega_2$-directed closed forcing. -Now, we add a generic normal $\omega_2$-tree by the forcing $\mathbb{Q}$ (the conditions of $\mathbb{Q}$ are the normal trees of successor ordinal height $<\omega_2$ and width $\omega_1$. $\leq_{\mathbb{Q}}$ is end extension). Let $V$ be the universe at this point. $V$ has Aronszajn trees and therefore also $\square(\omega_2,\omega_1)$ sequences. -The generic tree that was added by $\mathbb{Q}$ is Suslin, and it's easy to see that the forcing that adds a branch through this generic Suslin tree, $\mathbb{T}$, satisfies that $\mathbb{Q\ast T}$ has a $\omega_2$(-directed)-closed subset. Since the tree is Suslin, $\mathbb{T}$ is $\omega_2$.c.c.. -Let $\mathcal{C}$ be a $\square(\omega_2,\omega_1)$ sequence in $V$. Since $\mathbb{T}$ restore the tree property - it adds a thread to $\mathcal{C}$, which is a club $E$ such that for every $\alpha \in \text{acc }E$, $E\cap \alpha\in\mathcal{C}_\alpha$. But since $\mathbb{T}$ is $\omega_2$.c.c., this thread contains some club from $V$, and this club weakly threads $\mathcal{C}$.<|endoftext|> -TITLE: Equal signs with fancy marks -QUESTION [8 upvotes]: Some people use $\stackrel{\mathrm{def}}{=}$, $:=$ or $\stackrel{\Delta}{=}$ for definitions. -In more informal contexts, I have also seen $\stackrel{?}{=}$, for "I wish to prove this equality, which implies the thesis", used when working backwards, or even the less common $\stackrel{!}{=}$ (for which it is difficult to infer a precise meaning, but it's something like "we are imposing this equality", or "this is an Ansatz"). -It seems like the use of these symbols is not common in papers (apart from maybe the first ones for definitions); however, they seem useful to me as a way to convey more meaning than the usual equality symbols. For instance, in some sense the "?" superscript changes the set of "allowed operations": from $a=c$ and $b=d$ it follows that $a+b=c+d$, but the same implications wouldn't hold with $\stackrel{?}{=}$ signs. -Q1: has the use of these symbols ever been properly formalized? -Q2: are there contexts in which they are used in papers? Are they standard in your field? Are they used in lectures/seminars/papers? - -REPLY [6 votes]: In a slightly different direction, $\overset{\mbox{(2.3)}}{\ge}$ (the overset should be in a smaller font using \tiny, but for some reason that wouldn't render here) is a nice way to indicate that the inequality uses $(2.3)$ - this is especially convenient with a long string of inequalities in display-math where it's otherwise hard work to figure out which inequality uses which previous calculation.<|endoftext|> -TITLE: Identifying a special function from its power series -QUESTION [5 upvotes]: Here is a power series, which looks a bit like a Hypergeometric function series, but I don't think that it is. Has anyone any idea what it is? Here $n,p,r$ are integers with $n\ge 0$ and $p\ge r\ge 0$: -$$ -f_{n,p,r}(x)\,=\,\sum_{s=0}^p \frac{x^s}{s!}\ \frac{p!\,(2n+p+s+2)!\,(n+r+s+2)!}{(p-s)!\,(2n+r+s+3)!\,(n+s+2)!} -$$ -Originally this occurred as a $q$-factorial series, but if the $q=1$ case given here could be recognised, it would be a big help. -Thanks for the answer below. A bit before I declare this answered (if I remember how to do that!) I would like to sneak in another question. Is there any sensible way to sum this in the case $x=-1$? The answer is likely zero if $r -TITLE: Possible Choices for Cofinality of $\aleph_n$ without Choice -QUESTION [7 upvotes]: $\text{ZFC}$ proves that each $\aleph_{n}$ for $n\in \omega$ is a regular cardinal. But it seems without the Axiom of Choice there are many consistent possible choices for cofinality of such cardinals. -Please introduce some references for any known consistency result with $\text{ZF}$ about possible cofinalities of $\{ \aleph_n\}_{n\in \omega}$. - -REPLY [7 votes]: Here are some partial answers: -1) By a result of Gitik, all $\aleph_n$'s can have cofinality $\omega.$ -2) the paper "Cofinality and measurability of the first three uncountable cardinals" by Apter, Jackson and Löwe completely solves the problem for $\aleph_1, \aleph_2$ and $\aleph_3.$ The following is taken from their introduction: -"In this paper, we investigate all possible patterns of measurability and cofinality for [$\aleph_1, \aleph_2, \aleph_3$]. Combinatorially, there are exactly 60 such patterns of which 13 are impossible for trivial reasons (e.g., if $\aleph_1$ is singular, then $\aleph_2$ cannot have cofinality $\aleph_1$). In this paper, we prove that the remaining 47 patterns are all consistent relative to large cardinals.'' -3) The following is appeared in Ioanna Dimitriou's PhD thesis: -Theorem. Assume V is a model of ZFC which contains $\omega-$many strongly compact cardinals. For any function $f: \omega \rightarrow 2$ there is a model of ZF in which $\aleph_{n+1}$ is regular if $f(n)=1$ and singular if $f(n)=0.$ - -REPLY [5 votes]: Ioanna Dimitriou's Ph.D. thesis as well many papers by Arthur Apter would be a good start. -You may want to read the proof of the Feferman-Levy construction, there's a nice proof in Jech "The Axiom of Choice" as well Ioanna Dimitriou M.Sc. and Ph.D. theses. Also interesting is the results in - -Truss, John "Models of set theory containing many perfect sets." Ann. Math. Logic 7 (1974), 197–219. - -Also relevant are the results by Magidor that if $\operatorname{cf}(\omega_1)=\operatorname{cf}(\omega_2)=\omega$ then $0^\#$ exists (this can be found, I believe in Magidor's paper proving the covering lemma, but also in Jech "Set Theory" as an exercise for Chapter 18); and similar applications of the covering lemma which prove the necessity of large cardinals for some of the results. Related to that is Andres Caicedo's answer to my [very old] MSE question: Consistency of $\operatorname{cf}(\omega_1)=\operatorname{cf}(\omega_2)=\omega$. -On the same breath, one should mention - -Busche, Daniel; Schindler, Ralf "The strength of choiceless patterns of singular and weakly compact cardinals." Ann. Pure Appl. Logic 159 (2009), no. 1-2, 198–248. - -Which also deals with the question of many singular $\aleph_n$'s and its consistency strength. -From these papers and their reference, I think one should be able to cover all the known material.<|endoftext|> -TITLE: Do there exist non-isomorphic groups with the same cohomology? -QUESTION [17 upvotes]: For any group $G$, cohomology can be viewed as a functor -$$ -H^\ast(G,-): G{\sf\text{-}mod}\to {\sf GrAbGrp}, -$$ -where $G{\sf\text{-}mod}$ denotes the category of (left) $\mathbb{Z}[G]$-modules and ${\sf GrAbGrp}$ denotes the category of (non-negatively) graded abelian groups. -It seems that there are non-isomorphic groups $G_1$ and $G_2$ whose integral group rings $\mathbb{Z}[G_1]$ and $\mathbb{Z}[G_2]$ are Morita equivalent (this is a result of Roggenkamp and Zimmermann). One could then ask the following question: - -Do there exist non-isomorphic groups $G_1$ and $G_2$ for which there is an equivalence of categories $F: G_1{\sf\text{-}mod}\to G_2{\sf\text{-}mod}$ such that the functors $H^\ast(G_1,-)$ and $H^\ast(G_2,-)\circ F$ from $G_1{\sf\text{-}mod}$ to ${\sf GrAbGrp}$ are naturally isomorphic? - -This is my (possibly naive) attempt to formalise the question in the title. It may be that the answer is trivially "no" by looking at $H^0$, ie the functor of coinvariants. In that case, I would like to know if there are other formulations for which the question becomes interesting. - -REPLY [15 votes]: The answer would seem to be yes if I understood your question properly. The paper http://www.m-hikari.com/ija/ija-password-2009/ija-password9-12-2009/ladraIJA9-12-2009.pdf shows that if $\mathbb ZG$ is isomorphic to $\mathbb ZH$, then one can choose an isomorphism which is augmentation preserving. They then show that if $f\colon \mathbb ZG\to \mathbb ZH$ is the augmentation-preserving isomorphism, then viewing a $\mathbb ZH$-module as a $\mathbb ZG$-module gives isomorphisms in homology and cohomology. I didn't check if their proof gives naturality of the isomorphisms, but I would be surprised if it didn't. -The paper http://www.jstor.org/stable/3062112 shows there are nonisomorphic finite groups with isomorphic integral group rings.<|endoftext|> -TITLE: Surjective entire functions without critical points -QUESTION [7 upvotes]: It is easy to construct surjective locally univalent holomorphic functions $f: {\mathbb D}\to {\mathbb C}$, where ${\mathbb D}$ is the open unit disk. -I am pretty sure that the answer to the following is positive and well-known (to experts) and is somewhere in the literature: -Question. Are there (non-affine) surjective entire functions $f: {\mathbb C}\to {\mathbb C}$ without critical points? -Such functions (up to an additive constant) would have the form $f(z)=\int_0^z e^{h(w)} dw$, where $h(w)$ is another nonconstant entire function. I do not see, however, how to verify surjectivity of such $f$. Of course, it can miss at most one point in ${\mathbb C}$ and, hence, it would be astounding if indeed this was the case for arbitrary $h$. However, I do not see how to rule this out. Applying Cauchy's argument principle in this setting looks extremely messy. - -REPLY [7 votes]: The simplest surjective entire function without critical points is the "probability integral", $\int_0^z e^{-\zeta^2}d\zeta$. -Proof. Suppose it omits $a$. As it is of order 2, it must be of the form $a+e^P$, -where $P$ is a polynomial of degree $2$. But this function has critical points:-) -The argument extends to $\int e^Q$ for any polynomial of degree at least $2$.<|endoftext|> -TITLE: How do I efficiently find a sequence of Reidemeister moves between equivalent link diagrams? -QUESTION [7 upvotes]: In knot theory, two link diagrams are equivalent if and only if they can be related by performing a finite number of Reidemeister moves. But sometimes it is so confusing that I don't know which type move should I perform on link to get desired result. Is there an efficient procedure to relate one link diagram to another provided we know that the two links are equivalent? What is the computational complexity of determining knot equivalence? Is it NP complete? -Also, a related question: Is there any computer software which efficiently finds Reidemeister moves between equivalent diagrams? Because sometimes I find it is very difficult to visualise the link after some Reidemeister moves. - -REPLY [5 votes]: I think yanglee's answer gives the state-of-the-art as far as determining how many Reidemeister moves are needed to get between diagrams of equivalent knots. -It is not known that knot recognition is NP-complete. -For software to find equivalent diagrams, I would try out the program Gridlink written by Marc Culler, which allows one to experiment with grid diagrams of links, and perform the corresponding grid moves. If two projects of knots in grid position are equivalent by Reidemeister moves, then they will be equivalent by the grid moves. I think grid diagrams are easier to work with computationally and to render output on a computer. The Kirby calculator also allows certain Reidemeister moves to be performed, but is not specifically designed for this purpose.<|endoftext|> -TITLE: Influence of Yau's solution to the Calabi Conjecture on the field of PDEs -QUESTION [7 upvotes]: I remember reading a long time ago(I can't recall where, unfortunately) that Yau's solution of the Calabi-Yau conjecture introduced new techniques that were very important for the field of partial differential equations. -My question is essentially two-fold: first of all, what exactly were the breakthroughs that he made, and secondly, why do they have a wide applicability to other PDEs? My background is in geometry, so I'm well-aware of the importance of the confirmation of the Calabi conjecture, but I'm interested in the analytic side here. I know that he applied the continuity method, and used some very hard a priori bounds to show closedness of the set of solutions, the openness having been proved by Calabi, but not much more. - -REPLY [4 votes]: Probably not quite the answer to your question, nevertheless some historical remarks: the barrier function employed by Yau for deriving the estimates on the Laplacian was already used by Progorelov in his resolution of the Minkowski problem. -In that approach, one reduces the Laplacian estimate to the zero order estimates, that is, to $\sup |\phi|$. This was the missing piece in deriving the estimates. -The case of finding K\"ahler-Einstein metrics, when $c_1<0$, was in that respect easier as the first order estimates are pretty easy to derive. This was independently done by Aubin as well. But rather than the continuity method, Aubin uses a more difficult variational approach. -Yau's method for finding the zeroth order estimate was pretty involoved, which was later simplified by Kazdan and Bourguignon. Notice that if one wanted to solve the equation on a domain with boundary rather than a closed manifold, the $C^0$ estimate again comes for free. -The third order estimates are also not straightforward to derive. The quantity, $S$, that was considered by Yau for deriving the third order bound was intorduced by Calabi in his ingenuous calculation in the case of real Monge-Amp`ere equation, and If I'm not mistaken, Yau's method for bounding $S$ using the maximum principle to was due to Nirenberg. -The continuity method itself was already classical, know, probably since Sergej Bernstein's work.<|endoftext|> -TITLE: Do interpolation nodes have to be dense? -QUESTION [5 upvotes]: Let $f(x) = \exp(x)$ and $(\xi_i)_{i=0}^\infty, \, \xi_i \in (0,1)$ be a sequence of points from the unit interval. -For $n \in \mathbb{N}$ let $P_n$ be a polynomial of degree $n$ that interpolates $f$ at $\xi_0, \ldots, \xi_n$, i.e. -$$ -f(\xi_i)-P_n(\xi_i) = 0, \text{ for all } i=0,\ldots,n, -$$ -Question: -Does uniform convergence of the interpolation process on $(0,1)$, i.e. -$$\lim_{n \rightarrow \infty} \|f - P_n\|_\infty =0$$ - imply that the sequence $(\xi_i)_{i=0}^\infty$ is dense in $(0,1)$? -My intuition says it is but I failed to proof it. -Thank you! - -REPLY [2 votes]: Unfortunately not. Just look at the (confluent) limit case when all $\xi_i=0$: in this case $P_n$ is just the $n$-th Taylor polynomial for $f$. which clearly converges to $f$ not only on $(0,1)$, but on the whole complex plane, but $\xi_i$ are all at the origin. -A less extreme argument is: if we consider the sequence $q_n$ of best polynomial approximants to $f$ on a subinterval $(\alpha, \beta)$ of $(0,1)$, then (i) $q_n$ actually interpolate $f$ on $(\alpha, \beta)$ (but not outside), and (ii) $(f-q_n) \to 0$ in the largest ellipse with foci at $\alpha$ and $\beta$ where $f$ has an analytic continuation. Since $f$ is entire, convergence holds on the whole $\mathbb{C}$, so on $(0,1)$. -EDIT after the comments of Dirk and Qiaochu Yuan: -I don't think that the fact that nodes are different or form a sequence makes any difference. Still, what about this argument: the Cauchy formula for the interpolation error at a point $t$ is -$$ -(f-p_n)(t)=\frac{f^{(n+1)}(\theta)}{(n+1)!} \omega_n(t), \quad \omega_n(t)=\prod_{i=1}^{n+1}(t-\xi_i). -$$ -For $t\in (0,1)$, we have $\|\omega_n\|_\infty\leq 1$, and $f^{(n+1)}(\theta)\leq e$, REGARDLESS the distribution of $\xi_i$'s on $(0,1)$. In particular, they could live on $(0, 1/2)$ only, etc.<|endoftext|> -TITLE: Higher dimensional Rubik's cube group -QUESTION [7 upvotes]: Since "cubes" with higher dimension than three exist I think it's natural to ask for higher dimensional Rubik's cubes. These so called hypercubes don't seem to have been described from a group theoretic point of view. -Are there any papers on this? Is the group of the $3\times 3\times 3 \times 3$ cube a subgroup of a wreath product of another wreath product? -In case you don't know about the $3\times 3\times 3$ cube. Its group is a subgroup of a product of wreath products. The wreath products describing the corner pieces and the edge pieces and representing a permutation of them with its action in the respective orientation. That is why I conjecture that in the $3\times 3\times 3 \times 3$ case we might get a wreath product of the permutation of the faces, which are now 3d cubes itself, by the the wreath product of the permutation of the 2d faces of these by the groups of their orientations. - -REPLY [4 votes]: The 4-dimensional, i.e. $3 \times 3 \times 3 \times 3$, equivalent of the -Rubik's cube has 8 three-dimensional sides, each of which consists of -$3^3 = 27$ three-dimensional colored "squares". -Of these 27 "squares", the one in the center is fixed. Thus in total our -4-dimensional Rubik's cube has $8 \cdot 26 = 208$ movable "squares", -and the group of its sequences of moves under composition embeds therefore -into ${\rm S}_{208}$. -Now just as the usual 3-dimensional Rubik's cube has 2 kinds of movable cubies -(edge stones and corner stones), our 4-dimensional Rubik's cube has 3 distinct -kinds of movable cubies: - -The corner stones, of which there are $2^4 = 16$. -Each corner stone has 4 visible colored "squares". -The edge stones, of which there are $2 \cdot 12 + 8 = 32$. -Each edge stone has 3 visible colored "squares". -The face stones, of which there are $2 \cdot 6 + 12 = 24$. -Each face stone has 2 visible colored "squares". - -Therefore, analogous to the usual 3-dimensional case, we can finally say -that our 4-dimensional Rubik's cube group embeds into the following direct -product of wreath products: -$$ - {\rm A}_4 \wr {\rm S}_{16} \times {\rm S}_3 \wr {\rm S}_{32} \times - {\rm C}_2 \wr {\rm S}_{24}. -$$<|endoftext|> -TITLE: Some special complex tori -QUESTION [7 upvotes]: Let $T=V/\Gamma $ be a complex torus; so $V$ is a finite-dimensional complex vector space and $\Gamma$ a lattice in $V$. -Moreover I have a positive definite hermitian form $H$ on $V$ such that the real part of $H$ takes integral values on $\Gamma $. -If $\dim V=1$, the existence of such a form is equivalent to saying that the elliptic curve $T$ has complex multiplication. What can one say if $\dim V>1$? - -REPLY [4 votes]: Some minor disagreement has arisen in the comments as to `what' an elliptic curve is. Concretely, to perhaps an algebraic geometer the quotient of $\mathbb{C}$ by the lattice $\mathbb{Z}+\mathbb{Z} \sqrt{-D}$ is a very reasonable and simple elliptic curve. However I'm prone to dismiss it (or specifically, rescale it to have covolume one) since $diag(1, \sqrt{D})$ does not lie in $SL_2\mathbb{R}$ but rather $GSL_2$. Likewise, I am prone to only admit principally polarized abelian varieties, i.e. arising from $Sp_{2g}$ and not $GSp_{2g}$. This unfortunately invalidates my perspective on the OP's question, since he/she appears open to arbitrary covolume lattices and my comments below are (i see now) specific to being in the unimodular (normalized covolume one) setting. This confuses the discussion because the CM is not scale-invariant, e.g. the torus $\mathbb{Z} D^{1/4} + \mathbb{Z} iD^{3/4}$ has $\mathbb{Q}(i)$-CM, versus $\mathbb{Q}(\sqrt{-D})$-CM in the case of $\mathbb{Z}[\sqrt{-D}]$. -I'd like to make an extended comment. As I see it, the positive definite hermitian form arising from an elliptic curve (or, principally polarized abelian variety) has the form $H=g_j+i\omega$, where $\omega$ is the standard symplectic form $\omega(x,y)={}^txj_{std}y$ on $\mathbb{R}^{2g}$ and $g_j(x,y):=\omega(jx,y)$ for $j$ an $Sp_{2g}\mathbb{R}$ conjugate of $j_{std}$. In otherwords the data of a complex elliptic curve (or ppav) amounts to an $Sp_{2g}\mathbb{R}$-conjugate of the standard almost complex structure $j_{std}=\begin{pmatrix} 0 & -I_g \\ I_g & 0 \end{pmatrix}$. -In the special case of ppavs, the hypothesis that the real part of $H=H_j$ be integral on the lattice $\mathbb{Z}^{2g}$ (i.e. $j\mathbb{Z}^{2g} \subset \mathbb{Z}^{2g}$) is equivalent to having $j\mathbb{Q}^{2g} \subset \mathbb{Q}^{2g}$. This is because if $j\mathbb{Q}^{2g} \subset \mathbb{Q}^{2g}$, then $j\mathbb{Z}^{2g}$ is a sublattice of some $\frac{1}{N}\mathbb{Z}^{2g} \subset \mathbb{Q}^{2g}$ on which the symplectic form $\omega$ (i.e. imaginary part of $H=H_j$) is integral and unimodular. But $\mathbb{Z}^{2g}$ is the maximal submodule of $\mathbb{Q}^{2g}$ on which this occurs. In otherwords, we know $j$ is defined over $\mathbb{Q}$. -Now the claim that I want to make is this: if $j$ is defined over $\mathbb{Q}$ then the cyclic subgroup $\{ \exp j\theta\}_{\theta}$ of $Sp_{2g}\mathbb{R}$ is defined over $\mathbb{Q}$, where $\exp$ refers to matrix exponential. This would establish that a ppav has CM (in sense of hodge theory) if $j$ is defined over $\mathbb{Q}$. Now the converse of this statement is not clear (and i don't believe true). In an earlier version of this `answer' i wrongly said they were equivalent. In fact $Sp_{2g}\mathbb{R}$ has $\mathbb{Q}, \mathbb{R}, \mathbb{C}$-rank all equal to $g$, and $e^{j\theta}$ is merely a one-dimensional torus. This explains somewhat how the equivalence of $j$ rational and CM degenerates in higher dimension. -Hodge structures (in the sense, say, of Griffiths/Green/Kerr) on ppav's consist of nonconstant linear representations $\rho:\mathbb{S}^1 \to Sp_{2g}\mathbb{R}$. The image of $i \in \mathbb{S}^1$ actually determines the representation, with $\rho(i)$ giving an almost complex structure $j$ and extending to all of the unit circle via the exponential formula $\exp j\theta=(\cos\theta) I_{2g}+(\sin\theta) j$. I first learned of this formula right here on MO thanks to R. Bryant! The so-called Mumford-Tate group $MT(\rho)$ of the representation is the minimal algebraic subgroup of $Sp_{2g}\mathbb{R}$ defined over $\mathbb{Q}$ and containing the image of the representation $\rho(\mathbb{S}^1)$. From this point-of-view, CM is characterized as those representations (i.e. almost complex structures) whose MT-group is a (necessarily compact!) algebraic torus defined over $\mathbb{Q}$ in $Sp_{2g}\mathbb{R}$.<|endoftext|> -TITLE: Distribution and moments of ratio of two beta variables? -QUESTION [5 upvotes]: If $X$ and $Y$ are two Beta random variables, I am interested in the distribution of their ratio $X/Y$. More specifically, I am interested in the moment generating function of this ratio. There is a paper of Pham-Gia that apparently computes the distribution but I don't have access to it and I don't know how helpful it will be for determining the moment generating function. What is known about these? -Edit (copied from comment added Nov 7 '13): I've located the paper, so now the only question is: what is known about the moments of this ratio? - -REPLY [7 votes]: OP wrote: - -what is known about the moments of this ratio? - -I have not seen the paper ... but one does not even need to derive the distribution of the ratio in order to derive the moments of the ratio. In particular: -If $X$ ~ $Beta(a,b)$ and $Y$ ~ $Beta(c,d)$ are independent, then the joint pdf of $(X,Y)$ is, say, $f(x,y)$: - -(source) -Then, the $k$-th raw moment of the ratio $\frac{X}{Y}$ can be derived immediately as: - -(source) -where I am using the Expect function from the mathStatica add-on to Mathematica to automate the nitty-gritties for me (I am one of the developers of the former). If desired, one can express the solution slightly more neatly as: -$$\frac{B(a+k,b) B(c-k,d)}{B(a,b) B(c,d)}$$ -where $B$ denote the Euler beta function.<|endoftext|> -TITLE: Is $(G,K)$ a strong Gelfand pair? -QUESTION [6 upvotes]: Let $F$ be a $p$-adic field with ring of integers $\mathcal{O}$. When $G={\rm GL}_n$, it is a classical result that $(G(F),G(\mathcal{O}))$ is a Gelfand pair. Is it actually a strong Gelfand pair? I am particularly interested in the case $n=2$. - -REPLY [3 votes]: $\DeclareMathOperator\Res{Res}\DeclareMathOperator\Ind{Ind}\DeclareMathOperator\GL{GL}$You mean restriction of irreducible smooth, admissible representations of $G(F)$ to $G(o)$ decomposes with multiplicity one? Then yes for $n=2$. -Here are some more exact references: - -One-dimensional representations are obvious. - -For supercuspidal representations: Kristina Hansen. "Restriction to $\GL_2({\scr O})$ of supercuspidal representations of $\GL_2(F)$", Pacific J. Math. 130 (2) 327–349, 1987. https://projecteuclid.org/euclid.pjm/1102690181 - - --For principal series representations and Steinberg: Casselman - Restriction to $GL_2(o)$: https://doi.org/10.1007/BF01355984 -Note here that $\Res_{G(o)} \Ind_{B(F)}^{G(F)} \mu = \Ind_{B(o)}^{G(o)} \mu$, of which Casselman gives an explicit decomposition in the first lemma. The Steinberg as a quotient/submodule of some $\Ind_{B(F)}^G(F) \mu$ for some $\mu$ has then also the property. -I don't know a more conceptual proof. You need classification of all irreducible smooth admissible representations and then you need to look at the restriction, though. It's annoying. For $n>3$, we don't actually know the representation theory of $\GL_n(Z_p)$, so I think it's pretty much open. The corresponding questions for $\GL(n,R)$ or $\GL(n, C)$ seem to be wrong for $n>3$, but are right for $n=2$. I would only care about the types necessary to classify smooth admissible representations. There I think you have a positive answer meaning they occur with single multiplicity in irreducible admissible representations.<|endoftext|> -TITLE: Asymptotic behavior of the sequence $u_n = u_{n-1}^2-n$ -QUESTION [12 upvotes]: I am currently interested in the following sequence: -$$\begin{cases}u_0 & = & \alpha\\u_n & = & u_{n-1}^2-n\end{cases}$$ where $\alpha > C \approx 1.75793275... $ with $C$ being the Nested Radical Constant. -I'm interested to get the behaviour of this sequence as $n$ grows to infinity. -Furthermore, even though I would love to have an answer for any $\alpha$, you can for the sake of simplicity simply consider the case where $\alpha = 2$. In what follows, I consider only this case. -The OEIS know little about this sequence (http://oeis.org/A198959), and is not really helpful. -Now, I know (meaning I proved) very few about this sequence. It is obviously non-negative, and non-decreasing. Plus it goes to infinity. -Then, I feel like we should have $u_n \sim \lambda^{2^n}$ for some $1 < \lambda \leqslant \alpha$. The good news is that, when I compute the first hundred of terms, this equivalent seems pretty correct, with (in the case $\alpha=2$) $\lambda \approx 1.613590596957970...$. -Unfortunately, I am stuck here. I have been unable to prove this equivalent, or to find the following terms in the expansion of this sequence. And Plouffe's inverter (now Inverse Symbolic Calculator) does not recognize this number. -Do any of you have any ideas, clues, proofs, or links to papers related to this sequence? As it is just a matter of mathematical curiosity, I would take any relevant advance (a proof of the equivalent, a way to get the following terms, a close value for $\lambda$ (or an implicit formula, or involving $C$ itself, or an algorithm, etc.), a solution for the general problem with $\alpha \neq 2$, etc.). -Of course, as I did not proved the equivalent, it may be just false, in that case a counterargument would be welcomed too :-). -Thanks all. - -REPLY [6 votes]: Rather than defining a sequence in which the recursion depends on the number of iterations, it might be advantageous to simply consider this as iteration of the $2$-dimensional polynomial recursion -$$ (x,y)\longrightarrow (x^2-y,y+1)$$ -and look at the orbit of the point $(a,0)$ or $(a,1)$. There is a large literature on the dynamics of polynomial maps, although offhand I'm not sure how to search for this particular one. It sort of looks like a generalized Henon map, which are maps of the form -$$ (x,y)\longrightarrow (a_0 x^2 + a_1 y + a_2,b_0 x + b_1 y+b_2),$$ -but the fact that your $b_0$ is $0$ makes it more of a degeneration of a Henon map. Alternatively, it is an example of what is sometimes called a triangular map, which is a polynomial map of the form -$$ (x,y) \longrightarrow \bigl(F(x,y),G(y)\bigr). $$ -Again, there's lots in the literature on the dynamics of such maps.<|endoftext|> -TITLE: Accessible proofs of contemporary results in mathematics -QUESTION [11 upvotes]: Are there strong results in contemporary mathematical research (last 20 years) which have a proof which every mathematician (holding a PhD) can completely understand within a few days? -- If yes, please give examples. If not, please explain the phenomenon. - -REPLY [5 votes]: Manjul Bhargava's proof of the 15 theorem (in Quadratic Forms and Their Applications, Contemp. Math. 272, 1999) is strikingly simple, especially when contrasted with Conway and Schneeberger's original, intricate proof.<|endoftext|> -TITLE: Can an open manifold with positive Ricci curvature be non simply connected at infinity? -QUESTION [5 upvotes]: The question is in the title, I haven't been able to locate a discussion of these kind of properties. - -REPLY [9 votes]: For interiors of compact manifolds simply connected at infinity is equivalent to assuming that each boundary component is simply-connected. There are many such manifolds of positive Ricci curvature, e.g. the product of a circle and a high dimensional Euclidean space. See my paper with Guofang Wei for more examples of this kind. -On the other hand, the supply of known examples is limited and not much is known about topology of ends of complete manifolds of positive Ricci curvature.<|endoftext|> -TITLE: Expected supremum of average? -QUESTION [10 upvotes]: Is there either a closed form (in terms of the moments of $X_1$, say) or good bounds on -$$ -\mathbb{E} \sup_{k \leq n} \frac{1}{k} \sum_{i=1}^k X_i, -$$ -where $X_i$ are iid and arbitrarily nice? (In my specific application, $X_i$ are given by $(B_i - p)^2$, where $B_i$ are iid Bernoulli variables with mean $p$.) I am particularly interested in the correct functional dependence on $n$; e.g., is there a constant bound that holds for all $n$? - -REPLY [8 votes]: Here is an "arbirarily nice" example with closed form results. -Let $X_1,X_2,\ldots$ be i.i.d. real random variables with partial sums $S_k:=\sum_{i=1}^kX_i$ and let $M_n:=\sup_{k\leq n} \frac{S_k}{k}$, -$R_n:=\inf_{k\leq n} \frac{S_k}{k}= - \sup_{k\leq n} \frac{-S_k}{k}$, and let $M_\infty,R_\infty$ be the all-time supremum/infimum (possibly $\infty/-\infty$). -(1) The distribution of $M_n$ resp. $R_n$ can -be obtained as follows. -For $a\in \mathbb{R}$ let -$$T_a:=\inf\{k\geq 1\;:\;S_k-ka>0\} \mbox{ and } U_a:=\inf\{k\geq 1\;:\;S_k-ka\leq 0\}$$ -be first strictly ascendending resp. weakly descending ladder epochs of the random walk generated by the steps $X_i-a$, then clearly -$$\{M_n > a\}=\{ T_a \leq n\}\;\mbox{ and }\; \{R_n > a\}=\{ U_a > n\}\;\;, $$ -The fluctuation theory of random walks (see e.g. chapter XII in Feller II (1971)) gives for the generating functions -$g_a(z):= \mathbb{E}(z^{T_a}),\, h_a(z):=\mathbb{E}(z^{U_a}) $ that -$$\log\left(\frac{1}{1-g_a(z)}\right)=\sum_{n=1}^\infty \frac{z^n}{n} \mathbb{P}(S_n>na)\;\;\mbox{ (Sparre Andersen's theorem, Thm 1 in XII.7)}$$ -$$\mbox{ and } (1-g_a(z))(1-h_a(z))=1-z\;\;\mbox{ (duality, Thm 4 in XII.7).}$$ -Thus $\mathbb{P}(M_n>a)$ resp. $\mathbb{P}(R_n>a)$ can (in principle) be computed when only the probabilities $\mathbb{P}(S_k>ka), k=1,\ldots,n$ are known. -(2) -Now let the $X_i$ be $\exp(1)$-distributed. Then clearly $M_n, R_n\geq 0$ so only $a\geq 0$ need to be considered. Here it is known that for $a\geq 0$ -$$\mathbb{P}(T_a=n) = \frac{n^{n-1}}{n!} a^{n-1}e^{-an} \;\;\;\mbox{ for } n\geq 1, \mbox{ resp. that }$$ -$$g_a(z)=z\,e^{-a+T(zae^{-a})} \;\;\;\;(*)$$ -where $T(z):=\sum_{n=1}^\infty \frac{n^{n-1}}{n!} z^n$ denotes the "tree function". $(*)$ can e.g. be proved using Sparre Andersen's thm. and Lagrange inversion -(the series defining $T$ converges for $|z|\leq e^{-1}$ and represents the inverse of $z\mapsto z e^{-z}$ there). -Thus -$$\mathbb{P}( M_n>a) = \sum_{k=1}^n \frac{k^{k-1}}{k!} a^{k-1}e^{-ak}\;\;\mbox{ and }$$ -$$\mathbb{E} (M_n) = \int_0^\infty \mathbb{P} (M_n>a)\,da=\sum_{k=1}^n \frac{1}{k^2}$$ -Let $[z^n] F(z)$ denote the $n$-th cofficient of a (formal) power series $F(z)$. We have -\begin{align*} -\mathbb{P}(R_n >a) &=1-\mathbb{P}(U_a\leq n)\\ - &=[z^n]\frac{1-h_a(z)}{1-z}\\ - &=[z^n]\frac{1}{1-g_a(z)}\end{align*} -where in the last step the duality relation was used. -Expanding $\frac{1}{1-g_a(z)}$ into a geometric series and using Lagrange inversion on the individual terms now gives -$$\mathbb{P}( R_n>a) = \left(\sum_{k=0}^n \frac{(n-k)}{n}\frac{(na)^k}{k!}\right)e^{-na}\;.\;\mbox{ Thus }$$ -$$\mathbb{E} (R_n) = \int_0^\infty \mathbb{P} (R_n>a)\,da=\frac{1}{n^2}\sum_{k=0}^n (n-k)=\frac{1}{2}+\frac{1}{2n}$$ -(3) Passing to the limit -gives that -$\mathbb{P}(M_\infty >a)=e^{-a+T(ae^{-a})}$ (note that $\mathbb{P}(M_\infty > a)=1 \mbox{ for } a\in[0,1]$), and -$$\mathbb{E} (M_\infty) = \int_0^\infty e^{-a+T(ae^{-a})}\,da=\zeta(2)=\frac{\pi^2}{6}$$ -and that -$R_\infty $ is uniformly distributed on $[0,1]$: -we have that -$$\mathbb{P}( R_n >a) =\mathbb{P}(\mathrm{Poiss}(na)\leq n) -a\,\mathbb{P}(\mathrm{Poiss}(na)\leq n-1)$$ -By the strong law of large numbers for sums of i.i.d. $\mathrm{Poiss}(a)$-variables, both $\mathbb{P}(\mathrm{Poiss}(na)\leq n)$ and $\mathbb{P}(\mathrm{Poiss}(na)\leq n-1)$ tend to $0$ if $a>1$ and to $1$ if $01)=\mathbb{P}(\mathrm{Poiss}(n)=n)$ tends to zero (e.g. by Stirling's formula). Thus for $a>0$ $$ \mathbb{P}(R_\infty>a)= 1-\min(a,1)\;\;,$$ -that is, $R_\infty$ is uniformly distributed on $(0,1)$. -Further -$$\mathbb{E} (R_\infty) = \frac{1}{2}$$<|endoftext|> -TITLE: Centralizers of one parameter subgroups in semi-simple Lie groups -QUESTION [6 upvotes]: Suppose G is a connected semi-simple Lie group with finite center, and A, B are one parameter subgroups of the same Cartan subgroup. If the connected components of the identity of the centralizers of A and B are equal does it follow that the centralizers of A and B are equal ? - -REPLY [6 votes]: The answer is affirmative, and it is not necessary to assume that the center of $G$ is finite; discreteness (which is a consequence of semisimplicity of the Lie algebra) is sufficient. The crux of the matter is to carefully turn the analytic problem into an algebraic one (keeping track of connectedness issues on the analytic and algebraic sides of the story), even though the given Lie group and certainly the 1-parameter subgroups need not have any algebraic structure at all. The merit of the algebraic setting is that one has very general connectedness results for torus centralizers in connected linear algebraic groups over any field (such as $\mathbf{R}$). -To explain this, we first review the close connection between linear algebraic groups and the image of the adjoint representation of connected semisimple Lie groups, as well as a related algebraicity feature for Cartan subgroups. -Let $\mathfrak{g} = {\rm{Lie}}(G)$. The natural map ${\rm{Ad}}_G:G \rightarrow {\rm{GL}}(\mathfrak{g})$ induces ${\rm{ad}}_{\mathfrak{g}}$ on Lie algebras. This makes $\mathfrak{g}$ a perfect Lie subalgebra of $\mathfrak{gl}(\mathfrak{g})$. Bringing in the theory of linear algebraic groups with the Zariski topology, there is a unique (smooth) connected closed $\mathbf{R}$-subgroup $j:\mathbf{G} \hookrightarrow {\rm{GL}}(\mathfrak{g})$ such that ${\rm{Lie}}(\mathbf{G}) = \mathfrak{g}$ inside $\mathfrak{gl}(\mathfrak{g})$ (inclusion of $\mathfrak{g}$ via ${\rm{ad}}_{\mathfrak{g}}$); see Corollary 7.9 in Ch. II of Borel's textbook on algebraic groups. Since ${\rm{Lie}}(\mathbf{G})$ is semisimple, it follows by various means that $\mathbf{G}$ is semisimple in the sense of linear algebraic groups. -Consider the $\mathbf{R}$-homomorphism ${\rm{Ad}}_{\mathbf{G}}:\mathbf{G} \rightarrow {\rm{GL}}(\mathfrak{g})$. (Here we abuse notation in the usual manner, letting ${\rm{GL}}(V)$ to denote either the linear algebraic $\mathbf{R}$-group or the associated disconnected Lie group of $\mathbf{R}$-points, with the context making the intended meaning clear.) On Lie algebras this induces ${\rm{ad}}_{\mathfrak{g}}$, so ${\rm{Ad}}_{\mathbf{G}} = j$ due to Zariski connectedness of $\mathbf{G}$. In particular, $\ker {\rm{Ad}}_{\mathbf{G}} = 1$, which is to say $\mathbf{G}$ is of adjoint type. -The map ${\rm{Ad}}_G$ factors through the closed subgroup $\mathbf{G}(\mathbf{R})^0$ because it does so on Lie algebras (due to the compatibility of formation of Lie algebras with respect to passage from algebraic $\mathbf{R}$-groups to Lie groups of $\mathbf{R}$-points). -The resulting Lie group homomorphism $G \rightarrow \mathbf{G}(\mathbf{R})^0$ with discrete kernel $Z := Z_G$ (center of $G$) is an isomorphism on Lie algebras and hence identifies $\mathbf{G}(\mathbf{R})^0$ with $G/Z$. (Note also that $\mathbf{G}(\mathbf{R})^0$ always has finite index inside $\mathbf{G}(\mathbf{R})$; we will not use it). This is the "algebraicity" (up to disconnectedness!) of the image $G/Z$ of ${\rm{Ad}}_G$ inside ${\rm{GL}}(\mathfrak{g})$. -Likewise, if $\mathfrak{a} \subset \mathfrak{g}$ is a Cartan subalgebra then there is a (unique) maximal $\mathbf{R}$-torus $\mathbf{T} \subset \mathbf{G}$ with associated Lie algebra $\mathfrak{a}$ inside $\mathfrak{g}$ (as for connected reductive groups over any field of characteristic 0), ultimately because the group-centralizer of a semisimple element in the Lie algebra of a connected reductive group has reductive identity component (so we can perform dimension induction, using that the center of a connected reductive group has identity component that is a torus). Thus, $\mathbf{T}(\mathbf{R})$ has preimage in $G$ whose identity component $C$ maps onto $\mathbf{T}(\mathbf{R})^0$ with discrete kernel and hence is the Cartan subgroup ${\rm{exp}}_G(\mathfrak{a}) \subset G$ associated to $\mathfrak{a}$. This establishes a bijection between the sets of Cartan subgroups of $G$ and maximal $\mathbf{R}$-tori in $\mathbf{G}$. - -Now we address the question at hand. Let $f:\mathbf{R} \rightarrow G$ be a 1-parameter subgroup, and consider the induced homomorphism $\overline{f}:\mathbf{R} \rightarrow G/Z = \mathbf{G}(\mathbf{R})^0 \subset \mathbf{G}(\mathbf{R})$. Since $Z$ is discrete and central in $G$, $f$ is uniquely determined by $\overline{f}$. Hence, if $g \in G$ then $g$ centralizes $f$ if and only if its image $\overline{g} \in \mathbf{G}(\mathbf{R})$ centralizes $\overline{f}$. Thus, the centralizer $Z(f)$ of $f$ in $G$ is the full preimage under $G \rightarrow \mathbf{G}(\mathbf{R})$ of the centralizer $Z(\overline{f})$ of $\overline{f}$ in $\mathbf{G}(\mathbf{R})$, with $Z(f)$ mapping onto $Z(\overline{f}) \cap \mathbf{G}(\mathbf{R})^0$ (with discrete kernel). In particular, $Z(f)^0$ maps onto $Z(\overline{f})^0$. -Let $h:\mathbf{R} \rightarrow G$ be a second 1-parameter subgroup and assume $Z(h)^0 = Z(f)^0$, so $Z(\overline{h})^0 = Z(\overline{f})^0$. To prove that $Z(h) = Z(f)$ (centralizers in $G$) it suffices to prove that $Z(\overline{h}) = Z(\overline{f})$ (centralizers in $\mathbf{G}(\mathbf{R})$, which may of course be disconnected). -Assume $h$ and $f$ are valued inside a common Cartan subgroup $C$ of $G$. Clearly $\overline{h}$ and $\overline{f}$ are 1-parameter subgroups of $\mathbf{G}(\mathbf{R})$ valued in the group of $\mathbf{R}$-points of the associated maximal $\mathbf{R}$-torus $\mathbf{T}$ of $\mathbf{G}$. -Now comes the key idea: we consider Zariski closures of analytically-defined and typically non-algebraic subgroups of $\mathbf{G}(\mathbf{R})$. Let $\mathbf{S}, \mathbf{S}' \subset \mathbf{G}$ be the respective Zariski closures of the images of $\overline{h}$ and $\overline{f}$. In view of the "algebraicity" of the group law on $\mathbf{G}(\mathbf{R})$, $Z(\overline{h})$ is the group of $\mathbf{R}$-points of the algebraic group centralizer $Z_{\mathbf{G}}(\mathbf{S})$, and likewise $Z(\overline{f})$ is the group of $\mathbf{R}$-points of $Z_{\mathbf{G}}(\mathbf{S}')$. Thus, it suffices to show that $Z_{\mathbf{G}}(\mathbf{S}) = Z_{\mathbf{G}}(\mathbf{S}')$ inside $\mathbf{G}$. -The hypothesis that $Z(\overline{h})^0 = Z(\overline{f})^0$ implies that the closed $\mathbf{R}$-subgroups $Z_{\mathbf{G}}(\mathbf{S})$ and $Z_{\mathbf{G}}(\mathbf{S}')$ in $\mathbf{G}$ have the same Lie algebra inside $\mathfrak{g}$ because the Lie algebra of any algebraic $\mathbf{R}$-group is naturally identified with the analytically-defined Lie algebra of the identity component of its group of $\mathbf{R}$-points. But in characteristic 0, a connected (for Zariki topology!) smooth closed subgroup of a linear algebraic group is uniquely determined by its associated Lie algebra as a subalgebra of the ambient Lie algebra, so it suffices to prove that $Z_{\mathbf{G}}(\mathbf{S})$ and $Z_{\mathbf{G}}(\mathbf{S}')$ are connected for the Zariski topology. -By construction, $\mathbf{S}$ and $\mathbf{S}'$ are (smooth) closed $\mathbf{R}$-subgroups of the $\mathbf{R}$-torus $\mathbf{T}$, so they are tori provided that they are Zariski-connected. Moreover, it is a well-known fact in the algebraic theory that the centralizer of a torus in a (Zariski-)connected linear algebraic group over any field whatsoever is always (Zariski-)connected. This is really the whole point of this entire argument: in the algebraic setting, the disconnectedness problems that plague the consideration of centralizers in the analytic theory often do not arise. -So it suffices to prove that $\mathbf{S}$ and $\mathbf{S}'$ are connected for the Zariski topology. More generally, if $\mathbf{G}$ is an arbitrary linear algebraic group over $\mathbf{R}$ (not assumed to be connected for the Zariski topology) and if $H \subset \mathbf{G}(\mathbf{R})$ is an arbitrary subgroup (not necessarily closed!) that is connected for the subspace topology defined via the analytic topology on $\mathbf{R}$ (e.g., $\overline{h}(\mathbf{R})$ and $\overline{f}(\mathbf{R})$, with $\mathbf{G}$ the connected semisimple $\mathbf{R}$-group of adjoint type considered above) then we claim that the Zariski closure $\mathbf{H}$ of $H$ in $\mathbf{G}$ is Zariski-connected. In view of the generality of this setup, since $\mathbf{H}$ is certainly itself a linear algebraic group we may rename it as $\mathbf{G}$ to reduce to proving that if $\mathbf{G}(\mathbf{R})$ contains a subgroup that is connected for the analytic topology (though not necessarily closed for the analytic topology) and is also Zariski-dense in $\mathbf{G}$ then in fact $\mathbf{G}$ is Zariski-connected. We can replace $\mathbf{G}$ with $\mathbf{G}/\mathbf{G}^0$ to reduce to the case when $\mathbf{G}$ is finite etale, and then the only connected subgroup of the finite discrete group $\mathbf{G}(\mathbf{R})$ is the trivial one, so the identity point is Zariski-dense in the finite etale $\mathbf{R}$-group $\mathbf{G}$. This forces $\mathbf{G} = 1$, so we are done. -QED<|endoftext|> -TITLE: Higher commutators in E_n algebras and the Maurer--Cartan equation -QUESTION [7 upvotes]: Let $A$ be an associative algebra in $dgVect_k$. Then the commutator $[\cdot,\cdot]:A\otimes A\to A$ defined by $[x,y]=xy-(-1)^{|x||y|}yx$ gives $A$ the structure of a (dg-)Lie algebra. The Maurer--Cartan equation: -$$d\alpha+\frac 12[\alpha,\alpha]=0$$ -for $\alpha\in A$ (necessarily of degree $1$) plays an important role in deformation theory. - -Does anyone recognize the following generalization of the above construction from associative algebras (that is, $E_1$-algebras) to $E_n$-algebras? - -Let $A$ be an $E_n$-algebra in $dgVect_k$. The $E_n$-algebra structure gives a map $A\otimes A\otimes C_\bullet(Conf_2(D^n))\to A$ where $Conf_2(D^n)\simeq S^{n-1}$ is the configuration space of two distinct points in $D^n$. By picking a cycle in $C_\bullet(Conf_2(D^n))$ representing $[S^{n-1}]$, we get a pairing $[\cdot,\cdot]:A\otimes A\to A$ which I will think of as a sort of "higher commutator" (of course, this recovers the usual notion of commutator when $n=1$). -Now for some questions: - -Does this "higher commutator" endow $A$ with the structure of a dg-Lie algebra (or, more likely, an $L_\infty$-algebra), or is it something more exotic? -What is the significance of the solutions of the Maurer--Cartan equation (using the higher commutator) to the given $E_n$-algebra? - -REPLY [9 votes]: This operation appears prominently in the theory of $n$-fold loop spaces, where it is called a Browder operation and is related by suspension to the Samelson product. In characteristic $p$, this operation accompanies Dyer-Lashof operations, and these operations together give enough structure to compute $H_*(\Omega^n \Sigma^n X;\mathbf{F}_p)$ as an explicit functor of $H_*(X;\mathbf{F}_p)$ for any space $X$. This was part of Fred Cohen's 1972 PhD thesis and appears in "The homology of iterated loop spaces" http://www.math.uchicago.edu/~may/BOOKS/homo_iter.pdf. -This does not directly answer the questions, but I thought the history of these operations might be of some interest. The answer to question 1 in characteristic zero is well-known and is summarized in Section 5 of "Operads, algebras, and modules", http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf. The algebras over the homology of an $E_{n+1}$-operad $\mathcal{C}_{n+1}$ are $n$-braid algebras, which are commutative algebras and $n$-Lie algebras that satisfy the Poisson formula. When $n=1$, we see Batalin-Vilkovisky algebras. The free $n$-braid algebras are described explicitly in Theorem 5.6 op cit, where the description is deduced from topology.<|endoftext|> -TITLE: Admissible subcategories of $D^b(\mathbb{P}^n)$ -QUESTION [5 upvotes]: Recall that a triangulated subcategory $\mathcal{A}$ of a triangulated category $\mathcal{B}$ is called admissible if the inclusion functor has both left and right adjoints. - -Is it true that all admissible subcategories of $D^b(\mathbb{P}^n_k)$ (the bounded derived category of coherent sheaves on $\mathbb{P}^n_k$, for a field $k$) are generated by exceptional collections? - -Perhaps the right setting for the question is to consider admissible subcategories $\mathcal{A}$ of a $k$-linear triangulated category $\mathcal{B}$ generated by a (strong) exceptional collection rather than just $D^b(\mathbb{P}^n_k)$. - -REPLY [2 votes]: Reposting my comment as a slightly extended answer. -For $n=1$ this is folklore (it is I think a pleasant exercise using global dimension 1 and the description of coherent sheaves on curves), whilst for $n=2$ it was settled last year by Pirozhkov in his preprint Admissible subcategories of del Pezzo surfaces. He also gives the analogous statement for del Pezzo surfaces. -For $n\geq 3$ it's still open as far as I know.<|endoftext|> -TITLE: Consistency Strength of the Failure of Square on Singular Cardinals -QUESTION [5 upvotes]: Q1. What is the consistency strength of the failure of square on singular cardinals? -Q2. What are known as partial results in this direction? - -REPLY [4 votes]: One place to start reading is this: - -James Cummings and Sy-David Friedman, "$\square$ On the Singular Cardinals". The Journal of Symbolic Logic Vol. 73, No. 4 (Dec., 2008), pp. 1307-1314.<|endoftext|> -TITLE: Set of Positive Definite matrices with determinant > 1 forms a convex set -QUESTION [9 upvotes]: While reading a paper An Arithmetic Proof of John’s Ellipsoid Theorem by Gruber and Schuster, I have a question on their proof. - -Consider an $n\times n$ real symmetric and positive definite matrix $\mathbf A$. - -As this kind of matrix is symmetric, its $n(n+1)/2$ upper diagonal terms are enough to represent it. Hence, we can consider such a matrix as a point in $\mathbb R^{n(n+1)/2}$. -A conical combination of two positive definite matrices is also positive definite. Hence, the set of all symmetric positive definite matrices forms an open convex cone $\mathcal P\in\mathbb R^{n(n+1)/2}$ with apex on the origin. - - -Now they claim the following theorem without proving it. - -Theorem: The set $\ \mathcal D = \{\mathbf A \in \mathcal P: \det \mathbf A \geq 1\}$ is a closed, smooth, strictly convex set in $\mathcal P$ with non-empty interior. - -They just gave some hint that we can use Implicit Function Theorem and Minkowski's Determinant Inequality which states that - -For two $n\times n$ positive semidefinite Hermitian matrices $\mathbf X$ and $\mathbf Y$, - $$\det (\mathbf X + \mathbf Y)^{1/n}\geq \det(\mathbf X)^{1/n} + \det(\mathbf Y)^{1/n} $$ - -Any hint or suggestion on how to prove the above theorem about the set $\mathcal D$? - -REPLY [11 votes]: Here is a textbook level description of the above. I assume you know what a convex set and convex function on this set are. Given that, let us know prove that the determinant is strictly log-concave on hermitian positive definite matrices. -Claim. Let $A, B > 0$. Then, $\det\left(\frac{A+B}{2}\right) \ge \sqrt{\det(AB)}$ -Proof Consider $\phi(A) := \log\det(A)$. The first derivative of this is $A^{-1}$, while the second derivative may be identified with $-A^{-1}\otimes A^{-1}$, which is clearly negative definite if $A > 0$. This proves the desired concavity of $\phi(A)$, and therewith the claim above. -Note: Minkowski's determinant inequality is not the same as the above log-concavity. It is stronger, and enjoys a variety of different proof attempts. For a great list of these, have a look at the following much older MO question.<|endoftext|> -TITLE: coloring in lattice -QUESTION [5 upvotes]: This is a mathematical question raised from engineering and physics: -Is there some established mathematical approach in filling a physical lattice with some colored basis (black and white here)? For example, a triangular lattice can be filled with - to get -While alone cannot give any filled colored graph. Is there any general systematic mathematical approach in solving this problem? - -REPLY [6 votes]: The fact that Wang's procedure cannot theoretically work for arbitrary large tile sets does not render it useless for practical purposes. - -This is from the Wikipedia web page on Wang tiles. -Here is an algorithm for special cases that may (not certain) apply to your situation ($|T|=4$). -$T$ is a subset of $\mathbb{Z}^2$. - -Abstract. ... Here we present two algorithms, one for the case when $|T|$ is prime, and another for the case when $|T|=4$ ... -Szegedy, Mario. "Algorithms to tile the infinite grid with finite clusters." Foundations of Computer Science, 1998. Proceedings. 39th Annual Symposium on. IEEE, 1998. (IEEE link) - -And finally, a bit off the beaten path, but very interesting: - -Abel, Zachary, Nadia Benbernou, Mirela Damian, Erik D. Demaine, Martin L. Demaine, Robin Flatland, Scott D. Kominers, and Robert Schwelle. "Shape replication through self-assembly and RNase enzymes." In Proceedings of the Twenty-First Annual ACM-SIAM Symposium on Discrete Algorithms, pp. 1045-1064. Society for Industrial and Applied Mathematics, 2010. (ACM link):<|endoftext|> -TITLE: ${\rm Ext}^1$ and extensions of line bundles on a curve -QUESTION [5 upvotes]: I am confused about the following. I know that for two line bundles $L_1, L_2$ on an algebraic curve $C$ the vector space ${\rm Ext}^1(L_1,L_2)$ classifies isomorphism classes of rank two vector bundles on $C$ which are extensions of $L_2$ by $L_1$. My question is, does this mean that there is a "universal" rank two vector bundle $E$ on ${\rm Ext}^1(L_1,L_2) \times C$ ? -I.e. if I let $p_2 : {\rm Ext}^1(L_1,L_2) \times C \to C$ be the projection then does there exist a rank two vector bundle $E$ on ${\rm Ext}^1(L_1,L_2) \times C$ such that $0 \to p_2^*L_1 \to E \to p_2^*L_2 \to 0 $ is a short exact sequence and for all $x \in {\rm Ext}^1(L_1,L_2)$ if we restrict the above short exact sequence to $x \times C$ we get the exact sequence $0 \to L_1 \to E_x \to L_2 \to 0 $ of vector bundles on $C$ which corresponds to the extension class $x$ ? - -REPLY [8 votes]: I wanted to work this out for myself anyways, so here's a summary of the argument in Le Potier. -Suppose $X$ is a projective variety, and $E,G$ are locally free sheaves on $X$. Put $S = {\rm{Ext}}^1(G,E)$, and let ${\bf E},{\bf G}$ be the constant families on $S\times X$, namely ${\bf E} = p_2^\ast E$ and ${\bf G} = p_2^\ast G$. We want to construct a universal extension -$$0\to {\bf E} \to {\bf F} \to {\bf G}\to 0.$$ -We can specify $\bf F$ by giving an element of ${\rm Ext}^1({\bf G},{\bf E})$. -What is ${\rm Ext}^1({\bf G},{\bf E})$? Intuitively, we can specify an extension of ${\bf G}$ by ${\bf E}$ by specifying for each $s\in S$ an extension of $G$ by $E$; that is, it should be the space of morphisms $S\to S$; furthermore, if $f:S\to S$ and ${\bf F}(f)$ is the sheaf corresponding to $f$, then, ${\bf F}(f)_s$ is the extension of $G$ by $E$ corresponding to the extension class $f(s)$. More formally, considering the diagram -\begin{array}{ccc} S\times X &\stackrel{p_2}{\to} &X\\ \downarrow {p_1} & & \downarrow p_1'\\ -S & \stackrel{p_2'}{\to} & pt\end{array} -we find -\begin{align}{\rm Ext}^1({\bf G},{\bf E})&= H^1( {\mathcal H}om({\bf G},{\bf E})) \\&= H^0(R^1 p_{1*} {\mathcal H}om({\bf G},{\bf E})))\\ -&= H^0(p_2'^* R^1 p'_{1*} {\mathcal H}om(G,E))\\ -&= H^0(\mathcal O_S \otimes {\rm Ext}^1(G,E)),\end{align} -making use of various identities from Hartshorne III.6-9. The desired universal family corresponds to the distinguished identity morphism $S\to S$.<|endoftext|> -TITLE: Stacks over diffeologies -QUESTION [7 upvotes]: Konrad Waldorf shows in his paper one may realize a Grothendieck topology on the category of diffeological spaces. Is there any work exploring stacks over the category of diffeologies? - -REPLY [8 votes]: I will show that stacks over diffeological spaces are "the same" (in the sense of equivalence of 2-categories) as ordinary stacks on manifolds. -The Grothendieck pre-topology in question is the Grothendieck topology of "subductions". A map $f:X \to Y$ between diffeological spaces is a subduction if for every map $g:M \to Y$ with $M$ a manifold, the pullback $$X \times_Y M \to M$$ admits local sections. The pre-topology consists of singleton subductions as covers. -Notice that a map between smooth manifolds is a subduction if and only if it admits local sections. Given an open cover $U_\alpha$ of a manifold $M,$ $$\coprod\limits_\alpha U_\alpha \to M$$ is a subduction, and any subduction between manifolds can be refined by such a map. It follows that the restriction of the subduction pretopology to manifolds generates the same Grothendieck topology (i.e. the same covering sieves) as the standard open cover pretopology. -Denote, the full and faithful inclusion of manifolds into diffeological spaces by $$i:Mfd \hookrightarrow DiffSp,$$ where I allow arbitrary disjoint unions of manifolds to be considered manifolds. Consider a diffeological space $X.$ Let $$X_n=\coprod\limits_{f \in Hom\left(\mathbb{R}^n,X\right)} \mathbb{R}^n.$$ $X_n$ is a manifold and has a canonical map $$X_n \to X.$$ Putting all these together we get a map $$\pi_X:\tilde X=\coprod\limits_n X_n \to X,$$ and $\tilde X$ is a manifold. Since any point of a manifold has a neighborhood diffeomorphic to $\mathbb{R}^n,$ it follows that $\pi_X$ is a subduction. By the comparison Lemma (see SGA 4, III), it follows that $i$ induces an equivalence $$Sh\left(Mfd\right) \simeq Sh\left(DiffSp\right)$$ between the topoi of sheaves on manifolds and diffeological spaces respectively, where on the left we have open covers, and on the right we have subductions. Moreover, by applying the comparison lemma again, we can replace the category $Mfd$ with the subcategory $Man$ consisting of only embedded submanifolds of Euclidean spaces. A standard argument implies that in fact this equivalence extends to an equivalence -$$St\left(Man\right) \simeq St\left(DiffSp\right)$$ between their 2-categories of stacks. (E.g. you could model these as 1-truncated hypersheaves using the Jardine model strtucture, and the result follows from the ordinary comparison lemma). -In summary: -Stacks over diffeological spaces are "the same" as stacks over manifolds.<|endoftext|> -TITLE: A Question on 1, 2 ,3 Conjecture -QUESTION [25 upvotes]: The 1, 2, 3 conjecture is well-known: -If $G$ is a simple graph which is not $K_2$ then one can assign a number among $1, 2, 3$ to every edge such that if we label each vertex with the sum of the numbers of edges incident with it then we obtain a proper vertex coloring. -Question: Is the following weaker statement true? -For any simple graph $G$ there are three natural numbers $p_G, q_G, r_G$ such that if : -(a) We label any edge of $G$ with a number among $p_G, q_G, r_G$. -(b) We label any vertex of $G$ with the sum of numbers of the incident edges. -Then the vertex labels form a proper vertex coloring of the $G$. -Remark: In other words the question is about the truth of the $1, 2, 3$ conjecture when we replace global numbers $1, 2, 3$ with numbers localized to each given $G$. - -REPLY [19 votes]: If you choose $p_G$, $q_G$ and $r_G$, such that $p_G>\Delta~q_G>\Delta^2~r_G>0$, (with $\Delta=\Delta(G)$), then your question is equivalent to "neighbor distinguishing colorings by multisets". -As far as I know, the best known bound for this problem is proved here: -L. Addario-Berry, R. E. L. Aldred, K. Dalal, and B. A. Reed. Vertex colouring -edge partitions. J. Combin. Theory Ser. B, 94(2):237–244, 2005. -They prove that four different edge labels are sufficient, three should be open.<|endoftext|> -TITLE: A Problem on Linear Algebra -QUESTION [8 upvotes]: I'm trying to calculate an integral over the generalized Poincare upper half plane, then I find that I need to show the following identity: - -Let $X=(X_{i,j})\in\mathrm{GL}(n,\mathbb R)(n\geq 3)$ and $r$ be a positive integer such that $2\leq r\leq n-1$. For any $1\leq \ell_1<\ell_2<\cdots< \ell_r\leq n$, we define - \begin{eqnarray*} -\det(\ell_1,\ell_2,\cdots,\ell_r):=\det\left\{ -\left( - \begin{array}{cccc} - X_{n-r+1,\ell_1} & X_{n-r+1,\ell_2} & \cdots & X_{n-r+1,\ell_r} \\ - X_{n-r+2,\ell_1} & X_{n-r+2,\ell_2} & \cdots & X_{n-r+2,\ell_r} \\ - \vdots & \vdots & \cdots & \vdots \\ - X_{n,\ell_1} & X_{n,\ell_2} & \cdots & X_{n,\ell_r} \\ - \end{array} -\right)\right\}. -\end{eqnarray*} - Let - \begin{eqnarray*} -A&=&\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-1}\leq n-1}[\det(\ell_1,\cdots,\ell_{r-1})]^2\right)\times\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-1}\leq n-1}[\det(\ell_1,\cdots,\ell_{r-1},n)]^2\right)\\ -B&=&\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r}\leq n-1}[\det(\ell_1,\cdots,\ell_{r})]^2\right)\cdot\left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-2}\leq n-1}[\det(\ell_1,\cdots,\ell_{r-2},n)]^2\right)\\ -C&=& \left(\sum\limits_{1\leq \ell_1<\cdots<\ell_{r-1}\leq n-1}\det(\ell_1,\cdots,\ell_{r-1})\cdot \det(\ell_1,\cdots,\ell_{r-1},n)\right)^2. -\end{eqnarray*} - Show that $A-B=C$. - -The cases $n=3,4$ can be verified by Mathematical. -The case $r=2$ with any $n\geq 3$ can be verified easily by a direct calculation, but how to prove this identity for general $r$ ? Any comments and suggestions are welcomed! - -REPLY [3 votes]: The following proof is due to Yeping ZHANG: -Let $V$ be a $n$ dimensional Euclidean vector space. Let $(e_1,\cdots,e_n)$ be an orthogonal basis. -For any $k=1,\cdots,n$, let $\Lambda^k V$ the $k$-th exterior product of $V$. We equip $\Lambda^k V$ with a metric ${\lVert \cdot \rVert}_{\Lambda^k V}$ such that $(e_{l_1}\wedge\cdots\wedge e_{l_k})_{1\leq l_1 < \cdots < l_k\leq n}$ is an orthogonal basis. Let ${\langle \cdot,\cdot \rangle}_{\Lambda^k V}$ be the associated inner product. -Let $v_1,\cdots,v_k\in V$, whose coodinates with respect to $(e_1,\cdots,e_n)$ are given by -\begin{equation} -v_i = \sum_{j=1}^n t_{ij}e_j \;, -\end{equation} -where $i = 1,\cdots,k$. Let $T=(t_{ij})_{1\leq i \leq k, 1\leq j \leq n}$ be the associated matrix. For $1\leq l_1 < \cdots < l_k\leq n$, set $T(l_1,\cdots,l_k)=(t_{ij})_{1\leq i \leq k, j = l_1,\cdots l_k}$. Then -\begin{equation} -v_1 \wedge\cdots\wedge v_k = \sum_{1\leq l_1 < \cdots < l_k\leq n} \, \det \, T(l_1,\cdots,l_k) \, e_{l_1}\wedge\cdots\wedge e_{l_k} \;, -\end{equation} -in particular -\begin{equation} -{\lVert v_1 \wedge\cdots\wedge v_k \rVert}_{\Lambda^k V}^2 = \sum_{1\leq l_1 < \cdots < l_k\leq n} [ \, \det \, T(l_1,\cdots,l_k) \, ]^2 \;. -\end{equation} -With the above observation, we can interpret the question as the follows. -Set -\begin{equation} -x_i = \sum_{j=1}^n X_{ij}e_j \;, -\end{equation} -where $i=1,\cdots,n$. Let -\begin{align} -x_{n-r+2} \wedge\cdots\wedge x_n = \alpha + \beta \wedge e_n \;,\nonumber\\ -x_{n-r+1} \wedge\cdots\wedge x_n = \gamma + \delta \wedge e_n \;, -\end{align} -with $\alpha,\beta,\gamma,\delta$ generated by $e_1,\cdots,e_{n-1}$. -We have -\begin{align} -A & = {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^2 \cdot {\lVert \delta \rVert}_{\Lambda^{r-1} V}^2 \;,\nonumber\\ -B & = {\lVert \gamma \rVert}_{\Lambda^r V}^2 \cdot {\lVert \beta \rVert}_{\Lambda^{r-2} V}^2 \;,\nonumber\\ -C & = {\langle \alpha , \delta \rangle}_{\Lambda^{r-1} V}^2\;. -\end{align} -We have -\begin{equation} -x_i = y_i + X_{i,n}e_n \;, -\end{equation} -where $y_i = X_{i,1}e_1 + \cdots X_{i,n-1}e_{n-1}$. Then -\begin{align} -\alpha & = y_{n-r+2}\wedge\cdots\wedge y_n \;,\nonumber\\ -\beta & = \sum_{l=n-r+2}^n (-1)^{n-l} X_{l,n} y_{n-r+2}\wedge\cdots\wedge y_{l-1}\wedge y_{l+1}\wedge\cdots\wedge y_n \;,\nonumber\\ -\gamma & = y_{n-r+1}\wedge\cdots\wedge y_n \;,\nonumber\\ -\delta & = \sum_{l=n-r+1}^n (-1)^{n-l} X_{l,n} y_{n-r+1}\wedge\cdots\wedge y_{l-1}\wedge y_{l+1}\wedge\cdots\wedge y_n \;. -\end{align} -Case $1$. If one of $y_{n-r+2},\cdots,y_n$ is zero, we have -\begin{equation} -\alpha = \gamma = 0 \;, -\end{equation} -trivially, we get $A=B+C$. -Case 2. If $y_{n-r+1}=0$, we have -\begin{align} -\gamma & = 0 \;,\nonumber\\ -\delta & = (-1)^{r-1}X_{n-r+1,n}\alpha \;. -\end{align} -Thus -\begin{align} -A & = X_{n-r+1,n}^2 {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^4 \;,\nonumber\\ -B & = 0 \;,\nonumber\\ -C & = X_{n-r+1,n}^2 {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^4 \;. -\end{align} -We find $A=B+C$. -Case 3. We suppose that none of $y_{n-r+1},\cdots,y_n$ is zero. -Without changing $\alpha,\beta,\gamma,\delta$, we can replace $x_i$ by $x_i - tx_j$ for any $j>i$, any $t\in\mathbb{R}$. Thus we can suppose that -\begin{equation} -\langle y_i,y_j \rangle = 0 \;, -\end{equation} -for any $i \neq j$. Since $A,B,C$ are homogenious functions of $y_{n-r+1},\cdots,y_n$, we can suppose that $\lVert y_i \rVert = 1$. -With the above assumptions, we have -\begin{align} -& {\lVert \alpha \rVert}_{\Lambda^{r-1} V}^2 = 1 \;,\nonumber\\ -& {\lVert \beta \rVert}_{\Lambda^{r-2} V}^2 = \sum_{l=n-r+2}^n X_{l,n}^2 \;,\nonumber\\ -& {\lVert \gamma \rVert}_{\Lambda^{r} V}^2 = 1 \;,\nonumber\\ -& {\lVert \delta \rVert}_{\Lambda^{r-1} V}^2 = \sum_{l=n-r+1}^n X_{l,n}^2 \;,\nonumber\\ -& {\langle \alpha , \delta \rangle}_{\Lambda^{r-1} V}^2 = X_{n-r+1,n}^2 \;. -\end{align} -Hence $A = B + C$.<|endoftext|> -TITLE: For a set of matrices $S$, find $X$ such that the elements of $SX$ commute -QUESTION [6 upvotes]: Let $S := \{A_0, A_1, \dots, A_d\}$, where $A_k \in \mathbb{C}^{n \times n}$, be a set of (generally noncommuting) matrices. I am interested in finding a nonsingular $X \in \mathbb{C}^{n \times n}$ such that the elements of -$$SX = \{A_kX \colon k=0,1,\dots,d\}$$ -commute. In other words, I want a nonsingular $X$ such that -$$A_iXA_j = A_jXA_i, \quad \forall i,j \in \{0,1,\dots,d\}.\tag{*}$$ - -More precisely, depending on $d$, I am interested in: - -the conditions that $S$ has to fulfill so that such $X$ exists, - -an algorithm to find such matrix $X$, - -any structural properties that either $X$ or the elements of $SX$ might have. - - -Preferably, I'd like to keep within the matrices of order $n$, i.e., I want to avoid my problems to grow to order $nd$ or $n^2$. - -Note that it is perfectly O.K. to request nonsingular $X,Y$ such that the elements of $XSY$ commute, but this is equivalent to -$$XA_iYXA_jY = XA_jYXA_iY, \quad \forall i,j \in \{0,1,\dots,d\},$$ -which is the same as -$$A_i(YX)A_j = A_j(YX)A_i, \quad \forall i,j \in \{0,1,\dots,d\},$$ -so observing $XSY$ is equivalent to observing just $SX$. -Of course, $(*)$ is a system of linear equations, but its order is $nd$, and I'd like to avoid dealing with that. Also, I would like to be able if such $X$ exists before actually trying to find it (my point 1 above). -Searching for a way to solve this, I have found the (answered) question "Is there a name for the matrix equation $A X B + B X A + C X C = D$?", which looks a lot like my $d = 1$ case *I don't impose structural restrictions that are present there). However, I want to avoid using Kronecker product suggested in the most voted answer there, for two reasons: - -It is hard to determine if $X$ is nonsingular from $\operatorname{vec}(X)$, and the theoretical aspect (my point 1 above) is my primary interest. - -The matrices I obesrve may be quite large, so blowing them up from order $n$ to order $n^2$ is not acceptable. - - -Testing the case $d = 1$ on random generated matrices suggests that such $X$ (almost?) always exists, but I have no idea how to prove that. For $d > 1$, as one would expect, such $X$ sometimes exists, and sometimes does not, but I've managed to find no pattern on when it does. - -REPLY [2 votes]: This is not a complete solution by any means, but here are some ideas. -If one of $A_j$ (or their linear combinations) is invertible, then one can get -a necessary and sufficient condition. Namely, if $B_i=A_iX$ commute then so do $B_iB_j^{-1}=A_iA_j^{-1}$. So one can take $A_iA_j^{-1}$ and see if it commutes with $A_kA_j^{-1}$. In the other direction, if $A_iA_j^{-1}$ commute for all $i$ and fixed $j$ then you get what you want by picking $X=A_j^{-1}$. -This doesn't sound like a particularly practical criterion, because inverse calculation may be a mess. Also, even if $A_i$ are nice, say sparse, the matrices $A_iA_j^{-1}$ may not be sparse at all. Still, it's something. -It might be easier if one of $A_iA_j^{-1}$ has distinct eigenvalues. One can try to calculate eigenvectors of $A_iA_j^{-1}$ by finding the -roots of $det(A_i-\lambda A_j)$ and then solving the system (don't know how practical this is).<|endoftext|> -TITLE: Coloring $K_n$ via edge-weight sums -QUESTION [12 upvotes]: This is a question inspired by -and tangential to "A Question on 1, 2 ,3 Conjecture"—and certainly -much easier! -Suppose one assigns a random edge weight among $\{1,2,3,\ldots,k\}$ to each -edge of $K_n$, and then sums the incident edge weights to each vertex -and assigns those as vertex weights. -For example, here is $K_5$ with edge weights from $\{1,2,3,4,5\}$: -       -(The top vertex has weight $13=3+1+4+5$.) -In this case, the vertex weights do not form a proper coloring, because there -are two vertices assigned weight $14$. - -What is the probability that $K_n$ will be properly colored when the - edge weights are chosen uniformly from among $\{1,2,3,\ldots,k\}$? - -Empirically the function is well-behaved. Here is an accounting over $1000$ trials -on $K_5$ for each $k$ from $3$ to $20$: -       - -As requested, a plot of Lucia's $e^{-\frac{\sqrt{\frac{3}{\pi }} n^{3/2}}{2 k}}$ for $n=5$ (with $k$ on the horizontal axis): - -REPLY [7 votes]: Here's a heuristic which suggests that the right scale for $k$ is on the order of $n^{3/2}$, and predicts the actual probability for large $n$. More precisely, we conjecture the following: Let $n$ be large, and suppose $k$ is close to $t n^{3/2}$ and we have in mind that $t$ is of roughly constant size. Then the probability that one gets a proper coloring is (as $n$ goes to infinity) asymptotically $\exp(-\sqrt{3}/(2\sqrt{\pi}t))$. As $t$ goes to zero this is exponentially small, and as $t$ goes to infinity it tends to $1$ at a linear rate. -Here's the reasoning: If $X_1$, $\ldots$, $X_n$ are chosen uniformly (with replacement) from $1$ to $k$, then the sum $X_1+\ldots+X_n$ has mean $n(k+1)/2$ and variance $n(k^2-1)/12$, and for large $n$ it is approximately normal with this mean and variance. Therefore the sums for each of the $n$ vertices has mean $n(k+1)/2$ and wiggles around that on the scale of about $\sqrt{n}k$. Let us imagine that different vertices have independent sums -- this is not quite true because they share an edge, but I would expect that effect to be small (this is a heuristic, not a proof after all!). Then we have a situation like the birthday problem: Pick $n$ numbers from an interval of length about $\sqrt{n}k$, what is the chance of a coincidence. From the birthday problem we know that if $n^2$ is of size $\sqrt{n}k$ then there is a good chance of a coincidence. (Note that for the proper coloring we want there not to be a coincidence of sums.) This suggests that $k$ of size $n^{3/2}$ is the correct scaling. The actual formula was obtained by computing the analog of the birthday problem in this Gaussian setting -(which is a nice exercise). -The only thing preventing this from being a proof is the lack of independence for the sums at two vertices -- I think this can be fixed, but will need a better probabilist than me!<|endoftext|> -TITLE: Decomposition of a semi-definite matrix into sums of sparse semi-definite matrices -QUESTION [5 upvotes]: I'll first provide the background. -Let $x\in\mathbb{R}^N$ be decomposed into $n$ non-overlapping blocks of variables -$x^{(1)},\ldots,x^{(n)}$. -We say that $f:\mathbb{R}^N\rightarrow\mathbb{R}$ is partially separable of degree $w$, that it can be written in the form of -\begin{align} -f(x)=\sum_{J\in\mathcal{J}}f_J(x), -\end{align} -where $\mathcal{J}$ is a finite collection of nonempty subsets of $\{1,\ldots,n\}$, -$f_J$ are differential convex functions such that $f_J$ depends on blocks $x^{(i)}$ -for $i\in J$, and -\begin{align} -\lvert J\rvert\le w,\forall J\in\mathcal{J} -\end{align} -My problem is to find the minimum degree of partial separability of function -$f(x)=x^{\mathrm{T}}\mathbf{M}x$, $x\in\mathbb{R}^N$, $\mathbf{M}\succeq\mathbf{0}$. -That is to decompose $\mathbf{M}$ into sums of sparse semi-definite matrix -$\mathbf{M}_J$, while minimizing the number of entries of $\mathbf{M}_J$ which -has the maximum number of entries for all $J\in\mathcal{J}$ -\begin{align} -\min\max_{J\in\mathcal{J}}&\lVert\mathbf{M}_J\rVert_0\\ -\textrm{s.t. }\sum_{J\in\mathcal{J}}\mathbf{M}_{J}&=\mathbf{M}\\ -\mathbf{M}\succeq\mathbf{0},\mathbf{M}_J\succeq&\mathbf{0},\forall J\in\mathcal{J}\\ -\mathcal{J}\in2&^{\{1,\ldots,N\}} -\end{align} -where $\lVert\cdot\rVert_0$ represents the number of entries in a matrix. -This nonlinear optimization problem seems very difficult. -I don't even know if there is any feasible solution. - -REPLY [2 votes]: A heuristic approach may be to first construct a chordal embedding of the sparsity pattern, which automatically identifies the cliques. This defines the set $J$, and then the problem should be convex (tractable). -(See, for example, Chapter 4 in http://www.seas.ucla.edu/~vandenbe/publications/chordalsdp.pdf, or for a black box implementation, see symbolic.py in the Chompack package https://github.com/cvxopt/chompack.)<|endoftext|> -TITLE: Monstrous moonshine for $M_{24}$ and K3? -QUESTION [22 upvotes]: An important piece of Monstrous moonshine is the j-function, -$$j(\tau) = \frac{1}{q}+744+196884q+21493760q^2+\dots\tag{1}$$ -In the paper "Umbral Moonshine" (2013), page 5, authors Cheng, Duncan, and Harvey define the function, -$$H^{(2)}(\tau)=2q^{-1/8}(-1 + 45q + \color{blue}{231}q^2 + 770q^3 + 2277q^4 + 5796q^5+\dots)\tag{2}$$ -It was first observed in Notes on the K3 Surface and the Mathieu group M24 (2010) by Eguchi, Ooguri, and Tachikawa that the first five coefficients of $(2)$ are equal to the dimensions of irreducible representations of $M_{24}$. -Edit (Nov. 23) -For info, in the paper cited by J.Harvey below, in page 44, eqn(7.16) and (7.19), the authors missed a tiny but crucial + sign $n\in\mathbb{Z^{\color{red}{+}}}$ in the summation: -$$\begin{aligned}h^{(2)}(\tau)&=\frac{\vartheta_2(0,p)^4-\vartheta_4(0,p)^4}{\eta(\tau)^3}-\frac{24}{\vartheta_3(0,p)}\sum_{n\in\mathbb{Z^{\color{red}{+}}}}\frac{q^{n^2/2-1/8}}{1+q^{n-1/2}}\\ -&=q^{-1/8}(-1+45q+231q^2+770q^3+2277q^4+\dots)\end{aligned}$$ -where $q = p^2$, nome $p = e^{\pi i \tau}$, Jacobi theta functions $\vartheta_n(0,p)$, and Dedekind eta function $\eta(\tau)$. -Questions: - -Does anyone know how to compute the rest of the coefficients of $(2)$? (The OEIS only has the first nine.) -For $\tau=\tfrac{1+\sqrt{-163}}{2}$, is $H^{(2)}(\tau)$ algebraic or transcendental? Does it have a neat closed-form expression like $j(\tau) = -640320^3$? -Is the appearance of $\color{blue}{231}$ in $j\big(\tau) = -12^3(231^2-1)^3$ a coincidence? (Note also that its smaller sibling $j\big(\tfrac{1+\sqrt{-67}}{2}\big) = -12^3(21^2-1)^3$ and $M_{23}$ has dimensions $1,42,210$.) -Likewise, in $\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12} = -3^5 \left(231-\sqrt{2(-26679+2413\sqrt{3\cdot163})} \right)^{-2}$ where $\eta(\tau)$ is the Dedekind eta function. - -REPLY [11 votes]: I can answer your first question. In arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies -$H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$ -where $E_2(\tau)$ is the quasi modular Eisenstein series and -$F_2^{(2)}(\tau)= \sum_{r>s>0,r-s\ \mathrm{odd}} (-1)^r s\, q^{rs/2}$ which you can use to compute $H^{(2)}(\tau)$ to whatever order you desire. I have no idea concerning your questions 2,3, and 4.<|endoftext|> -TITLE: Is there a finite family of functions such that the max of any two functions can be dominated by a third? -QUESTION [37 upvotes]: Is it true that for every $t$ there is an $n$ and there exists a finite function -family, $\cal F$, whose members are from $[n] \to \mathbb N$ (taking all different -values) and for any $f_1, \ldots, f_t \in \cal F$ there is a $g \in \cal F$ such that $g > -\max(f_1,\ldots, f_t)$ on -(weak form:) more than $n/2$ inputs, -(strong form:) $(1-\epsilon)n$ inputs (for some small $\epsilon>0$)? -I already don't know the answer for $t=2$, while for $t=1$ it is easy to give such a family. -The question is an equivalent formulation of a problem regarding discrete voronoi games (see -http://arxiv.org/abs/1303.0523). -Update: Wow, no answers even after the bounty, quite surprising. -Meanwhile, Lev Borisov has observed in the comments that if the strong version is true, it is enough to prove it for $t=2$. -Seva has posed a weaker (?) version of the problem: "Circular" domination in ${\mathbb R}^4$. -The current best bound is due to Sam Zbarsky (see his answer). -Here are some somewhat related, potentially useful results: -https://arxiv.org/abs/1504.03602 -https://arxiv.org/abs/1601.04146 - -REPLY [19 votes]: For any $t$, we can get $p$ arbitrarily close to $\frac{2}{t+1}$ (in particular, for $t=2$, this gives $p=2/3-\epsilon$). -Take some large $N$ and let $n=(t+1)N$. For $a\in [t+1]$ and $b\in [N]$, define $h_{a,b}:[t+1]\times [n] \to \mathbb{N}$ by $h_{a,b}(c,d)=\big(a+c\pmod{t+1}\big)N+\big(b+d\pmod{N}\big)$ for any $d\in [N]$ and $c\in [t+1]$ (intuitively, $c$ and $a$ matter more, while $b$ and $d$ are used for tie-breaking). Let $\mathcal{F}=\{h_{a,b}\}$. -Given $f_1,\ldots,f_t$, let $a_i,b_i$ be such that $f_i=h_{a_i,b_i}$. A pigeonhole argument gives us that there must be some $a$ so that $a_i\not\equiv a+1 \pmod{t+1}$ for all $i$ and there is at most one $j$ with $a_j=a$. If there is no such $j$, we pick $g=h_{a,1}$. If such a $j$ exists, we pick $g=h_{a,b_j+1\pmod{N}}$. Then whenever $c=t+1-a$ or $c\equiv t-a\pmod{t+1}$ and $(c,d)\ne(t+1-a,N-b_j)$, we have $g(c,d)>\max(f_1(c,d),\ldots,f_t(c,d))$. Thus we get $p=\frac{2N-1}{(t+1)N}=\frac{2}{t+1}-\frac{1}{(t+1)N}$.<|endoftext|> -TITLE: Forcing over models of determinacy -QUESTION [6 upvotes]: Consider a ctm $\mathfrak{M}$ of $ZF+AD^+$. Is it possible to force over $\mathfrak{M}$ to get a model of ZFC which satisfies further the following: - -Every projectively definable family of sets of reals has an $OD_a$ member; -Every $OD_a$ set of reals has the property of Baire; -Where $a$ is any real parameter. - -Here statement 1 is as in my previous question (Projectively definable family of sets of reals). -PS: I am asking this question because there is a paper by Steel & Van Wesep "Two consequences of determinacy consistent with choice" which uses a similar technique. - -REPLY [10 votes]: $\newcommand\R{\mathbb{R}}\newcommand\OD{\text{OD}}$ -The answer is no. -First, notice that the family consisting of all subsets $R\subset\R^2$ that are a well-ordering of the reals is projectively definable in your sense, since we can say that $R$ is a well-ordering by quantifying only over countable objects: $R$ is reflexive, transitive, anti-symmetric and linear, every real is mentioned, and there is no infinite descending sequence (and we can code such $R$ canonically with a subset of $\R$, to avoid the need for $\R^2$ here). Thus, if statement 1 holds, we get an $\OD_a$ well-ordering of $\R$ for some real parameter $a$. But from any such well-ordering, we can define a set of reals, such as the Vitali set, that does not have the property of Baire, thereby violating property 2.<|endoftext|> -TITLE: What is the status of this strong form of Hedetniemi's conjecture? -QUESTION [6 upvotes]: In this question, a graph is a finite, undirected graph without loops or multiple edges, and a colouring of a graph is a proper vertex colouring. The product $G \times H$ of graphs $G$ and $H$ is the graph whose vertex-set is the product of the vertex-sets of $G$ and $H$ and whose edge-set is the product of the edge-sets of $G$ and $H$, with the obvious incidence relation. -Let $G$ and $H$ be graphs. Any $n$-colouring of $G$ gives rise to an $n$-colouring of $G \times H$: just paint $(x, y)$ the same colour as $x$. (Or, if you prefer, an $n$-colouring of a graph is just a homomorphism into the complete graph $K_n$, so we can compose the colouring $G \to K_n$ with the projection $G \times H \to G$ to obtain a colouring $G \times H \to K_n$.) Similarly, any $n$-colouring of $H$ gives rise to an $n$-colouring of $G \times H$. Let us say that a colouring of $G \times H$ arising in one of these two ways is obtained by projection. -The previous paragraph makes it clear that $\chi(G \times H) \leq \min\{ \chi(G), \chi(H) \}$, where $\chi$ means chromatic number. Hedetniemi's conjecture states that this is an equality. In other words, it says that there are no colourings of a product more economical than those obtained by projection. My question: - -Let $G$ and $H$ be graphs. Is every colouring of $G \times H$ with $\chi(G \times H)$ colours obtained by projection? - -The answer can't be known to be yes, unless I've missed some news, since that would imply Hedetniemi's conjecture. But perhaps the answer is no, or perhaps it's known that this apparently stronger conjecture would actually be implied by Hedetniemi's original conjecture. - -Edit  Assume the graphs are connected, otherwise the answer is trivially no. (E.g. consider $(K_2 \sqcup K_2) \times K_2$.) - -REPLY [5 votes]: The answer is no. In the comments, Gil Kalai suggested looking at products of two odd cycles, and indeed, this yields a counterexample. -Consider $C_3 \times C_5$, where $C_n$ denotes the $n$-cycle. This has chromatic number $3$, and there is a 3-colouring given by -$$ -\begin{pmatrix} -1&3&1&2&3\\ -1&2&1&2&3\\ -1&2&1&2&3 -\end{pmatrix} -$$ -(in what I hope is obvious notation). This is plainly not obtained by projection. -In fact, this is also a counterexample to a weaker form of this strong form of Hedetniemi's conjecture: that every colouring of $G \times H$ by $\chi(G \times H)$ colours is obtained by projection after some automorphism of $G \times H$. For any such colouring of $C_3 \times C_5$ has either 5 of each colour or 3 of one and 6 of each of the other two; but neither is the case for the colouring shown above.<|endoftext|> -TITLE: A problem related with 'Postage stamp problem' -QUESTION [5 upvotes]: A friend of mine taught me this question. I found that it is related with 'Postage stamp problem' (though it does not seem to be same). -Let $m,a_1\lt a_2\lt \cdots\lt a_n$ be natural numbers. Now let us consider the following condition : -Condition : For each $k\in\mathbb N$ which satisfies $1\le k\le m$, $k$ can be represented as $a_i, 2a_i$ or $a_i+a_j\ (i\not=j)$. -Letting $\min_m(n)$ be the min of $n$, then here are my questions. - -Question 1 : What is $\min_{m=100}(n)$ under the condition? -Question 2 : What is $\min_{m}(n)$ under the condition? - -Example : For $m=100$, $(1,2,\cdots, 8,9,10, 20, \cdots,80,90)$ is an obvious example with $n=18$. -Motivation : For $m=100$, it was almost by chance that I discovered the following example with $n=16$ : -$$(1,3,4,7,8,9,16,17,21,24,35,46,57,68,79,90)$$ -I expect that $\min_{m=100}(n)$ would be $16$, but I can't prove it. -For smaller $m$, we can get $\min_{m}(n)$ such as -$$\min_{m=1}(n)=\min_{m=2}(n)=1,\min_{m=3}(n)=\min_{m=4}(n)=2,$$$$\min_{m=5}(n)=\min_{m=6}(n)=\min_{m=7}(n)=\min_{m=8}(n)=3,\min_{m=9}(n)=4.$$ -Since we know that $(1,2,\cdots,k)$ is an example for $m=2k$, we get $\min_{m=2k}(n)\le k$ for any $k\in\mathbb N$. However, I don't have any good idea. Can anyone help? -Update : For $m=100$, I found an interesting example with $n=16$ : -$$(1,2,3,7,11,15,19,23,27,28,29,30,61,64,67,70)$$ -Using arithmetic progressions might be a key. - -REPLY [7 votes]: To my understanding, you are asking about the smallest size of a subset $A\subset[0,m]$ such that $[0,m]\subset 2A$ (where $2A:=\{a'+a''\colon a',a''\in A\}$ is the doubling of $A$). This is OEIS sequence A066063. There is no hope to give an explicit formula, but clearly, you need $\binom{|A|}2+|A|\ge m+1$, implying $|A|\ge(1+o(1)\sqrt{2m}$. On the other hand, one can get away with $|A|=(2+o(1))\sqrt m$: to see this, assume for simplicity that $m$ is a square, say $m=l^2$, and take $A=\{0,1,\ldots,l-1\}\cup\{l, 2l,\ldots,l^2\}$. -This problem has been studied and better bounds are known, but I do not have references handy.<|endoftext|> -TITLE: Is there a cohomology for magmas? -QUESTION [8 upvotes]: Is there a cohomology theory for magmas? Or cohomology theories for any class of non-associative algebras (other than Lie and maybe Jordan)? - -REPLY [3 votes]: Eilenberg S., Mac Lane S., Algebraic cohomology groups and loops, Duke Math. J. -14 (1947), 435-463. -Johnson K.W., Leedham-Green C.R., Loop cohomology, Czech. Math. J. 40 (1990), -182-194.<|endoftext|> -TITLE: Reduced special fiber implies reduced generic fiber for a projective morphism? -QUESTION [6 upvotes]: Let $f:X \to Y$ be a projective morphism between irreducible Noetherian schemes. If a fiber over a closed point of $Y$ is reduced is the generic fiber reduced? - -REPLY [19 votes]: Let's impose a flatness hypothesis (whose necessity is explained in Jason Starr's answer). The answer is still negative, but to explain the context for the counterexample it is instructive to first record some necessary features of any counterexample, so we know where to look. -In view of my above comment about geometric fibers, any search for a counterexample will necessarily have to involve a special fiber that is reduced yet not geometrically reduced. In other words, any counterexample will have to involve an imperfect residue field at the special point in the base. -Moreover, the answer is affirmative if the base $Y$ is Dedekind. Indeed, we can assume in such a case that the base $Y$ is Spec($R$) for a dvr $R$ with uniformizer $t$, so $X$ is a proper flat $R$-scheme having reduced special fiber $X_0$. If $N$ is the coherent sheaf of nilpotents inside $O_X$ then for any section $f \in N(U)$ over an open $U \subset X$ we know that $f_0 = f|_{U_0}$ vanishes since $X_0$ is reduced, so $f$ is a section of $tO_X$. In other words, $N \subset tO_X$. By $R$-flatness of $X$, it is clear (check!) that $N \cap tO_X = tN$, so $N \subset tN$. Hence, by Nakayama's Lemma along points of the special fiber, it follows that $N$ has vanishing stalks along the special fiber $X_0 \subset X$. The closed support of the coherent $N$ inside the $R$-proper $X$ is therefore disjoint from the special fiber. But every non-empty closed set in $X$ must meet $X_0$ due to $R$-properness of $X$ and locality of $R$. Thus, the support is empty, which is to say $N = 0$. Then $X$ is reduced, so its localization given by the generic fiber over the integral base $Y$ is also reduced. That settles the question affirmatively when $Y$ is Dedekind (and $X$ is any proper flat $Y$-scheme). -In view of the preceding observations, an "optimal" counterexample should involve a local base $Y = {\rm{Spec}}(R)$ where $R$ is a non-Dedekind 1-dimensional local noetherian domain having imperfect residue field with characteristic $p > 0$. And in fact there are counterexamples over such $R$, as we now construct. This shows that the EGA result with reducedness for geometric fibers is essentially "best possible". We will give counterexamples in equicharacteristic $p$. Maybe someone else can address the story with generic characteristic 0. (SEE THE END FOR SUCH AN EXAMPLE) -The idea is to make such an $R$ that is an order in a dvr of characteristic $p$ whose fraction field contains a certain $p$th root but for which $R$ does not contain that $p$th root. Let $k$ be a field of characteristic $p > 0$ and let $A = k(t)[x]_{(x)}$, a dvr with uniformizer $x$ and residue field $k(t)$ that is obviously not perfect. Let $F = {\rm{Frac}}(A) = k(t,x)$ and let $A' = A[T]/(T^p - t) = k(T)[x]_{(x)}$, so the dvr $A'$ with uniformizer $x$ is the integral closure of $A$ in $F' = F[T]/(T^p - t) = F(t^{1/p})$. Note that $A \rightarrow A'$ is an example of a finite extension of discrete valuation ring whose ramification degree is 1 but residual extension is not separable (so it is not "unramified"). -The residue field of $A'$ is $k(T)$ in which the element $t$ from the residue field of $A$ has image $T^p$ that is a $p$th power, but we can "fix" that (or rather, "ruin" it) by considering the order -$R = A + xA'$. Concretely, $R$ is the preimage of $k(t)$ under the reduction map $A' \twoheadrightarrow k(T)$. -Clearly $R$ is a 1-dimensional local noetherian domain whose fraction field is $F'$, residue field is $k(t)$, and normalization is $A'$. Note that the element $T = t^{1/p}$ in the fraction field $F'$ of $R$ does not lie in $R$. -Inside $\mathbf{P}^2_A$ with homogeneous coordinates $[U,V,W]$, consider the closed subscheme $Z$ defined by $U^p + t V^p$. This is $A$-flat since $A$ is Dedekind. Its special and generic fibers are the "same" projective scheme over the residue field $k(t)$ and fraction field $k(t,x)$ respectively, and as such these fibers are both reduced (as $t$ is not a $p$th power in either the residue field or fraction field of $A$, each of characteristic $p$) and are also both geometrically irreducible. The base change -$X = Z_R$ is certainly $R$-flat (since $Z$ is $A$-flat) and projective with special fiber that is the same as that of $Z$ since $R$ and $A$ have the same residue field by design. In particular, $X_0$ is reduced (though not geometrically so!). The generic fiber $X_{F'} = Z_{F'}$ is non-reduced since $t$ is a $p$th power in $F'$. Note however that the $R$-flat $X$ has irreducible generic fiber, so $X$ is also irreducible. Voila. - -EDIT: -Here is an even better example in the same spirit, but with a twist on the idea so that it works in generic characteristic 0. It will use an order in -$\mathbf{Z}[T]_{(p)}$ as I had been hoping. This example was pointed out to me by somebody "offline". -Let $A = \mathbf{Z}[t]_{(p)}$, a dvr with uniformizer $p$, residue field $\mathbf{F}_p(t)$, and fraction field $F = \mathbf{Q}(t)$. Let $A'= \mathbf{Z}[T]_{(p)}$ made into an $A$-algebra via $t \mapsto T^p$, so its fraction field $F'$ is $F(t^{1/p})$. -Let $R = A + pA'$, a 1-dimensional local noetherian domain with residue field $\mathbf{F}_p(t)$ and fraction field $F'$. -Consider the polynomial $f(U,V) = (U + T V)^p \in U^p + t V^p + pA'[U,V] -\subset R[U,V]$. -Obviously $f$ is a $p$th power over $F' = \mathbf{Q}(T)$, but it is not a (unit multiple of a) $p$th power over $R$ since the reduction of $f$ over the residue field of $R$ is $U^p + t V^p \in \mathbf{F}_p(t)[U,V]$, which is even irreducible. -Let $X \subset \mathbf{P}^1_R$ be defined by the vanishing of the homogeneous $f$. -(The equicharacteristic-$p$ example above ought to have been considered inside $\mathbf{P}^1$ just as well; I mistakenly thought being in $\mathbf{P}^2$ would provide better irreducibility properties, but those hold already with $\mathbf{P}^1$.) -The $R$-scheme is $R$-flat because each of $U^p$ and $V^p$ have coefficients that are units in $R$ (ensuring that after dehomogenization, the coordinate ring of the affine open chart of $X$ is a finite free $R$-module). In fact, we see that $X$ is finite flat over $R$, with degree $p$. -The special fiber of $X$ is the zero scheme of $U^p + t V^p$ in $\mathbf{P}^1$ over $\mathbf{F}_p(t)$, and this is visibly reduced. On the other hand, the generic fiber is $(U+TV)^p=0$ in $\mathbf{P}^1_{F'}$, and this is visibly not reduced (but is irreducible, and likewise $X$ is irreducible; in fact, $X_{\rm{red}} = {\rm{Spec}}(R)$!).<|endoftext|> -TITLE: Costa's minimal surface and the structure of lungs -QUESTION [7 upvotes]: Seeing this image of Costa's minimal surface -    -   (MathWorld image) -made me wonder if the fine-grained structure -of the human lung is somehow composed of pieces of minimal surfaces? -I could not find an ideal image of a real lung; -apologies in advance if you find this offensive! -          -            (Image link) -I searched around a bit without finding a connection. Which may mean none exists... -Does anyone know? Thanks! - -I since found one possibly relevant quote in -John Oprea's book -Differential Geometry and Its Applications. -(MAA, 2007), p.163: - -When the pressure difference is zero, the Young-Laplace equation leads to minimal surfaces. - -REPLY [5 votes]: I'll muse here a little about the shapes of the alveoli which are basic functional units of respiratory exchange in the lungs. Two caveats: (1) This isn't a mathematical answer, (2) I don't really know that much about the stuff below. - from http://en.wikipedia.org/wiki/File:Bronchial_anatomy.jpg -The alveoli appear to be roughly globular sacs surrounding cavities - to a very very rough approximation you might say spherical and hence constant mean curvature. -Minimal and constant mean curvature surfaces show up in nature often when one is considering the shape of an interface between two fluid phases, e.g. air and water (as in soap films) -- this is because they minimize area given some fixed pressure difference between the two phases. As such they are often a "go-to" for interesting mathematical surfaces. -However, lung tissue is made out of cells connected by an intricate network of proteins; see e.g. this wikipedia article. Thus their response to forces from air pressure is more like that of elastic sheets rather than fluid membranes; in particular there is an energy cost to bending as well as stretching and the interplay between these can lead to very diverse behavior. Such elastic sheets are described (in a certain idealized limit) by the Föppl–von Kármán equations rather than the minimal surface or constant mean curvature equations. No doubt there is also a spherical solution to these equations if one sets the forces and boundary conditions properly (corresponding to say a very slightly stretched balloon). -Though you could get a spherical shape both ways, the difference in mechanism is important if you want to understand how the shape changes as the pressure difference changes, or if there is a mutation affecting the density of the collagen network, etc. -Complicating matters is the fact that the cells are not static building blocks but grow and die over time. While this is unlikely to be important over the timescale of a single breath, I think an interesting and so-far unaddressed question (at least as far as my searches went) is how the mechanical forces from pressure during breathing affect the growth and development of the alveoli. -In particular it seems significant that the alveolar sacs do not seem to develop in detail until after birth (see here) suggesting that their development may be in response to mechanical stimuli. -There has even been some amount of attention in the media about studies showing that alveoli can regrow even in adults with speculation about future lung regeneration therapy, e.g. here and here. I think even the article that is the source for the second image in your question discusses the growth and development of alveoli in older rats (the older wisdom being that this process is halted after infancy). -While I couldn't find anything about alveoli in particular, there is a growing (pun intended) body of work on the interplay between mechanics and form in biological systems. Here's one recent very nice paper by Savin et al: "On the growth and form of the gut".<|endoftext|> -TITLE: Mapping class group and CAT(0) spaces -QUESTION [15 upvotes]: I hope the questions are not too vague. - -Is the mapping class group of an orientable punctured surface $CAT(0)$ ? -Is any of the remarkable simplicial complexes (curve complex, arc complex...) built on a punctured surface $CAT(0)$? -Is there any "nice" action (say, proper or cocompact) of the mapping class group on a $CAT(0)$ space? - -REPLY [17 votes]: (1) Bridson showed that if a mapping class group of a surface (of genus at least 3) acts on a CAT(0) space, then Dehn twists act as elliptic or parabolic elements. This implies that the mapping class groups of genus $\geq 3$ are not CAT(0) (Edit: as pointed out by Misha in the comments, this was originally proved by Kapovich and Leeb, based on an observation of Mess that there is a non-product surface-by-$\mathbb{Z}$ subgroup of the mapping class group of a genus $\geq 3$ surface). On the other hand, the mapping class group of a genus 2 surface acts properly on a CAT(0) space (this is not surprising, since it is linear). I think it's unresolved whether the mapping class group of genus 2 is CAT(0) though (this is essentially equivalent to the same question for the 5-strand braid group). -(2) The curve complex cannot admit a CAT(0) metric, since it is homotopy equivalent to a wedge of spheres. -(3) The mapping class group acts cocompactly by isometries on the completion of the Weil-Petersson metric on Teichmuller space, which is CAT(0). However, this metric is not proper (although as Bridson shows above, the action is semisimple, Dehn twists acting by elliptic isometries). -So I guess it's unresolved whether there is a proper action of the mapping class groups of genus $\geq 3$ on a CAT(0) space (where the Dehn twists act as parabolics). This is unsurprising, since it is unknown whether these groups are linear (a finitely generated linear group acts properly on a CAT(0) space which is a product of symmetric spaces and buildings).<|endoftext|> -TITLE: Are topoi and etale geometric morphisms locally small? -QUESTION [9 upvotes]: This question is related to this one: Local smallness and (higher) topoi which has not yet been answered. -The $2$-category of topoi and geometric morphisms is not locally small. (As I mentioned in the question above, for example, if $A$ is the classifying topos for abelian groups, the category of geometric morphisms from $Set$ to $A$ is equivalent to the category of all abelian groups, which is not small.) -**Question:**Is the $2$-category of topoi and only etale geometric morphisms locally small? -Note: I'm actually interested in the corresponding statement for infinity topoi. -Remark: Notice that this $2$-category is locally a topos, so, it is locally locally small (sorry for the weird wording). That is, if $E$ is a topos, the slice 2-cat over $E$ consisting of etale morphisms over $E$ is known to be equivalent to $E$ itself (and hence is really a 1-category). There is a corresponding result for infinity topoi as well. Perhaps there is a way to use this to prove the result. - -REPLY [8 votes]: There is only a set of isomorphism classes of étale geometric morphisms between any two Grothendieck toposes. In fact, the same is true for essential geometric morphisms. -Recall that Grothendieck toposes are locally presentable categories, and that the left adjoint of a $\kappa$-accessible functor between $\kappa$-accessible categories must send $\kappa$-compact objects in the domain to $\kappa$-compact objects in the codomain. Since there is only a set of isomorphism classes of $\kappa$-compact objects in a $\kappa$-accessible category, we deduce that there is only a set of isomorphism classes of $\kappa$-accessible right adjoints between $\kappa$-accessible categories. But if $\mathcal{E}$ and $\mathcal{F}$ are locally $\kappa$-presentable toposes and $f : \mathcal{F} \to \mathcal{E}$ is an essential geometric morphism, then the inverse image functor $f^* : \mathcal{E} \to \mathcal{F}$ is a $\kappa$-accessible right adjoint. Thus there is only a set of isomorphism classes of essential geometric morphisms $\mathcal{F} \to \mathcal{E}$. -I imagine the same argument works verbatim for $(\infty, 1)$-toposes, but I do not know the details well enough to be sure.<|endoftext|> -TITLE: What is the correct definition of the (derived) tensor product over a dg-algebra? -QUESTION [5 upvotes]: Let $A_\bullet$ be a dg-algebra over a field $k$. Let $M_\bullet$ (resp. $N_\bullet$) be a right (resp. left) $A_\bullet$-module. There is then a notion of the derived tensor product: -$$M_\bullet\otimes^L_{A_\bullet}N_\bullet=\bigoplus_{p\geq 0}M_\bullet\otimes(A_\bullet)^{\otimes p}\otimes N_\bullet$$ -(where $\otimes$ is $\otimes_k$) with differential given by the internal differential on each direct summand plus an alternating sum of all possible maps multiplying adjacent elements (decreasing $p$ by one). We can rewrite this as: -$$(M_\bullet\otimes^L_{A_\bullet}N_\bullet)_j=\bigoplus_{p\geq 0}(M_\bullet\otimes(A_\bullet)^{\otimes p}\otimes N_\bullet)_{j-p}$$ -But now (based on my limited understanding) it is not clear to me why we shouldn't instead take direct product instead of direct sum, and define: -$$(M_\bullet\otimes^L_{A_\bullet}N_\bullet)_j=\prod_{p\geq 0}(M_\bullet\otimes(A_\bullet)^{\otimes p}\otimes N_\bullet)_{j-p}$$ -Note that with this definition the differential still makes sense. -The definition with the direct sum seems to be standard. On the other hand, I have encountered a context where I want to write down a certain element in $M_\bullet\otimes^L_{A_\bullet}N_\bullet$, but it lies in the direct product (and not in the direct sum). Hence my question is: - -Which of the two definitions of $M_\bullet\otimes^L_{A_\bullet}N_\bullet$ given above is correct, and why? - -Note that when $A_\bullet$ is concentrated in nonnegative degrees, taking direct product instead of direct sum makes no difference, since for fixed $j$, the factors on the right become trivial for sufficiently large $p$. Hence this question is only non-vacuous when $A_\bullet$ is nontrivial in negative degrees. - -REPLY [2 votes]: In the case when $M=N=A$, your question reduces to the comparison between the usual direct sum Hochschild chain complex versus the direct product chain complex. It is important to use the direct sum chain complex in proving that Hochschild homology is a homotopy invariant of $A$. To see this, observe that the Hochschild differential decomposes to $d_1+d_2$ with $d_i$ induced by $m_i$ of $A$. In the case of direct sum, we can start with $d_1$ to compute Hochschild homology using spectral sequence method, which would yield the same complex if $A$ and $B$ are quasi-isomorphic. This argument fails in the direct product case.<|endoftext|> -TITLE: What is the structure of the group of rational points of an abelian variety over a Laurent series field? -QUESTION [9 upvotes]: Let $K = \mathbb{F}_q((t))$, and let $A_{/K}$ be a nontrivial abelian variety. Then $A(K)$ is a compact $K$-adic Lie group. What can be said about its structure? -By way of comparison, if $K/\mathbb{Q}_p$ is an extension of degree $d$ and $A$ has dimension $g$, then $p$-adic Lie theory shows that $A(K) \cong \mathbb{Z}_p^{dg} \oplus T$, where $T$ is a finite group. I am looking for a similar description in the positive characteristic case. -I've wondered about this off and on for years but all of a sudden I have a good reason to know: in particular, I would like to know the structure of $A(K)/p^aA(K)$, which I suspect is always an infinite group of exponent $p^a$. This is the case for e.g. $p$-adically uniformized abelian varieties, unless I am very much mistaken. -In particular, in the elliptic curve case it would be nice if the height of the formal group told the majority of the story, in the sense that e.g. if $E_1$ and $E_2$ were any two ordinary elliptic curves over $K$, then $E_1(K)$ and $E_2(K)$ would admit isomorphic finite-index subgroups. Is this true? -Added: After seeing Professor Lubin's answer I can be more precise. I would like a proof that $A(K) \cong \left(\prod_{i=1}^{\infty} \mathbb{Z}_p \right) \oplus T$, where $T$ is a finite group. - -REPLY [5 votes]: I’m feeling fairly fuzzy on this, but it seems to me that everything should be just what you expect, and without regard to the particular shape of the formal group. I’ll just deal with the case of elliptic curves over $R=k[[t]]$, and for the constant field $k=\mathbb F_p$. I’m sure you can do the mutatis mutandis. -We always have the exact sequence -$$ -0\longrightarrow \hat E(R)\longrightarrow E(K)\longrightarrow E(k)\longrightarrow0\,, -$$ -and I think you’re just asking about the shape of $\hat E(R)$, by which I mean the points of the formal group $\hat E$ of $E$ with values in $R$, and these have to be elements of the maximal ideal $tR$. -Seems to me that in the ordinary case, the story is exactly the same as for the group $1+tR$, the $R$-valued points of the formal multiplicative group. You get one more generator for each exponent prime to $p$, so the group has structure -$$ -\prod_iZ_i\,, -$$ -where each group $Z_i$ is a free $\mathbb Z_p$ module of rank one, generated by $t^i$. Remember that for $\alpha\in\mathbb Z_p$, $\alpha\cdot t^i=[\alpha](t^i)$, where the right-hand side refers to the $\alpha$-endomorphism of the formal group. And in the case of height one (ordinary), the indices $i$ are not all positive $i$, but just those prime to $p$. -I’m a little nervous about the height-two (supersingular) case, but since in the case $k=\mathbb F_p$, the endomorphisms of the formal group are exactly $\mathbb Z_p[\pi]$, where $\pi$ is the Frobenius endomorphism $x^p$, it’s not too bad there. In this case, the points of the formal group seem to be -$$ -\prod_iY_i\,, -$$ -where each $Y_i$ is a free $\mathbb Z_p$-module of rank two, with basis elements $t^i$ and $\pi(t^i)=t^{pi}$. Same indices as before. -It seems to me that the moral of the story is that in the category of topological groups, the points of the formal group are always the same.<|endoftext|> -TITLE: Hilbert space compression of the lamplighter group -QUESTION [6 upvotes]: What is the Hilbert space compression exponent of the standard lamplighter group $\mathbb{Z_{2}} \wr \mathbb{Z}$? For $\mathbb{Z} \wr \mathbb{Z}$ it is known to be $2/3$ by work of Austin, Naor and Peres, but I couldn't find a reference for the $\mathbb{Z_{2}}$ case. - -REPLY [3 votes]: A direct argument is the following. Fix a finite group $F$ of cardinality $c \geq 2$. - -Notice that the Diestel-Leader graph $DL(c)$ is the Cayley graph of $F \wr \mathbb{Z}$ with respect to some finite generating set. (See Wolfgang Woess' paper Lamplighters, Diestel-Leader graphs, random walks, and harmonic functions.) - -Explicitely, $DL(c)$ is the following graph. Take two $c$-regular trees $T_1$ and $T_2$, and fix two Busemann functions $\beta_1 : T_1 \to \mathbb{R}$ and $\beta_2 : T_2 \to \mathbb{R}$. Then $DL(c)$ is the graph whose vertices are the pairs of vertices $(x,y) \in T_1 \times T_2$ satisfying $\beta_1(x)+\beta_2(y)=0$ and such that two vertices $(x_1,y_1)$ and $(x_2,y_2)$ are linked by an edge if $x_1$ and $x_2$, and $y_1$ and $y_2$, are adjacent respectively in $T_1$ and $T_2$. - -The next step is to show that the canonical embedding $DL(c) \hookrightarrow T_1 \times T_2$ is quasi-isometric. -Finally, because a tree has Hilbert space compression one, it follows that the lamplighter group $F \wr \mathbb{Z}$ has also Hilbert space compression one. - -There is another approach in my PhD thesis, but which is essentially equivalent to the previous one. Define the graph of wreaths $\mathscr{W}$ as follows: -A vertex of $\mathscr{W}$ is an equivalence class $[(\varphi,a,b)]$, where $a -TITLE: Is Logic/Set Theory necessary for studying Topos Theory? -QUESTION [6 upvotes]: I have just completed a postgraduate course, in which I studied Category Theory, without having a background in Set Theory and Logic - this probably already sounds absurd to many. This did not seem to be a problem - at least until now. -As a (reasonable, I thought) progression to my studies, I am trying to teach myself Topos Theory. Until the notion of geometric morphisms, I thought it might be ok to just omit the Logic examples from my studying. And it worked fine, I could keep up with learning. But I have now arrived to the concept of a classifying topos. It feels that now my lack of knowledge in Logic is going to be a very big obstacle. It seems almost impossible to continue. I am worried that this was a big mistake from the beginning. So I am asking the Topos theorists that read this, for their opinion on how stupid they think I have been by thinking in this way, and maybe what is the best way/source to learn what is needed. Thanks for any help. - -REPLY [12 votes]: My first advice is to stop worrying about what progression you should be learning stuff. Just read things you find interesting, and if you find yourself reading something that makes no sense to you then you can start backtracking and looking at prerequisites. You don't need to plan ahead so much. Anyway, I second the recommendation of the book of Mac Lane and Moerdijk. -Also, I don't think it sounds absurd to study category theory without having a background in set theory/logic. Most people who know a lot of category theory do not, either; I would say most applications of category theory are in topology/algebraic geometry and related areas. But there are also logicians (as you mention), computer scientists, mathematical physicists, and many others that apply category theory in their work. -Finally, there's a big risk your question is going to closed. Don't take it personally, it's just a bit too discussion-y to be a great fit for the MO "format".<|endoftext|> -TITLE: Formulas in a Field and in a Field Extension -QUESTION [8 upvotes]: Let $\mathbb F$ be a field and let $a, b, c, d$ be fixed elements in the field $\mathbb F$. -Consider the formulas -1) $\exists\;x\;\;:\;\;x^2=-1.$ -2) $\exists\;x\;\;:\;\;(xa=c\land xb=d).$ -Formula $(1)$ can be false in $\mathbb F$ but true in a field extension of $\mathbb F$. For exemple, formula $(1)$ is false in the reals $\mathbb R$ but true in the complex numbers $\mathbb C$. -The same does not happen with the formula $(2)$ since formula $(2)$ is true if and only if the matrix -$$\left(\begin{array}{cc}a&b\\c&d\end{array}\right),$$ -is singular, and it is well known that the singularity of a matrix is not changed by an extension of the field that contains the elements of the matrix. -My question is: -Is there any characterization of the formulas, in a field, whose validity does not change by an extension of the field? -Remark: I put this question yesterday in https://math.stackexchange.com/questions/561178/formulas-in-a-field-and-in-a-field-extension. But was not answered. - -REPLY [4 votes]: This was intended to be a comment to SJR's post, but was too long. -Here is a proof that any formula that is preserved downwards in models of $T$ is equivalent in $T$ to a universal formula, from which we deduce that formulas preserved upwards are equivalent to existential formulas. -Let $T$ be an $\cal L$-theory. Suppose $\phi(\bar v)$ is preserved downwards in models of $T$. -Let $\Gamma(\bar v)=\{\psi(\bar v): \psi$ is universal and $T\models \phi(\bar v)\rightarrow \psi(\bar v)\}$. -We claim that $T+\Gamma(\bar v)\models \psi(\bar v)$. If that happens then there are $\psi_1,\dots,\psi_n\in\Gamma(\bar v)$ such that $T\models \phi(\bar v)\leftrightarrow (\psi_1(\bar v)\land\dots\land\psi_n(\bar v)$. -Suppose not. Then there is ${\cal M}\models T$ and $\bar a\in \cal M$ such that ${\cal M}\models \psi(\bar a)$ -for $\psi\in\Gamma$ and ${\cal M}\models \neg\phi(\bar a)$. -Let $T_1= T+ \phi(\bar a) + $ atomic diagram of $\cal M$. If $T_1$ is satisfiable there is ${\cal N}\models T$ -with ${\cal M}\subset {\cal N}$ and ${\cal N}\models \phi(\bar a)$, contradicting the fact that $\phi$ is preserved downward. Thus $T_1$ must be unsatisfiable. Thus there is a quantifier free formula $\theta(\bar v,\bar w)$ -and $\bar b\in \cal M$ such that $$T\models \theta(\bar a,\bar b)\rightarrow \neg\phi(\bar a).$$ Since the parameters $\bar a$ and $\bar b$ don't occur in $T$, -$$T\models \forall \bar v\ (\phi(\bar v)\rightarrow\forall \bar w\ \neg\theta(\bar v,\bar w))$$ -But then $\forall \bar w\ \neg\theta(\bar v,\bar w)\in \Gamma$ contradicting the fact that ${\cal M}\models -\theta(\bar a,\bar b)$.<|endoftext|> -TITLE: How did Gauss and contemporaries think of modular forms? -QUESTION [14 upvotes]: Accounts of modular forms say that they were studied in the early 19th century, but then define modular forms using terminology that didn't exist until the 20th century. How did the earliest mathematicians to investigate modular forms define them? What motivated their exploration? - -REPLY [13 votes]: There is also the book of F. Klein, Development of Mathematics in XIX century, vol. I, which has a large chapter on Gauss which describes his work on modular forms. This was written in XX century, but Klein was essentially a XIX century mathematician, so you can see from this book "how did they think".<|endoftext|> -TITLE: Global sections of determinant bundle of symmetric powers -QUESTION [7 upvotes]: Let $E$ be a globally generated vector bundle of rank $r$ on a normal irreducible projective variety $X$. Suppose that $E$ induces a finite map -$$X \to \mathbb{G}r (H^0(E), r)$$ -to the Grassmannian of rank r quotients. -Can we say that a symmetric power of $S^mE$ for $m>>0$ induces an immersion -$$X \hookrightarrow \mathbb{G}r (H^0(S^mE), R)$$ -where $R = \binom{m+r-1}{m} = \mathrm{rk} S^m E$? -In case $r=1$ it is straightforward, as such a line bundle would be ample, -here we ae not supposing $E$ to be ample, the hypothesis imply only that $E$ is nef and $\det E$ is ample. -A positive answer would be equivalent to say, under the same hypothesis for $E$, that for $m >>0$ the image $V$ of the determinant map -$$ \Lambda^R H^0(S^m E) \twoheadrightarrow V \subseteq H^0 (\det S^m E)$$ separate points and tangent vectors of $X$. - -REPLY [7 votes]: The suggestion in my comment turns out to be impossible, and this leads to counterexamples to the OP's question, already for $\text{dim}(X)=2$. Let $U$ be a $2$-dimensional vector space over a field $k$. Let $\mathbb{P}(U^\vee)$ denote $\text{Proj} \text{Sym}^\bullet_k U$, i.e., $\mathbb{P}^1_k$. To be very precise, there is a short exact sequence on $\mathbb{P}(U^\vee)$, -$$ 0 \to (\bigwedge^2_k(U^\vee))^\vee\otimes_k \mathcal{O}(-1) \xrightarrow{\alpha^\dagger} U\otimes_k \mathcal{O}_{\mathbb{P}(U^\vee)} \xrightarrow{\beta} \mathcal{O}(+1) \to 0,$$ -and the surjection $\beta$ is universal in the usual sense. -Let $V$ be $\text{Syt}_k^2(U)$, the $k$-subspace of $U\otimes_k U$ of symmetric tensors (the same as the symmetric quotient $\text{Sym}_k^2(U)$ except in characteristic $2$). Then $V$ is a $3$-dimensional $k$-vector space. Let $\mathbb{P}(V^\vee)$ be $\text{Proj}\text{Sym}_k^\bullet V$, i.e., $\mathbb{P}^2_k$. To be very precise, on $\mathbb{P}(V^\vee)$ there is a short exact sequence, $$ 0 \to F^\vee \xrightarrow{\gamma^\dagger} V\otimes_k \mathcal{O}_{\mathbb{P}(V^\vee)} \xrightarrow{\delta} \mathcal{O}(1),$$ -and the surjection $\delta$ is universal in the usual sense. -Consider $X=\mathbb{P}(U^\vee) \times_k \mathbb{P}(U^\vee)$, i.e., $\mathbb{P}^1_k\times_k \mathbb{P}^1_k$, with its two projection $\pi_1,\pi_2: X \to \mathbb{P}(U^\vee)$. There is a surjection, -$$ \pi_1^*\beta\otimes \pi_2^*\beta: (U\otimes_k U)\otimes_k \mathcal{O}_X \twoheadrightarrow \pi_1^*\mathcal{O}(+1)\otimes_{\mathcal{O}_X}\pi_2^*\mathcal{O}(+1).$$ Computing locally, the restriction of $\pi_1^*\beta\otimes\pi_2^*\beta$ to the subsheaf $$\text{Syt}^2_k(U)\otimes_k \mathcal{O}_X\subset (U\otimes_k U)\otimes_k \mathcal{O}_X$$ is still surjective, i.e., there is an induced surjection, -$$ -\text{Syt}^2\beta:\text{Syt}^2_k(U)\otimes_k \mathcal{O}_X \twoheadrightarrow \pi_1^*\mathcal{O}(+1)\otimes_{\mathcal{O}_X}\pi_2^*\mathcal{O}(+1).$$ -Thus, there exists a unique morphism $f:X\to \mathbb{P}(V^\vee)$ such that $f^*\delta$ equals $\text{Syt}^2\beta$. Of course this is nothing other than the usual degree $2$ morphism $\mathbb{P}_k^1\times_k \mathbb{P}_k^1 \to \mathbb{P}_k^2$, but with due attention paid in characteristic $2$. -Now consider the dual of the short exact sequence on $\mathbb{P}(V^\vee)$, i.e., -$$ 0 \to \mathcal{O}(-1) \xrightarrow{\delta^\dagger} V^\vee\otimes_k \mathcal{O}_{\mathbb{P}(V^\vee)} \xrightarrow{\gamma} F \to 0. $$ -The surjection $\gamma$ is also universal, thus identifying $\mathbb{P}(V^\vee)$ with the Grassmannian, $\mathbb{G}r(2,V)$. Define $E$ to be $f^*F$ on $X$. Consider the restriction of $F$ to points and lines in $\mathbb{P}(V^\vee)$. -For a point of $\mathbb{P}(V^\vee)$ corresponding to a rank $1$ quotient, $V\twoheadrightarrow Q_1$, the fiber $F|_{[Q_1]}$ is simply the dual of the $2$-dimensional kernel of the quotient. For a pointed line in $\mathbb{P}(V^\vee)$ corresponding to a flag of quotients, $V\twoheadrightarrow Q_2\twoheadrightarrow Q_1$, the restriction of $F$ to the line is isomorphic to $\mathcal{O}\oplus \mathcal{O}(1)$, but where the fiber of the positive summand at the marked point $[Q_1]$ corresponds to the annihilator of the kernel of the surjection to $Q_2$. In particular, this depends on the choice of the line containing the specified point $[Q_1]$. As we hold $Q_1$ fixed and vary the line, this image in $F|_{[Q_1]}$ of the positive summand varies over all $1$-dimensional subspaces of the $2$-dimensional fiber. -Claim. For every integer $m\geq 0$, the locally free sheaf -$F_m := \text{Sym}^m_{\mathcal{O}}(F)\otimes_{\mathcal{O}}\mathcal{O}(-1)$ on $\mathbb{P}(V^\vee)$ has only the zero section. -Proof of Claim. -Since the restriction of $F$ to a pointed line is isomorphic to $\mathcal{O}\oplus \mathcal{O}(1)$, also -the restriction of $F_m$ to every pointed line $(L,[Q_1])$ is isomorphic to a direct sum, $$\mathcal{O}(-1)\oplus \mathcal{O}\oplus \mathcal{O}(+1)\dots \oplus \mathcal{O}(m).$$ In particular, the subsheaf spanned by global sections is a locally free subsheaf whose quotient is an invertible sheaf. For a specified point $[Q_1]$ of the line, the fiber at $[Q_1]$ of the "spanned subsheaf" is a codimension $1$ linear subspace $S_L$ of the fiber $F_m|_{[Q_1]}$. By the same analysis as in the previous paragraph, for fixed $[Q_1]$, for every nonzero element $v$ of $F_m|_{[Q_1]}$, there is a choice of line $L$ containing $[Q_1]$ such that $v$ is not contained in $S_L$. Since the fiber $v$ of a global section of $F_m$ at $[Q_1]$ is contained in $S_L$ for every choice of $L$, it follows that every global section of $F_m$ vanishes at $[Q_1]$. Since this holds for every point, and since $\mathbb{P}(V^\vee)$ is reduced, etc., it follows that $F^m$ has only the zero global section. -QED Claim. -Now consider the locally free sheaf on $X$, $f^*\text{Sym}^m_{\mathcal{O}}(F) \cong \text{Sym}^m_{\mathcal{O}}(E)$. The global sections on $X$ of $f^*\text{Sym}^m_{\mathcal{O}}(F)$ are the same as the global sections on $\mathbb{P}(V^\vee)$ of $f_*(f^*\text{Sym}^m_{\mathcal{O}}(F))$. -By the projection formula, this is canonically isomorphic to $\text{Sym}^m_{\mathcal{O}}(F)\otimes_{\mathcal{O}}f_*\mathcal{O}_X$. However, $f_*\mathcal{O}_X$ is canonically isomorphic to $\mathcal{O}_{\mathbb{P}(V^\vee)}\oplus \mathcal{O}(-1)$, where the summand $\mathcal{O}_{\mathbb{P}(V^\vee)}$ is just the image of $f^\#$. Since, as proved in the claim, $\text{Sym}^m_{\mathcal{O}}(F)\otimes_{\mathcal{O}}\mathcal{O}(-1)$ has only the zero global section, -it follows that the pullback map on global sections, -$$ f^*: H^0(\mathbb{P}(V^\vee),\text{Sym}^m_{\mathcal{O}}(E)) \to H^0(\mathbb{P}(U^\vee)\times_k \mathbb{P}(U^\vee), \text{Sym}^m_{\mathcal{O}}(f^*E)),$$ -is an isomorphism. Therefore, for every induced morphism as in the OP's question, the morphism factors through $f$. So the induced morphism can never be a closed immersion.<|endoftext|> -TITLE: Are the Kahler Identities for a Holomorphic Vector Bundle Actually Interesting? -QUESTION [5 upvotes]: For a Kahler manifold $M$, and a smooth vector bundle $E$ over $M$, let us denote by $A^{(p,q)} := \Omega^{(p,q)} \otimes E$ the bundle of forms with values in $E$. Now with respect to a choice of connection on $E$, we can extend $d$ to a mapping d$ _E: A^{(p,q)} \to A^{(p+1,q+1)}$. Moreover, we can extend $\partial$, and $\overline{\partial}$, to mappings $\partial_E: A^{(p,q)} \to A^{(p+1,q)}$, and $\overline{\partial}_E: A^{(p,q)} \to A^{(p,q+1)}$ respectively. As is well-known, if we also assume that $E$ is holomorphic, then we can choose a connection such that $\overline{\partial}_E: A^{(0,\bullet)} \to A^{(0,\bullet)}$ is a chain complex. However, the same cannot be said for the sequences ${\partial}_E: A^{(\bullet,0)} \to A^{(\bullet,0)}$, and d$ _E: A^{(\bullet,\bullet)} \to A^{(\bullet,\bullet)}$, meaning that in general we have no analogue for holomorphic Dolbeault, or de Rham, cohomologies. We do have, however, natural analogues of the Kahler identities. What I would like to know is, are these still of any real interest, given that they can no longer be used to prove equality of holomorphic, anti-holomorphic, and de Rham cohomologies? For example, in Voisin and Huybrecht's books they do not even appear. - -REPLY [6 votes]: They can be used to prove the vanishing theorems (and more). Kodaira-Nakano vanishing theorem and Kodaira embedding theorem follow from these identities. One proves that the difference of the $\partial$ and $\bar\partial$-Laplacians is a commutator of Hodge -$\Lambda$ operator and the curvature. This commutator happens to be positive or negative when the bundle is positive or negative; vanishing of cohomology follows immediately, because the Laplacians themselves are positive operators.<|endoftext|> -TITLE: Asymptotic Weyl Character Formula -QUESTION [10 upvotes]: Let $G$ be a complex semi-simple group along with a chosen pair of opposite Borel subgroups (so we get all the root-theoretic data we need). Let $\lambda$ be a dominant weight, and let $V(\lambda)$ be the corresponding highest weight representation. I'm interested in a formula for the character of $V(n \lambda)$ as $n \rightarrow \infty$. Of course to have this limit converge, I need to normalize things. So I consider the character of $V(n\lambda)$ as a measure on the Lie algebra of the torus, which I rescale to live on precisely the convex hull of the weights appearing on $V(\lambda)$, and which I scale by a fixed power of $n$. Then I consider the weak limit of this measure. This measure is also the Duistermaat-Heckman measure of the flag variety corresponding to the line bundle $\mathcal{O}(\lambda)$ (perhaps I should restrict to $\lambda$ regular to be correct). -The Fourier transforms of these measure appearing in the limit are precisely the character in the usual sense (given by the Weyl character formula), so the Fourier transform of the limiting measure should be calculable. So I am asking for the Fourier transform of this measure. -In the case of $SL_2$, for $\lambda$ equal to the fundamental weight $\omega$, this is just the Duistermaat-Heckman measure of $\mathbb{P}^1$. This is just the Lebesgue measure supported on the interval $[-\omega,\omega]$, which clearly agrees with what we know about characters of $SL_2$ representations. And it's Fourier transform is (up to some constants) the sinc function $\frac{\sin(x)}{x}$, which can also be seen from the Weyl character formula since $\frac{\sin(x)}{x} = \frac{e^{ix} - e^{-ix}}{2ix}$ However, I do not know how to write down such a formula in other types. I expect that there should be some formula with roughly the form of the Weyl character formula. Does anyone know any such formula? - -REPLY [8 votes]: I personally consider a "character" to be function on the group, whereas you speak about it living on the convex hull of the weights, which is in the dual of the Lie algebra of the torus. -Anyway, the point is that you already understand Heckman's asymptotics of the weight multiplicity function on the weight lattice, as a measure on the vector space containing it, and you're asking for the corresponding limit of the Fourier transforms. -This is easy to get from the WCF directly: the relevant limit is $1 - \exp(-x) \sim x$. The limits of the Weyl character (thought of as a function on $T$) in its two forms -$$ \sum_{w\in W} w \cdot \frac{t^{\lambda}}{\prod_{\beta \in \Delta_+}(1 - t^{-\beta})} = \frac{\sum_{w\in W} (-1)^w t^{w\cdot(\lambda+\rho)-\rho}}{\prod_{\beta\in\Delta_+}(1-t^{-\beta})}$$ -are the functions on $\mathfrak t$ -$$ \sum_{w\in W} w \cdot \frac{\exp(-\lambda)}{\prod_{\beta\in \Delta_+} \beta} = \frac{\sum_{w\in W} (-1)^w \exp(-w\cdot\lambda)}{\prod_{\beta\in \Delta_+} \beta}$$ -The $\rho$-shift disappears entirely! -Instead of thinking of this as a limiting process, you could compare [Atiyah-Bott '65] localization in K-theory to [Atiyah-Bott '84] localization in cohomology, applied to the $\lambda$ Borel-Weil line bundle on the flag manifold or the Chern character thereof. -As for the $\sin(x)$ formula you want, that's special to the case $\lambda$ a multiple of $\rho$. Check out Why do flag manifolds, in the P(V_rho) embedding, look like products of P^1s?<|endoftext|> -TITLE: Model structure for cooperads and for coalgebras -QUESTION [13 upvotes]: I am studying the homotopy theory of (algebraic) operads and I came up with several questions I am unable to answer to. I would like to stress that I don't have applications in mind, I just would like to understand the "state of art". -Initial assumption. In what follows, every operad is assumed to be reduced. -Let me recall a few well-known facts: - -fix a closed symmetric monoidal model category $(\mathcal E, \otimes, I)$, which is moreover cofibrantly generated. It is shown in Spitzweck's thesis [1] that the collection of operads in $\mathcal E$ has always a J-semi-model structure; moreover, this becomes a model structure whenever $\mathcal E$ has a cocommutative Hopf interval in the sense of [2]; -Following [2], we say that an operad $P$ is admissible if $P\textrm{-} \mathrm{Alg}$ obtains a model structure which is transferred from $\mathcal E$. Again in [1] it is shown that if $(\mathcal E, \otimes, I)$ satisfies the monoid axiom of [2], any cofibrant operad (w.r.t. the semi model structure) is admissible; -more generally, if $P$ is an operad endowed with a map $P \to P \otimes Q$ (where this is the Hadamard product) and the category $\mathcal E$ has an interval endowed with a $Q$-coalgebra structure, then $P$ is admissible. As corollary, we obtain that if there exists a cocommutative coassociative interval, then every operad is admissible. This is shown in [2]. - -Point 3. seems really interesting to me: it relates algebras and coalgebras in a sort of pairing. Let me be more precise: if we knew that there is a map of operads $P \to P \otimes Q$ and we knew moreover that the category $Q \textrm{-} \mathrm{coAlg}$ of $Q$-coalgebras has a model structure where weak equivalences and cofibrations are defined via the forgetful functor, then the condition stated in 3. would hold trivially. However I cannot hope to use the transfer principle, because the adjunction (forgetful,cofree) for coalgebras goes in the wrong direction. My first question is therefore: - -Question 1. Is there in the literature a notion of "coadmissible operad", i.e. an operad $P$ such that $P\textrm{-}\mathrm{coAlg}$ has a model structure such that the forgetful functor to $\mathcal E$ preserves weak equivalences and cofibrations? - -Changing somehow approach and flavour, I began to study B. Vallette's Homotopy Theory of Homotopy Algebras [4]. I learned that when $\mathcal E = \mathrm{dgVect}_k$, if $P$ is Koszul then conilpotent $P^¡$-coalgebras do have a model structure satisfying Question 1 (this makes sense because coalgebras over a cooperad are simply coalgebras over the linear dual operad). This model structure is obtained via the bar-cobar adjunction induced by the obvious twisting morphism $\kappa \colon P^¡ \to P$, but it is not a simple application of the transfer principle: once again, the direction of the adjoint functors is the wrong one and one needs to work a lot with spectral sequences (using conilpotency) to get the result. Therefore I am led to the following question: - -Question 2. Is there in the literature a notion of "admissible twisting morphism"? Let us say that this could mean the following: a twisting morphism $\alpha \colon C \to P$ from a cooperad to an operad is said to be admissible if the category of conilpotent $C$ coalgebras have a model structure where cofibrations are degreewise monomorphisms and a map $f$ is a weak equivalence iff $\Omega_\alpha(f)$ is a weak equivalence. Here $\Omega_\alpha$ is the bar construction, part of the bar-cobar adjunction $\Omega_\alpha \colon \text{conil } C\textrm{-}\mathrm{coAlg} \leftrightarrows P \textrm{-}\mathrm{Alg} \colon B_\alpha$. - -Remark. I am not too strict on the above definition of admissible twisting morphism. My question is: "has someone studied something looking similar?". -Finally, the last question is: - -Question 3. Is there a treatment of cooperads similar to the one given in [2]? Is there a general principle comparable with "cofibrant operads are admissible"? If not, can someone explain me the obstructions? - -Bibliography -[1] M. Spitzweck, Operads, Algebras and Modules in General Model Categories, avaiable at http://arxiv.org/abs/math/0101102 -[2] C. Berger, I. Moerdijk, Axiomatic homotopy thoery for operads. -[3] S. Schwede, B. Shipley, Algebras and modules in monoidal model categories. -[4] B. Vallette, Homotopy theory of homotopy algebras, available at http://math.unice.fr/~brunov/HomotopyTheory.pdf - -REPLY [7 votes]: Joey Hirsh and I have given an answer to question 2. We assume our operads and cooperads are $\Sigma$-split outside characteristic zero for the usual technical reasons and additionally require a weight grading on the cooperad so that we can use Bruno's results. Then any twisting morphism is admissible in the sense of the question. -These structures are functorial in both cooperad and operad and promote the bar and cobar functors to Quillen adjoints, which (in characteristic zero, at least) are Quillen equivalences if and only if the twisting morphism is Koszul.<|endoftext|> -TITLE: How to check whether a positive integer can be written as linear combination of given others, where all coefficients are positive? -QUESTION [11 upvotes]: Let $n$, $k$ and $m_1, \dots, m_k$ be positive integers. Which is the most efficient -algorithm to find out whether there are positive integers $a_1, \dots, a_k$ such that -$n = \sum_{i=1}^k a_i m_i$? -To make things nontrivial, think of $k$ being in the hundreds, and of $n$ and the $m_i$ -having hundreds of decimal digits, each. -- -Clearly if we would remove the requirement that the $a_i$ are positive, the -Chinese Remainder Theorem would tell us the answer -- but we do require them to be -positive. - -REPLY [4 votes]: The problem can be thought of as a coin problem. There are $k$ coins with denominations $m_1,\dots,m_k$ and you want to express an amount $n$ with these coins. As states, the problem is an integer programming question which is NP-complete when $k$ is part of the input. It is in P (with exponential dependence on $k$) when $k$ is fixed by an algorithm by Lenstra. -The problem is closely related the Frobenius/Sylvester coin problem - to find the minimum $n$ so that every larger integer has such a representation. See here and here. A polynomial algorithm when $k$ is bounded was achieved by Ravi Kannan. (The dependence on $k$ is double-exponential.) -These two problems (finding a representation for fixed $n$ and finding the value of $n$ above which a representation always exists) represent the first two levels in Presburger Hierarchy. An important open problem here is to find a P-algorithm for higher order problems in the Presburger Hierarchy. -Of course, another important question is how to solve such questions in practice. I suppose other people can answer that better than me. One method that certainly comes to mind is to consider the linear programming relaxation (i.e. to allow rational $a_i$s) and then apply some rounding and "local" improvement. -The range proposed by the OP where $k$ - (the number of coins) is in the hunderds is interesting. I don't know if current algorithms can scratch this value.<|endoftext|> -TITLE: Expression of a non-orthogonal projection in a $C^*$ algebra via an orthogonal one -QUESTION [6 upvotes]: A paper I'm currently reading uses the following fact. If $A$ is a unital $C^*$-algebra, $P=P^2\in A$, then there are $T, F\in A$ s.t. $F$ is an orthogonal projection ($F=F^*=F^2$) and -$$P=F+FT(1-F).\tag{1}$$ -The question: - -Since no proof or further reference is given, I assume, that this fact - is obvious from some point of view. Is it? What do I have to learn, so - the similar facts would be obvious for me? - -P.S. Here is a proof I came up with after some trial and error. -We would like to define $F=f(PP^*)$, where $f(x)=0$ if $x=0$ and $f(x)=1$ otherwise. To check, that $F$ is well-defined we need to check, that $f$ is continuous on the spectrum of $PP^*$. To do this it is enough to show, that the open interval $(0,1)$ doesn't intersect with its spectrum or, equivalently, that $(2PP^*-1)^2\geq 1$. Expanding we see, that this is equivalent to $PP^*PP^* \geq PP^*$. -Define $A=P-P^*$. Then $A^*A\geq 0$. On the other hand expanding we get $A^*A=PP^*+P^*P-P-P^*$, so $PP^*+P^*P\geq P+P^*$. Multiplying by $P$ on the left and by $P^*$ on the right we get $PP^* + PP^*PP^*\geq 2 PP^*$ or $PP^*PP^*\geq PP^*$. Thus $F=f(PP^*)$ is indeed well defined. -Since $f(x)$ is real valued and $f(x)=f(x)^2$ we have $F=F^*=F^2$. To show, that $FP=P$ notice, that $f(x)x=x$ for $x\geq 0$. Therefore $FPP^*=PP^*F=PP^*$ and -$(FP-P)(FP-P)^*=(F-1)PP^*(F-1)=0(F-1)=0$. Thus indeed $FP=P$. -Thus we have -$$P=F+(P-F)=F+F(P-1).$$ -Finally, notice, that -$(P-1)F=(P-1)PP^*F=0P^*F=0$, so $P-1=(P-1)(1-F)$ and -$$P=F+F(P-1)(1-F).$$ - -REPLY [3 votes]: Just a few remarks in addition to David Handleman's answer and your work. To make sense of ranges and nullspaces below, you can assume your $C^*$-algebra is sitting in $B(H)$. I just added that hoping it makes things more "obvious". But the whole thing works of course in an abstract $C^*$-algebra. Even the notion of range projection. - -If such an $F$ exists, it must be the projection (self-adjoint idempotent) onto the range of the idempotent $P$ (this is often called the range projection of $P$), since $PF=F \iff \mbox{im} F\subseteq \ker (I-P)=\mbox{im} P$ and $FP=P\iff \mbox{im} P\subseteq \ker (I-F)=\mbox{im} F$. Altogether, these two conditions are equivalent to $PA=FA$. -Now for every idempotent $P$, the range projection of $P$ is given, for instance, by the formula $$F=P(P+P^*-I)^{-2} P^*$$ where the existence of the central factor follows from the fact that $(P+P^*-I)^2=I-(P-P^*)^2=I+H$ with $H=(P-P^*)^*(P-P^*)$ is a positive element. Simple algebraic manipulations show that this formula yields the range projection of $P$: that is $P^2=P=P^*$, $PF=F$, and $FP=P$. The formula also shows that $F$ lies in the $C^*$-algebra generated by $P$. -To understand what $T$ could be, it is helpful to know about Peirce decomposition with respect to an orthogonal decomposition of the unit: here $I=F\oplus (I-F)$. In the latter, we have -$$ P=\pmatrix{A& B\\C&D} \quad A=FPF=F\quad\;B=FP(I-F)\quad C=(I-F)PF=0\quad D=(I-F)P(I-F)=0$$ -So the most natural choice is $T=P$ as we have $P=F+FP(I-F)$. -Hopefully, now things should be as obvious as in the following particular case. Take a rank one idempotent $P$ on $\mathbb{C}^2$. Then in $\mathbb{C}^2=\mbox{im} P\oplus (\mbox{im} P)^\perp$, we have -$$ -P=\pmatrix{1& b\\ 0&0}\quad F=\pmatrix{1& 0\\ 0&0}\quad P=\pmatrix{1& 0\\ 0&0}+\pmatrix{0& b\\ 0&0}=F+FP(I-F) -$$<|endoftext|> -TITLE: Hodge structure versus Weight structure -QUESTION [9 upvotes]: This is a naive question. -One is told that, somehow, Hodge theory for varieties over complex numbers, is an analog of weight theory for varities over finite fields. In weight theory, one considers eigenvalues of Frobenius and so on. Hodge theory should capture symmetries of the Galois group of complex numbers over real numbers. -My question is: It is confusing; For a variety over a finite field, it makes sense that the Galois group will act on different things and give some information etc.; But in Hodge theory, we begin with a variety over the complex numbers, not over the real numbers; So what is the logic by which the Galois group of complex numbers over real numbers enters the picture? What is the logic behind the words "hidden symmetries" in the book by Gelfand and Manin? -i.e., when we have a variety over the algebraic closure of a finite field, we do not claim to have "weight theory"; We need structures to be defines over the finite field.. -I hope my naive question is clear, -Thank you, -Sasha - -REPLY [9 votes]: I don't know how helpful this will be either. But let me modify your question a bit to ask what is the correct analogue of the Galois group (of say a finite field or number field) in Hodge theory? It is certainly true that when $X$ is a smooth complex projective variety, -$Gal(\mathbb{C}/\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$ acts on $H=H^i(X,\mathbb{C})$, -by virtue of the fact that it is obtained by complexifying real de Rham cohomology. However, this is rather weak information; certainly too weak to recover the Hodge structure. Fortunately, one can do better. Deligne's torus $\mathbb{S}$ is the real algebraic group whose real points $\mathbb{S}(\mathbb{R})=\mathbb{C}^\ast$. -So that $\mathbb{S}(\mathbb{C})=\mathbb{C}^\ast\times \mathbb{C}^\ast $. Note that we have natural cocharacter $\mathbb{G}_m\to \mathbb{S}$ corresponding to the inclusion $\mathbb{R}^\ast\subset\mathbb{C}^\ast$ on real points. -$H$ becomes a real representation of this group, by letting $(z_1, z_2)\in \mathbb{S}(\mathbb{C})$ act on $H^{pq}\subset H$, by $z_1^pz_2^q$. This is a better answer to the question, because this action completely determines the real Hodge structure on $H$. The weight $i$ is gotten by restricting the action to $\mathbb{G}_m$, and observing that the characters of $H$ are just $i$th twists of the standard character. -If you look at the rational Hodge structure, then there is a bigger group called the Mumford-Tate group $MT(H)$, that I won't attempt to define, that acts on $H_{\mathbb{Q}}= H^i(X,\mathbb{Q})$ -and competely determines the Hodge structure on it. When $X$ is defined over a number field $K$, then conjecturally $MT(\mathbb{Q}_\ell)$ should be (essentially) the same as the image of the action of $Gal(\bar K/K)$ on $\ell$-adic cohomology $H^i(\bar X_{et},\mathbb{Q}_\ell)$.<|endoftext|> -TITLE: Original proof of Gödel's completeness theorem compared to Henkin's proof -QUESTION [7 upvotes]: May I have some clarification about original proof of Gödel's Completeness Theorem compared to "standard" Henkin's proof based on Model Existence Lemma ? -My understanding of Gödel's original proof is this : -1) Perform some syntactical transformation, in order to reduce the general problem to a particular class of wffs -2) Build an assignment of object to te free vars of the "reduced" formula in order to satisfy it. The assignment is built up using n-uples of natural number corresponding to index of the free vars. -3) The last step is made using some sort of König's Lemma. -From a "philosophical" perspective, Gödel made two assumptions (in line with his "platonism") : the existence of natural number and some properties of infinite sets (König's Lemma). -The Henkin proof needs (if I remember well) Zorn's Lemma and build up the model using only "syntactical objects" : the constants of the language as witnesses. -There are also other proofs (based on Ultrafilters) but they are (for me) less transparent then Henkin's one. -So it seems to me that the theorem needs a "big amount" of set theory. Is it true ? If so, this fact gives us some information about the foundational issue regarding first-order logic and higher-order logic (according to Quine : not "real" logic at all, but set theory in disguise) ? -thanks - -REPLY [4 votes]: With regard to the amount of set theory required to prove the completeness theorem, the wikipedia page on The completeness theorem asserts: - - -When considered over a countable language, the completeness and compactness theorems are equivalent to each other and equivalent to a weak form of choice known as weak König's lemma, with the equivalence provable in RCA0 (a second-order variant of Peano arithmetic restricted to induction over Σ01 formulas). Weak König's lemma is provable in ZF, the system of Zermelo–Fraenkel set theory without axiom of choice, and thus the completeness and compactness theorems for countable languages are provable in ZF. However the situation is different when the language is of arbitrary large cardinality since then, though the completeness and compactness theorems remain provably equivalent to each other in ZF, they are also provably equivalent to a weak form of the axiom of choice known as the ultrafilter lemma. In particular, no theory extending ZF can prove either the completeness or compactness theorems over arbitrary (possibly uncountable) languages without also proving the ultrafilter lemma on a set of same cardinality, knowing that on countable sets, the ultrafilter lemma becomes equivalent to weak König's lemma.<|endoftext|> -TITLE: $\Sigma_1$ Statements and Forcing Extensions -QUESTION [5 upvotes]: Assuming consistency of $\text{ZFC}$ (with some large cardinal axiom), is the following statement consistent with $\text{ZFC}$? -Any $\Sigma_1$ statement with parameters $\omega_1,\omega_2$ which holds in a $\omega_1,\omega_2$ preserving forcing extension holds in $V$. - -REPLY [9 votes]: The answer is yes. Your theory holds in any model of the Maximality Principle for $(\omega_1,\omega_2)$-preserving forcing. So indeed, one can get a stronger result. -Specifically, this version of the maximality principle asserts that whenever a statement $\sigma$ is forceable by $(\omega_1,\omega_2)$-preserving forcing, in such a way that it remains true in all such further extensions, then it is already true in $V$. Thus, it can be expressed modally as the scheme of assertions -$$\Diamond\square\sigma\to\sigma,$$ -where $\Diamond\psi$ means that $\psi$ is forceable by $(\omega_1,\omega_2)$-preserving forcing, and $\square\psi$ means that $\psi$ holds in all such forcing extensions. The main point here, in connection with your question, is that the forceable $\Sigma_1$ assertions are of course forceably necessary, since once you force them, they remain true in all further extensions. Thus, your theory is generalized by the theory asserting that all forceably necessary statements are already true. -Theorem. The maximality principle $\text{MP}(\omega_1,\omega_2\text{-preserving})$ is equiconsistent with $\text{ZFC}$. -Proof. -Suppose ZFC is consistent, and let $T$ be the theory of ZFC together with all instances of this maximality principle. I claim that this theory $T$ is consistent. To see this, consider any finitely many assertions from $T$. This subtheory mentions only finitely many assertions $\sigma_0,\ldots\sigma_n$. Start from any model $M\models\text{ZFC}$, and then form extensions $M[G_0]\cdots[G_n]$, where at stage $k$ we force $\square\sigma_k$ over $M[G_0]\cdots[G_{k-1}]$, if this is possible to do while preserving $\omega_1$ and $\omega_2$, and otherwise use trivial forcing. The final model $M^+=M[G_0]\cdots[G_n]$ will satisfy all the instances in the subtheory, since if $\square\sigma_k$ is forceable over $M^+$ while preserving $\omega_1$ and $\omega_2$, then it would have been forced already at stage $k$ and hence would be true in $M[G_0]\cdots[G_k]$ and hence also in $M^+$. So every finite subtheory of $T$ is consistent, and so the theory is consistent, as desired. QED -Your theory follows from $\text{MP}(\omega_1,\omega_2\text{-preserving})$, since every forceable $\Sigma_1$ assertion is forceably necessary, because once you force it, it remains true in all further extensions. The parameters $\omega_1$ and $\omega_2$ are definable, by definitions that are absolute under $(\omega_1,\omega_2)$-preserving forcing, and so one doesn't need to have them as parameters, once we loosen the $\Sigma_1$ restriction to the forceably necessary statements. -This kind of argument is used extensively in my paper A simple maximality principle. -(Click on the edit history to see my original answer, which had some remarks on the Levy absoluteness theorem, which implies that $\Sigma_1$ sentences cannot be affected by forcing.)<|endoftext|> -TITLE: Can you form the "spectrum" of a sheaf of algebras? -QUESTION [9 upvotes]: If X is a scheme, we know there is a one-one correspondence between quasi-coherent -sheaves of $\mathcal O_X$-algebras on X and affine morphisms $Y \longrightarrow X$ -But what about arbitrary (not necessarily quasi-coherent) sheaves of $\mathcal O_X$-algebras? Do they correspond to schemes $Y\longrightarrow X$? -It seems to me that given a morphism $f:Y\longrightarrow X$ of schemes, -for any $U\subseteq X$, the association $U\mapsto \mathcal O_Y(f^{-1}(U))$ -defines a sheaf of $\mathcal O_X$-algebras. That's one direction. Does it not work in the other direction for some reason? - -REPLY [9 votes]: You could ask about this in the category of locally ringed spaces, rather than just schemes. For every locally ringed space $(X,\mathcal{O}_X)$, and for every sheaf $\mathcal{A}$ of $\mathcal{O}_X$-algebras, there is a locally ringed space, $(S,\mathcal{O}_S)$, a morphism of locally ringed spaces, $$(\pi,\pi^\#):(S,\mathcal{O}_S)\to (X,\mathcal{O}_X),$$ and a morphism of $\mathcal{O}_S$-algebras, $\phi:\pi^*\mathcal{A} \to \mathcal{O}_S$, that represents the contravariant functor from the category of locally ringed spaces over $X$ to the category of sets that associates to each morphism of locally ringed spaces, $(f,f^\#):(Y,\mathcal{O}_Y)\to (X,\mathcal{O}_X)$, the set of morphisms of $\mathcal{O}_Y$-algebras, $\psi:f^*\mathcal{A}\to \mathcal{O}_X$. Of course when $(X,\mathcal{O}_X)$ is a scheme and $\mathcal{A}$ is quasi-coherent, $(S,\mathcal{O}_S)$ is isomorphic (over $X$) to $\textbf{Spec}_X(\mathcal{A})$ as constructed in Hartshorne's book. In this sense, the locally ringed space $(S,\mathcal{O}_S)$ deserves to be called "Spec".<|endoftext|> -TITLE: Maximal centralizer in full matrix ring -QUESTION [5 upvotes]: I will be so thankful if someone can help me with the following question. -Is it possible to obtain all maximal centralizers in the full matrix ring, $M_n(F)$, for an arbitrary finite field $F$? Here, by maximal centralizer I mean: $C=C_R(x)$ is maximal if $C\subseteq C_R(y)$, then $C=C_R(y)$ or $y\in Z(R)$. -In Akbari et al., Linear Alg. App. 390 (2004) 345-355, lemma 3 determines all centralizers with maximum dimension. So some of the maximal centralizers are determined. Is it true that the set of centralizers with maximum dimension and the set of maximal centralizers are equal? -Hamid. - -REPLY [5 votes]: Let $x \in M_n(F)$. If the characteristic polynomial of $x$ has distinct prime factors in the ring $F[t]$, then there is some idempotent that commutes with everything that commutes with $x$, hence, if the centralizer is maximal, then it is the centralizer of that idempotent. The centralizer of an idempotent is clearly maximal. So this gives one kind of maximal subgroup. -In the remaining kind, the characteristic polynomial has the form $f(t)^k$ for some irreducible polynomial $f(t)$. Here we split into two cases - either $x$ is semisimple or it isn't. If $x$ is semisimple, its centralizer is a representation $M_k(F')$ where $F'$ is an extension of $F$. The elements which commute with this ring are just the elements of $F'$, and the only ones that have a larger centralizer are the subextensions. So this algebra is maximal if and only if $F'$ is an extension of prime degree. This gives a second kind of maximal subgroup. -If $x$ is not semisimple, we can take the Jordan decomposition $x=x_{ss}+x_n$, and the centralizer of $x$ is contained in the centralizer of $x_n$, but $x_n$ is not in the center of $M_n(k)$, so we may assume $x=x_n$. Then by taking a power we may assume $x^2=0$. In this case, we can check by looking explicitly at the Jordan block that anything that commutes with the centralizer of $x$ must be a linear combination of $1$ and $x$, and so the centralizer of $x$ is maximal. -So there are three cases: -The algebra is the centralizer of an idempotent, the product of two matrix algebras, as in Amitanshu's example. -The algebra is the algebra of matrices over a prime-degree extension of $F$. -The algebra is the centralizer of a nilpotent $x$ satsifying $x^2=0$.<|endoftext|> -TITLE: Automorphisms of $P(\Bbb N)$ -QUESTION [15 upvotes]: I believe I've proved that the power semigroup of non-negative integers with addition has a trivial automorphism group. The proof is a bit long, completely elementary and rather unremarkable (as the fact itself probably) so I won't post it here. However, I spent a long time thinking it up. Probably way too long. So I would like to know whether the fact - -is actually a fact; -is completely trivial; -follows immediately from a theorem that's maybe not that trivial; -has any significance. - -Added: For a semigroup $(S,\star)$, the power semigroup of $S$, denoted by $P(S)$ is the semigroup of all non-empty subsets of $S$ with the operation $A\star B=\{a\star b\,|\,a\in A,b\in B\}.$ -Added later: I'm confused by the answers given so far. The answerers seem to be showing that any automorphism of $P(\Bbb N)$ has to fix every singleton. Showing this was the first step in my proof, but I couldn't see how it showed immediately that such an automorphism has to fix everything. That's why my proof was getting longer and longer. I kept showing that an automorphism had to fix more and more things, until it had to fix all things. If that was unnecessary, I would like to understand it. Even if it is completely obvious, please do spell it out for me if possible. My mathematical spirits haven't been above sea level for a very long time, and being unable to understand something that people seem to consider obvious always brings me down. -Edit: I only really started feeling like I might get anywhere with this when I realized it was enough to work on sets containing zero. This is because the idempotent elements in $P(\Bbb N)$ are exactly the submonoids of $\Bbb N$ and $\Bbb N$ is the only idempotent $E$ such that for any other idempotent $F$ we have $E+F=E$. (In fact more is true: $E\supseteq F\iff E+F=E,$ which is to say that inclusion is the reversed standard partial order on idempotents of a semigroup: $e\leq f\iff ef=fe=e.$) This means $\Bbb N$ is fixed by any automorphism, and a set $X$ contains $0$ iff $X+\Bbb N=\Bbb N.$ So sets containing zero must go to sets containing zero. And additionally any set can be written uniquely as the sum of a singleton and a set containing zero, so it's enough to show any automorphism has to fix any such set. And these are a lot nicer because there are all kinds of relationships between adding such sets and inclusion. -Further on my proof is just ugly grinding so it doesn't make any sense to post it here I think. - -REPLY [2 votes]: Cannot make a remark so write here: every automorphism of $P(\mathbb{N})$ may be extended to an endomorphism of $P(\mathbb{Z})$, which in fact may be even turns out to be an automorphism of $P(\mathbb{Z})$, and may be then you can reason in a much simpler way, since by thesis of Peter Gallagher you know what are the indecomposables in the group power-semigroups. -The only little warning i would like to make: you see, power-semigroups is a tricky topic to work on, since it leads eventually to lots of frustration (as the whole semigroup theory in fact). I used to study at bachelor with a really genius guy so much naturally fitting to Maths and he did his phd on power-semigroups, it just happenned so, but since the topic is quite weird, he would feel not to finish his phd and went into industry and Maths lost a very brilliant person. So be careful!<|endoftext|> -TITLE: Frobenius-Perron eigenvalue and eigenvector of sum of two matrices -QUESTION [9 upvotes]: Suppose that I have two positive matrices, $A$, and $B$, and I know their Frobenius-Perron eigenvalues ($\lambda_A$, $\lambda_B$) and eigenvectors ($v_A$, $v_B$). I'm interested in what I can say about the Frobenius-Perron eigenvalue and eigenvector of $C=A+B$. I'm sure that this question must have been considered before, but I haven't found a good source. -Here are two particular questions: -(1) One clear bound on $\lambda_C$ is that $\max(\lambda_A,\lambda_B) \le \lambda_C$. Is there a nice upper bound? Also, if $v_A=v_B$, then $\lambda_C = \lambda_A + \lambda_B$. But can we do better than this if we know $v_A$ and $v_B$? -(2) What can we say about $v_C$? I expect it to be "somewhere between" $v_A$ and $v_B$, but that's just vague intuition. (Assume all the eigenvectors are normalized.) Is there a simple bound on the elements of $v_C$ given knowledge of $\lambda_A$, $\lambda_B$, $v_A$, and $v_B$? -I'd prefer not to assume that $A$ and $B$ are symmetric, but if the question has only been solved for the symmetric case, that's fine. -Edit: I realized that I should clarify that all of the $v$'s are right eigenvectors. If it helps to know the left Frobenius-Perron eigenvectors $u_A$ and $u_B$, then feel free to use this too. -Edit 2: In light of David's answer, then I've corrected my "intuitive" bound on $\lambda_C$ which was completely wrong. - -REPLY [9 votes]: First, the definition of $positive$ should be clarified. It could mean all entries strictly positive, or merely nonnegative, or that the matrix is primitive (all entries nonnegative and some power strictly positive). -I will use $\rho(M)$ to denote the spectral radius of a matrix $M$, which in case $M$ is nonnegative, is the Perron eigenvalue. -Note a weird example: if $M$ is strictly upper triangular with all entries above the diagonal strictly positive, then the spectral radius is $0$, and the same for $M^T$. However, $M+M^T$ is primitive, so its spectral radius exceeds zero. We can modify this to obtain strictly positive $M$ with $\rho(M)$ small, but $\rho(M+M^T)$ big, simply by adding a tiny amount to all the zero entries. In this case, $\rho(C)$ can be made much larger than $\rho(A) + \rho(B) $. -A well-known result ($well-known$ means I can't give a reference off the top of my head) is that if $C \geq A$ entrywise, $C \neq A$, and $A$ is strictly positive (or merely primitive), then $\rho(C) > \rho (A)$. (But the argument is easy: use an H-transform to convert $A$ to one with all column sums equal; adding anything will increase the spectral radius.) So if $A$ and $B$ are primitive, then $\rho(A+B) > \max \{\rho(A), \rho(B) \}$. -I doubt there's much of anything beside the obvious that can be said about the Perron eigenvectors. -Perhaps you have matrices in special forms in mind?<|endoftext|> -TITLE: For which $n$ is there only one group of order $n$? -QUESTION [35 upvotes]: Let $f(n)$ denote the number of (isomorphism classes of) groups of order $n$. A couple easy facts: - -If $n$ is not squarefree, then there are multiple abelian groups of order $n$. -If $n \geq 4$ is even, then the dihedral group of order $n$ is non-cyclic. - -Thus, if $f(n) = 1$, then $n$ is a squarefree odd number (assuming $n \geq 3$). But the converse is false, since $f(21) = 2$. -Is there a good characterization of $n$ such that $f(n) = 1$? Also, what's the asymptotic density of $\{n: f(n) = 1\}$? - -REPLY [17 votes]: The original question has already been answered, so I thought I would provide a slightly more general version. -The short paper http://www.math.ku.dk/~olsson/manus/three-group-numbers.pdf describes those orders for which there are precisely 1, 2 or 3 groups of the given order.<|endoftext|> -TITLE: Why aren't operator semigroups studied from a dynamical perspective? -QUESTION [7 upvotes]: Often times one talks about iterating a continuous map to get discrete topological dynamics, or having a 1-parameter family of continuous maps to get continuous topological dynamics. -When studying Feller processes, or in general semigroups of operators defined on a Banach space, it seems the only notion of dynamical systems that appears is "ergodicity" and it is really about the asymptotic behavior of the family. Why aren't things like fractals in Banach spaces, and non-asymptotic objects such as the orbit and Hausdorf dimension of the orbit and so on studied? For instance, in the Feller case "ergodicity" means that there is exactly one stationary distribution for the semigroup and it is attracting in some sense. All other properties studied (keep in mind I study it from the perspective of a probabilist) seem to be functional analytic. - -REPLY [6 votes]: Jack Hale had important work in generalizing concepts of dynamical systems to the infinite dimensional setting, see his monographs: -Asymptotic behavior of dissipative systems -It turns out that, under compactness assumtions, it is possible to generalize a lot of the finite dimensional concepts. -The work of Temam, Foiaș and their group connected to the Navier-Stokes equations is also extremely important, see -Infinite dymensional dynamical systems -On more recent work of the chaotic behavior (as mentioned b Uwe in his comment) you should consult the excellent monograph -Linear Chaos -where there is an account on recent results. -But of course the main problem is, as remarked by Uwe, that we are still struggling with the proper notions of generalization of some of the concepts mentioned.<|endoftext|> -TITLE: Nonmetrizable compact totally disconnected spaces without isolated points -QUESTION [8 upvotes]: Brouwer proved that a topological space is homeomorphic to the Cantor set if and only if -1) it's non-empty, -2) it's compact, -3) it's totally disconnected, -4) it has no isolated points, and -5) it's metrizable. -What are some nice examples of spaces that meet conditions 1)-4) but not 5)? -Perhaps it's worth noting that a compact Hausdorff space is metrizable if and only if it's second-countable, by a spinoff of the Urysohn Metrization Theorem. I suppose I'd prefer my examples to be Hausdorff. If so, they must be non-second-countable. Such spaces might thus be seen as 'like the Cantor set, but bigger'. -[1] L. E. J. Brouwer, Over de structuur der perfekte puntverzamelingen, Amst. Ak. Versl. 18 (1910) 833-842. - -REPLY [7 votes]: The boolean algebra of clopen subsets of a space satistying conditions 1,2 and 3 is always a base for the topology (i.e. the space is zero-dimensional). So the space is homeomorphic to the Stone space of that boolean algebra. The space will satisfy condition 4) if and only if the algebra is atomless. Non-metrizability corresponds exactly to the algebra being uncountable. So Shane's comment actually gives a characterization of such spaces. -There is only one countable atomless boolean algebra and that is why 1)-5) characterizes a single space which happens to be Cantor's space. -The space $2^\kappa$ in Nates's comment correspond to the free boolean algebra in $\kappa$ generators. A space satisfies 1)-4) if and only if it is homeomorphic to a perfect subspace of $2^\kappa$ for some $\kappa$ (the smallest $\kappa$ for which this happens coincides with the weight of the space, and such space will be metrizable if and only if it has countable weight).<|endoftext|> -TITLE: Mumford-Tate groups of products of Hodge structures -QUESTION [6 upvotes]: Let $V_1$, $V_2$ be two polarised simple $Q$-Hodge structures which are non-isomorphic. -I am assuming that the Mumford-Tate groups of $V_1$ and $V_2$ are semi-simple adjoint. -Is it true in this case that $MT(V_1 \times V_2) = MT(V_1) \times MT(V_2)$? -(I can easily see that this is not true when $MT(V_1)$ and $MT(V_2)$ are tori !) - -REPLY [6 votes]: The answer is no. Namely, let $E$ be an elliptic curve without complex multiplication and $V=H^1(E,Q)$ its first rational cohomology group, which is a 2-dim'l rational Hodge structure of weight 1. Its Hodge (resp. Mumford-Tate) group is $SL(V) \cong SL(2)$ (resp. $GL(V)$). Let $m$ be a positive integer and $V_m=Sym^{2m}(V)$ be the $2m$-th symmetric power of $V$, which is the (absolutely) irreducible rational Hodge structure of weight $2m$ with Hodge group $PSL(V)\cong PSL_2$. (The irreducibility follows from the representation theory of $SL_2$.) Now its twist $\tilde{V}_m=V_m(-m)$ is the irreducible rational Hodge structure of weight 0 with ``the same" (adjoint) Hodge/Mumford-Tate group $PSL(V)$; the Hodge structures $\tilde{V}_m$ are not isomorphic for different $m$, because they have different dimension, namely, $1+2m$. Now take two distinct positive integers $m$ and $n$. Then $MT(\tilde{V}_m \times \tilde{V_n})$ is still $PSL(V)$, which is strictly less than $PSL(V)\times PSL(V)$. This provides a desired counterexample.<|endoftext|> -TITLE: A New Continuum Hypothesis (Revised Version) -QUESTION [16 upvotes]: Define $N_n$ as $n$ th natural number: $N_0=0, N_1=1, N_2=2, ...$. -What happens after exponentiation? -We have the following equation: $2^{N_n}=N_{2^{n}}$. -(Which says: For all finite cardinal $n$ we have: $2^{n~\text{th finite cardinal}}=2^{n}~\text{th finite cardinal}$). -What this means? -The gap between $N_n$ and $2^{N_n}$ is rapidly increasing in exponential speed. -Now look at the $\text{GCH}$. It says that the gap between an infinite cardinal $\kappa$ and $2^{\kappa}$ is just a constant number $1$ in cardinals. Even in models for total failure of $\text{GCH}$ we usually have a finite gap between $\kappa$ and $2^{\kappa}$. Now if we look at the infinite cardinals as generalization of natural numbers it seems we should restate continuum hypothesis with more acceleration for the function $\kappa \mapsto 2^{\kappa}$ in order to uniform the behavior of exponentiation function in finite and infinite cardinals. Note to the following statement: -For all cardinal $\kappa$ we have $2^{\kappa~\text{th infinite cardinal}}=2^{\kappa}~\text{th infinite cardinal}$. -This is a direct generalization of the equation $\forall n\in \omega~~~~~2^{N_n}=N_{2^n}$ to the following form: -Natural Continuum Hypothesis (NCH): $~~~\forall \kappa\in Card~~~~~2^{\aleph_{\kappa}}=\aleph_{2^{\kappa}}$ -Unfortunately $\text{NCH}$ is contradictory by Konig's lemma because assuming $\text{NCH}$ we have: $\aleph_{\aleph_0} -TITLE: A special tessellation -QUESTION [12 upvotes]: Let $P$ be a convex $n$-gon. Suppose that we have an infinite number of $P$s, and that each of them is colored either red or blue. Here, let us consider the following operations : -Operation 1 : Place a red $P$ on a plane. -$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -Operation 2 : Place $n$ blue $P$s around the red $P$ such that each of the blue $P$s and the red $P$ are laid symmetrically with each $E_i\ (i=1,2,\cdots,n)$ where $E_i$ is an edge of the red $P$. -$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -Operation 3 : Place $n$ red $P$s around every blue $P$ in the same way as operation 2. -$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -Operation 4 : Place $n$ blue $P$s around every red $P$ in the same way as operation 2. -Operation 5 : Repeat operation 3 and 4. -Here, let us consider the following conditions : -Condition 1 : These $P$s are plane tessellation figures. -Condition 2 : There exists no place where both red $P$ and blue $P$ are placed. -Note that a regular hexagon, for example, does not satisfy the condition 2. See the figure below. The blue $P$ on which a letter $P$ is written, for example, will be colored red by the next operation. -$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ -Then, here is the first question. - -Question 1 : Is the following true? -$P$ satisfies these two conditions $\iff$ $P$ is either "a $45–45–90$ triangle", "a $30–60–90$ triangle", "an equilateral triangle" or "a rectangle". - -I reached this conjecture by considering the inner angles of $P$. The followings are what I've thought : Every inner angle, say $\alpha$, of $P$ has to satisfy $2m\alpha=360^{\circ}$ where $m\ge 2\in\mathbb N$. Hence, $\alpha$ has to be any of the positive divisors of $180$ except $180$. This leads $n\le 4$ and so on. -Then, here is the second question. - -Question 2 : Letting $P$ be a convex polyhedron, how about the three dimensional version of this question? Suppose that we consider the plane symmetry instead of the line symmetry. - -In the two dimensional version, I think we can consider the inner angles of $P$. However, I don't have any good idea for the higher dimensional version. Can anyone help? - -REPLY [2 votes]: This is in response to questions posed in the comments, but is too long for a comment. -The use of the reflection trick in mathematical billiards, and detailed results for regular billiards are contained in - -S. Tabachnikov Geometry and billiards (AMS, 2005). - -including the classification of regular 2D billiards already given in another answer here. The connection between higher dimensional billiards and Coxeter groups was made in - -A. Y. Plakhov and A. M. Stepin Scattering and Coxeter groups Russ. Math. Surv. 53 401-402 (1998). - -Both conditions 1 and 2 are needed for regular polytope billiards. The red and blue polytopes then correspond to regions where the mapping back to the original space is a translation or a reflection. The condition that angles in a polygon are of the form $\pi/n$ is exactly what is needed for corners to be regular, in other words, parallel billiard paths that are close but pass on opposite sides of a corner remain close after reflections near the corner. The example of a hexagon, which tesselates the plane but does not satisfy condition 2, has corners not of this form and so parallel billiard paths on opposite sides of the corner emerge at different angles. I am not aware of results which distinguish this case from the very popular more general case of billiards with rational angles. These are described by translation surfaces of higher genus, for example see - -V. Delecroix, P. Hubert and S. Lelievre Diffusion for the periodic wind-tree model arxiv:1107.1810<|endoftext|> -TITLE: Strength of some claims about finitely additive measures on infinite sets? -QUESTION [6 upvotes]: Assume ZF. Consider the claim: -(1) For any infinite set $\Omega$, there is a finitely additive probability measure $\mu:2^\Omega\to[0,1]$ with $\mu(A) = 0$ whenever $|A|<|\Omega|$. -Then (1) is implied by Hahn-Banach and the claim that the union of two sets of cardinality less than $|\Omega|$ has cardinality less than $|\Omega|$ when $\Omega$ is infinite. (Let $\scr B$ be the boolean algebra $2^\Omega$ and let $\scr I$ be the ideal of all subsets of $\Omega$ of smaller cardinality. HB is equivalent to the claim that every boolean algebra has a f.a. probability measure, so $\scr B/I$ has a f.a. probability measure, which lifts to a f.a. probability measure on $\scr B$ with the requisite property.) -[Deleted Question 1 as it was predicated on something false.] -Question 2: Is anything known about the strength of (1)? -A related claim is: -(2) Every set has a f.a. probability measure on its powerset which is not supported on any set of smaller cardinality (i.e., if $|A|<|\Omega|$ then $P(A)<1$). -Clearly (1) implies (2) (at least in the infinite case, but the finite case of (2) is trivial). -Question 3: Does (2) imply (1)? -Updates: The answers below so far show: (1) is equivalent to AC. ZF+DC is not strong enough to show (2). It's still open in this discussion whether (2) is equivalent to AC (I doubt it). - -REPLY [3 votes]: A quick note that (2) doesn't follow from ZF (assuming ZF is consistent) or even ZF+DC. -(2)+Countable Choice for Finite Sets (CC(fin)) implies that every uncountable set has a non-principal (finitely additive, non-trivial) measure. Let $\Omega$ be uncountable. Let $\mu$ be as in (2). Let $A = \{ x : \mu(\{ x \}) > 0 \}$. The standard proof that a finite measure has only countable many atoms only uses CC(fin) (there are at most $n$ elements $x$ such that $\mu(\{ x \}) \in [1/n, 1/(n-1))$). So $A$ is countable. So $\mu(A)<1$. Then $\nu(B) = \mu(B-A)$ defines a non-principal measure and is non-trivial since $\nu(\Omega)=1-\mu(A)>0$. -But Blass gave a model of ZF+DC with no non-principal measures ("A model without ultrafilters", Bull. Acad. Polon. Sci. 25 (1977), 329–331; I haven't read the paper, but Pincus and Solovay say that it proves this). Since DC implies CC(fin), (2) is false in that model. -A different way to see that (2) doesn't follow from ZF is this. Take a model where there is an infinite Dedekind-finite set but CC(fin) still holds (the Consequences of AC website lists a bunch of models where Form 10 is true and Form 9 is false). -Now if $\Omega$ is an infinite Dedekind-finite set and $\mu$ satisfies (2), then every point of $\Omega$ will have to be an atom of $\mu$ (or else $\mu(\Omega - \{ x \}) = \mu(\Omega)$). But by CC(fin), the set of atoms is countable, which is impossible for a Dedekind-finite set. -I'm curious if (2) implies CC(fin).<|endoftext|> -TITLE: On the obstruction of a sequence of simplicial spaces that is levelwise a fibration -QUESTION [5 upvotes]: Given a sequence of simplicial spaces (actually bisimplicial sets) -$$F\to E\to B$$ -that is level-wise a fibration, then the geometric realisation does not necessarily have to be a fibration. -If I understand it correctly then a sufficient criterion is that all spaces are group-complete H-spaces and that the map of simplicial sets $\pi_0(E)\to \pi_0(B)$ is surjective. -Now consider a map of simplicial group complete H-spaces $f:X\to Y$, the canonical maps $p_X:X\to \pi_0(X)$, $p_Y:Y\to \pi_0(Y)$ and the induced map $g:\pi_0X\to \pi_0 Y$. -Let me denote by $\operatorname{hofib} |f|$ the homotopy fibre of the map $f$ and by $|\operatorname{hofib} (f)|$ the geometric realisation of the level-wise homotopy fibres, similarly for $g$. -Then I have induced maps $F: |\operatorname{hofib} (f)|\to \operatorname{hofib} |f|$ and $G:|\operatorname{hofib} (g)|\to \operatorname{hofib} |g|$. -Now $\operatorname{hofib}(F)$ and $\operatorname{hofib}(G)$ measure the defect how far the geometric realisation of the level-wise fibres is away from the actual fibre. By the sufficient condition in the beginning this obstruction should be completely captured by the behaviour on the $\pi_0$. But by construction this is the same for $f$ and $g$. Can I somehow conclude that -$$\operatorname{hofib} (F)\simeq \operatorname{hofib} (G)?$$ -Edit: The space $|\operatorname{hofib}(g)|$ as written above doesn't really make sense. In the end I am talking about maps of sets here and over every connected components (i.e. every point) the fibre will be just the preimage of this point. What really should be there is $|\pi_0 \operatorname{hofib}(f)|$. I am not sure if this makes the question easier or harder. - -REPLY [2 votes]: Working with spaces meaning spaces, I gave a reasonably strong result along these lines in -Theorem 12.7 of The Geometry of Iterated Loop Spaces http://www.math.uchicago.edu/~may/BOOKS/geom_iter.pdf. -If $p\colon E\to B$ is a simplicial Hurewicz fibration such that $B$ is proper (Reedy cofibrant in more modern terminology) and each $B_q$ is connected, then $|p|\colon |E|\to |B|$ is a quasifibration with fiber $|F|$.<|endoftext|> -TITLE: Does Gromov's Waist Inequality imply Borsuk-Ulam? -QUESTION [15 upvotes]: I'm curious if anyone can see a route to get the Borsuk-Ulam theorem from Gromov's waist inequality. For the sake of notation, here's the inequality: -Let $S^n$ denote the round unit sphere in $\mathbb{R}^{n+1}$. It carries the spherical measure which we denote $vol(\cdot)$ and a Riemannian metric which induces a distance function $d(\cdot,\cdot)$. We write $S^{n-k}$ for the equatorial $(n-k)$-sphere inside $S^n$: - $$ S^{n-k} = \{ (x_1, \dots, x_{n+1}) \in S^n : x_i = 0, i>n-k+1\} $$ - For a continuous map $f : S^n \rightarrow \mathbb{R}^k$ we write $S_z = \{s \in S^n : f(s) = z\}$ for the level sets of $f$. Write $U_\epsilon(S) = \{\sigma \in S^n : d(\sigma, S) < \epsilon\}$ for an $\epsilon$-neighbourhood of a subset $S$. - -Theorem (Gromov): If $f : S^n \rightarrow \mathbb{R}^k$ is continuous then there is a point $z \in \mathbb{R}^k$ such that: $$ vol(U_\epsilon(S_z)) \geq vol(U_\epsilon(S^{n-k})) $$ for all $\epsilon > 0$. - -This inequality is quite strong. It implies the spherical isoperimetric inequality. It also implies topological invariance of domain. For more background, see Guth's survey The Waist Inequality in Gromov's Work. -Getting back to my question: If $n=k$ and the level set given by the inequality contains two points then taking $\epsilon = \pi/2$ gives the Borsuk-Ulam theorem. Can one arrange for this to always happen and hence derive Borsuk-Ulam? - -REPLY [14 votes]: Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started. -Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that -$$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ -It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. -Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. -This gives the following (weaker) version of a Bursuk Ulam type result: - -If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$. - - $U_{\pi/2}$ for two points which are not antipodal. -In the original paper Gromov is more careful. He writes: - -If $k=n$, and $\operatorname{card}(f^{-1}(z))\leq 2$, $z\in\mathbb{R}^n$, then [the waist of the sphere theorem] applied to $\pi/2$ amounts to the Borsuk–Ulam theorem: some level $f^{-1}(z)$ of $f\;:S^n\to \mathbb{R}^k$ equals a pair of opposite points. - -Of course $f^{-1}(z)$ is just a different notation of $S_Z$. So the important detail is: "and $\operatorname{card}(f^{-1}(z))\leq 2$". -In general $\operatorname{card}(f^{-1}(z))\leq 2$ is not true for arbitrary maps. And even if you perturbe $f$ a little bit this condition won't be fullfilled. Of course any two maps from $S^n$ to $\mathbb{R}^n$ are homotopic, so in particular $f$ will be homotopic to a map with $\operatorname{card}(f^{-1}(z))\leq 2$ for all $z$, but this doesn't really give you anything. -To summarize the answer is: Gromov's Waist Inequality almost implies Borsuk-Ulam.<|endoftext|> -TITLE: Chevalley Groups over an arbitrary ring. -QUESTION [6 upvotes]: My question is simply about the Chevalley groups over rings. In many books, including Carter's book on "Simple groups of Lie types", the groups are considered over fields. I have checked the computations and I noticed that the computations works over any commutative $\mathbb{Z}$-algebra. Why these groups are not introduced in the general format? Of course if these groups are defined over rings then they are not necessarily simple. Is it the reason why most people want to define them over fields? -Just to give a general definition of Chevalley groups over rings let me review the construction. -Let $L$ be a finite dimension complex simple Lie algebra with a Chevalley basis -$$\{ e_r: r\in\Pi; e_r: r\in\Phi \},$$ -where $\Pi$ is basis for the root system $\Phi$. Pick $\zeta\in\mathbb{C}$. Therefore -$$ -\exp(\zeta ad_{e_r}), -$$ -is an Lie-algebra automorphism of $L$. One can observe that the entries of the matrix associated to $\exp(\zeta ad_{e_r})$, denoted by $A_r(\zeta)$, with respect to a Chevalley basis, are of the form $a\zeta^l$ where $a\in\mathbb{Z}$ and $l\in\mathbb{Z}_{\geq 0}$. Now let $B$ be a commutative $Z$ algebra with the structure map $\rho: \mathbb{Z}\to B$. Pick $b\in B$, then we consider the matrix $\overline{A_r(b)}$ which is obtained by transforming the entry of $a\zeta^l$ of the matrix $A_r(\zeta)$ into $\rho(a)b^l$. Then One can consider the linear transformation, denoted by $\overline{x_r(b)}$, obtained by $\overline{A_r(b)}$ on $L(B):=L(\mathbb{Z})\otimes_{\mathbb{Z}}B$. One can show that $\overline{x_r(b)}$ is indeed a Lie-algebra automorphism of $L(B)$. Then the Chevalley group $G_{ad}(B,\Phi)$, is defined by the subgroup of $GL(L(B))$ generated by $\overline{x_r(b)}$. -Is there a problem in this construction that I am not considering? -Many thanks for your answers. - -REPLY [6 votes]: A couple of further clarifications, to supplement the extensive answer by marguax and the many comments: - -Chevalley's influential 1955 paper was mainly concerned with finding a uniform approach to most of the known simple finite groups of Lie type (supplemented soon afterward by the introduction of twisted groups as well as the groups of Suszuki and Ree). The original "Chevalley group" was defined as a group of automorphisms of a certain Lie algebra, generated by unipotent elements; this group is simple over almost all fields (except a few very small ones). The Lie algebra here is obtained by first reducing mod $p$ a Chevalley $\mathbb{Z}$-basis of a simple Lie algebra over $\mathbb{C}$, then tensoring with a field of prime characteristic $p$. - -A subtle point here is that the "Chevalley Lie algebra" obtained over an algebraically closed field is actually the Lie algebra of the corresponding simply connected algebraic group, whereas the Chevalley groups themselves are closer to the adjoint groups. -Anyway, Steinberg in his 1967-68 Yale lectures broadened the notion of "Chevalley group" by using other faithful Lie algebra representations, to include special linear groups and the like. - -In fact, there is a huge amount of literature about Chevalley groups over (commutative) rings. Here the ideal structure of the ring contributes to normal subgroup structure in the group, so the groups are typically non-simple. But they do come up naturally in algebraic K-theory, including the study of the congruence subgroup problem. N.A. Vavilov and many others have written extensively about such groups. There is less textbook literature along these lines, except for a few books on algebraic K-theory including one by Hahn and O'Meara. And the entire subject becomes quite technical. - -ADDED: Concerning the more sophisticated viewpoint of Chevalley-Demazure group schemes, there is an important recent paper by Lusztig (not yet freely available online) Study of a $\mathbf{Z}$-form of the coordinate ring of a reductive group. The arXiv preprint is here.<|endoftext|> -TITLE: Fundamental group of the moduli stack of ordinary generalized elliptic curves -QUESTION [6 upvotes]: Let $M$ be the moduli stack of ordinary but possibly nodal elliptic curves over the field $\overline{\mathbf{F}_p}$. Then $M$ has a $\mathbb{Z}_p^{\times}$-torsor over it, given by the moduli scheme of "trivialized" elliptic curves: that is, elliptic curves equipped with an isomorphism between the formal group and $\widehat{\mathbf{G}_m}$. A theorem of Igusa states that this cover (or rather, inverse system of covers) is connected. -Away from $p$-primary information, is this the universal cover of $M$, or is there an additional portion of the fundamental group? (In other words: what is the fundamental group of Katz's ring of $p$-adic modular forms?) - -REPLY [5 votes]: For $p>13$, there are at least two supersingular $j$ invariants, say $a$ and $b$, and adjoining $\sqrt[N]{\frac{j-a}{j-b}}$ is always an etale cover for $N$ prime to $p$. This gives an additional portion of the fundamental group. -To compute the full fundamental group, we can first take an etale cover that kills the elliptic points. To do this, if $a \neq 1728$ is a supersingular $j$ invariant, adjoin $\sqrt{ \frac{j-a}{j-1728}}$ and if $a \neq 0$ is a supersingular $j$ invariant, adjoin $\sqrt[3]{\frac{j-a}{j}}$. If $1728$ or $0$ is supersingular, just ignore that steps. -Now our stack is basically just a scheme - an algebraic curve minus finitely many points. We can compute its fundamental group (or the prime-to-$p$ part?) the usual way. Then there is an extra central extension by $\mathbb Z/2$, coming from the automorphism $y \to -y$ shared by all elliptic curves.<|endoftext|> -TITLE: Localizations of hereditary rings -QUESTION [5 upvotes]: It is known that if a commutative Noetherian ring $R$ is hereditary then for any maximal ideal $M$ the localization $R_M$ is also hereditary. Is the Noetherian assumption necessary? - -REPLY [5 votes]: You do not need to suppose your ring to be noetherian. In fact, the following stronger statement is true: - -If $R$ is a hereditary commutative ring and $S\subseteq R$ is a subset, then the ring $S^{-1}R$ is hereditary. - -In order to prove this we use the following two facts: -1) A commutative ring $R$ is hereditary if and only if any epimorphism of $R$-modules with injective source has injective target (i.e., quotients of injectives are injective). (See e.g. T.Y. Lam, Lectures on modules and rings, Theorem 3.22.) -2) If $R$ is a commutative ring, $S\subseteq R$ is a subset an $M$ is an $S^{-1}R$-module, then $M$ is injective if and only if it is so considered as an $R$-module by means of scalar restriction. (See e.g. M.P. Brodmann, R.Y. Sharp, Local cohomology, Lemma 10.1.12 (where the noetherian hypothesis is not used), or E.C. Dade, Localization of injective modules, J. Algebra 69 (1981), 416--425.) -Now, suppose that $R$ is a hereditary commutative ring and consider a subset $S\subseteq R$. Let $M\rightarrow N$ be an epimorphism of $S^{-1}R$-modules with injective source. By means of scalar restriction to $R$ we can consider this as an epimorphism $M\rightarrow N$ of $R$-modules with injective source by 2). Hence, its target is injective by 1), and therefore the $S^{-1}R$-module $N$ is injective by 2). It follows that $S^{-1}R$ is hereditary.<|endoftext|> -TITLE: Is there a function defined on real numbers which is continuous from the left, but not from the right, everywhere -QUESTION [23 upvotes]: I am teaching Mathematical analysis. A student asked this question. I think this is a good question, but don't know the answer. - -REPLY [14 votes]: Let $\omega(f,A):= \sup \{|f(x)-f(y)| : x,y \in A\}$ be the oscillation of $f$ on $A$, and let $O_n$ be the union of all open sets $A$ s.t. $\omega(f,A) < 1/n$. Then $O_n$ is open and its complement is countable because every uncountable closed set contains a point that is a limit point from the left of the set, while for every real $x$ there is a $y -TITLE: What is a good introduction to branching rules in representation theory? -QUESTION [12 upvotes]: I'm looking for a book or introductory article, that explains branching rules in representation theory of real Lie groups. -When a Lie group has a set of irreducible representations, I'd like to know how these representations decompose into irreducible representations of a subgroup. -I heard of "Symmetry, representations, and invariants" by Goodman and Wallach and "Representation Theory" by Fulton and Harris, but I couldn't get an account on the special cases I'm interested in, which are $U(1) \to SU(2)$ and $SO(4) \to SO(5)$. I know that $U(1) \cong \operatorname{Spin}(2, \mathbb{R})$ and $SU(2) \cong \operatorname{Spin}(3, \mathbb{R})$, but Goodman/Wallach and Fulton/Harris only seem to treat $\operatorname{Spin}(n, \mathbb{C})$ and $SO(n, \mathbb{C})$. - -REPLY [2 votes]: T. Kobayashi (recently joint with Birgit Speh), has written many papers giving details about branching for classical groups/algebras. Just "google" "T. Kobayashi, branching laws". Much of it is on-line.<|endoftext|> -TITLE: Fantastic properties of Z/2Z -QUESTION [53 upvotes]: Recently I gave a lecture to master's students about some nice properties of the group with two elements $\mathbb{Z}/2\mathbb{Z}$. Typically, I wanted to present simple, natural situations where the only group satisfying the given constraints is $\mathbb{Z}/2\mathbb{Z}$ (also $\mathbb{Z}/2\mathbb{Z}$ as a ring or as a field could qualify, but I'd prefer to stick to the group if possible). Here are some examples of theorems that I proved to the students : - -Let $G$ be a nontrivial group with trivial automorphism group. Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. -Let $G$ be a nontrivial quotient of the symmetric group on $n>4$ letters (nontrivial meaning here different from 1 and the symmetric group itself). Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. -Let $k$ be an algebraically closed field and let $k_0$ be a subfield such that $k/k_0$ is finite. Then $k/k_0$ is Galois and $G=\text{Gal}(k/k_0)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. (Moreover $k$ has characteristic $0$ and $k=k_0(i)$ where $i^2=-1$.) This is a theorem of Emil Artin and I actually did not prove it because my students did not have enough background in field theory. -Let $k$ be a field with the following property: there exists a $k$-vector space $E$ of finite dimension $n>1$ and an isomorphism $E\simeq E^*$ between the space and its linear dual which does not depend on the choice of a basis, i.e. is invariant under $\text{GL}(E)$. Then $k=\mathbb{Z}/2\mathbb{Z}$, $n=2$ and the isomorphism $E\simeq E^*$ corresponds to the nondegenerate bilinear form given by the determinant. - -I am looking for some more fantastic apparitions of $\mathbb{Z}/2\mathbb{Z}$. Do you know some? - -REPLY [7 votes]: The automorphism group of the category of categories is $\mathbb{Z}/2$. -That is, the group of invertible functors $F: \mathrm{Cat} \to \mathrm{Cat}$ is $\mathbb{Z}/2$.<|endoftext|> -TITLE: Quasi-reflexive spaces which are not isometric to dual spaces -QUESTION [6 upvotes]: My question may sound weird and I have no deep motivation behind it other than curiosity. -As is well-known, quasi-reflexive spaces have the Radon-Nikodym property hence their balls have lots of extreme points (they even have the so-called Krein-Milman property). However, can one give me an example of a quasi-reflexive space which is not isometric to a dual space? Of course, every quasireflexive space is isomorphic to a dual space. -I suspect that a clever renorming of the James space should do the job. - -REPLY [6 votes]: Every non-reflexive Banach space can be equivalently renormed so as not to be isometrically isomorphic to a dual space. -$$ -$$ -Davis, William J.; Johnson, William B. -A renorming of nonreflexive Banach spaces. -Proc. Amer. Math. Soc. 37 (1973), 486–488.<|endoftext|> -TITLE: Permuting collinear points on a curve -QUESTION [14 upvotes]: Let $C \subset {\bf CP}^2$ be an irreducible algebraic smooth (projectively) planar curve over the complex numbers of degree $d$ (we allow finitely many points to be deleted from $C$ to make it smooth). Then a generic complex line in ${\bf CP}^2$ intersects $C$ in $d$ distinct points $z_1,\ldots,z_d$. - -Question 1: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, is it possible to continuously deform the $d$ points $z_1,\ldots,z_d$ to $z_{\sigma(1)},\ldots,z_{\sigma(d)}$ while keeping all the points collinear, distinct, and on $C$ at all times? In other words, does there exist continuous maps $z_i: [0,1] \to C$ with $z_i(0) = z_i$, $z_i(1) = z_{\sigma(i)}$, and the $z_1(t),\ldots,z_d(t)$ distinct and collinear for all $t\in [0,1]$? - -I believe the answer to this question is yes, because one can transpose any two of the $z_i,z_j$ while keeping the other $z_k$ unchanged by bringing them close together near a generic point of $C$ (and making the collinear line close to the line of tangency of this point to $C$, which generically will not be tangent to anywhere else in $C$), performing the transposition, and then returning back to the original position. One can also phrase the question as follows: if we let $S \subset C \times Gr(2,1)$ be the set of incidences $(p,\ell)$ with $p \in C$ and $\ell \in Gr(2,1)$ a line through $p$, then (after passing to an open dense subset of $Gr(2,1)$), $S$ is a covering space over (most of) $Gr(2,1)$ whose fibre has $d$ points, and the question asserts that the fundamental group of the base acts completely transitively on the fibre (in that every permutation of the fibre shows up); equivalently, the question asserts that the $d^{th}$ exterior power of $S$ over $Gr(2,1)$ is irreducible. -However, I'm having more trouble with proving the following generalisation of Question 1. Keep the same setup as before, but now suppose we also have another irreducible curve $C'$ covering $C$ with fibres of cardinality $d'$. Thus, above a generic $z_i$ in $C$ we have $d'$ points $w_{i,1},\ldots,w_{i,d'}$ in $C'$. - -Question 2: for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_d,k_1,\ldots,k_d \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d$ points $w_{1,j_1},\ldots,w_{d,j_d}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d),k_d}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct? - -Like Question 1, one can phrase this question in terms of the transitivity properties of the action of the fundamental group of a suitable open dense subset of $Gr(2,1)$ on some fibre, but this formulation becomes rather complicated to state and I won't give it here. -I believe I can resolve this problem in the affirmative if $C$ has a smooth closure, in which case I can use ad hoc arguments to move around each of the $j_i$ independently, but encountered a problem if $C$ has a singular point in its closure, and $C'$ ramifies above this point, as then the points $z_i$ appear to become entangled with each other near this point in a manner that I was not able to analyse by ad hoc topological arguments. -[My motivation for this problem was to understand arcs in a finite plane ${\bf F}_p^2$, that is to say sets that do not contain any collinear triples, and to classify when these arcs can be constructed as the ${\bf F}_p$-points of a low-degree algebraic curve $C$ (or as the projection of the ${\bf F}_p$-points of a low-degree algebraic curve $C'$). Using a Lefschetz principle argument, one can transfer the problem (in the large $p$ limit) to a problem in the complex plane related to the questions above.] - -REPLY [3 votes]: While both of my original questions have been answered by David, I wanted to record the fact that a slightly weaker version of Question 2 is in fact true (and good enough for my particular application), in which one allows one of the points to move freely: - -Question 2': for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j_1,\ldots,j_{d-1},k_1,\ldots,k_{d-1} \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d-1$ points $w_{1,j_1},\ldots,w_{d-1,j_{d-1}}$ to $w_{\sigma(1),k_1},\ldots,w_{\sigma(d-1),k_{d-1}}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct? - -We sketch the proof as follows. Let $V \subset C^d$ be the set of all $k$-tuples of distinct collinear points $(z_1,\ldots,z_d)$ in $C$; by the positive solution to Q1, this is irreducible. The embedding $V \subset C^d$ induces a homomorphism $\pi_1(V) \to \pi_1(C)^d$; by abstract nonsense, Q2' follows from (and is basically equivalent to) the assertion that the resulting image of $\pi_1(V)$ in $\pi_1(C)^d$ projects surjectively onto $\pi_1(C)^{d-1}$. -Now, since any two points determine a line, $C^2$ lifts to $V$, which implies that the projection of $\pi_1(V)$ to $\pi(C)^2$ is surjective. Also, the image of $\pi_1(V)$ in $\pi_1(C)^d$ is symmetric with respect to permutations of the $d$ coordinates. This is enough to imply the surjectivity onto $\pi_1(C)^{d-1}$ by elementary group theory (this is a trick I learned from a paper of Furstenberg and Weiss). Indeed, for any $g \in \pi_1(C)$, the surjectivity onto the final $\pi_1(C)^2$ implies that the image of $\pi_1(V)$ in $\pi_1(C)^d$ contains an element of the form $(g_1,\ldots,g_{d-1},g,1)$ for some $g_1,\ldots,g_{d-1}$; by symmetry it also contains $(g_1,\ldots,g_{d-1},1,g)$,and by dividing it thus contains $(1,\ldots,1,g,g^{-1})$. Thus the projection onto $\pi_1(C)^{d-1}$ contains $(1,\ldots,1,g)$; permuting and composing we obtain the required surjectivity.<|endoftext|> -TITLE: Which fields have multiplicative group isomorphic to additive group times Z/2Z? -QUESTION [20 upvotes]: Let $K$ be a field, $K_{*}$ its multiplicative group and $K_{+}$ its additive group. As Richard Stanley notes in this answer, the only field for which $K_{+} \simeq K_{*} \times \mathbb{Z}/2\mathbb{Z}$ is $K=\mathbb{Z}/2\mathbb{Z}$. Which fields have $K_{*} \simeq K_{+} \times \mathbb{Z}/2\mathbb{Z}$? For instance, $K=\mathbb{R}$ is an example. - -REPLY [5 votes]: Here's another unorderable example. -Consider nonstandard models; let $p$ be an infinite prime such that $(p-1)/2$ is relatively prime to every standard prime, and let $F$ be the internal finite field $\mathbf{F}_p$. The additive group is cyclic of order $p$ and the multiplicative group is the product of $\mathbb{Z} / 2 \mathbb{Z}$ and the cyclic group of order $(p-1)/2$. -By the four square theorem, this field has elements $a,b,c,d$ such that $a^2 + b^2 + c^2 + d^2 \equiv -1 \pmod{p}$. -Viewed externally, the additive group and the multiplicative group of squares are both uniquely divisible abelian groups of the same cardinality, and assuming we can arrange for the cardinality to be non-pathological, we conclude that $F_+ \cong F_\times \times \mathbb{Z} / 2 \mathbb{Z}$. -Furthermore, since $-1$ is a sum of squares, $F$ can't be ordered.<|endoftext|> -TITLE: Quotients of $\ell_\infty$ by separable subspaces -QUESTION [13 upvotes]: Given a (closed) separable subspace $M$ of $\ell_\infty$, I am interested in conditions implying that the quotient $\ell_\infty/M$ is isomorphic to a subspace of $\ell_\infty$. -It is not difficult to see that being $M$ reflexive is sufficient, and Bourgain proved that $\ell_\infty/c_0$ does not admit an equivalent strictly convex norm (while $\ell_\infty$ does). So $\ell_\infty/c_0$ is not isomorphic to a subspace of $\ell_\infty$. - -REPLY [14 votes]: While I do not have a complete answer to the OP’s question, I made enough observations that I think it is worthwhile to record them as an answer. -(1) If $X$ and $Y$ are isomorphic (closed) subspaces of $\ell_\infty$, then $\ell_\infty/X$ embeds into $\ell_\infty$ iff $\ell_\infty/Y$ embeds into $\ell_\infty$. -Indeed, it is clear that $\ell_\infty/X$ embeds into $\ell_\infty$ iff -$(\ell_\infty \oplus \ell_\infty)/(X \oplus \{0\})$ embeds into $\ell_\infty$, so without loss of generality we can assume that $\ell_\infty$ embeds into both $\ell_\infty/X$ and into $\ell_\infty/Y$. We then get from [LR, [Theorem 3(i)] that every isomorphism from $X$ onto $Y$ extends to an automorphism of $\ell_\infty$. -(2) Suppose that $Y$ is a subspace of $\ell_\infty$ and $X$ is a subspace of $Y$. Assume that $\ell_\infty/Y$ embeds into $\ell_\infty$. Then $\ell_\infty/X$ embeds into $\ell_\infty$ iff $Y/X$ embeds into $\ell_\infty$. -The “only if” part is clear because $Y/X$ embeds into $\ell_\infty/X$. The other direction is an easy consequence of Lindentrauss’ observation that if $U\subset V$ and both $U$ and $V/U$ embed into $\ell_\infty$, then $V$ embeds into $\ell_\infty$. To prove this observation, let $T$ be an isomorphism from $U$ into $\ell_\infty$ and $S$ an isomorphism from $V/U$ into $\ell_\infty$. The space $\ell_\infty$ is $1$-injective (this is immediate from the Hahn-Banach theorem), so $T$ extends to a bounded linear mapping (which we also denote by $T$) from $V$ into $\ell_\infty$. Denoting the quotient map from $V$ to $V/U$ by $Q$, we see that $T \oplus SQ: x\mapsto (Tx, SQx)$ defines an isomorphism from $V$ into $\ell_\infty \oplus_\infty \ell_\infty \equiv \ell_\infty$. -(3) Suppose that $X$ is a subspace of of $\ell_\infty$ and $X$ is isomorphic to $Y^*$ for some separable $Y$. Then $\ell_\infty/X$ embeds into $\ell_\infty$. -Indeed, $Y$, being separable, is isomorphic to the quotient space $\ell_1/W$ for some subspace $W$ of $\ell_1$, and $W^\perp$ in $\ell_\infty$ is isomorphic to $X$. But $\ell_\infty/{W^\perp}$ is isometric to $W^*$, which embeds into $\ell_\infty$ because $W$, being separable, is a quotient of $\ell_1$. Hence by (1), the space $\ell_\infty/X$ also embeds into $\ell_\infty$. -(4) Suppose that $X$ a subspace of of $\ell_\infty$ and $X$ embeds into a separable conjugate space. Then $\ell_\infty/X$ embeds into $\ell_\infty$. -This follows from (3), (2), and the fact that every separable space embeds into $\ell_\infty$. -(5) $\ell_\infty/{c_0}$ does not embed into $\ell_\infty$. Hence if $X$ is a separable subspace of of $\ell_\infty$ and $X$ contains a subspace isomorphic to $c_0$, then $\ell_\infty/X$ does not embed into $\ell_\infty$. -The second statement follows from the first and (1), (2). The first statement is another observation that I think is due to Lindenstrauss; namely, that $\ell_\infty/{c_0}$ contains an isometric copy of $c_0(\Gamma)$ with $|\Gamma|=2^{\aleph_0}$ (consider the image in $\ell_\infty/{c_0}$ of characteristic functions of $2^{\aleph_0}$ infinite subsets of the natural numbers all of whose pairwise intersections are finite). The space $c_0(\Gamma)$ does not embed into $\ell_\infty$ because the unit vector basis for the space is a non separable set that has weakly compact closure, and every weakly compact subset of the dual to a separable space is separable. -(6) If $X$ is a subspace of $\ell_\infty$ that is isomorphic to $L_1(0,1)$, then $\ell_\infty/X$ embeds into $\ell_\infty$. Hence if $X$ is a subspace of $\ell_\infty$ that embeds into $L_1(0,1)$, then $\ell_\infty/X$ embeds into $\ell_\infty$. -The second statement follows from the first by what is now standard reasoning. For the first statement, note that $L_1(0,1)$ embeds into $C[0,1]^*$ as a complemented subspace, and by (3), if $Y$ is a subspace of $\ell_\infty$ isomorphic to $C[0,1]^*$, then $\ell_\infty/Y$ embeds into $\ell_\infty$. Now use (2) and (1). -[LR] J. Lindenstrauss and H. P. Rosenthal, Automorphisms in $c_0$, $\ell_1$, and $m$, Israel J. Math. 7 (1969), 227–239.<|endoftext|> -TITLE: Must a weak homotopy equivalence induce an isomorphism between stable homotopy groups? -QUESTION [15 upvotes]: I'm confused by the following question: -$f:X\to Y$ is a weak homotopy equivalence, that is $f_*:\pi_*(X)\to \pi_*(Y)$ is an isomorphism for any dimensional homotopy groups. However, for the stable homotopy groups, is the homomorphism $f_*:\pi_*^s(X)\to \pi_*^s(Y)$ still an isomorphism? -Any comments are welcome! Many Thanks! - -REPLY [10 votes]: As Peter points out, stable homotopy groups are usually defined using the reduced suspension, which requires $X$ and $Y$ to be based. Let's talk about both situations. -As in the comment above, let $Y$ be the subspace $\{1,1/2,1/3,1/4,\ldots,0\}$ of $\mathbb{R}$, and let $X$ be $\mathbb{N}$ with the discrete topology. There is a weak equivalence $X \to Y$. -Let's take 0 to be the basepoint and start taking suspensions. This map becomes a map from a countable wedge of spheres to a higher-dimensional Hawaiian earring that features in a famous paper of Barratt-Milnor (you can prove that this is homeomorphic by the standard "bijection from compact to Hausdorff" argument). -The paper "Homotopy and homology groups of the $n$-dimensional Hawaiian earring" by Eda and Kawamura essentially shows that for $n > 1$, the map $X \to Y$, on $\pi_n$, becomes the embedding $\oplus \mathbb{Z} \to \prod \mathbb{Z}$ from a countable direct sum to a countable product. Therefore, the map on stable homotopy groups is not an isomorphism. (This would be some very small manifestation of Tom Goodwillie's result from the comments.) -However, you can also define stable homotopy groups using iterated unreduced suspension (the groups only become well-defined after a couple of suspensions due to basepoint issues). The unreduced suspension is more homotopically well-behaved, and in particular preserves weak equivalence because it only collapses along cofibrations.<|endoftext|> -TITLE: About a solid which satisfies $\sum_{i=1}^{n}x_i=0, |x_i|\le1\ (i=1,2,\cdots,n)$ -QUESTION [9 upvotes]: For $n\ge 2\in\mathbb N$, let $S_n$ be the volume of a $(n-1)$ dimensional solid which satisfies -$$\sum_{i=1}^{n}x_i=0, |x_i|\le1\ (i=1,2,\cdots,n).$$ -Then, here is my question. - -Question : Can we represent $S_n$ by $n$ ? - -Remark : This question has been asked previously on math.SE without receiving any answers. -Motivation : I've been interested in this simple question. I've got the followings : -$$S_2=2\sqrt 2, S_3=3\sqrt 3, S_4=\frac{32}{3}.$$ -$S_4$ is the volume of a regular octahedron, whose edge length is $\sqrt 8$, which passes through the following six points : -$$(1,1,-1,-1),(1,-1,1,-1),(1,-1,-1,1),(-1,1,1,-1),(-1,1,-1,1),(-1,-1,1,1).$$ -However, I don't have any good idea for $n$ in general. Can anyone help? - -REPLY [5 votes]: Section 2 in the paper http://arxiv.org/abs/math/0503115 expresses this volume in terms of sinc integrals $$\sigma_n=\frac2\pi\int_0^\infty \left(\frac{\sin x}{x}\right)^n\,dx$$ -Namely, it is shown that the volume you are interested in is equal to $2^{n-1}\sqrt{n}\sigma_n$.<|endoftext|> -TITLE: Computing $L$-rank (constructible universe) -QUESTION [5 upvotes]: I posted this question on Math StackExchange but did not get a full answer. I hope it's not a problem if I ask again here. -Is there a way to compute explicitly the $L$−rank $\rho(\bigcup x)$ of $\bigcup x$ in terms of $\rho(x)$? I know it's necessarily $\rho(\bigcup x)\leq\rho(x)$ and that both $\rho(\bigcup x)<\rho(x)$ and $\rho(\bigcup x)=\rho(x)$ can hold, but what else can be said, maybe distinguishing between $\rho(x)$ limit ordinal and successor ordinal? For example, is it possible that equality holds if $\rho(x)$ is a successor ordinal? If so, could you give an example? - -REPLY [9 votes]: There is no way to compute explicitly $\rho(\bigcup x)$ in terms of $\rho(x)$, in any meaningful fashion: e.g. (working in $L$) for arbitrarily large countable $\alpha$ there are reals $x\subseteq \omega$ with $\rho(x)$ a limit or successor greater than $\alpha$ but with $\rho(\bigcup x) =\omega$ of course. (Because for arbitrarily large such $\alpha$ there is a new real $x$ definable over $L_\alpha$. If $\alpha$ is not a successor, then there will always be a new real definable over $L_{\alpha +1 } $ too; for example the real that codes $L_\alpha$; similarly if $\alpha$ is not a limit there will in any case be a new real definable over $L_{\alpha + \omega}$.) Thus the gap between the two ranks under either assumption can be as wide as it could conceivably be. -If $\rho(x)=\gamma+1$ then $\rho(\bigcup x)= \rho(x)$ can happen. Work in $L$: Suppose $x\subseteq \omega$ has $\rho(x)=\gamma +1$. Let $y = \{ \{u\} \mid u \in x \}$. Then $\rho(y)\leq\gamma +1 $; $\rho(\bigcup y) =\rho(x)$, but $\rho(y)\leq\gamma$ would imply that $\rho(x)\leq\gamma$.<|endoftext|> -TITLE: Measure theory in nuclear spaces -QUESTION [10 upvotes]: Much of the literature on measure theory in linear spaces focuses on the case of normed linear spaces (e.g., the outstanding book by Vakhania, or its sequel). However, nuclear linear spaces "as far from being normed as possible" [nLab], and I haven't been able to find a reference in this setting. -Is there a good reference on measure theory in nuclear spaces? - -REPLY [3 votes]: The fourth volume of I. M. Gel'fand's "Generalized Functions", subtitled "Applications of Harmonic Analysis" (written together with N. Ya. Vilenkin and published by Academic Press in 1964, now recently reissued by AMS-Chelsea) discusses this topic abstractly and in depth. It is, in fact, the standard reference for the subject as cited in other books such as J. Glimm's and A. Jaffe's "Quantum Physics - A Functional Integral Approach" (2nd. ed., Springer-Verlag, 1988) and L. Schwartz's book cited in alpha's answer. -It must be said, though, that cylinder set measures as they appear e.g. in the important Bochner-Minlos theorem are actually built over topological duals of nuclear locally convex vector spaces (lcvs).<|endoftext|> -TITLE: Algebra structure $Tor(A,A)$ -QUESTION [7 upvotes]: This is a question i asked on math.stackexchange but i didn't get any answer. -Let $A$ be algebra over commutative ring $k$ and $P_{\bullet}=(P_i,d_i)\rightarrow A$, $k$ projective resolution. Then we have obvious lift of multiplication $f:A\otimes A\rightarrow A$ to $F:P\otimes P \rightarrow P$. Of course there is no reason for $F$ to be associative so we can't claim that $P_{\bullet}$ has algebra structure. Consider the map -$Tor_*(A,A)\otimes Tor_*(A,A)=H_*(P_{\bullet}\otimes P_{\bullet})\otimes H_*(P_{\bullet}\otimes -P_{\bullet})\rightarrow H_*(P_{\bullet}\otimes P_{\bullet}\otimes P_{\bullet}\otimes P_{\bullet})\rightarrow H_*(P_{\bullet}\otimes P_{\bullet})=Tor_*(A,A)$ -I was wondering if this map gives graded algebra structure on $Tor_*(A,A)$. Do i need any assumptions? -Edit: -Can you give me references to any papers about those algebras? - -REPLY [5 votes]: More generally, if $A,B,C,D $ are $k$-algebras, there's a multiplication -$$Tor_m^k(A,B)\otimes_k Tor_n^k(C,D)\rightarrow Tor^k_{m+n}(A\otimes_k C,B\otimes_k D)\qquad(1)$$ -If you take $C=A$ and $B=D$, this becomes -$$Tor_m^k(A,B)\otimes_k Tor_n^k(A,B)\rightarrow Tor^k_{m+n}(A\otimes_A A,B\otimes_k B)$$ -and you can compose with the maps induced by multiplication on $A$ and $B$ to get -$$Tor_m^k(A,B)\otimes_k Tor_n^k(A,B)\rightarrow Tor_{m+n}^k(A,B)$$ -The resulting map is commutative up to a sign (namely $(-1)^{m+n}$). -To get (1), you can represent elements of $Tor_m^k(A,B)$ and $Tor_n^k(A,B)$ as cycles in he appropriate complexes, then tensor these cycles together to get a cycle in the total tensor product complex. -You can find the details, among other places, in Eisenbud's book on commutative algebra, where he mentions that a lot of work has been done on the structure of this algebra in the case where $k$ is a local ring and $A=B$ is the residue field. (Here's where I really wish you'd called your ring $R$ instead of $k$!)<|endoftext|> -TITLE: Order of magnitude of $\sum \frac{1}{\log{p}}$ -QUESTION [15 upvotes]: Question: What is the order of magnitude of the following sum? -$$ \sum_{\substack{p -TITLE: Fields aren't group objects in Ab, so what are they? -QUESTION [12 upvotes]: This might be a vague question, but I am troubled by the fact that fields do not admit a nifty categorical definition. An obvious attempt such a definition would be to say that fields are commutative groups in $\mathcal{Ab}$, using that rings are monoids. This fails because the tensor product isn't the categorical product and thus doesn't have diagonals. -Can this be fixed using a more general definition of a group object [1] -or can the definition of a field be relaxed in a sensible way? -Is this issue with definability related to why the category of fields is badly behaved? (eg. no limits / colimits, lack of a "free field") [2] - -REPLY [8 votes]: Perhaps it is better to think of fields as particularly nice local rings. Local rings can be defined internal to a topos, and indeed this is (EDIT: related to) why stalks of sheaves of rings field-valued functions (EDIT: on schemes) turn out to be local rings. As far as a category-theoretic view on fields, there are several ways to look at them, as discussed at the nLab entry field in the section 'constructive notions'; note that here one can essentially ignore the constructivism, and think of these as different ways to specify a field object in a category (with appropriate structure/properties). The categories of various types of internal fields can given by the categories of models for various limit-colimit sketches, with the different notions given by different sketches. Group and ring objects are given by models of Lawvere theories, which are sketches using only cones over discrete diagrams - too simple to give the full definition of a field.<|endoftext|> -TITLE: Why not a Roadmap for Homotopy Theory and Spectra? -QUESTION [40 upvotes]: MO has seen plenty of roadmap questions but oddly enough I haven't seen one for homotopy theory. As an algebraic geometer who's fond of derived categories I would like some guidance on how to build up some background on homotopy theory. -Does the analogue of Hartshorne exist? Are there any must-reads for stable homotopy theory and spectra? What would your advice for a beginning graduate student be? -Just to set a starting point, I would ask for suggestions for someone who's familiar with a good chunk of most concepts covered in Hatcher and some differential geometry, but not much more. -I guess in the back of my mind is trying to understand some of this "brave new algebraic geometry", ie geometry over an $E_\infty$-ring spectrum. - -REPLY [23 votes]: There have been several questions previously in this vein. This one asks for an advanced beginners book. The consensus seemed to be that it was difficult to find a one-size-fits-all text because people come in with such diverse backgrounds. Peter May's textbook A Concise Course in Algebraic Topology is probably the closest thing we've got. If you like that, then you can also read More concise algebraic topology by May and Ponto. I also recommend Davis and Kirk's Lecture Notes in Algebraic Topology. I think these would be a very reasonable place for a beginning grad student to start (assuming they'd already studied Allen Hatcher's book or something equivalent). -Another question asked for textbooks bridging the gap and got similar answers. Finally, there was a more specific question about a modern source for spectra and this has a host of useful answers. Again, Peter May and coauthors have written quite a bit on the subject, notably EKMM for S-modules, Mandell-May for Orthogonal Spectra, and MMSS for diagram spectra in general. Another great reference is Hovey-Shipley-Smith Symmetric Spectra. On the more modern side, there's Stefan Schwede's Symmetric Spectra Book Project. All these references contain phrasing in terms of model categories, which seem indispensible to modern homotopy theory. Good references are Hovey's book and Hirschhorn's book. -Since you mention that you're especially interested in $E_\infty$ ring spectra, let me also point out Peter May's survey article What precisely are $E_\infty$ ring spaces and $E_\infty$ ring spectra?<|endoftext|> -TITLE: Is the tree of large cardinals linear? -QUESTION [10 upvotes]: Kanamori in the introduction of his book "The Higher Infinite" says: -"The investigation of large cardinal hypotheses is indeed a mainstream of modern set theory, and they have been found to play a crucial role in the study of definable sets of reals, in particular their Lebesgue measurability. Although formulated at various stages in the development of set theory and with different incentives, the hypotheses were found to form a linear hierarchy reaching up to an inconsistent extension of motivating concepts." -My question simply is: -Question: Is the tree of large cardinals linear? Precisely: Are there four large cardinal axioms like $\text{A}, \text{B}, \text{C}, \text{D}$ such that at least one of the following diagrams be true? - -Note that in the current tree of large cardinals one can find some diamonds like above diagrams but we usually interpret this phenomenon as "unknown" situation not a possible "incomparablility" between large cardinal axioms. For example note to the following diagram: - -REPLY [10 votes]: Let me just add to Andres excellent answer that The $\Omega$-conjecture of Hugh Woodin implies that all large cardinals are wellordered under the relation "implies the consistency of". This is a major line of investigation in Inner Model Theory. I recommend Woodin's "Suitable extender Models" (the introduction provides motivation to this problem).<|endoftext|> -TITLE: Is every connected space equivalent to some B(Aut(X))? -QUESTION [14 upvotes]: Given a connected space $B$, is there always some space $X$ with $B \simeq \mathbf{B}(\mathrm{Aut}(X))$? -Here by space I mean simplicial set, by $\mathrm{Aut}(X)$ I mean the simplicial monoid of auto-equivalences of $X$ (not just strict automorphisms), and by $\mathbf{B}$ I mean the classifying space of this; but I’d also be interested in answers for other reasonable interpretations of these terms. -Edit: A little background — this simply arose out of curiosity, not out of any desired application. Most of the people I’ve mentioned it to have a strong first impulse that the answer should be “no”, but none of us have been able to substantiate this. - -REPLY [12 votes]: Here is the $1$-type case. I assume all spaces are of the homotopy type of CW. Let me write $haut(X)$ (resp. $haut_*(X)$) for the monoid of self-equivalences (resp. pointed ones) to avoid posible confusion with the group-theoretic notation. These spaces have the correct homotopy type by our assumption. -From All Groups are Outer Automorphism Groups of Simple Groups by Droste, -Giraudet and Göbel, for every discrete group $G$ we can write $G\cong out(S)$ where $S$ is simple. Recall the exact sequence $0\to Z(S)\to S\to aut(S)\to out(S)\to 0$. Since (thanks to Ricardo Andrade for pointing this) the $S$ appearing in the theorem is in fact non-abelian, $Z(S)=0$. -Looping down the universal fibration with fiber $BS$ we have the homotopy principal fiber sequence -$S\to haut_*(BS)\to haut(BS)$ and thus $haut_*(BS)//S\simeq haut(BS)$. But $haut_*(BS)\simeq aut(S)$ and $S\to aut(S)$ is injective so the homotopy quotient $haut_*(BS)//S$ is the (ordinary) quotient $aut(S)/S\cong out(S)\cong G$. -Hence, $G\simeq haut(BS)$.<|endoftext|> -TITLE: Checking if a matroid is binary(Detecting $U^2_4$ minor in a matroid) -QUESTION [5 upvotes]: I am wondering what is the (computationally) best way to tell if a matroid of size $n$ and rank $r$ is binary(or whether it has a $U^2_4$ minor) given either one of these: -1) An independence oracle -2) Its rank vector -3) A basis oracle -I want to implement $U^2_4$/binary detection(Most likely given a basis oracle which will be in form of the reverse-lexicographic encoding in [3]). -What I know(or what I think I know): -1) can't be solved by asking polynomial number of questions to the oracle which is due to Seymour[1]. -2) and 3) One can write a brute force implementation which would compute and check all minors of size 4. -Also, I considered following ideas: -a) For problem (3), given a matroid $M$ one can start with an $r\times r$ Identity matrix and assign its columns to elements of some basis say $B^*$. Now we expand this matrix by adding columns that correspond to $e\in E(M)\setminus B^*$. We now fill r entries in this new column with $0$ or $1$. For filling $i^{th}$ entry in the column, we ask the basis oracle if $(B^*\setminus i)\cup e$ is a basis. If it is, we put a $1$ in that entry, $0$ otherwise. When we are done constructing all $n-r$ columns this way we have the only binary matrix that correctly reflects if an $r$-subset S of $E(M)$ with $|S\cap B^*|=r-1$ is a basis or not. As for other $r$-subsets, we can test their rank in the matrix and ask basis oracle whether each of them is a basis. If the two results don't match for any $r$-subset, we conclude that the matroid is not binary representable. This way seems better than the brute-force 'compute and check all minors' in terms of complexity. -b) One can compute all $r-1$ flats of $M$ from bases(Using an algorithm again due to Seymour [2]). Then compute all $r-2$ flats. Then use scum theorem(i.e. check if any $r-2$ flat is contained in more than three $r-1$ flats). -EDIT -Some extra info: Here is what I am really trying to do: Much on the lines of [3] and [4] I am trying to enumerate(list) matroids, but only the binary ones using single element extensions(by simply not extending any non-binary matroids produced). This list(scalar binary codes) and more lists created after pairing of bits(vector binary codes) are then to be used for computer assisted achievability proofs of rate regions of multisource network coding problems. Matsumoto [3] was very generous to share his c++ code which I am parallelizing using OpenMPI(My C $U^2_4$ check which is currently brute-force seamlessly goes into this scheme). Since the binary matroids are so less in number[5] compared to all matroids, I am expecting to go much farther. I have to care about computational efficiency because there are millions of matroids to be tested just at ground set size 14. -Given all this, although I appreciate and have been following sage-matroid library, I am not much of a Python guy I am not very sure how sage's code will go into my OpenMPI-C++ code(well, I won't exclude the possibility of existence of some hack that will let me do this). Hence, what I am looking for is pointers to best algorithms around which Dr. Royle's answer does tell. -[1] Seymour, P. D.; Walton, P. N. (1981), "Detecting matroid minors", Journal of the London Mathematical Society, Second Series 23 (2): 193–203 -[2] P.D. Seymour, A Note on Hyperplane Generation, Journal of Combinatorial Theory, Series B, Volume 61, Issue 1, May 1994, Pages 88-91 -[3] Yoshitake Matsumoto, Sonoko Moriyama, Hiroshi Imai, and David Bremner. Matroid enumeration for incidence geometry. Discrete & Computational Geometry, 47(1):17–43, 2012 -[4] Dillon Mayhew and Gordon F. Royle. Matroids with nine elements. Journal of Combinatorial Theory, Series B, 98(2):415 – 431, 2008 -[5] Marcel Wild: The Asymptotic Number of Binary Codes and Binary Matroids. SIAM J. Discrete Math. 19(3): 691-699 (2005) - -REPLY [3 votes]: This is actually implemented (as well as a host of other features) in the latest version of Sage. This is the culmination of 2 and a half years of hard work by Stefan van Zwam and Rudi Pendavingh, together with help from Michael Welsh and Gordon Royle. See this page from the Matroid Union Blog to get started. For your particular question, it is easy to construct $U_{2,4}$ via the Sage command -Sage: N = matroids.Uniform(2,4) -To test if an input matroid M has an N-minor you can run the Sage command -Sage: M.has_minor(N) -As you can see, Stefan and Rudi have worked hard to make the syntax easy to understand. Of course this is a very generic approach, so I am not sure how optimal it will be. Feel free to contact Stefan if you have any questions, or (better yet) want to develop for the package (Sage is open-source).<|endoftext|> -TITLE: Self-dual surfaces in $\mathbb P^3$ with isolated singularities -QUESTION [5 upvotes]: I am aware of the following examples of normal surfaces in $\mathbb P^3$ that are projectively isomorphic to their dual varieties: - -the smooth quadric; -Kummer surfaces; -The surface with the equation $x_0^3=x_1x_2x_3$ (in homogeneous coordinates). - -What else is known? The base field is algebraically closed of characteristic zero. -Thank you in advance, -Serge - -REPLY [6 votes]: In his paper [Some invariants for conics and their applications, Publ. RIMS (Kyoto Univ.) 19 (1983), 1139-1151] Naruki gives an example of a self-dual quartic surface in $\mathbb{P}^3$ with three singular points of type $A_3$ and seven points of type $A_1$ (i.e., ordinary double points). -The paper can be dowloaded here.<|endoftext|> -TITLE: Determining the Lie algebra elements exponentiating to the center of a Lie group -QUESTION [5 upvotes]: For a semi-simple compact Lie group $G$ with center $Z(G)$, one can characterize the preimage of $Z(G)$ in the Cartan subalgebra under the exponential map as the nodes of the Stiefel diagram (see for instance V.7.16 of "Representation of compact Lie groups", by Bröcker and tom Dieck). -Is there any generalization of this result, for instance for non-compact Lie groups, or for classes of infinite dimensional Lie groups? -Update: A look at $SL(2,\mathbb{R})$ shows that the preimage of the center consists in elements of the form -$ -\left(\begin{array}{cc} a & b \\ -(k^2\pi^2 + a^2)/b & -a \end{array}\right) \;, \quad a \in \mathbb{R} \;, \quad b \in \mathbb{R}^\ast \;, \quad k \in \mathbb{N} -$ -One complication in this case is that the Cartan subalgebras are not all conjugate, and it looks indeed that the intersection with $\log Z(G)$ depends on the choice of Cartan subalgebra. - -REPLY [2 votes]: I have only some simple remarks here, valid also for non-compact Lie groups. -Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. A first remark is that for $X\in Z(\mathfrak{g})$ and $Y\in \mathfrak{g}$, $\exp (X)$ and $\exp(Y)$ commute. Hence we have $\exp(Z(\mathfrak{g}))\subseteq Z(G)$. However, $\exp$ need not be injective, even if $G$ is simply connected. Also, $\exp$ need not be surjective (but there is a classification of all simple Lie groups with surjective exponential). There are some special cases, where we can say which elements do not exponentiate to the centre of $G$. As an example, the following is true: -Lemma: Let $G$ be a real or complex connected Lie group whose Lie algebra $\mathfrak{g}$ is linear and centerfree. If $X ∈\mathfrak{g}$ is nilpotent and $\exp(X) \in Z(G)$, then -$X =0$. -Proof: By assumption $\exp({\rm ad} X)Y=Y$ for all $Y\in \mathfrak{g}$, so that $[X,Y]=0$ for all $Y\in \mathfrak{g}$. It follows $X=0$ because $Z(\mathfrak{g})=0$.<|endoftext|> -TITLE: Artin representations in Serre's book 'local fields' -QUESTION [6 upvotes]: Let $K$ be a complete local field with discrete valuation, and let $L/K$ be a finite Galois extension. Use $G=Gal(L/K)$ to denote the Galois group. -In Serre's book 'local fields', chapter 6, a linear representation of $G$, called Artin representation, is defined by identifying its character. Explicit description of such representations is discussed in the case of algebraic curves in section 4 of that chapter. -But I'm more interested in the case when $L,K$ are $p$-adic fields, and I wanna know the following questions: -(1) How to construct such representations explicitly in $p$-adic fields case? -(2) What's its role in ramification theory, or more general in (local) class field theory? -(3) What other properties do we know about such representations? Like when will it be irreducible? -Thank you very much for any answer or references for any of the above questions. - -REPLY [3 votes]: Serre (Annals, 1960) proves that the Artin representation corresponding to $L|K$ is rational over $\mathbf{Q}_l$ for every prime $l$ distinct from the residual characteristic $p$ and gives an example where it is not rational over $\mathbf{R}$. -Fontaine (Annales ENS, 1971)) gives a general criterion for the Artin representation to be rational over a given field of characteristic $0$, and proves that it is rational over the ring $W(k)$ of Witt vectors (if the residue field $k$ of $K$ is perfect). It might be a good idea to start with these classic papers to learn more about the Artin representation. -Serre 1960 -Fontaine 1971<|endoftext|> -TITLE: Reducing 12th degree eqns (12T179) to an 11th degree eqn -QUESTION [5 upvotes]: I always wondered if the fact that the quartic can be solved by a cubic can be generalized to other even degrees $n$, namely if there is an ordering of the roots $x_i$ of form $x_1x_2+x_3x_4+\dots+x_{n-1}x_{n} = y$ such that $y$ is reduced to an algebraic number of deg $n-1$. It seems it can be if its Galois group is of a certain kind. -Call the resolvent polynomial formed by $y$ as $P_{n}(y)$. (The labeling of the transitive groups below is from Kluener's database of number fields.) -The 3-deg $P_4(y)$ has a 3th deg factor for the general quartic. -The 15-deg $P_6(y)$ has a 5th deg factor if $6T12, 6T14$. -The 105-deg $P_8(y)$ has a 7th deg factor if $8T25, 8T36, 8T37, 8T48$ (the first two are solvable groups). -The 945-deg $P_{10}(y)$ has an 9th deg factor if (none?), -The 10395-deg $P_{12}(y)$ has an 11th deg factor if $12T179, \text{(and?)}$ -For example, given, -x^12-2x^11+17x^10-28x^9+138x^8-157x^7+549x^6-382x^5+1099x^4-77x^3+1016x^2+9x+1912 = 0 -(I've left it unformatted for easier copy-paste) which has group $L(2,11)$ and discriminant $D_{12} = 1831^6$. I use Mathematica, and its root ordering is, -$$x_1, x_2 = -0.822\mp0.744 i$$ -$$x_3, x_4 = -0.801\mp2.308 i$$ -$$x_5, x_6 = -0.452\mp2.271 i$$ -$$x_7, x_8 = 0.800\mp1.760 i$$ -$$x_9, x_{10} = 0.949\mp1.080 i$$ -$$x_{11}, x_{12} = 1.324\mp2.123 i$$ -Define, -$$y = x_1x_7+x_2x_8+x_3x_6+x_4x_5+x_9x_{10}+x_{11}x_{12}=15.60720071\dots$$ -then $y$ is a root of the monic 11-deg, -$$14155013839 + 4895349263 y + 3076830095 y^2 + 520581160 y^3 - 43394407 y^4 - 3261079 y^5 + 376457 y^6 + 33595 y^7 - 42 y^8 - 162 y^9 - 17 y^{10} + y^{11} = 0$$ -which has discriminant $D_{11} = 23^6(1831^4)$. -Questions: - -What other transitive groups for 12-deg eqns will have a $P_{12}(y)$ that has an 11-deg factor? -Does it follow that there are 14-deg eqns such that $P_{14}(y)$ has an 13-deg factor? (I suspect $14T30, 14T39$ which have groups $PSL(2,13), PGL(2,13)$, respectively.) - -P.S. My thanks to the Spearman, Watanabe, Williams paper PSL(2,5) Sextic Fields with a Power Basis which was the clue. - -REPLY [9 votes]: There are no other examples of degree $12$ besides the one you found. Moreover, there are no examples of degrees $10$ or $14$, and in fact the only additional degrees up to $32$ in which there exist examples are $16$, $28$, and $32$, where the groups in the latter cases are precisely (in Klueners/GAP/MAGMA notation) 16T447, 16T777, 16T1079, 16T80, 16T1329, 16T1508, 16T1653, 16T1654, 16T1753, 16T1840, 16T1906, 28T165, 32T35272, 32T397065, 32T2795174. Of these, only the first three degree-$16$ groups and the first two degree-$32$ groups are solvable; those degree-$16$ groups are the unique solvable transitive degree-$16$ groups of orders $240$, $480$, and $960$, and the solvable degree-$32$ groups are the unique transitive degree-$32$ groups of orders $992$ and $4960$. The biggest nonsolvable degree-$16$ group on this list is $\text{AGL}_4(2)$, the degree-$28$ group is $\text{P}\Gamma\text{L}_2(8)$, and the -nonsolvable degree-$32$ group is $\text{AGL}_5(2)$. -More generally, let $f(X)$ be a separable irreducible polynomial over a field $K$, and suppose that $n:=\text{deg}(f)$ is even. Suppose in addition that, if $\{C_1,C_2,\dots,C_{n/2}\}$ and $\{D_1,D_2,\dots,D_{n/2}\}$ are any two distinct partitions of the roots of $f(X)$ into $n/2$ two-element sets, then -$$ -\sum_{i=1}^{n/2} \prod_{x\in C_i} x \ne \sum_{i=1}^{n/2}\prod_{x\in D_i} x. -$$ -(Note that this condition will hold for any "randomly selected" polynomial having any prescribed Galois group.) Under these hypotheses, if $G$ denotes the Galois group of $f(X)$ over $K$, then the following are equivalent: - -there is an ordering $x_1,\dots,x_n$ of the roots of $f(X)$ such that $x_1x_2+x_3x_4+\dots+x_{n-1}x_n$ has degree $n-1$ over $K$. -there is an index-$(n-1)$ subgroup $H$ of $G$, and a pair of distinct roots $(x_1,x_2)$ of $f(X)$, such that every element of $H$ maps $\{x_1,x_2\}$ to either itself or to a disjoint set of roots, but every subgroup of $G$ which properly contains $H$ must contain an element which maps $\{x_1,x_2\}$ to a set having exactly one element in common with $\{x_1,x_2\}$. - -This can be used to test your condition for specific values $n$, and perhaps group-theoretic results can be used to prove theorems for infinitely many $n$. For instance, I will show later that $G=\text{AGL}_d(2)$ has the required property for $n=2^d$. -To see this equivalence, first assume that the first condition holds; then we can let $H$ be the set of elements in $G$ which fix $y:=x_1x_2+x_3x_4+\dots+x_{n-1}x_n$. This $H$ will be an index-$(n-1)$ subgroup of $G$, and every element of $H$ permutes the collection of 2-element sets $\{\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}\}$. Thus, each element of $H$ maps $\{x_1,x_2\}$ to some $\{x_i,x_{i+1}\}$ with $i$ odd. Moreover, since $f(X)$ is irreducible, $G$ acts transitively on $\{x_1,\dots,x_n\}$, so $[G:G_{x_1}]=n$; but since $[G:H]=n-1$ is coprime to $n$, we have $[G:G_{x_1}\cap H]=n(n-1)$ and thus $[H:G_{x_1}\cap H]=n$, so $H$ is transitive. Any subgroup $J$ of $G$ which properly contains $H$ must contain an element $j$ which does not fix $y$, so this element must not permute the collection $\{\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}\}$; by multiplying $j$ on both sides by suitable elements of $H$, we obtain an element of $J$ which maps $\{x_1,x_2\}$ to a set having one element in common with $\{x_1,x_2\}$. -Conversely, assume that the second condition holds. As above, $H$ is transitive. It follows that the images of $\{x_1,x_2\}$ under $H$ consist of $n/2$ pairwise disjoint two-element sets, which we can write as $\{x_1,x_2\}, \{x_3,x_4\}, ..., \{x_{n-1},x_n\}$. Then $y:=x_1x_2+x_3x_4+...+x_{n-1}x_n$ is fixed by $H$, but not by any larger subgroup of $G$, so $[K(y):K]=n-1$. This completes the proof of the equivalence. -Added later: here is a proof that $G:=\text{AGL}_d(2)$ has the required property for $n=2^d$. The group $G$ acts on the $\mathbf{F}_2$-vector space $V:=(\mathbf{F}_2)^d$, and $\text{GL}_d(2)$ consists of the elements fixing $0$. Pick some nonzero $c\in V$, and let $J$ be the stabilizer of $c$ in $G$. Let $H$ be the subgroup of $G$ generated by $J$ and the translations $x\mapsto x+u$. I claim that $H$ and $\{0,c\}$ have the required properties. Note that $\text{GL}_d(2)$ acts transitively on $V\setminus\{0\}$, so $[\text{GL}_d(2):J]=2^d-1$. Since the group of translations has order $2^d$, and is normalized by $\text{GL}_d(2)$ (and hence by $J$), we have $\#H=2^d\cdot\#J$ so $[G:H]=2^d-1$. Every translation maps $\{0,c\}$ to a set $\{b,b+c\}$, and any such set either equals $\{0,c\}$ or is disjoint from $\{0,c\}$. Finally, let $K$ be a subgroup of $G$ which properly contains $H$. Since $K$ contains all the translations, and every element of $G$ is a translation times an element of $\text{GL}_d(2)$, it follows that $K$ contains an element of $\text{GL}_d(2)$ which is not in $H$. Any such element maps $c$ to some $b\notin\{0,c\}$, and hence maps $\{0,c\}$ to the set $\{0,b\}$ which has exactly one element in common with $\{0,c\}$.<|endoftext|> -TITLE: Mapping class groups of small Seifert-fibred 3-manifolds -QUESTION [13 upvotes]: Are computations of the mapping class groups of small Seifert-fibred 3-manifolds recorded in some convenient location? -For most Seifert manifolds working out the mapping class group is easy-enough (at least, reducing it to a 2-dimensional mapping class group is easy) since diffeomorphisms usually are isotopic to fibre-preserving diffeomorphisms. But for the small Seifert-fibred manifolds that isn't always the case. -For lens spaces and general spherical 3-manifolds this was worked out by Darryl McCullough and others, about 10 years ago. - -McCullough, Darryl. Isometries of elliptic 3-manifolds. J. London Math. Soc. (2) 65 (2002), no. 1, 167–182. - -For the purpose of this question, "mapping class group" of a 3-manifold means the diffeomorphism group modulo the subgroup of diffeomorphisms isotopic to the identity, i.e. $\pi_0 Diff(M)$. -If this is in one of the standard references like Orlik's book, please let me know. I looked over it briefly but it looked to me like Orlik's book does not cover this, at least not explicitly. I imagine one could derive the computation from any suitably-detailed proof of the uniqueness of Seifert-fiberings, but I haven't worked out the details myself. - -REPLY [15 votes]: The determination of mapping class groups of small Seifert manifolds was completed by M. Boileau and J.-P. Otal in a paper in Invent. Math. 106 (1991), 85-107. They give references for cases previously done: - -Lens spaces by Bonahon and Hodgson-Rubinstein. -Multiple fibers of orders (2,2,n) by Asano and Rubinstein. -Multiple fibers of orders (2,3,4) by Birman-Rubinstein. -Multiple fibers of orders (p,q,r) in most of the cases with infinite fundamental group by Scott. - -This left only the cases (2,3,p) other than (2,3,4), and the cases (3,3,p). These are the cases treated in the Boileau-Otal paper. In particular this includes the Poincaré homology sphere, where the mapping class group happens to be trivial. -All this was done in the 1980s. (Boileau and Otal announced their results in a C.R. note in 1986.)<|endoftext|> -TITLE: Deep theorems and long proofs -QUESTION [22 upvotes]: I ran across this discussion by Daniel Shanks, - -"Is the quadratic reciprocity law a deep theorem?." - Solved and Unsolved Problems in Number Theory. Vol. 297. AMS, 2001. p.64ff. - -which made me wonder: - -Q. Is there a theorem in some formal system whose proofs are known to be - necessarily "long" in some sense, perhaps in the Kolmogorov-complexity sense? - -I know it has been established that there are relatively "short" theorems that have -only enormously "long" proofs (apparently[?] due to Gödel,"On the length of proofs"), -but I'm asking if it is known that some -particular theorem only has long proofs? -(This is in some sense the obverse of the MO compendium, -"Quick proofs of hard theorems.") -Obviously this is a naive question! - -REPLY [20 votes]: If you're willing to accept a contrived statement, then it's not hard to get an explicit example, but this may not be the sort of example you're looking for. -Following Gödel's approach, you can make a statement of the form "I have no formal proof of less than a hundred pages." (More precisely, you should use implicit self reference, along the lines of "There is no formal proof of less than a hundred pages for this string followed by its quotation: 'There is no formal proof of less than a hundred pages for this string followed by its quotation:'") -Assuming your formal system is consistent, this statement cannot be proved in less than a hundred pages. In that case it actually is provable, because you can prove it by enumerating all possible proofs of less than a hundred pages and checking that they don't work. Thus, we have a relatively short theorem that has no short formal proof. Sadly, the theorem is of no intrinsic interest. -Note that the reason it didn't take us a hundred pages to prove the theorem is because we assumed the formal system was consistent. By Gödel's second incompleteness theorem, the system can't prove its own consistency, so this approach is not available within the system. In fact, this argument proves the second incompleteness theorem: if consistency had a short enough proof, then we could carry out this argument to prove the theorem in under a hundred pages. There's nothing special about a hundred pages in this argument, so there can be no proof of consistency at all. This approach to the second incompleteness theorem is fundamentally the same as Gödel's original proof. -Here I've been measuring proofs by length, rather than the Kolmogorov complexity mentioned in the question. That's not a useful measure for proofs, because you can reconstruct the shortest proof of theorem X in system S from little more than X and a proof checker for S (which you apply to do a brute force search for the shortest proof). This means if you fix S, the Kolmogorov complexity of the shortest proof is no more than an additive constant greater than that of the theorem itself.<|endoftext|> -TITLE: An order type $\tau$ equal to its power $\tau^n, n>2$ -QUESTION [21 upvotes]: (This is a re-post of my old unanswered question from Math.SE) -For purposes of this question, let's concern ourselves only with linear (but not necessarily well-founded) order types. -Recall that: - -$0, 1, 2, \dots$ — unique linear order types for each finite cardinality, that can be identified with natural numbers. -$\omega$ — the order type of $\mathbb{N}$ ordered by magnitude, the smallest infinite ordinal. -$\eta$ — the dense countable order type of rational numbers $\mathbb{Q}$ ordered my their magnitude, which is also an order type of any dense denumerable linear order without first and last elements (e.g. the set of positive algebraic numbers). - -The sum and product of order types are natural generalizations of (and are consistent with) the sum and product of ordinals, which we consider well-known. Let's agree that a positive integer power of an order type is just a syntactic shortcut for repeated multiplication. -There are some order types satisfying $\tau^2=\tau$, for example: $0,\,1,\,\eta,\,\omega\,\eta$ and $\omega^2\eta$. - -Question: Is there a linear order type $\tau$ such that $\tau^2\ne\tau$, but $\tau^n=\tau$ for some integer $n>2$? - -REPLY [6 votes]: [This partial answer was posted Nov. 19 to Math Stack Exchange but seems to have been ignored there. It seems to be a more concise version of the easy case (no endpoints) of Garrett Ervin's answer.] -Such questions have been considered in the past. From W. Sierpiński's Cardinal and Ordinal Numbers, second edition revised, Warszawa 1965, p. 235: "We do not know so far any example of two types $\varphi$ and $\psi$, such that $\varphi^2=\psi^2$ but $\varphi^3\ne\psi^3$, or types $\gamma$ and $\delta$ such that $\gamma^2\ne\delta^2$ but $\gamma^3=\delta^3$. Neither do we know any type $\alpha$ such that $\alpha^2\ne\alpha^3=\alpha$." Also, from p. 254: "We do not know whether there exist two different denumerable order types which are left-hand divisors of each other. Neither do we know whether there exist two different order types which are both left-hand and right-hand divisors of each other." Of course, if $\tau^n=\tau$ for some integer $n\gt2$, then $\tau^2$ and $\tau=\tau^2\tau^{n-2}=\tau^{n-2}\tau^2$ are both left-hand and right-hand divisors of each other. -For what it's worth, here is a partial answer to your question, for a very special class of order types. By "order type" I mean linear order type. An order type $\xi$ is said to have a "first element" if it's the type of an ordered set with a first element, i.e., if $\xi=1+\psi$ for some $\psi$; the same goes for "last element". -Proposition. If $\alpha$ is a countable order type, and if $\alpha\xi=\alpha$ for some order type $\xi$ with no first or last element, then $\alpha\beta=\alpha$ for every countable order type $\beta\ne0$. -Corollary. If $\tau$ is a countable order type with no first or last element, and if $\tau^n=\tau$ for some integer $n\gt1$, then $\tau^2=\tau$. -The corollary is obtained by setting $\alpha=\beta=\tau$ and $\xi=\tau^{n-1}$ in the proposition. -The proposition is proved by a modified form of Cantor's back-and-forth argument. Namely, let $A$ be an ordered set of type $\alpha=\alpha\xi$, and let $B$ be an ordered set of type $\alpha\beta=\alpha\xi\beta$. Since $A$ and $B$ are countable sets, let's fix an enumeration of each set. -An isomorphism between $A$ and $B$ will be constructed as the union of a chain of partial isomorphisms $f_k$ of the following form. The domain of $f_k$ is $I_1\cup I_2\cup\dots\cup I_k$, where $I_1,\dots,I_k$ are intervals in $A$ of order type $\alpha$; $I_1\lt\dots\lt I_k$; the interval in $A$ between $I_j$ and $I_{j+1}$ ($1\le j\lt k$), as well as the interval to the left of $I_1$ and the interval to the right of $I_k$, have order types which are nonzero right multiples of $\alpha$. The range of $f_k$ is $J_1\cup\dots\cup J_k$ where $J_1,\dots,J_k$ are intervals in $B$ of type $\alpha$, etc. etc. etc., and $f(I_1)=J_1,\dots,f(I_k)=J_k$. -Here is an informal description of the induction step. Suppose we have defined $f_k$ etc. as described above. Let $x$ be the first point in the enumeration of $A$ which is not in the domain of $f_k$, say $I_1\lt x\lt I_2$. The gap between $I_1$ and $I_2$ is of order type $\alpha\gamma$ for some order type $\gamma\gt0$. Since $\alpha\xi=\alpha$, we have $\alpha\gamma=\alpha\xi\gamma=\alpha\theta$ where $\theta=\xi\gamma$ is a nonzero order type with no first or last element. That is, the gap between $I_1$ and $I_2$ is the linear sum of a $\theta$-type set of intervals of type $\alpha$, and $x$ is in one of those intervals, let's call it $I$. Note that the gap between $I_1$ and $I$, as well as the gap between $I$ and $I_2$, has an order type which is a nonzero right multiple of $\alpha$. Choose an interval $J$ between $J_1$ and $J_2$ in the same way, except that there is no particular point to be covered. Extend $f_k$ to $f_{k+1}$ so that $I$ is mapped isomorphically onto $J$.<|endoftext|> -TITLE: What motivated Rademacher's contour along the Ford circles? -QUESTION [22 upvotes]: Apologies if this question isn't suitable for MathOverflow; I posted it on MSE here but it didn't get a response and it felt like it was on the cusp of being suitable for here. -After Ramanujan and Hardy found the infinite sum representation of the partition function $p(n)$, Rademacher went about simplifying their proof; the form generally seen involves integrating $\frac{P(q)}{q^{n+1}}$ along a circle centered at $0$, where $P(q)=\sum_{n \geq 0} p(n) q^n$. We fix an integer $N$ and consider the Farey sequence of fractions with denominator at most $N$; we then split the circle into parts, "centering" each part at a given element of this Farey sequence; as $N$ goes to infinity (and our radius goes to $1$), this gives us the infinite sum we want. -Later, Rademacher rewrote this proof using a different contour along the upper half plane (which can be sent to the unit disc by $z \mapsto e^{2\pi i z}$). He integrated along the Ford circles, starting at $i$ and going to $i+1$ with a sequence of paths, each path going further 'down' the Ford circles than before; he constructed the ford circles for the $N$th Farey sequence, started at the top of the largest circle, followed an arc until he reached the last tangency with another circle, followed that circle to the right, etc. This ultimately leads to a cleaner proof, I'd guesstimate about a third the length of the original. However... -It's not clear at all how he thought to use this method. Why did Rademacher choose to integrate along the Ford circles? Is it just because they're a geometric way of looking at the Farey fractions (which are key in the circle method), so he said "well why don't I give it a shot"? Why does it work so magically in leading to quicker and cleaner integrals and approximation? What he's essentially doing is, instead of having each bit of the contour be an arc small enough that we integrate 'less' near the most prominent singularities, allowing the contour to approach the various singularities (and getting naturally closer to the ones with greater height - the 'less important' ones - faster). It's clear that this contour provides better bounds in proving the infinite sum form of $p(n)$ but in no way is it clear to me why, or how Rademacher first came up with the idea. - -REPLY [14 votes]: In 1954 Hans Rademacher gave a series of lectures in which he described his reasoning step by step, see Lectures on Analytic Number Theory, page 113 and following. This should give some insight into his creative process, of which he himself says: "The path [of integration] is complicated and was not invented in one sitting."<|endoftext|> -TITLE: Entire functions with a null real escaping set -QUESTION [6 upvotes]: Let $f$ be a entire function (stable on $\mathbb{R}$), and $E_{\mathbb{R}}$ its real escaping set : $$E_{\mathbb{R}} = \{ x \in \mathbb{R} : f^{(k)}(x) \rightarrow_{k \to \infty} \infty \} $$ -We put the Lebesgue measure on $\mathbb{R}$. - -Question : If $E_{\mathbb{R}}$ is a (measurable) null set, is it also empty ? - -REPLY [5 votes]: For every continuous function $g$ on the real line and for every positive continuous function $\epsilon$ on the real line, there exists an entire function $f$ such that -$|f(x)-g(x)|<\epsilon(x)$ for all real $x$ (This is due to Carleman). So if you can construct a real continuous -function with your property then you can also construct an entire one. But construction -of a real continuous function with this property does not seem to be difficult. -Here is a sketch. -First, let us make the negative semi-axis invariant: $x\leq 0$ implies $f(x)\leq 0$. -On the positivde ray, let $f(x)\leq 0$ except some small disjoint intervals $I_k$ -tending to $+\infty$. Let $E=\{ x:f(x)\leq 0\}$ be the complement of these intervals. On the intervals we arrange like this: Let $J_k\subset I_k$ be a smaller -interval near the middle of $I_k$, where our function is large and has a local max, -but this $J_k$ is mapped to $E$, and on the two subintervals $I_k\backslash J_k$ -has very large derivative (by absolute value) and the image of these two subintervals -contains only one interval of the $I_j$, namely $I_{k+1}$. -Then the escaping set consists of points whose orbits $x_k$ are in the intervals for $k$ -large enough. It is clear that this is not empty, and on each interval the set of -escaping points is a Cantor set, which can be easily made of measure $0$ by -making the derivative very large on the "side" subintervals.<|endoftext|> -TITLE: Exponentiation in finite simplicial sets -QUESTION [12 upvotes]: A finite simplicial set is a simplicial set having only a finite number of non degenerate simplicies. My question is: if $A$ and $B$ are finite simplicial sets, does this imply that the simplicial set $A^B$ is also finite? - -REPLY [7 votes]: I think the statement is false for $ X = A = \Delta_4/\partial \Delta_4$ and $B = \Delta_1$: -As Charles Rezk already mentioned, it is enough to consider the growth of $f_{X^{\Delta_1}}(n) = |Hom(\Delta_1\times \Delta_n, X)|$. Writing the prism $\Delta_1 \times \Delta_n$ as a coequalizer over its $(n+1)$- simplices, I think we can identify $Hom(\Delta_1\times \Delta_n, X)$ with tuples $(x_0, \dotsc, x_n)$ where $x_i \in X_{n+1}$ and $ d_{i+1} x_i = d_{i+1} x_{i+1}$ whenever this makes sense. Now, we can restrict our attention to tuples where even $d_{i+1} x_i = d_{i} x_i = [\partial \Delta_4]$ holds (and for simplicity $x_n = [\partial \Delta_4]$ where $[\partial \Delta_4]$ denotes the unique degeneracy of the basepoint). -These definitely satisfy the gluing conditions, and the number of them grows exponentially: -We can choose the $x_i$ independently, and for $n$ large enough there are at least 2 possibly choices for any $i< n$: If we denote the unique non-degenerate simplex in $X$ by $\iota$, then we can consider morphisms in the $\Delta$-category $\sigma : [n] \rightarrow [4]$ which are surjective and have the property that $\sigma^{-1}(\sigma(i))$ and $\sigma^{-1}(\sigma(i+1))$ are singleton sets. Then $\sigma^* \iota$ is a possible choice for $x_i$. -This gives exponential growth as a lower bound, so $X^{\Delta_1}$ can't be finite.<|endoftext|> -TITLE: Consistency strength of the failure of Shelah's Strong Hypothesis (SSH) -QUESTION [7 upvotes]: Some known facts about SSH (Shelah's Strong Hypothesis): -i) "$0^\sharp$ does not exist" implies SSH. -ii) SSH implies SCH (Singular Cardinal Hypothesis). -iii) The failure of SCH is equiconsistent with -$\exists \kappa (o(\kappa) = \kappa^{++})$. -What else is known about the consistency strength of the failure of SSH? - -REPLY [5 votes]: Additional informations from - -Pierre Matet. Large cardinals and covering numbers, - Fundamenta Mathematicae 205 (2009), 45-75. - doi:10.4064/fm205-1-3 - -on the first page: - -And, as Moti Gitik pointed out to the author, one obtains a model of - "$u (\omega_1 , \omega_{\omega}) > {\omega_{\omega}}^+$ - (and hence $\neg$SSH) + SCH" by adding $\aleph_{\omega + 1}$ - Cohen reals to a model of - "for every infinite cardinal $\nu$, $2^{\nu}$ equals $\nu^{++}$ - if $\nu = \omega_{\omega}$, and $\nu^+$ otherwise". - -and - -It is not known whether the failure of SSH is equiconsistent with that of SCH.<|endoftext|> -TITLE: Why do convex polytope options constrict with dimension, rather than expand? -QUESTION [10 upvotes]: There are an infinite number of regular polygons in the plane, -five regular polyhedra, -six regular polytopes in $\mathbb{R}^4$, -and then three regular polytopes in every dimension $d > 4$. -There are eight convex deltahedra (all faces equilateral triangles), -five in $\mathbb{R}^4$ (see, "Convex deltahedra in higher dimensions"), -and then three deltatopes in every dimension $d > 4$. - -Q. Is there some intuitive reason why freedom is removed in higher dimensions? - -I ask because, if I didn't know better, I would think the opposite. -There is vastly much more "room" in higher dimensions, and one might think forms proliferate, even under constraints. Concerning "much more room," -think of the severe contraints on planar graphs vs. -the fact that every graph can be realized as embedded in $\mathbb{R}^3$. -I seek a corrective to my faulty intuition. Thanks! - -REPLY [13 votes]: This is a well known "dimension curse" phenomenon. It is easier to explain for spherical polyhedra. Let $P$ be a regular spherical convex polyhedron of dimension $d$. Starting with dimension $d=3$, all spherical convex polyhedra are rigid, so they are determined by their links (intersection of a vertex cones with an $\epsilon$-sphere). The latter are themselves regular (spherical) polyhedra. Therefore, for $d\ge 4$ these links are rigid themselves and regular polyhedra are uniquely determined by their links (which are all congruent, of course, by regularity). Thus, as dimension grows the combinatorial types of regular polyhedra can disappear but new cannot appear. -For Euclidean polyhedra the same type of analysis works, but goes through the links which are spherical polyhedra. Therefore, as regular spherical polyhedra disappear so do Euclidean ones, but this is less intuitive perhaps. I haven't seen "deltatopes" before, but it's the same story there as well, if I understood definitions correctly.<|endoftext|> -TITLE: Consecutive non squarefree integers -QUESTION [12 upvotes]: Question: Is there a function $f(n) \rightarrow \infty$, such that infinitely often the interval $[n,n+\frac{f(n) \log(n)}{\log{\log(n)}}]$ does not contain a squarefree integer? -Additional information: If we consider the simultaneous congruences: -$$ x \equiv 0 \ (\text{mod} \ 2^2)\\ x \equiv 1 \ (\text{mod} \ 3^2) \\ \cdots \\ x \equiv k-1 \ (\text{mod} \ p_{k}^2) $$ -Which has a solution not bigger than $ \Pi_{i=1}^{k} p_i^2 \leq e^{2(1+\varepsilon)k\log k} $. -This way we can see that there is a constant such that there are no squarefree numbers in the interval $[n,n+c\frac{\log(n)}{\log{\log(n)}}]$ infinitely often. We can optimize the above equations, for example on the right hand side we don't have to consider numbers divisible by $4$ after the first equation. Or numbers congruent to $1$ modulo 9 after the second equation etc. but that just gives a better constant! - -REPLY [12 votes]: Erdos has mentioned this lower bound in several places, adding always that he's never been able to improve it. For example see http://renyi.hu/~p_erdos/1951-13.pdf (page 107; in fact he gives an explicit constant here that he says he cannot improve), and page 8 of http://hsb.org.hu/~p_erdos/1981-21.pdf . -A standard Borel-Cantelli type heuristic suggests that the gaps should be bounded by some constant times $\log n$, analogous to the Cramer conjectures for gaps between primes. I don't know if anyone has written down such a conjecture in this context. But I did find a paper by Kevin McCurley where he considers the least square-free number in an arithmetic progressions, and formulates a Borel-Cantelli type conjecture in this context. In analytic number theory, the situations of arithmetic progressions and short intervals are usually very similar, and one could adapt McCurley's argument to write down conjectures in the short interval case. McCurley's paper is here: http://www.ams.org/journals/tran/1986-293-02/S0002-9947-1986-0816304-1/S0002-9947-1986-0816304-1.pdf .<|endoftext|> -TITLE: Shelah's book on "Classification Theory" -QUESTION [10 upvotes]: As we know one of the most important and fundamental books in stability, simplicity, forking and ... classification theory, is Shelah's "Classification Theory" where lots of original ideas of the subject could be found. -I am studying model theory and wanted to start reading that book in order to be able to understand the current literature of the research in the areas of stability and ..., and of course as it is a book that a model theorist must better read. -I ask some of the people around me and they prevent me from reading the book, by the reason that "it is old fashioned and could not enable you to be involved in the current literature of research." -Is this book still recommended or there are some more new resources covering that topics (including the original ideas and the core concepts)? - -REPLY [14 votes]: I think Shelah's book is very good, but it is very difficult to read, in part intrinsically because of its subject matter, and in part because it was written by Shelah. Recently, Richard Elwes reminded us of the review by Lascar (I do not agree with portions of the review, but it gives you a good idea of what you are up against): - -Unfortunately, the book is very difficult to read. This is undoubtedly due in part to the difficulty of the topic itself, but it also has to do with the way the book has been written. The reader should be warned about the numerous misprints and inaccuracies he will have to detect and the time he will need to work out the numerous "easy'' or "left to the reader'' proofs. He will have to understand that statements made under impossibly complicated hypotheses for the sake of generality, and often in an axiomatic setting, may conceal, as a special case, a simple and important fact. Also, he should not expect a rigorous structure: notions are usually introduced when needed, and it will be up to him to guess how important they may be in the rest of the book. -All of these features do not make for easy reading, and we shall certainly not advise anyone (except perhaps the author himself) to "devote himself to reading and solving the exercises till he knows the book by heart'' (cf. Introduction). We do not recommend it as a textbook, and it will be difficult to use it as a reference book, considering how hard it is to find any particular result. But we do think that nobody involved in research in model theory can avoid studying it. We are unable to recommend any particular angle of attack. We did find the opening remarks in each chapter rather helpful (with the exception of Chapter III). - -For modern work in the subject, you probably want to read eventually about Geometric Stability Theory, as in Pillay's book. An introduction to stability that would better prepare you for this would probably be best. Pillay's introductory book on Stability Theory is certainly an option. Other suggestions are Marker's book, Poizat's and Hodges's. The truth is, recently there have been quite a few nice texts published, so it should not be difficult to find a good text with which to start. -(But I would still suggest to have Shelah's book as a reference and to consult it as you learn more of the theory.) -On a vein different from the algebraic geometric approach, you may be interested in how set theory interacts with the subject. Shelah's book has plenty of this while constructing non-isomorphic models, but there is significantly more in this direction, in the context of abstract model theory and abstract elementary classes. I would recommend Grossberg's text as an introduction.<|endoftext|> -TITLE: Continuous functions $f$ with $f(A)$ linearly independent when $A$ is independent -QUESTION [27 upvotes]: Is there any characterization of continuous functions $f : \Bbb{R}\longrightarrow \Bbb{R}$ such that for any linearly independent set $A$ (over the rationals) $f(A)$ is also linearly independent ? - -REPLY [9 votes]: Trevor Wilson has reduced the possibilities to $\alpha x^\beta$ with $\alpha\ne 0$ and $\beta\ge 0$. The following argument should reduce the possibilities to $\beta=0$ or $\beta=1$, modulo a lemma that I can't quite seem to prove. Perhaps some other MO reader can plug the gap. -The lemma I need is: - -For any $q_1, q_2, q_3\in\mathbb{Q}$, not all zero, and any positive $\beta\ne1$, the equation $$q_1+q_2x^\beta+q_3(1+x^{\beta^2})^{1/\beta} = 0$$ has at most countably many real solutions. - -This lemma implies that for any positive $\beta\ne1$, there exists a constant $c>0$ such that the three numbers $x_1:=1$, $x_2:=c$, and $x_3:=(1+c^{\beta})^{1/\beta}$ are linearly independent over $\mathbb{Q}$. On the other hand, $\alpha x_1^\beta + \alpha x_2^\beta = \alpha x_3^\beta$, so their images under the function are linearly dependent.<|endoftext|> -TITLE: Uniform incentre of collection of quasi convex subspaces in hyperbolic spaces -QUESTION [5 upvotes]: I'm reading a paper of Wise on cubulations and the following fact is used: -Let $H$ be a quasi-convex subgroup of a $\delta$-hyperbolic group and let $H_i, i\in I$ be a finite family of translates of $H$ such that they are pairwise at bounded distance $D$. Then there is a constant $C$ and a point $x$ such that $d(x,H_i) -TITLE: Which positive definite symmetric matrices have solvable characteristic polynomial? -QUESTION [5 upvotes]: I am interested in the structure of the space of $n \times n$ positive definite symmetric matrices with rational entries whose characteristic polynomials are solvable (i.e. the Galois group is solvable). Is this an algebraic variety, for instance? I can't find any characterization of such matrices, but I wouldn't necessarily know where to look. - -REPLY [6 votes]: Here is a small observation. As a subspace of the space $\mathbb{Q}^n$ of monic polynomials of degree $n$ with rational coefficients, the solvable polynomials are dense (and so in particular are not contained in an algebraic or even semialgebraic subset $\mathbb{Q}^n$). To see this it suffices to observe that any such polynomial is a product of real linear or quadratic polynomials and that we can approximate these by rational linear or quadratic polynomials. The corresponding products are clearly solvable.<|endoftext|> -TITLE: Is definability of a basis for $\mathbb{R^N}$ independent of ZFC? -QUESTION [6 upvotes]: By $\mathbb{R^N}$ I mean the real vector space with the natural componentwise addition and scalar multiplication. Certainly ZFC+(V=L) gives definable bases, but does ZFC? - -REPLY [4 votes]: The answer, if I understand the question correctly, is negative. That is, if we understand "definable" as "can be defined from ordinals [and a real number]", or in simpler words, sets which are in $\sf HOD(\Bbb R)$. Here is a sketch of an argument I suspect is correct. -Consider Solovay's model which we get from collapsing an inaccessible cardinal to $\aleph_1$. In that model every set of real numbers definable from an ordinal and a real number has the Baire property. Since I am not going to care about the forcing, $V$ is going to be the model after the collapse. -Now suppose that $B$ was a basis and was definable, since $\sf HOD(\Bbb R)$ and $V$ both agree that $B$ must have the cardinality of the real numbers, they also agree that it has more than $2^{\aleph_0}$ permutations. Each of those permutations extends uniquely to an automorphism of $\Bbb{R^N}$. Therefore in $\sf HOD(\Bbb R)$ there are more than $2^{\aleph_0}$ automorphisms of the space. -However in $\sf HOD(\Bbb R)$ we have automatic continuity for Polish groups, and therefore every automorphism is continuous. This is a contradiction since there can only be $2^{\aleph_0}$ continuous automorphisms of a Polish space. -Therefore $B$ cannot be definable to begin with. One can also get away from the inaccessible cardinal by considering Shelah's model in which not all sets are Lebesgue measurable, but all sets have the Baire property. - - -Solovay, Robert M., "A model of set-theory in which every set of reals is Lebesgue measurable." Ann. of Math. (2) 92 1970 1–56. -Shelah, Saharon, "Can you take Solovay's inaccessible away?" Israel J. Math. 48 (1984), no. 1, 1–47. -Raisonnier, Jean, "A mathematical proof of S. Shelah's theorem on the measure problem and related results." Israel J. Math. 48 (1984), no. 1, 48–56.<|endoftext|> -TITLE: Long gaps between primes -QUESTION [8 upvotes]: Zhang's celebrated result established the existence of bounded gaps between primes, that is, there exists a constant $B$ (in Zhang's paper $B$ can be taken to be $7 \times 10^7$, and this was almost immediately improved significantly so that one may take $B= 4680$) such that there are infinitely many primes $p_n$ such that $p_n - p_{n-1} < B$. -My question considers the opposite case. Do there exist, infinitely often, gaps that are much larger than average? The prime number theorem implies that the average gap between $p_n$ and $p_{n+1}$ is about $\log p_n$. -Thus, for which constant $C > 1$ can one establish the existence of infinitely many primes $p_n$ such that $p_n - p_{n-1} > C\log p_n$? And what is the best known constant to date? -Edit: in view of the wikipedia entry on this topic, it seems that the correct thing to look at is -$$\displaystyle p_n - p_{n-1} > \frac{c \log n \log \log n \log \log \log \log n}{(\log \log \log n)^2}.$$ -This seems like a rather unnatural function, but for now it is not known if the constant $c$ in the above inequality may be taken to be arbitrarily large. Is this inequality expected to be the right order of magnitude? -Thanks for any insight. - -REPLY [4 votes]: Erdos's problem, which asks whether or not the bound -$$\displaystyle p_n - p_{n-1} > \frac{c \log n \log \log n \log \log \log \log n}{(\log \log \log n)^2}$$ -holds with $c$ arbitrarily large, has been answered in the positive independently by Ben Green, Kevin Ford, Sergei Konyagin, and Terry Tao and James Maynard. However, it seems that they have decided to write a single paper on the subject, which is found here: http://arxiv.org/abs/1412.5029<|endoftext|> -TITLE: Are ramified covering of negatively curved manifolds negatively curved? -QUESTION [5 upvotes]: Gromov and Thurston proved in "Pinching constants for hyperbolic manifolds" that any finite ramified covering of a compact hyperbolic manifold, along a codimension $2$ totally geodesic submanifold, can be endowed with a riemannian metric of negative sectional curvature. Pansu mentions in a survey (4/TSG_1985-1986_4_101_0/TSG_1985-1986_4_101_0.pdf">http://archive.numdam.org/ARCHIVE/TSG/TSG_1985-1986_4/TSG_1985-1986_4_101_0/TSG_1985-1986_4_101_0.pdf) that he doesn't know how to make Gromov and Thurston's calculation in the general case. Is it possible to generalize ? -Namely let $M$ be a compact manifold endowed with a riemannian metric of negative sectional curvature. Suppose $X \longrightarrow M$ is a ramified covering of finite degree, along a compact submanifold of codimension $2$. Can $X$ be endowed with a riemannian metric of negative sectional curvature ? Can a counterexample be found when the locus of ramification is not totally geodesic ? Can one hope a pinching of the curvature like Gromov and Thurston ? Is the covering, in any reasonable sense, "more negatively" than the base ? - -REPLY [5 votes]: In the case of 3-dimensional negatively curved manifolds, this result was proved by Yong Hou (at least when the submanifold is totally geodesic; as Misha says, this is the most natural interpretation of your question). The statement is a generalization of Theorem 3.2 of his paper, -which applies as stated only to cyclic covers branched over a null-homologous geodesic; however it's clear that the proof generalizes to arbitrary branched covers since it is purely local. The idea of his proof is to show that in a tubular neighborhood of the metric, one may make a deformation in which the core geodesic has a neighborhood with constant negative sectional curvature (this sort of argument goes back to Gao). Then one may perform the change in curvature exactly as in the case of Gromov-Thurston. -So one could try to emulate Hou's proof in higher dimensions, but I don't know if this has been carried out.<|endoftext|> -TITLE: How difficult will it be for me to switch fields (details below) after my Ph.D. in pure mathematics? -QUESTION [10 upvotes]: I'm a first year postdoctoral researcher, working in pure areas of Riemann surfaces and differential geometry, after just finishing my Ph.D. in 2013. Recently I've also started taking interest in applied fields relating to the above areas: like medical imaging and computer vision. I've no prior knowledge in these areas, indeed in my department there's nobody I can talk to about these subjects. So I've recently started contacting people asking for notes and their current research so that I can at least begin in an amateur way. The notes/research I'm studying relate to use of quasiconformal maps, Riemann surface, differential geometry (for example, Ricci flow) to imaging and computer vision. -My plan is to apply for a postdoc next year in these areas, thereby switching from 'pure' to 'applied'. But the thing that concerns me is my zero-experience in this field. While some people did reply to my e-mail saying I've sufficient background for research in those areas, some other forwarded my CV etc. to the committee in case there's been a vacancy already, even though I didn't actually formally apply there, but it got rejected. -So, to you who specially work in theoretical/computational mathematical/medical imaging or computer vision problems, what will be your opinion on hiring somebody as postdoc who has lot of related pure mathematics experience and willingness to learn the skills related to these areas, but have almost zero experience in the fields themselves? How difficult will it be for me to switch fields now? - -REPLY [6 votes]: I switched from the theory of Banach spaces, to fluid dynamics. I made the change after I got tenure. I am really glad that I waited until after tenure, because it had a huge negative impact on my career. I am glad I did it, because I love my new applied area. But I am also glad I waited until I had tenure and didn't have to worry so much about publish or perish. -My situation isn't quite the same as yours, so I don't really know how to advise you. But without a track record, and also without the connections that your Ph.D. adviser or other colleagues can provide, I think you are looking to go a hard way. -Continuing in your research area and waiting until you get tenure: is that an option? Also, you might be quite surprised at the extent to which your experience in pure mathematics will give you insights that are quite unique.<|endoftext|> -TITLE: consequence of Novikov conjecture -QUESTION [6 upvotes]: Novikov conjeture is a famous open problem in Geometric topology.It predicts that higher signature is oriented-homotopy invariant. -http://en.wikipedia.org/wiki/Novikov_conjecture -I am a student interested in geometric topology.My question is: -What are the possible consequences of this famous conjecture? Thank you! - -REPLY [7 votes]: The Novikov conjecture is known for some groups, and open for other groups. Often Novikov conjecture is implied by a more general conjecture (e.g. Borel, Farrell-Jones, or Baum-Connes) which in my amateur view are more intersting. -All these conjectures are true e.g. for $G=\mathbb Z^n$, $n>4$ and in many other cases. -Out of those the Borel conjecture is the easiest to state, it says: a homotopy equivalence of closed $K(G,1)$ manifolds is homotopic to a homeomorphism. There is also a relative Borel conjecture: the homotopy equivalence of compact aspherical manifolds that is a homeomorphism on the boundary is homotopic to a homeomorphism. -Here is a common way in which the conjectures get used: obstructions to various topological problems lie in groups associated with $G$ such as the Whitehead group (is an h-cobordism trivial?), projective class group (can one attach a boundary to an open manifold?) etc, which vanish if the appropriate Novikov-type conjecture hold for $G$. -There is a friendly textbook "The Novikov conjecture'' by Kreck and Lueck which starts as a slow introduction and ends with a survey.<|endoftext|> -TITLE: A special c.c.c forcing notion and adding minimal generic reals -QUESTION [7 upvotes]: This question is related to my question "Forcing with c.c.c forcing notions, Cohen reals and Random reals". -A natural way to answer Prikry's conjecture is to build a c.c.c. forcing notion which adds a minimal generic real. -I arrived to the idea of constructing such a forcing notion by considering the following: -1) Mathis forcing at $\omega$ is essentially the same as Prikry's forcing at some measurable cardinal, -2) Recently, a minimal Prikry type forcing at a measurable cardinal is constructed by Koepke-Räsch-Schlicht (see "A minimal Prikry-type forcing for singularizing a measurable cardinal, J. Symbolic Logic Volume 78, Issue 1 (2013), 85-100." and also Is Prikry forcing minimal? ). -So a natural idea is to define a version of the above forcing for $\omega$. It turns out to me that such a forcing is constructed many years ago by Judah-Shelah in "Forcing minimal degree of constructibility" and there it is shown that under $CH$ we can choose nice ultrafilters so that the resulting forcing adds a minimal generic real which is minimal. -Let me describe the forcing I have in mind: Let $(A_n: n<\omega)$ be a partition of $\omega$ into infinite sets and for each $n$ let $U_n$ be a non-principal ultrafilter over $A_n$. Let $\mathbb{P}$ consists of all pairs $(t, T)$ where $T \subseteq [\omega]^{<\omega}$ -is a tree with trunk $t$ and for all $t \unlhd u\in T$ (where $\unlhd$ denotes end-extension relation) $Suc_T(u) \in U_{max(u)}.$ We define $(t, T) \leq (s, S)$ iff $T \subseteq S$ and $(t, T) \leq^* (s, S)$ iff $T \subseteq S$ and $t=s.$ The following can be proved easily: -1) $(\mathbb{P}, \leq)$ satisfies the c.c.c., -2) $(\mathbb{P}, \leq, \leq^*)$ satisfies the Prikry property, -3) Let $G$ be $\mathbb{P}-$generic over $V$ and let $f_G=\bigcup \{t: \exists T, (t, T)\in G \}.$ Then $f_G$ is a real, -4) Suppose $d \subseteq f_G$ is infinite, then $V[d]=V[f_G]$ (the proof is the same as in the proof of Proposition 4.4 in Koepke-Räsch-Schlicht paper). -Question. Is it consistent with $ZFC$ that there is no sequence of ultrafilters as above such that the corresponding forcing defined above adds a minimal generic real? -Note that in such a model $CH$ or $MA+\sim CH$ should fail. -Remark 1. If we can show that the forcing does not add a minimal generic real when all of our ultrafilters are on the same class of near-coherence ultrafilters, then the answer to question 1 will become positive, as it is consistent that there is just one class of near-coherence non-principal ultrafilters. -Remark 2. As pointed out by Dorais, if all of the ultrafilters are RK equivalent, then the resulting forcing does not add a minimal generic real. - -REPLY [5 votes]: Groszek studied this and related forcings in Combinatorics on ideals and forcing with trees (JSL 52 (1987), 582–593; MR0902978). (Note that Groszek allows each node $\sigma$ to have a different ideal, so yours is a special case of what Groszek denotes $L(\Sigma^+)$. Also note that $L(\Sigma^+) = L(\Sigma^1)$ when all ideals are maximal.) -Theorem 7 gives a sufficient condition — the simultaneous tree property — for the generic $L(\Sigma^+)$-real to have minimal degree. I'm not sure the exact strength of the existence of a $\Sigma$ with the simultaneous tree property is but it feels like it should be possible to construct such a thing under CH. (The paper contains a construction under V = L.)<|endoftext|> -TITLE: Homotopy of quivers -QUESTION [5 upvotes]: The matrix ring $k^{n\times n}$ can be realized in many ways as a quotient of a path algebra: For example choose the quiver $1\leftrightarrows 2 \leftrightarrows \cdots \leftrightarrows n-1\leftrightarrows n$ and impose the relations $i\to i\pm 1\to i = e_i$ where $e_i$ is the idempotent associated to the vertex $i$. So in this sense loops are trivial. The isomorphism is given by $e_i\mapsto E_{ii}$ and $(i\to i\pm 1) \mapsto E_{i,i\pm 1}$. -In fact one can show that every path algebra over a strongly connected finite quiver has a matrix ring as a quotient: Factoring out all "loops" (i.e. imposing the relation $i_1\to i_2 \to \ldots \to i_k \to i_1=e_{i_1}$ for every closed loop in the quiver) gives a matrix algebra where the vertex idempotents become the diagonal idempotents in the matrix ring. -It is not even necessary to quotient out all loops as can be seen be the example with the linear quiver. One has to quotient out just enough loops that all other loops lie in the ideal generated by that. -This feels like a homotopy condition: Consider the quiver as one dimensional CW-complex and glue in a 2-cell to some loops you want to get rid of. If the resulting complex is simply connected, the corresponding path algebra should be a matrix ring. -Now my questions: - -This naive form of the statement seems not to be true because it does not take into account the orientations of the loops. Is there a way to make this homotopy feeling precise? (If the quiver is symmetric in the sense that edges come in pairs of opposite orientation then I think I have proved the statement.) -Is there a more general version of this line of thinking? Maybe some result that says that if two quivers are homotopy equivalent and the relations are in compatible in some sense then the corresponding path algebra quotients are morita equivalent? - -REPLY [10 votes]: Once upon a time I noticed roughly this but didn't know what to do with it. I would rephrase as follows. -Any category $C$ has an associated "category algebra" $k[C]$ spanned by the morphisms of $C$ where the product of two morphisms is $0$ if they can't compose and their composition otherwise. This algebra is unital iff $C$ has finitely many objects; in this case it can be thought of as the endomorphism algebra of a formal direct sum of all of the objects of $C$. If $C$ is the one-object category associated to a group $G$, then $k[C] \cong k[G]$. Somewhat more generally the following is true. - -Theorem: Let $C$ be a groupoid with finitely many objects. Suppose $C$ is the disjoint union of connected groupoids $C_i$ with $n_i$ objects, each of which is isomorphic, and each of which has automorphism groups $G_i$. Then -$$k[C] \cong \bigoplus_i M_{n_i}(k[G_i]).$$ -In particular, the Morita equivalence class of $k[C]$ only depends on the homotopy type (by which I mean equivalence class under equivalence of categories) of $C$, and $k[C]$ is a matrix algebra over $k$ iff $C$ is contractible (by which I mean equivalent to the terminal groupoid). - -I don't have a conceptual explanation of this, though. By a conceptual explanation I mean one would want to upgrade $C \mapsto k[C]$ to a $2$-functor from the $2$-category $\text{Cat}$ of categories, functors, and natural transformations to the $2$-category $\text{Bimod}$ of rings, bimodule categories, and bimodule homomorphisms (equivalence in this $2$-category is Morita equivalence), but as far as I can tell $C \mapsto k[C]$ is not even a functor! (The problem is that functors can cause morphisms which were previously not composable to become composable.) Perhaps one should instead talk about the free $k$-linear category on $C$... - -Edit: Aha! We should definitely instead be talking about the free $k$-linear category on $C$, which I will confusingly also denote $k[C]$. Sorry about that. -But now $C \mapsto k[C]$ is a $2$-functor. It takes values in the $2$-category of $k$-linear categories, functors, and natural transformations, and so in particular it sends equivalent categories to equivalent $k$-linear categories. Now the problem reduces to understanding why equivalent $k$-linear categories give rise to Morita equivalent rings. It will be helpful, but not necessary, to be aware that Morita theory naturally generalizes to $k$-linear categories as follows. - -Theorem: Let $R, S$ be $k$-linear categories (thought of as generalizations of $k$-algebras, which we reduce to when $R, S$ have one object). The categories $\text{Mod}(R), \text{Mod}(S)$ of functors $R^{op} \to k\text{-Mod}, S^{op} \to k\text{-Mod}$ (thought of as either analogues of presheaves or generalizations of right modules) are equivalent if and only if $R$ and $S$ have equivalent Cauchy completions. - -Here the Cauchy completion of a $k$-linear category $R$ is obtained from $R$ by first adjoining formal direct sums and then splitting idempotents. It is a slight generalization of the Karoubi envelope, which figures in the corresponding theorem for ordinary categories. -Example. Let $R$ be the one-object $k$-linear category associated to a ring which I will, again confusingly, also denote by $R$. Adjoining formal direct sums gives us the category of finitely generated free right $R$-modules, while splitting idempotents gives us the category of finitely generated projective right $R$-modules. -Now we just need one more observation. -Observation: Let $R$ be a $k$-linear category with finitely many objects. Then $\text{Mod}(R)$ is equivalent to the module category of the endomorphism ring of the formal direct sum of the objects of $R$. -You can just verify this directly or, a little more abstractly, observe that the identity morphisms of the objects of $R$ become idempotents in the endomorphism ring, and roughly speaking splitting these idempotents gives us our original objects back. -So, to recap: - -Suppose $C$ and $D$ are equivalent groupoids with finitely many objects. -Then the free $k$-linear categories $k[C]$ and $k[D]$ are equivalent, and hence have equivalent module categories. -Since $k[C]$ and $k[D]$ have finitely many objects we can replace them with the endomorphism rings of the formal direct sum of all their objects, and the result is two Morita equivalent rings.<|endoftext|> -TITLE: Is there a $K(0)$-local Rezk logarithm? -QUESTION [10 upvotes]: If $R$ is a $K(n)$-local $E_\infty$-algebra, then a construction of Rezk gives a natural transformation -$$ \mathfrak{gl}_1(R) \to R,$$ -by using the equivalence (arising from the Bousfield-Kuhn functor) $L_{K(n)}\mathfrak{gl}_1(R) \simeq R$. This construction makes sense for $n \geq 1$. Is there an equivalent natural transformation (at the level of model or $\infty$-categories) for $n = 0$? (In Rezk's paper on the logarithm, it is noted that one can define a map on the level of cohomology theories by using the usual formula for the logarithm, so one gets a functor at the level of homotopy categories.) - -REPLY [8 votes]: You could try to build an analogue of the Bousfield-Kuhn functor rationally. -Given a pointed space $X$, associate to it a rational spectrum $P(X)$, with the property that for simply connected $X$, $\pi_*P(X)\approx \pi_*X_{\mathbb{Q}}$. Roughly, define $P(X)$ to be the derived primitives of the cocommutative coalgebra $C_*(X;\mathbb{Q})$ of rational chains on $X$. Since I want to get a spectrum, it would be better to think of the rational chains as $H\mathbb{Q}\wedge \Sigma^\infty X$, so that $P(X)$ is a rational spectrum equipped with a map -$$ -P(X) \to H\mathbb{Q}\wedge \Sigma^\infty X. -$$ -If $X=\Omega^\infty Y$ for a spectrum $Y$, then there is a natural map -$$ -\alpha\colon P(\Omega^\infty Y) \to H\mathbb{Q}\wedge Y. -$$ -One should show that alpha is an equivalence if $Y$ is $0$-connected. -Once you have this (and I don't claim to know how to do this properly myself, but surely the know-how to do it exists), then you can define a logarithm. It will have the form -$$ -\mathrm{gl}_1(R)\langle 1\rangle \to R_{\mathbb{Q}}; -$$ -the domain is the connected cover of the units spectrum.<|endoftext|> -TITLE: Are there trees for $(\Sigma^2_1)^{\text{uB}}$? -QUESTION [9 upvotes]: If there is a proper class of Woodin cardinals, then Woodin showed (using stationary towers) that $(\Sigma^2_1)^{\text{uB}}$ statements are generically absolute, where $\text{uB}$ denotes the pointclass of universally Baire sets of reals. -This generic absoluteness result has a more local version: if $\lambda$ is a limit of Woodin cardinals, then -$(\Sigma^2_1)^{\text{uB}_\lambda}$ statements are generically absolute for posets of size less than $\lambda$, where $\text{uB}_\lambda$ denotes the pointclass of $\lambda$-universally Baire sets of reals (or what some people would call $\mathord{<}\lambda$-universally Baire sets of reals.) -The "local" generic absoluteness for $(\Sigma^2_1)^{\text{uB}_\lambda}$ can be explained in terms of trees (although the proof uses stationary towers instead.) -More precisely, let $\varphi(v)$ be a formula in the language of set theory expanded by a unary predicate symbol. For every limit $\lambda$ of Woodin cardinals there is a tree $T_{\varphi,\lambda}$ such that in every generic extension $V[g]$ by a poset of size less than $\lambda$ we have -$$ V[g] \models p[T_{\varphi,\lambda}] = \{x \in \mathbb{R} : \exists A \in \text{uB}_\lambda\,(\text{HC}; \mathord{\in},A) \models \varphi[x]\}.$$ -This tree is obtained from the scale property for the pointclass $\Sigma^2_1$ of the derived model of $V$ at $\lambda$. -Given these trees $T_{\varphi,\lambda}$, the "local" generic absoluteness follows by a standard argument using the absoluteness of well-foundedness. -My question is, can the "global" generic absoluteness for $(\Sigma^2_1)^{\text{uB}}$ also be explained in terms of trees, assuming that there is a proper class of Woodin cardinals? More precisely, is there a single proper-class-sized tree $T_\varphi$ such that in every generic extension $V[g]$ we have -$$ V[g] \models p[T_{\varphi}] = \{x \in \mathbb{R} : \exists A \in \text{uB}\,(\text{HC}; \mathord{\in},A) \models \varphi[x]\}?$$ -I can think of two possible approaches, both with apparently serious problems. - -Consider the "derived model at $\text{Ord}$." Problem: this doesn't really exist. -Define $T_\varphi$ as the amalgamation of the trees $T_{\varphi,\lambda}$ for various $\lambda$, e.g. all limits of Woodin cardinals, or all limit of Woodin cardinals above some point. Problem: I don't see any way to show that the projection of such an amalgamated tree in some generic extension $V[g]$ is not too large. - -REPLY [4 votes]: The answer is yes. Hugh Woodin showed me the following argument, which I post here with his permission. -Let $\varphi(v)$ be a formula in the language of set theory expanded by a unary predicate symbol. Given a pair of ordinals $(\alpha, \beta)$, -working in $V^{\text{Col}(\omega,\alpha)}$ -we let $B$ be a universally Baire set of reals having Wadge rank $\beta$ -in the model $L(B,\mathbb{R})$, which satisfies $\mathsf{AD}^+$. Note that this model depends only on $\beta$ and not on $B$, and also that every set of reals in $L(B,\mathbb{R})$ is universally Baire because $B^\sharp$ exists and is universally Baire. Let $T_{\alpha,\beta}$ be the tree of a $(\Sigma^2_1)^{L(B,\mathbb{R})}$-scale on the set -$$\{x \in \mathbb{R} : \exists C \in L(B,\mathbb{R})\, (\text{HC}; \in, C) \models \varphi[x]\}.$$ -By the homogeneity of $\text{Col}(\omega,\alpha)$ this tree is is independent of the choice of generic filter and we have $T_{\alpha,\beta} \in V$. -Let $T$ by the amalgamation of all the trees $T_{\alpha,\beta}$, so that $T$ is a tree on $\omega \times \text{Ord}$ and $p[T] = \bigcup_{\alpha,\beta \in \text{Ord}} p[T_{\alpha,\beta}]$ in every generic extension of $V$. -We claim that -$$ V^{\text{Col}(\omega,\alpha)} \models p[T] = \{x \in \mathbb{R} : \exists C \in \text{uB}\, (\text{HC}; \in, C) \models \varphi[x]\},$$ -for every ordinal $\alpha$. The right-to-left inclusion follows immediately from the definition of the trees $T_{\alpha,\beta}$, -so it remains to prove the left-to-right inclusion. -Let $G \subset \text{Col}(\omega,\alpha)$ be a $V$-generic filter and let $x \in p[T]^{V[G]}$, say $x \in p[T_{\alpha',\beta'}]$ for ordinals $\alpha'$ and $\beta'$. We want to show -\begin{equation*}\tag{$*$} -\exists C \in \text{uB}^{V[G]}\, (\text{HC}^{V[G]}; \in, C) \models \varphi[x]. -\end{equation*} -If $\alpha' = \alpha$, this is easy. -There are two remaining cases to consider: - -$\alpha' > \alpha$. -$\alpha' < \alpha$. - -In case (1), we have ($*$) by $(\Sigma^1_2)^{\text{uB}}$ generic absoluteness for $\text{Col}(\omega,\alpha')$. -In case (2), we use the fact that if -$B \in V[G \restriction \alpha']$ is a universally Baire set as in the definition of the tree $T_{\alpha', \beta'}$, then $B^\sharp$ exists and is universally Baire, so there is an elementary embedding -$$ j : L(B, \mathbb{R}^{V[G \restriction \alpha]}) \to L(B^{V[G]}, \mathbb{R}^{V[G]}),$$ -and we have $j(T_{\alpha', \beta'}) = T_{\alpha, \beta}$ where $\beta$ is the Wadge rank of $B^{V[G]}$. Considering the pointwise image of a branch witnessing $x \in p[T_{\alpha',\beta'}]$, we have $x \in p[T_{\alpha,\beta}]$ . Therefore ($*$) is witnessed by a set of reals $C \in L(B^{V[G]}, \mathbb{R}^{V[G]})$.<|endoftext|> -TITLE: Is equivalence of functions built from nested exponentiations a decidable problem? -QUESTION [11 upvotes]: Let $\mathcal{E}$ be the minimal set of symbolic expressions (without any predefined meaning) such that - -The symbol $x$ is in $\mathcal{E}$, and -If expressions $P,Q\in\mathcal{E}$, then the superscript expression $(P)^{(Q)}\in \mathcal{E}$. - -For every expression $S\in\mathcal{E}$, define the interpretation of $S$ as the function $\mathbb{R}^+\to\mathbb{R}^+$ given by $x\mapsto S$, where each occurrence of $x$ in $S$ is considered as the function parameter, and superscript expressions are considered as exponentiations. Note that we concern ourselves only with positive values of $x$. -We can make an observation that interpretations of two structurally distinct expressions can be the same function, e.g. the expressions -$$\left((x)^{(x)}\right)^{\left((x)^{(x)}\right)},\ \left((x)^{\left((x)^{(x)}\right)}\right)^{(x)}$$ - both have the same function $x\mapsto x^{x^{x+1}}$ as their interpretation. - -Questions: - -Is there an algorithm that for every pair of expressions from $\mathcal{E}$ can decide whether they have the same interpretation? -If yes, can we give an explicit example of such algorithm? - -REPLY [6 votes]: This is a justification of the algorithm suggested in Dan Turetsky's comment. -Every expression in $E$ reduces to an expression in $E'$ which is the minumum language such that $x\in E'$ and $x^{(p_1*\dots*p_n)}\in E'$ whenever $p_1,\dots,p_n\in E'$. Let us define a linear order on $E'$ as follows. -First, every expression in $E'$ has a level, which is the maximum number of nested exponentiations. More precisely, the level of $x$ is 0, and the level of $x^{(p_1*\dots*p_n)}$ is the maximum of the levels of $p_i$'s plus one. Let $E_n$ denote the set of expresions of level at most $n$. -We define our linear order on $E_n$ by induction. First of all, $x$ is the minimal element of our order. Let $f,g\in E_n$ and $f,g\ne x$. Then $f=x^{p_1*\dots*p_m}$ and $g=x^{q_1*\dots*q_k}$ where $p_i$'s and $q_j$'s are from $E_{n-1}$. Let us sort $p_i$'s and $q_j$'s in non-ascending order with respect to the (already defined) linear order on $E_{n-1}$. Then we say that $f>g$ iff the sequence $(p_i)$ is lexocographically bigger than $(q_j)$, i.e. $p_l>q_l$ where $l$ is the first index of mismatch (or $(q_j)$ is an initial subsegment of $(p_i)$). Clearly this definition is consistent with its previous stage, so we have a well-defined linear order on $E'$. This order is easy to compute. -I claim that, if $f>g$ in this linear order, then $f$ dominates $g$ at infinity, i.e. $f(x)>g(x)$ for all sufficiently large $x\in\mathbb R_+$. Moreover, $f$ dominates $g^a$ for any $a\in\mathbb N$. The latter claim is easy to prove by induction. Indeed, $x^x$ dominates $x^a$, and the induction step goes as follows. Let $f=x^{p_1*\dots*p_m}$ and $g=x^{q_1*\dots*q_k}$ and $f>g$. We need to prove that $f$ dominates $g^a$, or, equivalently, that the product $p_1\dots p_m$ dominates $aq_1\dots q_k$. By removing common initial terms we may assume that $p_1>q_1$ in our order. (If all $q_j$'s cancel out completely, it remains to prove that $p_1$ dominates the constant $a$ and this is trivial.) Now by induction hypothesis, $p_1$ dominates $q_1^{k+1}$, which dominates $q_1^2q_2\dots q_k$, which dominates $aq_1\dots q_k$, q.e.d.<|endoftext|> -TITLE: Morita theorem for simplicial rings -QUESTION [7 upvotes]: My question is the following: is there an analog of Morita theorem in the simplicial setting? -I mean, we can define two simplicial rings $A,B$ to be simplicially Morita equivalent is the categories of simplicial modules $Mod(A)$ and $Mod(B)$ are equivalent (or maybe equivalent as simplicial categories?). Is there a result similar to the classical Morita theorem which gives a (simple) criterion for two rings to be Morita equivalent? -I tried to google, but I couldn't find anything useful. So I think maybe such an extension is just some trivial formality? Maybe it can easily be obtained from some well-known general result? -I have asked this question on math.stackexchenge, but didn't receive an answer (though comments there suggest that something like that should be true). -Thank you very much! - -REPLY [4 votes]: Recall that if $A$ is a ring and $\mathcal{C}$ is a cocomplete $\mathsf{Ab}$-category, then cocontinuous $\mathsf{Ab}$-functors $F : \mathsf{Mod}(A) \to \mathcal{C}$ correspond to left $A$-module objects $(M,\theta)$ in $\mathcal{C}$ (i.e. $M \in \mathcal{C}$ and $\theta : A \to \mathrm{End}(M)$ is a ring homomorphism). You can either prove this directly (define $M=F(A)$ etc.), or using that $\mathsf{Mod}(A)$ is the free $\mathsf{Ab}$-enriched cocompletion of the one-object $\mathsf{Ab}$-category $A$. The Theorem of Eilenberg-Watts is the special case $\mathcal{C}=\mathsf{Mod}(B)$ for another ring $B$. The special case of equivalences of categories is the Morita Theorem. -But free cocompletions work for an arbitrary cosmos. For instance we can take the category of simplicial abelian groups $\mathsf{simpAb}$. It follows that if $A$ is a simplicial ring and $\mathcal{C}$ is a cocomplete $\mathsf{simpAb}$-category, then cocontinuous $\mathsf{simpAb}$-functors $\mathsf{Mod}(A) \to \mathcal{C}$ correspond to $A$-module objects in $\mathcal{C}$. We get Eilenberg-Watts and then also Morita Theorem for simplicial rings: $\mathsf{Mod}(A) \simeq \mathsf{Mod}(B)$ as simplicial categories iff there is a simplicial $(A,B)$-bimodule $M$ and a simplicial $(B,A)$-bimodule $N$ such that $M \otimes_B N \cong A$ and $N \otimes_A M \cong B$.<|endoftext|> -TITLE: Large gaps between consecutive irreducible polynomials with small heights -QUESTION [6 upvotes]: For a prime gap of length at least $n$, a trivial upper bound for its first occurrence is $N=n!$ or $N=lcm(2,\dots,n)$. A bit better is $N=p_1\cdots p_n$ where $p_k$ is the $k$th prime, as then $N+2,\dots,N+(p_{n+1}-1)$ are all composite. In “real life” however, the first big gaps will occur much earlier, e.g. between $113$ and $127$, which is even below $ 2\cdot3\cdot5\cdot7=210$ (see http://oeis.org/A002386). -I won’t ask for better bounds for early occurrences. I rather have a corresponding question concerning irreducible polynomials. It started with a question on Math SE here (but note that I have 'reversed' the polynomials here for convenience of notation.) -For given $k$, we'll deal with integer polynomials $P=P(x)=a_nx^n+\cdots+a_1x$ such that $P,P+1,…,P+k-1$ are all reducible. Call such a polynomial a $k$-gap. -A trivial construction, corresponding somewhat to the initial one for prime gaps above, is $P=x(x+1)\cdots(x+k-1)+x$, so $P+j$ has a factor $(x+j)$. The downside of this construction is that the coefficients, i.e. the Stirling numbers, are growing much faster than $k$. -It seems however that $k$-gaps with very small (absolute) coefficients do exist, at least for $k=3,4,5,6$. For example, there is - a $4$-gap $P=x^4+x^3-3x^2-2x$, - a $5$-gap $P=x^5+x^4-4x^3-3x^2+4x$, - a $6$-gap $P=x^{10}-x^9-2x^8-2x^7-3x^6-x^5-x^4+x^3+2x^2+x$. -Note that I make no restriction on the degree of such a polynomial. I haven't yet found a $6$-gap with smaller degree. -For $k=3,4,5$ we even have $k$-gaps with all coefficients in $\lbrace-1,0,1\rbrace$. Examples: -a $3$-gap $P= x^5+x$ -a $4$-gap $P= x^7-x^6-x^4+x^2+x$ -a $5$-gap $P= -x^{12}+x^{11}-x^8-x^6+x^2+x$ -(no such $6$-gap found yet.) -To ask a hopefully more feasible question than that: - -Is there a construction that yields for any given $k$ a $k$-gap with relatively small coefficients, say $|a_j|< k$ for all $j$? - -REPLY [8 votes]: At the cost of having the degree be very large you can always choose a $k$-gap with coefficients in $\lbrace 0,1\rbrace$. Pick a large $n$ so that $n\equiv -j\pmod{p_j}$, for all $1\le j\le k$. Where $p_j$ is the $j$th prime. Let $P(x)=x^{a_1}+\cdots+x^{a_n}$ and choose the exponents $a_i$ by the chinese remainder theorem, so that they are all distinct and so that - -Exactly $\frac{n+1}{2}$ of the $a_i$'s are odd, so that $P(x)+1$ is divisible by $x+1$. -Exactly $\frac{n+2}{3}$ of $a_i$'s are $1\pmod{3}$, and similarly for $2\pmod{3}$, so that $P(x)+2$ is divisible by $x^2+x+1$. -Similarly for other $j\le k$, have the exponents be equally distributed $\pmod {p_j}$ with $\frac{n+j}{p_j}$ in each non-zero residue, so that $P(x)+j$ is divisible by $x^{p_j-1}+\cdots +1$.<|endoftext|> -TITLE: Pre-images of unipotent elements in $\operatorname{SL}_{n}(A)$ -QUESTION [7 upvotes]: The starting point of this question is the (presumably) well-known theorem (the proof I know is from Abelian $\ell$-adic representations and elliptic curves from J-P.Serre in which it is a lemma for $n=2$ and an exercise for $n>2$; which suggests that the result was already classical in the 60s). -Theorem I: If $G$ is a closed subgroup of $\operatorname{SL}_{n}(\mathbb Z_{p})$ which surjects on $\operatorname{SL}_{n}(\mathbb F_{p})$ and if $p≥5$, then $G=\operatorname{SL}_{n}(\mathbb Z_{p})$. -The theorem is optimal with respect to all hypotheses in the sense that there exists proper subgroups of $\operatorname{SL}_{2}(\mathbb Z_{p})$ mapping onto $\operatorname{SL}_{2}(\mathbb F_{p})$ when $p=2,3$ (a fact that played a role in the original proof of the modularity of semi-stable elliptic curves by A.Wiles, if I am not mistaken) and in the sense that for all $n≥2$ and all prime $p$, there exist discrete valuation rings $A$ of mixed characteristic $(0,p)$ such that $\operatorname{SL}_{n}(A)$ contains proper subgroups mapping onto $\operatorname{SL}_{n}(A/\mathfrak m)$ (just take a completely ramified $A$ over $\mathbb Z_{p}$ and consider $\operatorname{SL}_{2}(\mathbb Z_{p})$ inside $\operatorname{SL}_{2}(A)$). -A slightly less well-known fact is that the theorem admits the following generalization, due to N.Boston. -Theorem II: Let $A$ be a complete local noetherian ring with finite residual characteristic $p\neq 2$. If $G$ is a closed subgroup of $\operatorname{SL}_{n}(A)$ which surjects on $\operatorname{SL}_{n}(A/\mathfrak m^2)$, then $G=\operatorname{SL}_{n}(A)$. -This is in the appendix of On p-adic analytic families of Galois representations. -Compositio Math. 59 (1986), no. 2, 231–264 by B.Mazur and A.Wiles and again, this theorem is optimal in the sense that there exists a proper subgroup of $\operatorname{SL}_{2}(\mathbb Z_{2})$ surjecting on $\operatorname{SL}_{2}(\mathbb Z/4\mathbb Z)$. -Now my actual question. Let $A$ be a complete local noetherian ring (UPDATE: domain, actually) of mixed characteristic $(0,p)$ with $p≠2$ UPDATE: which one can assume to be a discrete valuation ring if necessary. Suppose $G$ is a closed subgroup of $\operatorname{SL}_{n}(A)$ which surjects on $\operatorname{SL}_{n}(A/\mathfrak m)$. - -Among the pre-images in $G$ of non-identity unipotents elements in $\operatorname{SL}_{n}(A/\mathfrak m)$, is it true that there exists a unipotent element (that is to say an element $\sigma$ such that $\sigma-1$ is nilpotent)? - -Granted theorem II, this is obviously true if $G$ maps onto $\operatorname{SL}_{n}(A/\mathfrak m^2)$. Without this hypothesis, it looks dubious to me but nevertheless, the obvious counterexamples to theorem I coming from ramified rings do not provide counterexamples to this claim and I confess that I don't quite know how to construct other counter-examples. Taking this into account, another more general point of view on the question would be the following. - -What are the subgroups of $\operatorname{SL}_{n}(A)$ which do not map onto $\operatorname{SL}_{n}(A/\mathfrak m^2)$ but which do map onto $\operatorname{SL}_{n}(A/\mathfrak m)$? - -Perhaps group cohomology of $\operatorname{SL}_{n}$ would help then. Any positive result, even in the case $n=2$ and $p>3$ would already be of interest to me. -UPDATE: Jim Humphreys asks in comments what I mean by a unipotent element in a matrix group. I meant an element $\sigma$ such that $\sigma-1$ is nilpotent, but I realize now that this might not be so great a definition when the ring of coefficients is not a domain, or at least reduced. So perhaps it is better to take $A$ a domain in the above, or even a discrete valuation ring. Also, even though Tim Dokchister answered the first question in the negative, I would be interested to know if someone has something to say for infinite closed subgroup $G$. - -REPLY [4 votes]: For the first question, at least for $p=5, n=2$ there is a counterexample.: -The group $\text{SL}_2({\mathbb F}_5)$ has a 2-dimensional symplectic representation with character in ${\mathbb Q}(\sqrt 5)$ and Schur index 2, so it is realizable over $A={\mathbb Z}_5(\zeta_5)$. So $\text{SL}_2({\mathbb F}_5)$ injects in $\text{SL}_2(A)$ and reduces onto $\text{SL}_2(A/{\mathfrak m})$. But it is just a finite group, so it has no non-identity unipotent elements. -I don't know whether it generalizes to higher $p$ and $n$ though, and whether there are examples with `interesting' infinite closed subgroups of $\text{SL}_2(A)$.<|endoftext|> -TITLE: Strata of the Affine Grassmannian -QUESTION [7 upvotes]: Let $G$ be a connected, simply-connected complex semisimple linear algebraic group, and denote by $\mathcal{G}$ its affine Grassmannian. Fix a maximal torus $T\subseteq G$. We know that $\mathcal{G}$ has a natural stratification $\{\mathcal{G}^{\lambda}\}_{\lambda\in\Lambda^+}$, where $\Lambda^+$ is the collection of dominant coweights of $T$. I am looking for a detailed formulation of the result that $\mathcal{G}^{\lambda}$ is an affine bundle over a partial flag variety. I would appreciate any and all references. - -REPLY [5 votes]: One can also argue as follows: -$G(\mathcal{O})$ operates transitivly on each stratum $\mathcal{G}^\lambda$. The isotropy group at $t^\lambda$ is $P^a_\lambda:= (t^\lambda)^{-1}G(\mathcal{O})t^\lambda\cap G(\mathcal{O})$. Let $P_\lambda$ denote the parabolic subgroup of $G$ with Levi factor $\lambda$. Then the evaluation at zero homomorphism $ev_0:G(\mathcal{O})\to G$ restricts to an epimorphism $P^a_\lambda \to P_\lambda$ with pro-unipotent kernel $U$. More over $ev_0$ induces a map $G(\mathcal{O})/P_\lambda^a\to G/P_\lambda $ with fiber $(\ker ev_0)/U$ which is easily seen to be affine.<|endoftext|> -TITLE: To what extent has the Haar measure been generalized? -QUESTION [12 upvotes]: It is known that all locally compact groups, and therefore compact groups, have a left-invariant Haar measure which is unique up to scalar constant, also a right-invariant one. Is there a strictly wider class of groups that has such a measure? What about weakening of these measures? What about hypergroups? -Thank you. HTTT. - -REPLY [12 votes]: For hyperfinite groups - ultraproducts $G$ of sequences of finite groups $G_n$ - one has the Loeb probability measure, which is analogous to Haar measure in that it is a bi-invariant probability measure. The catch though is that the Loeb measure is not measurable with respect to a Borel sigma algebra (indeed, there is no natural topology to place on a hyperfinite group), but instead on the Loeb sigma algebra generated by applying the Caratheodory construction to the Boolean algebra of internal sets (the ultraproduct of subsets $E_n$ of $G_n$, which have measure equal to the (standard part of the) ultralimit of $|E_n|/|G_n|$). -One way to think of this is that while a hyperfinite group $G$ is not a topological space, it is a sigma-topological space - it has a collection of "open" sets (countable union of internal sets) which obey a weakened version of the topology axioms in which one has closure only under countable unions rather than arbitrary unions. Loeb measure is then the analogue of Haar measure with respect to this sigma-topology. I discuss this a bit in my paper with Bergelson. -One can also generalise from the hyperfinite case to ultraproducts of compact groups, or even locally compact groups if one normalises things properly.<|endoftext|> -TITLE: Lattice points and convex bodies -QUESTION [14 upvotes]: Given are two convex bodies $K, L \subset \mathbb{R}^n$ that contain the origin as an interior point. Assume the number of integer points contained in $\lambda K$ equals the number of integer points contained in $\lambda L$ for every $\lambda > 0$. Is $K$ equal to $L$ modulo unimodular transformations? -If not, in what way is $K$ similar to $L$ besides the well-known fact that they must have the same volume? -Addendum: Anton's example shows that $K$ and $L$ are not necessarily equal modulo unimodular transformations even when they are taken to be two integral polygons in the plane and Abhinav Kumar points out in the comments that in the case $K$ is a integral polytope the number of integer points inside $n K$ ($n$ a natural number) is the value of the Ehrhart polynomial of $K$ at $n$. -Added 21/11/2013: For $K$ and $L$ ellipsoids this problem translates to "what can we say about isospectral flat Riemannian tori?". Therefore, by a famous remark by Milnor there exist ellipsoids $K$ and $L$ in $\mathbb{R}^{16}$ that are not unimodularly equivalent and for which the number of integer points contained in $\lambda K$ equals the number of integer points contained in $\lambda L$ for every $\lambda > 0$. - -REPLY [6 votes]: In this neat paper by C. Haase & T. McAllister (which appeared in Cont. Math. 451 (2008)) you can find many examples, but as already mentioned in Anton's answer, the question of equidecomposability is open (and in fact, the authors offer a conjecture about this).<|endoftext|> -TITLE: Effective Camille Jordan -QUESTION [10 upvotes]: It is a well-known (and frequently used) theorem of C. Jordan that a proper subgroup $H$ of a finite group $G$ can not intersect every conjugacy class. This is used most frequently for $G=S_n.$ Now, the question: for $H$ a subgroup of $S_n$ what are the possible numbers of cycle types $H$ can have? One might conjecture that there is $A_n$ which has a lot of cycle types, and then there is a big drop, and then there are some sort of further "phase transitions". There might also be big difference between transitive and intransitive and primitive and imprimitive subgroups. Is there? - -REPLY [6 votes]: It's difficult to come up with precise results for general subgroups. For example $S_{n-1}$ has lots of cycle types! -You can say more for transitive groups, and even more for primitive groups. Jordan proved in 1873 that, if $G \le S_n$ is primitive and contains a $p$-cycle with $p$ prime and $p216$).<|endoftext|> -TITLE: Action of $\pi_1(S)$ on its commutator subgroup -QUESTION [5 upvotes]: Let $G$ be a group. It acts canonically on its derived subgroup by conjugation. Can on describe the orbits of this action when $G$ is the fundamental group of a compact orientable surface of genus $g \geq 2$ ? Especially, is the number of orbits finite ? -$\textbf{Edit : Another question has arisen : what about the dynamic of $\pi_1(S)$ on $D\pi_1(S) / D^2\pi_1(S)$ ?} $ - -REPLY [4 votes]: I'm answering the edit: what is the action of $G=\pi_1(S)$ on $G'/G''$? A first thing to notice is that $G'$ acts trivially on $G'/G''$, so the action factors through to an action of $G/G'$ on $G'/G''$. The $\mathbb{Z}[G/G']$-module $G'/G''$ is known as the first Alexander module of $G$. (If $G$ is the fundamental group of $S^3-K$ for a knot, then this module is where the Alexander polynomial comes from.) -Topologically, $G/G'=H_1(S)$ and $G'/G''=H_1(\overline{S})$, where $\overline{S}$ is the universal abelian cover of $S$. That is, it is the cover associated to the abelianization homomorphism $G\to H_1(G)$, namely the cover with $\pi_1(\overline{S})=G'$ by the correspondence between subgroups and covers. A way to imagine the action of $G/G'$ on $G'/G''$ is that $H_1(S)$ acts on $H_1(\overline{S})$ as the group of deck transformations. We will characterize $H_1(\overline{S})$ as a $\mathbb{Z}[H_1(S)]$-module. -If $S$ is a compact surface with nonempty boundary, its fundamental group is the free group $F_k$ on some number of generators $k$. If $F_k=\langle g_1,g_2,\dots,g_k\rangle$ is the free generating set, the abelianization is a map $F_k\to \mathbb{Z}^k$ sending $g_i$ to a generator $t_i$. The ring $\mathbb{Z}[\mathbb{Z}^k]$ is isomorphic to $\mathbb{Z}[t_1^{\pm 1},\dots,t_k^{\pm 1}]$, the Laurent polynomial ring in $k$ variables with integer coefficients. I can't really explain the following calculations in enough detail here, other than what I am doing is calculating with $C_i(\overline{X_G})\cong \mathbb{Z}[H_1(G)]\otimes C_i(X_G)$ for a presentation complex $X_G$ of $G=\pi_1(S)$. For example, $t_ig_1$ represents the copy of $g_i$ (a distinguished lift of the $1$-cell for $g_i$ to $\overline{X_G}$) shifted by the deck transformation for $t_i$. (I might post a link to some notes about this at some point.) The complex only has $0$- and $1$-cells, so we need only identify the kernel of $\partial_1$, where $\partial_1(g_i)=t_i-1$. Since these are all pairwise coprime polynomials, $\ker\partial_1$ is minimally generated by $(t_j-1)g_i-(t_i-1)g_j$ for all $1\leq i -TITLE: Stable matchings when switches have costs -QUESTION [12 upvotes]: The Gale-Shapley algorithm finds a stable matching in the complete bipartite graph, for any preference matrix. It's also well-known that stable matchings don't always exist in the complete graph (roommates problem), and it seems to be still open whether they almost surely don't exist in a uniformly random instance of the latter. See -http://arxiv.org/pdf/cond-mat/0509221.pdf -Given an instance of the marriage or roommates problem, call a matching \emph{$t$-stable} if there is no unmatched pair $\{x,y\}$ such that $x$ ranks $y$ at least $t$ places higher than his match, and similarly for $y$. There are a couple of obvious motivations for considering this extension: (i) imagine that there is a cost associated with switching partners (e.g.: the time and money it takes to physically move, or the "social cost" incurred to one's reputation as an unreliable roommate, or the cost imposed by the laws of the society - in Ireland, where I'm from, divorce was illegal up to the mid-1990s), (ii) maybe each person has a bounded ability to distinguish beteen alternative partners. Hence, someone won't switch unless the alternative is significantly better. -Now there are some obvious questions one can ask, in particular in the roommates problem, where $1$-stable matchings don't always exist. For example: -Question 1: What is the smallest function $t=t(n)$ such that every instance of the roommates problem on $n$ nodes has a $t$-stable matching ? -Question 2: Determine a threshold function $t=t(n)$ such that a random instance of the roommates problem on $n$ nodes almost surely has a $t$-stable matching when $t \gg t(n)$, and almost surely doesn't when $t \ll t(n)$. -Question 3: Given $t,n$, say $n$ even, describe an algorithm which decides whether an instance of the roommates problem on $K_n$ has a $t$-stable matching or not, and finds one if it has. -I have not been able to locate anything at all on this notion in the literature. Remarks ? By the way, regarding Question 1, I think I can construct an instance on $K_{n^2}$ where there is no $n$-stable matching. Is this best-possible ? Regarding Question 2, since it's not even known if $1$-stable matchings a.s. don't exist, finding the right threshold is probably hard. But can we bound it from above ? Regarding Question 3, Irving's algorithm for $1$-stable matchings doesn't seem to extend trivially to $t$-stability. - -REPLY [2 votes]: As a first approach to answering Question 3 we can consider the following integer programming formulation: -$\sum _{j}x_{i,j}=1\; \; \; \forall i,$ -$x_{i,j}+\sum _{l:l<_{j}i}x_{l,j}+\sum _{l:l<_{i}j}x_{i,l}\leq 1\; \; \; \forall i,j,$ -$x_{i,j}\in \left \{ 0,1 \right \}\; \; \; \forall i,j.$ -See: http://www.columbia.edu/~js1353/marriage.ps p.18 -The only difference between the IP formulation of the stable roommates problem (SRP) and its $t$-stable variant is in an interpretation. The expression "$l<_{j}i$" is interpreted as “$j$ prefers $i$ to $l$” or equivalently: “$i$ is higher, in the $j$'s preference list, than $l$” in the SRP. In the $t$-stable variant of the SRP it should be interpreted as “$i$ is at least $t$ positions higher, in the $j$'s preference list, than $l$". -There are some similarities between $t$-stable matchings and weakly stable matchings (as defined in Irving, R. W. and Manlove, D. F. (2002) The stable roommates problem with ties). It is known that deciding whether an instance of stable roommates problem admits a weakly stable matching is NP-complete. Thus, the above IP formulation may not be as bad as it seems at first.<|endoftext|> -TITLE: Is there a practical criterion to determine whether the limit of a diagram of real chain complexes is also a homotopy limit? -QUESTION [10 upvotes]: Consider a diagram D: I→ChR of real connective chain complexes. -In the example I have in mind all chain complexes are concentrated in some fixed degree n. -There is a canonical map lim D → holim D from the limit of D to the homotopy limit of D. -I would like to have a practical criterion for determining when this map is a quasiisomorphism. -Of course, a sufficient condition for this would be the injective fibrancy of D. -However, injective fibrancy seems to be difficult to check in general. -Thus I'm wondering if the fact that we are in the category of real chain complexes, which has rather nice properties (e.g., every epimorphism splits) -might yield a more practical criterion. -Is there a practical way to check whether the canonical map lim D → holim D for a diagram D: I→ChR of real chain complexes is a quasiisomorphism? -Motivation for this question comes from a desire to compute the homotopy mapping space of simplicial presheaves on the site of smooth manifolds -of the form Hom(F,BnR). -Here Hom denotes the homotopy mapping space (i.e., the simplicial mapping space with the source cofibrantly replaced and the target fibrantly replaced), -BnR denotes the Eilenberg—MacLane object corresponding to the representable presheaf R of real numbers -(note that R here is not discrete), -and F is the smooth singular set of some smooth manifold X, i.e., Fn(U) := {Δn × U → X}. -Although the simplicial presheaf BnR is fibrant in the local projective model structure, -the simplicial presheaf F is not cofibrant, the primary obstacle being the fact that the corresponding simplicial components Fn -are infinite-dimensional (they are the mapping spaces Map(Δn,X)), whereas cofibrancy in the projective structure -requires them to be coproducts of retracts of representables, in particular they must be finite-dimensional. -One can try to circumvent this problem by observing first that any simplicial presheaf is equivalent to the homotopy colimit of its simplicial components Fn: -Hom(F,BnR) = Hom(hocolimn Fn, BnR) = holimn Hom(Fn, BnR). -Since R is contractible, it is plausible to expect that the higher cohomology of Fn with coefficients in R must vanish, -in particular we hope to have Hom(Fn, BnR) = Bn Hom(Fn, R) = Bn C∞(Fn). -To calculate Hom(Fn,BnR) we can use the above trick one more time and observe that every presheaf of sets is equivalent to the homotopy colimit -of its elements, hence we have Hom(Fn, BnR) = Hom(hocolimU→Fn U, BnR) = holimU→Fn Hom(U, BnR) = Γ holimU→Fn C∞(U)[n], -the latter equality coming from the fact that U is representable and therefore cofibrant. (The functor Γ is the Dold—Kan functor that sends a chain complex -to the corresponding simplicial set.) -If the latter homotopy limit coincides with the corresponding limit, then we have holimU→Fn C∞(U)[n] = limU→Fn C∞(U)[n] = C∞(Fn)[n], which answers the original question. - -REPLY [4 votes]: Since all your complexes are concentrated in a fixed degree $n$, the cohomology of the homotopy limit is $H^{n+r}(\operatorname{holim}D)=\lim^rD^n$, hence a criterion would be the vanishing of higher derived limits $\lim^rD^n=0$, $r>0$. This holds for all finite direct categories $I$, for instance, but in general it is complicated to know. I guess this doesn't help much.<|endoftext|> -TITLE: An extreme point of the ball of the space of compact operators -QUESTION [5 upvotes]: It is very easy to see that the unit ball of $c_0$ has no extreme points. I was trying to spot any extreme points in the unit ball of the space of compact operators on a Hilbert space (a non-commutative version of $c_0$) but without success. - -Does the unit ball of $K(\ell_p)$ have any extreme points for some $p\in (1,\infty)$? - -REPLY [8 votes]: For $p\neq 2$ there are plenty of extreme points in $K(\ell_p)$. -J. Hennefeld, Compact extremal operators, Il. J. Math. 21 (1977) 61-65.<|endoftext|> -TITLE: First order Elliptic operator -QUESTION [7 upvotes]: Assume that there exists a first order elliptic operator $D$ acting on functions from $\mathbb{R}^n$ to some vector space $V$. What can we conclude about $V$? -For example, is the dimension of $V$ always greater than $[n/2]$? Is $D$ necessarily some kind of modified Dirac operator? Can we somehow construct a pointwise Clifford module structure on $V$? -To sum it up, I would like to know if there is any truth to the statement "all first order elliptic operators are basically Dirac operators". - -REPLY [10 votes]: Here is a simple way of producing first order elliptic operators $\newcommand{\bR}{\mathbb{R}}$ $C^\infty(\mathbb{R}^n, W)\to C^\infty(\bR^n, W)$ with constant coefficients. Denote by $L(W)$ the space of linear operators $W\to W$. Consider a map $\newcommand{\si}{\sigma}$ -$$\si :\bR^n\to L(W), $$ -such that $\si(x)$ is invertible for any $x\in\mathbb{R}^n\setminus 0$. Denote by $e_1,\dotsc, e_n$ the canonical basis of $\bR^n$. Now set $\si_k=\si(e_k)$ and define $\newcommand{\pa}{\partial}$ -$$ D=\sum_{k=1}^n \si_k \pa_k: C^\infty(\mathbb{R}^n, W)\to C^\infty(\bR^n, W). $$ -This is an elliptic operator because its symbol is the map $\si: \bR^n\to L(W)$, up to a multiplication by $\sqrt{-1}$. Let me point two things. - -All elliptic first order p.d.o.'s with constant coefficients are obtained in this fashion and -The symbol at a point $p \in\bR^n$ of any first order p.d.o. $C^\infty(\mathbb{R}^n, W)\to C^\infty(\bR^n, W)$ is a map $\si_p:\bR^n\to L(W)$ with the above property. - -The problem is then to find the lowest dimensional vector space $W$ such that there exists a linear map $\si: \bR^n\to L(W)$ with the property that $\si(x)$ is invertible for any $x\in\mathbb{R}^n\setminus 0$.If $W$ is real, the minimal dimension is computed by the so called Radon-Hurwitz numbers. For details, I refer to the book Topological Geometry, by I.R. Porteous. At the end of Chapter 13 of the book you can find a discussion on these numbers and the above problem. In any case, the solution is not simple, it is related to the problem of vector fields on spheres and was elucidated by J.F. Adams using $K$-theoretic techniques. -Atiyah has a discussion where he explains that the index theorem for Dirac operators implies the index operators for arbitrary operators. I'll add the reference once I find it. -I have posted on MO a similar question myself a while ago. -Note. In the paper Topology of elliptic operators, Proc. Sympo.Pure. Math., vol. 16, Amer. Math. Soc 1970, p 101-119, Atiyah essentially answers your first question in the complex case, again referring to the work of Adams. -In the paper The index of elliptic operators, Colloquium Lectures, Amer. Math. Soc. Dallas, 1973 Atiyah gives a brief explanation why the index theorem for Dirac operators implies the index theorem in general. Essentially, it has to do with the Bott periodicity theorem. You can find both papers in the 3rd volume of Atiyah's Collected Works.<|endoftext|> -TITLE: If $f$ is $C^{\infty}$ and $f^2$ is analytic, then $f$ is analytic -QUESTION [9 upvotes]: Assume that $f:\mathbb{C}^n\rightarrow \mathbb{C}$ is a $C^{\infty}$ function such that $f^2$ is (complex) analytic. Then one can show that $f$ is analytic. (Note: Liviu Nicolaescu and Alexandre Eremenko have given elegant proofs of this below; my own proof involved the Weierstrass preparation theorem.) -Now, when we consider the question for functions in some quasi-analytic Denjoy-Carleman class these proofs don't carry over: -Question: If $f^2$ is a function in some quasi-analytic Denjoy-Carleman class, then $f$ is quasi-analytic belonging to the same class? -Weierstrass preparation theorem doesn't hold, for $n\geq2$, in quasi-analytic Denjoy-Carleman classes, and it is an open problem whether a $C^\infty$ function that belongs to a quasi-analytic Denjoy-Carleman class along every line belongs to that class. This is part of the difficulty of this problem for $n\geq2$. -Another open question in Denjoy-Carleman classes is about whether ideals are closed. For principal ideals this is related to solving for $f$ in $gf=h$, where $g$ and $h$ are known to belong to the Denjoy-Carleman class. The ideal generated by $g$ would not be closed if we can find a smooth $f$ that doesn't belong to the class such that it gets pushed into the class by the multiplication by $g$. In this way, the question above is about understanding whether a smooth function, not in the class, can be pushed into the class by multiplying by itself. If the square appears composed with $f$ on the other side then it is known to occur. This is, $f(x^2)$ may be in a quasi-analytic Denjoy-Carleman class while $f$ is not. -Note: I put in the tags model-theory and o-minimal because people that work in those areas sometimes have also worked with quasi-analytic functions enough to maybe have an idea for proving it. Subject classification in mathematics doesn't play the same role for exposition than for finding proofs. -Note: Joris' theorem states that if $f^2$ and $f^3$ are smooth functions then $f$ is smooth. This holds for several variables functions. Also, for the quasianalytic case, in the case of one variable, it is again easy to show that if $f$ is smooth and $f^2$ and $f^3$ belong to some quasianalytic Denjoy-Carlmenan class, then $f$ also belong to the same class. For several variables ... who know!? The problem with just $f^2$ is also a subproblem steaming form asking if Joris' theorem is true for quasianalytic DC classes. - -REPLY [22 votes]: Here is a simple proof for complex-analytic case. -If restrictions of $f$ on all complex lines are analytic, then $f$ is analytic. -This reduces the problem to the case $n=1$. Now $f^2$ is analytic so near every point, so -it has a representation $f^2(z)=(z-a)^mg(z),$ where $g(a)\neq 0$. If $m$ is even, -we obtain analyticity of $f$. If $m$ is odd, $f$ cannot be $C^\infty$; some derivative blows up. -The argument also works for real-analytic, as a real-analytic function extends to complex -analytic in some complex neighborhood. -EDIT. This proof extends to quasianalytic functions of $DC$ (Denjoy-Carleman) class. One needs two facts: - -If $f\in DC$ and $f(a)=0$ then $f(x)=(x-a)g(x)$ with $g\in DC$. The proof is based -on the formula -$$g(x)=\int_0^1f'(tx)dt.$$ -By induction, this gives $f(x)=(x-a)^mg(x)$, for every $f\in DC$ and every $a$, -with some non-negative integer $m$ and $g\in DC$, $g(a)\neq 0$. -If $f\in DC$ and $F$ is analytic on the range of $f$, then $F\circ f\in DC$. This follows from the great theorem of E. M. Dyn'kin (J.d'Analyse 60 (1993)) that characterizes -$DC$ in terms of pseudoanalytic extension to a complex semi-neighborhood of a real interval. - -In the proof above, $F$ is a branch of the square root. -Validity of facts 1 and 2 was explained to me by MO user fedja.<|endoftext|> -TITLE: Does the following operation on modular forms yield something modular? -QUESTION [6 upvotes]: Let $f(z) = \sum_{n=0}^\infty a_nq^n$ be the fourier expansion of a (quasi-)modular form (with $q = e^{2\pi i z}$). Consider the following related functions: -$$f_{m,k}(z) = \sum_{n=0}^\infty a_{mn + k}q^n$$ -$$g_{m,k}(z) = \sum_{n=0}^\infty a_{mn + k}q^{mn+k}$$ -for, naturally, $0 \leq k < m$. -My question: are these in any way modular? For some congruence subgroup, etc. etc.? -What if we put some restrictions on $f(z)$? Or shift the powers of $q$ somewhat? -As an example of when we do retain (quasi-)modularity, consider the case $f(z) = E_2(z), m = 2$. Then it is not too hard to show that -$$ -E_2(z) = g_{2,1}(z) + 3E_2(2z) - 2E_2(4z) -$$ -and so it follows that $g_{2,1}(z)$ is indeed quasi-modular for $\Gamma_0(4)$. Since then $g_{2,0} = E_2(z) - g_{2,1}(z)$, this is also quasi-modular. -So is this in general true? - -REPLY [3 votes]: I know nothing about quasimodular forms, so let me answer the question for modular forms in the classical sense. -First, your $f_{m, k}$. If $k = 0$ this is essentially the image of $f$ under the Hecke operator $U_m$, so it is always going to be modular of some level (and if f has level $\Gamma$, then $f_{m, 0}$ will have level $ \Gamma \cap \Gamma_0(m)$ or something like that). If $k \ne 0$ then $f_{m, k}$ will not be modular, because $f_{m, k}(mz) = q^{-k} g_{m, k}(z)$, $g_{m, k}$ is modular, and dividing by a power of $q$ destroys modularity. -Now $g_{m, k}$. This is just a linear combination of the forms $f(z + j/m)$ for $j = 0, ..., m$ by finite Fourier theory, and all these are modular, so $g_{m, k}$ is modular (although if $g$ is modular of level $\Gamma_1(N)$, then $g_{m, k}$ will have level something like $\Gamma_1(mN) \cap \Gamma_0(m^2 N)$ in general). -The forms $g_{m, k}$ are closely related to the forms -$$ g_{\chi} = \sum_n \chi(n) a_n q^n $$ -for Dirichlet characters $\chi$ modulo $m$, which are very interesting (and well-studied) objects; if $g$ is a Hecke eigenform then so are the forms $g_{\chi}$ (although the $g_{m, k}$ generally aren't).<|endoftext|> -TITLE: Complexity of computing the Galois group -QUESTION [5 upvotes]: There has been some discussion of computing the Galois group of a polynomial over the integers, but I can't seem to find any results, or even a question of what the complexity of this might be. For example, I assume that this is NOT in NP, and IS NP-hard (at least), but am not aware of any reduction... - -REPLY [4 votes]: According to a paper Upper bounds on the complexity of some Galois Theory problems by Arvind and Kurur here, a theorem of Landau gives an exponential upper bound in the size of the polynomial under a certain definition of size. More precisely, her theorem gives a polynomial bound in terms of the size of the Galois group and the size of the polynomial. But the size of the group can be exponential in the size of the polynomial. They show if the Galois group is solvable, then the order can be computed by a randomized polynomial time algorithm with an NP oracle. -This paper gives some complexity bounds on computing the order (they show this is in $P^{\# P}$ see the complexity zoo). Google seems to say there is a paper proving that nilpotence of the Galois group is decidable in polynomial time.<|endoftext|> -TITLE: Knots indistinguishable by HOMFLY -QUESTION [6 upvotes]: Is there any list (incomplete of course) of knots, that have similar HOMFLY polynomials? I am mainly interested in torus knots. - -REPLY [7 votes]: There is some list on Thistlethwaite's knot page, I think, see "the knot atlas". Kanenobu has shown that there are infinitely many distinct knots with the same HOMFLY polynomial: -Kanenobu, T. "Infinitely Many Knots with the Same Polynomial." Proc. Amer. Math. Soc. 97, 158-161, 1986. -For knots with "similar" HOMFLY polynomial, in particular the Alexander polynomial, the Conway polynomial and the Jones polynomial coincide. For torus knots there are easy formulas, e.g., the Jones polynomial for a $(m,n)$-torus knot is -$$ -V(t)=\frac{t^{{(m-1)(n-1)}/2}(1-t^{m+1}-t^{n+1}+t^{m+n})}{1-t^2}, -$$ -and the Alexander polynomial is -$$ -A(t)=\frac{(t^{nm}-1)(t-1)}{(t^m-1)(t^n-1)}. -$$ -This shows that the polynomials are different (except for obvious symmetries). -In general, the HOMFLY polynomial for torus knots is computed in section 3 of the paper "The HOMFLY polynomial for torus links from Chern-Simons gauge theory", by J.M.F. Labastida and M. Marino (see formula $3.1$).<|endoftext|> -TITLE: Moduli space of motives vs moduli space of varieties -QUESTION [14 upvotes]: A (projective) abelian variety $A$ over the complex numbers is determined by $H^1(A,\mathbb{Z})$ together with its Hodge structure and polarization. This miracle means that one can parametrise polarized abelian varieties (which sounds like a hard geometric problem) by parametrising their H^1's with the extra structure (which sounds like an easier algebraic problem, solved via the theory of Shimura varieties). -If we move away from abelian varieties, the miracle won't occur, and indeed attempting to understand a general variety (a complicated non-linear object) via its cohomology groups (linear objects) is presumably not going to work in general -- the linearization will lose information. -I have two related questions about what is going on in general. Let me talk about K3 surfaces below, although my real confusion has nothing to do with K3 surfaces in particular -- I could easily be saying "Calabi-Yau 8-folds" or "curves of genus 23" or indeed any random type of smooth projective variety. -Q1) I am confused about why the power of functors and representability theorems do not give me everything. This probably just reflects my lack of real understanding of what is going on. For example, let me just naively consider the functor sending a scheme S (over the complexes, if you like) to the set of isomorphism classes of polarized K3 surfaces over S (or the groupoid of polarized K3 surfaces). I now want to vaguely mutter that a big machine the likes of which I don't really understand says that this functor satisfies some basic continuity properties and hence (perhaps by some theorem of M. Artin) is representable by some algebraic stack. My understanding of the abelian variety situation is that this method is one way to prove the existence of things like Siegel modular varieties (perhaps even over $\mathbb{Z})$. Why does this method fail for more general families of varieties? I am guessing that I am perhaps being too sloppy with my polarizations but I don't really have a precise feeling for what goes wrong. -Q2) This general functor nonsense must surely fail, so here's a second approach to constructing moduli spaces of certain types of variety. Again let me stick to K3's for concreteness. I understand H^2 of a K3 surface quite well and presumably there is a Shimura variety parametrising the types of Hodge structures showing up as H^2 of a K3. I am wondering how far this Shimura variety would be from the "moduli space of K3 surfaces" which I am naively assuming exists. I can see the issue -- if the moduli space of K3's exists, there will be a map to the Shimura variety, but there is no reason to expect either injectivity or surjectivity on the face of it (I am losing information by linearising, and the linear stuff is too naive to know whether it comes from geometry). Let's forget about injectivity for a moment -- injectivity is an issue whose answer will depend on which type of variety I am trying to parametrise (e.g. for curves I have Torelli etc). But what about surjectivity? Because I'm not really interested in K3's, I'm interested in the general picture of "moduli spaces of varieties of type X" and understanding this "space" via Hodge structure, I am led to the following question, the answer to which is presumably well-known: -If $H$ is a polarizable $\mathbb{Z}$-Hodge structure of weight $n$, does one expect $H$ to be the singular cohomology of a pure motive? Perhaps more concretely, does one expect there to exist some smooth projective algebraic variety $X$ over the complexes such that $H$ is a subspace of $H^i(X,\mathbb{Z})(j)$ for some $i,j$, and even a subspace cut out by correspondences? - -REPLY [10 votes]: Let me expand my (and ulrich's) comment slightly concerning your last question. Let $D$ be the period domain of all Hodge structures with fixed Hodge numbers and polarization. For the sake of simplicity, let's say the weight is $2$. Suppose that $H\in D$ is a summand of $H^2$ of some smooth projective variety, then by weak Lefschetz, it's a summand of $H^2$ of a smooth projective surface. All surfaces embed into $\mathbb{P}^5$. -So using a Hilbert scheme argument, the set of surfaces is parametrized by a countable union of quasi projective varieties $\bigcup T_i$. With a bit more work, we -can find $\bigcup T_i$, such that $t\in T_i$ parametrize - pairs $(S_t, C_t)$, where $S_t$ is a surface and $C_t\subset S_t\times S_t$ is a correspondence yielding a motivic sub Hodge structure -$$H^2([S_t,C_t]):=im(p_{1*}[C_t]\cup p_2^*-):[H^2(S_t)\to H^2(S_t)]$$ After throwing away some components $T_i$ we can assume that $H^2([S_t,C_t])\in D$ for all $i$ and $t\in T_i$. So we get holomorphic period maps $f_i:\tilde T_i\to D$ from the universal covers. Griffiths tranversality says that -for any tangent vector $v$ of $\tilde T_i$, $df_i(v):F^p\subset F^{p-1}$, and this is typically (although not always) a nontrivial constraint. In such a typical case, this -forces the image $f_i(T_i)\subset D$ to be proper. By Baire category $\cup f_i(T_i)\subset D$ is a proper subset as well. This shows that a generic element of such a typical $D$ does not come from an effective motive. -Of course this argument is highly nonconstructive. -In fact, I don't know of a single explicit example of a polarizable Hodge structure which not motivic!<|endoftext|> -TITLE: "Abnormal" manifold -QUESTION [6 upvotes]: We often assume manifolds to be paracompact Hausdorff. Clearly, this implies normal. -However, there is a manifold (I mean locally Euclidean Hausdorff space) which is not paracompact. Without paracompactness, they are still regular because manifolds are locally compact, but does it imply normal? -The only example of non paracompact manifold which I know is the "long line". However, I hear this is normal. I want to know whether manifold implies normal or not. - -REPLY [5 votes]: There are even non-normal separable complex manifolds. The paper Complex analytic manifolds without countable base by Calabi and Rosenlicht gives examples $M,S$ of a non-normal separable complex manifolds of complex dimension 2 and hence real dimension 4. It was proven by Rado that every Riemann surface is paracompact, so the dimension of these manifolds is optimal.<|endoftext|> -TITLE: 4 squares almost in an arithmetic progression -QUESTION [9 upvotes]: Does there exist infinitely many coprime pairs of integers x,d such that x, x+d, x+2d, x+4d are all square numbers? -One example would be 49,169,289,529. This is the only example I have found so far and I believe it may be the only one. - -REPLY [29 votes]: The answer is yes. Here is another example, -$$ -x=2021231^2, \qquad d=82153503191760. -$$ -Then -$$ -x+d=9286489^2, \qquad -x+2d=12976609^2, \qquad -x+4d=18240049^2. -$$ -Basically, write $x=X^2$, $x+d=Y^2$, $x+2d=Z^2$, and $x+4d=W^2$. Eliminating $x$ and $d$ reduces to the two equations: -$$ -X^2-2Y^2+Z^2=0, \qquad -3X^2-4Y^2+W^2=0. -$$ -This pair of equations defines a smooth curve $C$ of genus one in $\mathbb{P}^3$, which we can put into Weierstrass form by sending the point $(1,1,-1,1)$ (say) to the point at infinity. The elliptic curve you obtain is -$$ -y^2 = x^3 - x^2 - 9x + 9 -$$ -with Cremona reference 192A, and Mordell--Weil rank $1$. The point $(5,8)$ has infinite order, and corresponds to your solution (up to changes in sign). Taking multiples of this point and mapping back to $C$ gives infinitely many coprime pairs $x$, $d$ such that $x$, $x+d$, $x+2d$, $x+4d$ are squares.<|endoftext|> -TITLE: Do level sets always correspond to even graphs? -QUESTION [8 upvotes]: Suppose I have a level set of some function $f\colon\mathbb{R}^n\rightarrow\mathbb{R}^m$, say $L:=\{x:f(x)=c\}$. Let $S$ denote the points in $L$ at which $L$ is locally diffeomorphic to an open interval on the real line. Now define a set of vertices by $V:=(\overline{S}\setminus S)\cup\{\infty\}$, where bar denotes closure. Let's say two vertices $u,v\in V$ are adjacent if there is a continuous path $\gamma\colon[0,1]\rightarrow\mathbb{R}^n\cup\{\infty\}$ such that $\gamma(0)=u$, $\gamma(1)=v$, and $\gamma(t)\in S$ whenever $t\in(0,1)$. -Question: Is this graph necessarily even if $f$ is polynomial? analytic? -As an example, the following depicts the $(x,y)$'s such that $(x^2+y^2-2)(x^3-y^2)(xy-1)=0$, along with the corresponding graph: - -In the graph, I put the point at infinity on the bottom right. The other vertices are $(0,0)$, $(1,1)$, $(1,-1)$ and $(-1,-1)$, and all of these have even degree. -Note that the graph is not necessarily even if $f$ is $C^\infty$, since the non-analytic smooth function in this article, namely -$$ -f(x):=\left\{\begin{array}{ll}e^{-1/x}&\mbox{if }x>0\\0&\mbox{otherwise}\end{array}\right. -$$ -has a level set $\{x:f(x)=0\}=(-\infty,0]$ whose graph is a path of two vertices (each of degree $1$). - -REPLY [11 votes]: This is true for real analytic functions $f:R^2\to R$. Let such a function be $f(x,y)$ and -we consider the level set $f(x,y)=0$, that is an analytic curve. If at some point $f_x$ or $f_y$ is different from $0$, this point is non-singular and belongs to your set $S$. -Now we show that a singular point is a common endpoint of an even number of branches of -the curve. Let this singular point be $O=(0,0)$. -According to Newton and Puiseux, the curve near $O$ is given by finitely many series -of the form $y=(c+o(1))x^{p/q}$, where $p$ and $q$ are positive integers, $c\neq 0$. -Each such series -defines an even number of branches. If one of $p$, $q$ is even, the principal term -is invariant either under $x\mapsto-x$ or $y\mapsto -y$. If both are odd, -use $(x,y)\mapsto (-x,-y)$ So we have an even number of branches converging at each -singular point. (For a reference, you can look at "Puiseux series" entry in Wikipedia, -or any book which has "algebraic curves" in the title. There is no difference between -algebraic and analytic curves in this respect). -EDIT. I asked the specialists in singularities, and they told me that this is true in any dimension, as a corollary of a much more general fact proved by Sullivan in: -MR0278333 -Sullivan, D. -Combinatorial invariants of analytic spaces. 1971 Proceedings of Liverpool Singularities—Symposium, I (1969/70) pp. 165–168 Springer, Berlin. -EDIT2. -Another reference: Real algebraic geometry" by Bochnak, Coste, -Roy, Corollary 11.2.4. It is stated for algebraic case but actually there is no difference, according to Thm 8.6.12 of the same book.<|endoftext|> -TITLE: $\aleph_2$ Suslin Hypothesis -QUESTION [7 upvotes]: Is it still open whether ZFC+GCH is consistent with the statement that there are no $\aleph_2$-Suslin trees? - -REPLY [6 votes]: From http://www.users.muohio.edu/larsonpb/kly_guessing.pdf (pdf 9 of 16): - -Note that the consistency of GCH + "no $\omega_{2}$-Suslin-trees" is - still an open question. ...this old question is now more open than - ever. - -As best as I can tell, the matter has not been settled in the six years since the published paper above. -Citation: -König, B., Larson, P., & Yoshinobu, Y. (2007). Guessing clubs in the generalized club filter. Fundamenta Mathematicae, 195(2), 177-191.<|endoftext|> -TITLE: A question on verbal subgroups of free groups -QUESTION [6 upvotes]: Are there any verbal subgroups in a rank 2 free group $F(a,b)$ arising as normal closures $\langle\langle r \rangle\rangle$ of a (nontrivial) element $r \in F(a,b)$ other than the commutator subgroup $[F,F]=\langle\langle [a,b] \rangle\rangle$? -More generally, are there any subgroups of the given type $\langle\langle r \rangle\rangle$ containing a verbal subgroup? - -REPLY [8 votes]: 1) The answer is no: namely, suppose that $n\ge 2$, $r\in F_n$ and $N=\langle\langle r\rangle\rangle$ is verbal, then either $r=1$, or $n=2$ and $r$ is conjugate to $[x_1,x_2]$. Indeed $F_n/N$ is a 1-relator group; it was established; if $r\neq 1$ it satisfies a law, and Magnus [a,b] proved that this only happens when $F_n/N$ is a solvable Baumslag-Solitar group. On the other hand, if $N$ is verbal then so is $N[F_n,F_n]$, and this implies that the abelianization of $F_n/N$ is the square of some finite group. This excludes all solvable Baumslag-Solitar groups except $\mathbf{Z^2}$. -2) The answer is yes, and precisely (by the above argument) the only examples are when $n=2$ and the relator defines a solvable Baumslag-Solitar group (in which case $N$ contains the second derived subgroup of $F_n$, which is a verbal subgroup). -[a] Magnus W. Das Identitätsproblem für Gruppen mit einer definierenden Relation, Math. Ann. (1932) 106. 295–307. -[b] Moldavanskii D. I. On a theorem of Magnus (in Russian) Uch. zap. Ivanovsk. gos. ped. inst. 1969. T.44. S.26–28.<|endoftext|> -TITLE: Understanding the intermediate field method for the $\phi^4$ interaction -QUESTION [6 upvotes]: In Rivasseau's and Wang's How to Resum Feynman Graphs, on page 11 they illustrate the intermediate field method for the $\phi^4$ interaction and represent Feynman graphs as ribbon graphs. I had to read up about ribbon graphs as I've never heard of them before and still don't fully understand their use for the intermediate field method. For the $\phi^4$ interaction, we have: - -In that case each vertex has exactly four half-lines. There are exactly three ways to pair these half-lines into two pairs. Hence each fully labeled (vacuum) graph of order $n$ (with labels on vertices and half-lines), which has $2n$ lines can be decomposed exactly into $3^n$ labeled graphs $G'$ with degree $3$ and two different types of lines - -the $2n$ old ordinary lines -$n$ new dotted lines which indicate the pairing chosen at each vertex (see Figure 5). - - -The extension illustrated in Figure 5 looks as follows: - -I'm not sure I correctly understand the way to create such "3-body extensions". I understand it in that way that we "split up" the given vertex into two vertices and add an edge between them. Then, each vertex has two additional half-edges and there are three ways to pair them together. Joining the upper half-edges together, we get the second term on the RHS and joining the two half-edges on one side together, we will get the first term on the RHS (the dashed line would be the new edge between the two vertices). The only case left is the one of joining the upper left and the lower right (and the upper right and lower left) half-edge together. However, when drawing that specific case, I don't see any way to stretch or rotate it such that the third term on the RHS results. -Am I misunderstanding the way 3-body extensions are intended to get or am I just missing a way to stretch the edges to get the third term on the RHS? Also, where exactly do we need the cyclic ordering characterising ribbon graphs? And why are there two dashed lines in the third term on the RHS? - -REPLY [3 votes]: The half-edges incident to a 4-valent vertex can be labeled 1,2,3,4. There are three ways to split them into pairs: 12-34, 13-24, and 14-23. Another way to think of the three pairings is by considering the 6 edges of a tetrahedron (with vertices labeled 1,2,3,4), and matching them into three skew pairs of edges. -The twisty shape in the picture indicates that you should join the pairs of half-edges that are diagonal to each other. I think the illustrator chose to move the dotted line to the side of the crossing to improve visibility.<|endoftext|> -TITLE: Reference for Wang Tile -QUESTION [10 upvotes]: I am working on projects in solving ground state of generalized ising models. One recent work involves tiling with basic tiles that filled the whole lattice. For example, we could obtain results: - -every row corresponds to one constructible structure with basic tiles being "Generating configurations of the structure". -The resulting structures could be shown as followed: - -9 and 12 are more complicated (can be periodic or aperiodic) shown as followed: - - -I realize (with the help of community), this is actually related to Wang Tiles. So the first few steps I wish to approach this problem are looking into all established theorems on Wang Tiles. Could someone suggest some relevant reference? -(PS:From Engineering respective, I think we (in engineer department) commonly would agree that as long as we could keep tiling the plane to a sufficiently large extent( e.g use 100*100 tiles without conflict), then we blindly think these few basics tiles could tile the plane. So the undecidability is not too crucial...) - -REPLY [4 votes]: Original references (Wang's "Proving theorems by pattern recognition" and Berger's "The undecidability of the domino problem") are not very friendly. You may find some reviews around, though. -I'll suggest you to take a look at Jarkko Kari's website where you will find some introductory material and these slides for some basic notions and references.<|endoftext|> -TITLE: When does the derived subgroup of $G(F)$ contains the $F$-points of unipotent subgroups of $G$ -QUESTION [5 upvotes]: Let $F$ be a local field of characteristic $0$ and $G$ a connected split reductive group over $F$. -Let's look at the derived groups. We have $(G(F),G(F)) \subset (G,G)(F)$ and this inclusion is of finite index according to this MO question. -My question is : do we have (maybe under stronger assumptions) $U(F) \subset (G(F),G(F))$ for any unipotent subgroup $U \subset G$ ? -If no, same question for a given unipotent subgroup $U \subset G$. - -REPLY [7 votes]: Yes, since in characteristic 0 all unipotent groups are connected and split. More generally, for any field $k \ne \mathbf{F}_2$ and any connected reductive $k$-group $G$ and split unipotent smooth connected closed $k$-subgroup $U \subset G$, necessarily $U(k) \subset (G(k), G(k))$. (This is false for ${\rm{SL}}_2(\mathbf{F}_2) = S_3$, whose commutator subgroup $A_3$ has order 3.) -In effect, this will reduce to the analogue for the non-reductive $H := \mathbf{G}_m \ltimes \mathbf{G}_{\rm{a}}$ with semi-direct product via either the usual scaling action or its precomposition with squaring on $\mathbf{G}_m$, taking $U$ to be $\mathbf{G}_{\rm{a}} = \mathscr{R}_u(H)$. In such cases, $(t,0)(1,x)(t,0)^{-1}(1,x)^{-1} = (1, (t^e-1)x)$ with $e = 1, 2$. But as $x$ varies through $k$ and $t$ varies through $k^{\times}$, $(t^e-1)x$ varies through $k$ when we can choose $t \in k^{\times}$ with $t^e - 1 \ne 0$; i.e., $k \ne \mathbf{F}_2$ when $e = 1$ and $k \ne \mathbf{F}_2, \mathbf{F}_3$ when $e = 2$. To deal with $\mathbf{F}_3$ below, we'll need to do a bit more work. (Note that ${\rm{SL}}_2(\mathbf{F}_3) = S_4$ does have commutator subgroup which contains the 3-Sylow subgroups.) -In the initial setup, every $U$ lies in $\mathscr{R}_u(P)$ for a minimal parabolic $k$-subgroup $P$ of $G$, so it suffices to treat such unipotent radicals (which are $k$-split). Of course, if $G$ has no proper parabolic $k$-subgroup (i.e., if $P = G$) then the only such $U$ is 1 and there is nothing to do, so we may assume $P \ne G$ and hence $G$ has a non-central split $k$-torus. -For a split maximal $k$-torus $S$ in $P$, the unipotent radical of $P$ is the direct product variety (under multiplication in $G$) given by $\prod_{c \in \Phi^+_{\rm{nd}}} U_c$ with the root groups associated to non-divisible elements in a positive system of roots in the relative root system $\Phi(G,S)$ (here ${\rm{Lie}}(U_c)$ is the $c$-weight space in ${\rm{Lie}}(G)$ unless $c$ is multipliable, in which case it is the span of the weight spaces for $c$ and $2c$). The hypothesis that $P \ne G$ ensures that the root system is non-empty (otherwise there is nothing to do). -It suffices to treat the $U_c$'s separately, so by replacing $G$ with $\mathscr{D}(Z_G(S_c))$ for the codimension-1 subtorus $S_c = (\ker c)^0_{\rm{red}}$ killed by $c$ we may focus on the case of semisimple $G$ with $k$-rank 1. -For the purpose of handling $k = \mathbf{F}_3$, the following additional reduction steps are also convenient and make sense with arbitrary $k$. We may replace $G$ with its simply connected central cover $\widetilde{G}$ (as the preimage of $P$ in $\widetilde{G}$ is a minimal parabolic $k$-subgroup whose unipotent radical maps isomorphically onto $U$), so $G$ is simply connected. Hence, $G = \prod_i {\rm{R}}_{k_i/k}(G_i)$ for a finite separable extensions $k_i/k$ and absolutely simple connected semisimple $k_i$-groups $G_i$ that are simply connected. The $k$-rank 1 conditions forces all but one $G_i$ to be $k_i$-anisotropic, so we can drop those factors and thereby get to the case where there's a single $i = i_0$. Then $P = {\rm{R}}_{k_{i_0}/k}(P_{i_0})$ and $U = {\rm{R}}_{k_{i_0}/k}(U_{i_0})$, so on $k$-points we can pass to $(G_{i_0}, U_{i_0})$; i.e., we can assume $G$ is also absolutely simple over $k$. -If the relative root system is A$_1$ then $U_c$ is a direct sum of copies of the 1-dimensional representation $\mathbf{G}_{\rm{a}}(c)$ of $S = \mathbf{G}_m$ via a nontrivial character $c$ that is at worst 2-divisible in the character group (since $\langle c, c^{\vee} \rangle = 2$). If the relative root system is BC$_1$ then $U_c$ is an extension of a direct sum of copies $\mathbf{G}_{\rm{a}}(c)$ by a direct sum of copies of $\mathbf{G}_{\rm{a}}(2c)$ where $c$ is a basis of the character group, and this extension of $k$-groups remains short exact on $k$-points due to additive Hilbert 90. -Hence, it now suffices to show that if $H := \mathbf{G}_m \ltimes \mathbf{G}_{\rm{a}}(\chi)$ for the character $\chi$ of $\mathbf{G}_m$ given by $\chi(t) = t$ or $\chi(t) = t^2$ then $(H(k), H(k))$ contains $U(k)$ where $U = \mathscr{R}(H)$ is $\mathbf{G}_{\rm{a}}(\chi)$. The computation near the start handles this when $|k| > 3$. What if $k = \mathbf{F}_3$? This is where the above passage to the absolutely simple and simply connected case (not yet used) is useful. The only such cases over a finite field are ${\rm{SL}}_2$ and ${\rm{SU}}_3$. For the first case, by inspection ${\rm{SL}}_2(\mathbf{F}_3) = S_4$, whose commutator subgroup $A_4$ contains the 3-Sylow subgroups. In the other case, a direct argument shows that $G(k)$ is perfect whenever $|k| > 2$.<|endoftext|> -TITLE: The combinatorial interpretation of an identity found in "Primes in tuples I" -QUESTION [6 upvotes]: I am currently reading the paper -D.A. Goldston, J. Pintz, C.Y. Yildirim, $\textit{Primes in tuples I}$, Annals of Mathematics $\textbf{170}$ (2009), 819-862 -and in particular I found equation (8.16), which records the identity -$$\displaystyle \frac{1}{u!} \sum_{i=0}^u (-1)^i \binom{u}{i} \frac{d(d+1)\cdots(d+i-1)}{(v+d+i)!} = \frac{1}{(u+v+d)!} \binom{u+v}{u}$$ -to be interesting. This identity itself admits a rather elegant proof via the Chu-Vandermonde identity. -I am interested in a suitable combinatorial interpretation for this identity, preferably an elementary counting argument. -Any insight would be appreciated. - -REPLY [13 votes]: Since the terms aren't integers we can't find a combinatorial interpretation directly. -If we multiply both sides by $(u+v+d)!$ and rearrange, we can rewrite the identity as -$$ -\sum_{i=0}^u (-1)^i \binom{u+v+d}{u-i}\binom{d+i-1}{i} =\binom{u+v}{u}, -$$ -which we can interpret combinatorially. Let $U$, $V$, and $D$ be disjoint sets of sizes $u$, $v$, and $d$, and suppose that $D$ is totally ordered. Then $(-1)^i\binom{u+v+d}{u-i}\binom{d+i-1}{i}$ counts ordered pairs $(X,Y)$ where $X$ is a $(u-i)$-subset of $U\cup V \cup D$ and $Y$ is an $i$-multisubset of $D$ (i.e., a selection of $i$ elements of $D$ with unlimited repetition), with sign $(-1)^i$. The right side counts such pairs with $i=0$ in which $X$ contains no elements of $D$. All other such pairs can be canceled by a sign-reversing involution: In any other pair $X$ or $Y$ contains an element of $D$. Let $\delta$ be the least element of $D$ occurring in $X$ or $Y$. If $\delta$ occurs in $X$, move it from $X$ into $Y$. If $\delta$ does not occur in $X$, move one copy of $\delta$ from $Y$ into $X$. -Alternatively, we may multiply both sides by $u!\, (v+d)!$ and rewrite the identity as -$$ -\sum_{i=0}^u (-1)^i \binom u i \frac{(d)_i}{(d+v+1)_i} = \frac{(v+1)_u}{(d+v+1)_u}, -$$ -where $(a)_j = a(a+1)\cdots (a+j-1)$. A probabilistic proof of this identity, using inclusion-exclusion and the Pólya-Eggenburger urn model can be found in section 5 of my paper Symmetric Inclusion-Exclusion, Séminaire Lotharingien de Combinatoire, B54b (2005), 10 pp.<|endoftext|> -TITLE: Growth of the "denominator" of powers of an algebraic number -QUESTION [19 upvotes]: Let $z \in \overline{\bf Q}$ be an algebraic number. Define the "denominator" of z to be the least natural number $n$ such that $nz$ is an algebraic integer. -By a rather ad hoc argument (playing with the minimal polynomial of $z$ to compute high powers of $z$ in terms of low powers of $z$, and measuring the coefficients obtained p-adically), I can show the following: - -If $z$ is an algebraic number that is not an algebraic integer, then the denominator of $z^m$ grows exponentially in $m$ in the limit $m \to \infty$ (thus there is $c>1$ such that the denominator is at least $c^m$ for all sufficiently large $m$). - -This is obvious in the case when $z$ is rational, from the fundamental theorem of arithmetic, but I couldn't find a similarly quick proof in the general case; presumably, unique factorisation into prime ideals for a suitable number field is the key, but (somewhat to my embarrassment) my algebraic number theory is too rusty to figure out how to exploit this here. So I am posing the question here to see if anyone can find an easy proof of this fact. - -REPLY [24 votes]: There is some number field $K$ containing $z$, and there we have a prime factorization -$$ -(z)=\frac{\mathfrak{p}_1\ldots\mathfrak{p}_a}{\mathfrak{q}_1\ldots\mathfrak{q_b}} -$$ -of fractional ideals, where no $\mathfrak{p}_i$ is equal to a $\mathfrak{q}_j$. If $nz^m$ is an algebraic integer, then $(\mathfrak{q}_1\ldots\mathfrak{q_b})^m$ divides $(n)$ (as ideals of $\mathcal{O}_K$). This means -$$ -N(\mathfrak{q}_1\ldots\mathfrak{q_b})^m|N(n)=n^{[K:\mathbb{Q}]}, -$$ -so that -$$ -n\geq \left(N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]}\right)^m. -$$ -We can take -$$ -c=N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]} -$$<|endoftext|> -TITLE: A question on eigenvalues -QUESTION [8 upvotes]: Let $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$, $A_{5}$ be linearly independent Hermitian matrices in the the space of $6$ by $6$ Hermitian matrices as a vector space over $\mathbb{R}$. Does there always exist a non-zero matrix $A_{0}\in\mbox{span}_{\mathbb{R}}\left(A_{1},A_{2},A_{3},A_{4},A_{5}\right)$ such that at least four of the eigenvalues of $A_{0}$ are $0$? - -REPLY [11 votes]: Unless I'm mistaken, the answer to your question is negative. -In the paper "On matrices whose real linear combinations are nonsingular", by J. F. Adams, P. D. Lax, R. S. Phillips, (Proc. AMS, 16 (1965), the authors show that $2b+1$ is the largest number of Hermitian matrices that one could pick so as to ensure that all of their nontrivial linear combinations are nonsingular. -Here, $b$ is implicitly defined as follows. For $n\times n$ matrices, where $n=(2a+1)2^b$, $b=c+4d$, where $a,b,c,d$ are integers with $0\le c < 4$. In our case $n=4$, so we use $a=0$, $b=2$, which completes our conclusion.<|endoftext|> -TITLE: History question: Roth's theorem on approximating algebraic numbers...before Roth -QUESTION [9 upvotes]: Roth's theorem has two universal quantifies, over irrational algebraic numbers $\alpha$ and over real $\epsilon>0$. Of course the theorem asserts in each instance that the inequality -$$|\alpha-\frac{p}{q}|<\frac{1}{q^{2+\epsilon}}$$ -has only finitely many solutions in integers $(p,q)$. -I seek confirmation (or not) for the following two historical claims: -Prior to Roth's paper... -1) the statement was not known to hold even for a single specific algebraic irrationality of degree $>2$ (but for all $\epsilon$); -2) the statement was not known to hold for single specific $\epsilon$ no matter how large (but for all $\alpha$). -The most famous precursors (Liouville--Thue--Siegel--Dyson) have $\epsilon$ depending on the degree of $\alpha$, so I feel pretty sure about 2), but less confident about 1). - -REPLY [7 votes]: I believe that both statements are correct. The ineffective results of Thue, Siegel, Gel'fond and Dyson do not lead to irrationality measures arbitrarily close to $2$, while earlier effective results (Thue, Siegel, etc) for classes of algebraic numbers give measures as close to $2$ as you like, but not of the $2 + \epsilon$ variety. If this last statement seems to be inconsistent, note that Thue proved irrationality measures for numbers like $\sqrt[3]{1+1/N}$, of the shape $2+c/\log N$ for some constant $c>0$. Even nowadays, we do not have an effective $2+\epsilon$ irrationality measure for even a single algebraic number of degree $3$ or higher.<|endoftext|> -TITLE: Is a unitary representation always semisimple? -QUESTION [7 upvotes]: I have been reading the online lecture notes by Fiona Murnaghan -http://www.math.toronto.edu/murnaghan/courses/mat1197/notes.pdf -The first lemma in p.35 says that every unitary representation of locally compact group $G$ is semisimple. In her notes, she defines a semisimple representation to be a representation which is a direct sum of irreducible representations. -On the other hand, people often say that the right regular representation of $G$ on $L^2(G)$, which is unitary, does not decompose into a direct sum of irreducible representations but it is a direct integral. -But if the above lemma by Murnaghan is correct, it seems that $L^2(G)$ must be a direct sum of irreducible representations. -I read the proof of the lemma in her note carefully, and noticed that even though she stated the lemma under the assumption that the group $G$ is $p$-adic and the representation is smooth, the proof goes through for any locally compact group and any uniterizable representation. Indeed, the proof only uses the property that any subrepresentation has the orthogonal complement and Zorn's lemma. -What am I missing? - -REPLY [5 votes]: Probably the version of "semi-simplicity" relevant in the context in those notes refers to a repn of $G$ that is admissible with respect to a compact subgroup $K$, in the sense that it is assumed to decompose with finite multiplicities over $K$. This assumption holds for repns induced from cuspidal repns on the compact itself, for example, as observed by Jacquet already in 1971. In such a context, the otherwise-too-glib remarks about complete-reducibility still reach a correct conclusion, even without completeness in a Hilbert-space sense. -The more down-to-earth situations, like $L^2(\mathbb R)$, or subspaces generated by rough functions on the circle, do not satisfy suitable "admissibility" conditions, perhaps-oddly.<|endoftext|> -TITLE: The origin of Discrete `Liouville's theorem' -QUESTION [25 upvotes]: It is known that discrete Liouville's theorem for harmonic functions on $\mathbb{Z}^2$ was proved by Heilbronn (On discrete harmonic functions. - Proc. Camb. Philos. Soc. , 1949, 45, 194-206). - -If a bounded function $f : \mathbb Z^2 \rightarrow \mathbb{R}$ satisfies the following condition - $$ -\forall (x, y) \in \mathbb{Z}^2,\quad f(x,y) = \dfrac{f(x + 1, y)+f(x, y + 1) + f(x - 1, y) +f(x, y - 1)}{4}$$ - then f is constant function. - -(A stronger version of discrete “Liouville's theorem” also follows from Heilbronn's proof.) -But recently I knew from Alexander Khrabrov that there is an older article of Capoulade with almost the same result (Sur quelques proprietes des fonctions harmoniques et des fonctions preharmoniques, - Mathematica (Cluj), 6 (1932), 146-151.) -Unfortunately the last article is not available for me. Does anybody have an access to this paper? Is this really the first proof discrete Liouville's theorem? -(This question is inspired by Liouville Theorem from Mathematics) - -REPLY [15 votes]: Mathematica (Cluj), 6, 146-151 (1932) - -In a recent article from this journal, Mr. Bouligand has indicated the - possibility of a modified proof of a theorem of Mr. Picard: "a - harmonic function that is positive in the whole space is a constant". - The purpose of this article is to continue along the same lines to - prove an analogous result for preharmonic functions. - -So this seems to be the required theorem. If help is needed with the French text of the paper, let me know.<|endoftext|> -TITLE: Finding the min of $m$ such that $k = \pm 1^n \pm 2^n \pm 3^n \pm \cdots \pm m^n$ for a given pair $(n,k)$ -QUESTION [7 upvotes]: Erdös-Suranyi theorem states that - -For each natural number $k$ there is an $m$ and an appropriate choice of $+$ and $-$ signs (which we write as $±$ in short) such that $k = \pm 1^2 \pm 2^2 \pm 3^2 \pm \cdots \pm m^2$. - -By the way, we can generalize this theorem as the following $(\star)$ (I'm going to write the proof at the end of this question): - -Letting $n\in\mathbb N$, for each natural number $k$ there is an $m$ and an appropriate choice of $+$ and $-$ signs (which we write as $±$ in short) such that $k = \pm 1^n \pm 2^n \pm 3^n \pm \cdots \pm m^n$. - -Here, for a given pair $(n,k)$, let $M(n,k)$ be the min of such $m$. -Then, here are my questions. - -Question 1 : What is $M(2,k)$ ? Can we represent it by $k$ ? Or can we estimate it from above? -Question 2 : What is $M(n,k)$ ? Can we represent it by $k$ ? Or can we estimate it from above? - -Remark : This question has been asked previously on math.SE without receiving any answers. -Motivation : I've been able to solve the $n=1$ case. Letting $A_m=\{k|k=\pm 1 \pm 2 \pm 3 \pm \cdots \pm m,k\in\mathbb N\}$, we get -$$A_1=\{1\}, A_2=\{1,3\},A_3=\{2,4,6\},A_4=\{2,4,6,8,10\},\cdots.$$ -These lead that $M(1,k)$ is the min of $m$ such that -$$\text{When $k$ is odd}\ \ \ k\le m(m+1)/2\ \ \ m\equiv 1,2\ (\text{mod $4$}),$$ -$$\text{When $k$ is even}\ \ \ k\le m(m+1)/2\ \ \ m\equiv 3,0\ (\text{mod $4$}).$$ -For the $n=2$ case, I've been able to find a way to represent $k$ by about $\sqrt k$ terms. Precisely speaking, we can represent $k$ by $m\ (\ge 10)$ terms such that -$$\text{When $k$ is odd}\ \ \ k\le (m+1)^2\ \ \ m\equiv 1,2\ (\text{mod $4$}),$$ -$$\text{When $k$ is even}\ \ \ k\le (m+1)^2\ \ \ m\equiv 3,0\ (\text{mod $4$}),$$ -(Note that $m\ge 10$.) -In the following, I'm going to explain this idea briefly. -We know that when $m=10$ we can represent every odd $k$ from $1$ to $121={11}^2$ such as -$$1={10}^2-9^2+8^2-7^2-6^2+5^2-4^2-3^2+2^2-1^2,$$ -$$121={10}^2-9^2+8^2+7^2+6^2-5^2-4^2-3^2+2^2-1^2.$$ -Then, we know that we can represent every even $k$ from $0$ to $144={12}^2$ such as -$${11}^2-121=0, {11}^2-119=2,\cdots,{11}^2-1=120, \cdots, {11}^2+23=144={12}^2.$$ -Then, we know that we can represent every even $k$ from $0$ to $169={13}^2$ such as -$${12}^2-144=0, {12}^2-142=2, \cdots, {12}^2-0=144,\cdots, {12}^2+24=168.$$ -Then, we know that we can represent every odd $k$ from $1$ to $196={14}^2$ such as -$${13}^2-168=1, {13}^2-166=3, \cdots, {13}^2-0=169, \cdots,{13}^2+26=195$$ -and so on. -However, I don't have any good idea for finding $M(n,k)$. Can anyone help? -Proof for $(\star)$ : Considering the $n$-th difference sequence, we can get -$$\begin{align}\pm (m+1)^n\pm (m+2)^n\pm\cdots\pm (m+2^n)^n=n!\times 2^{n(n+1)/2}\qquad(1)\end{align}$$ -Signs are determined when we make difference sequences. Letting $l$ be this value, we know that $-l$ can be represented as the left hand side of $(1)$. In the following, let us consider in mod $l$. -Since we know -$$-(al+1)^n-(al+2)^n-\cdots -(al+l)^n+(al+l+1)^n+\cdots+(al+l+l)^n\equiv 0,$$ -we also get -$$(al+1)^n-(al+2)^n-\cdots -(al+l)^n+(al+l+1)^n+\cdots+(al+l+l)^n\equiv 2.$$ -Letting $f(a)$ be the left hand side of this equation, we get -$$\begin{align}f(0)+f(2)+\cdots+f(2b)\equiv 2(b+1)\qquad(2)\end{align}$$ -Since the left hand side of $(2)$ is -$$1^n-2^n-\cdots -l^n+(l+1)^n+\cdots+(2l)^n+(2l+1)^n-\cdots+\{(2b+2)l\}^n,$$ -we know that this is in our form. Adding $\{(2b+2)l+1\}^n$ to this equation keeps the form as -$$\begin{align}f(0)+f(1)+\cdots +f(2b)+\{(2b+2)l+1\}^n\equiv 2(b+1)+1\qquad(3)\end{align}$$ -$(2)$ shows that every even in system of residues of mod $l$ can be represented in our form. Also, $(3)$ shows that every odd in system of residues of mod $l$ can be represented in our form. Hence, by using $(1)$ many times, we now know that every natural number can be represented in our form. - -REPLY [7 votes]: [Upon request from the OP, I have deleted my comment to re-post it as an answer.] -For related work, see the following source and references contained therein: Bleicher, M. N. (1996). On Prielipp's Problem on Signed Sums of kth Powers. Journal of Number Theory, 56(1), 36-51. -The article should be available for free here; an image of the abstract can be found below. -(Don't be misled by the typo in line one!) - -Also of interest may be a couple of the papers that have cited the aforelinked: GoogleScholar<|endoftext|> -TITLE: Do quantum "Sure-Shor separators" have a natural Veronese/Segre classification? (question inspired by Gil Kalai and Aram Harrow) -QUESTION [11 upvotes]: Aram Harrow asked: "Is there any place this is written up?" - -Update  Partly in answer to Aram's question, the thermodynamical properties of varietal dynamical systems now are written-up in our Soldier Healing Seminar's notes for 2013; and various motivations for caring about these properties are surveyed in a post to Dick Lipton and Ken Regan's Gödel's Lost Letter weblog. For engineers, the practical question at-issue is whether thermodynamics "just works" on varietal state-spaces, not only at regular points, but at singular points. - -Introduction  In considering the question "Are quantum states exponentially long vectors?" (arXiv:quant-ph/0507242) Scott Aaronson provocatively argued that: - -The real reason to study quantum computing is not to learn other people’s secrets, but to unravel the ultimate Secret of Secrets: Is our universe a polynomial or an exponential place? … -For me, the main weakness in the arguments of quantum computing skeptics has always been their failure to suggest an answer to the following question: What criterion separates the quantum states we’re sure we can prepare, from the states that arise in Shor’s factoring algorithm? - -In a (wonderfully interesting!) subsequent debate, the thesis that our universe is (in Aaronson's phrase) "an exponential place" has been ably defended by Aram Harrow, whereas Gil Kalai has marshalled cogent reasons — for example, in a UW/PIMS colloquium (Friday November 23; details here) — "Why quantum computers cannot work." -The questions asked  The questions asked concern the classification of secant varieties of Segre varieties of Veronese varieties (SSVs). Viewed as dynamical manifolds immersed in an embedding Hilbert state-space, SSVs find broad applications in quantum simulation problems. -The questions asked concretely concern a dynamical Hilbert space of 27 dimensions, which is viewed as a 26-dimensional projective space $\mathbb{P}^{26}$. Numerical studies indicate that the following five varieties are immersed submanifolds of dimension 25, that is, the following five varietal state-spaces are Hilbert-deficient by precisely one dimension: -$$\begin{alignedat}{2} -\sigma_4\big(\,&\text{Seg}(\,\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^2\,)\big)\\ -\sigma_5\big(\,&\text{Seg}(\,\mathbb{G}^2\times\mathbb{P}^2\times\mathbb{P}^2\,)\big)\\ -\sigma_7\big(\,&\text{Seg}(\,\mathbb{G}^2\times\mathbb{G}^2\times\mathbb{G}^2\,)\big)\\ -\sigma_9\big(\,&\text{Seg}(\,\mathbb{G}^8\times\mathbb{P}^2)\,\big)\\ -\sigma_{13}\big(\,&\text{Seg}(\,\mathbb{G}^{26}\,)\big) -\end{alignedat}$$ - -The Four Questions -Q1  In what mathematical respects (if any) do these five varietal spaces differ from one another, in regard to their topology, singularity structure, metric structure, and symplectic structure? Do they all pushforward/blowdown to the same varietal submanifold of Hilbert space? Or do they represent five essentially different "foamy" algebraic/metric/symplectic geometries? -Q2  By what physical measurement processes (if any) can dynamical trajectories on these manifolds be distinguished from one another, and from trajectories on the (flat, singularity-free) Hilbert space in which they are embedded? -Q3  In what quantum mechanical respects (if any) are trajectory unravelings on these manifolds sufficiently deficient in any and all forms of "quantum goodness" as to invalidate their claim to be "Sure-Shor separators" (in the sense of Aaronson)? -Q4  What classification theorems (if any) exist to help quantum systems engineers in exploiting these varietal spaces for purposes of large-scale quantum dynamical simulation? - -Physical remarks  The state-spaces are stipulated to inherit by pullback the standard metric, symplectic, Hamiltonian, and Lindbladian structures of the immersing Hilbert space, such that (from a Harrow-esque perspective) any quantum experiment envisioned to be performed in the Hilbert space, has an algebraically natural pullback/blowup/resolution of singularities on the varieties named, such that simulated quantum trajectories unraveled on the varietal state-space are a reasonable approximation to the "true" quantum state-space. -Alternatively (from a Kalai-esque perspective), any quantum experiment envisioned to be performed in the varietal state-space, has an algebraically natural pushforward/blowdown onto a fictitious Hilbert space of exponentially many dimensions, that is a maximally symmetric (but computationally intractable) approximation to the "true" varietal state-space. -In a nutshell  These varietal spaces are candidate manifolds for Aaronson-style "Sure-Shor separators", and the difficulty of distinguishing geometric perturbations of dynamical trajectories on these "foamy" algebraic manifolds from noise perturbations helps us appreciate the depth and subtlety of Gil Kalai's reasons "why quantum computers cannot work." -Notational remarks  The question adopts and slightly generalizes the notation of Landsberg's Tensors: Geometry and Applications as follows: $\sigma_r$ is the usual rank-$r$ secant join, $\text{Seg}$ is the usual Segre embedding, and $\mathbb{P}^k$ is the usual projective space -$$\mathbb{P}^k(\,\boldsymbol{\xi}\,) = [\xi_0,\xi_1,\dots,\xi_k]$$ -The projective space $\mathbb{G}^k$ is (what mathematicians call) a Veronese embedding -$$ \mathbb{G}^k(a,b) = [\,{% -\textstyle{{k}\choose{0}}^{1/2}a^k b^0,% -{{k}\choose{1}}^{1/2}a^{k-1} b^1,% -\dots,% -{{k}\choose{k-1}}^{1/2}a^{1} b^{k-1},% -{{k}\choose{k}}^{1/2}a^{0} b^{k}% -}\,]\hookrightarrow \mathbb{P}^k.$$ -Physicists regard this Veronese embedding as the space of coherent states; the "$\mathbb{G}$" of the above notation is a mnemonic for Roy Glauber, who pioneered the application of coherent states to broad-ranging practical problems, first in quantum optics and nowadays in many branches of science and engineering; the connexion between the physics viewpoint and the algebraic geometry viewpoint is surveyed by Dorje Brody and Eva-Maria Graefe's recent Coherent states and rational surfaces; the complex numbers $a,b$ are associated to the quaternionic rotation coordinates $\boldsymbol{q} = \{q_0,q_1,q_2,q_3\}$ that are favored by engineers by -$$ -\begin{aligned} -a(\boldsymbol{q}) &= q_0 + i q_3\\ -b(\boldsymbol{q}) &= i q_1 - q_2\ ; -\end{aligned} -$$ -and finally, the binomial weights ${{k}\choose{0}}^{1/2}$ etc. that appear in the definition of $\mathbb{G}^k$ commonly are omitted in the mathematical literature. The binomial weights ensure that the Veronese embedding is a metric and symplectic isometry, as is desirable to preserve various aspects of "quantum goodness" for simulation purposes; Katsumi Nomizu (1976) seems to have been among the first to appreciate their rationale. -Motivation  This question is constructed as a tribute to Aaronson's fine "Sure-Shor Separator" question, and as a tribute to the outstanding Kalai/Harrow debate that Aaronson's question helped stimulate. For engineers, Gil Kalai's proposition "Why quantum computers CAN'T work" belongs to a broad class of strategically crucial propositions that include "How large-scale quantum simulation DOES work" and "How atomic-resolution microscopes CAN work", and include too unanswered fundamental physics questions like "Is Nature's quantum dynamical state-space EXPERIMENTALLY static and flat?" -Acknowledgements  Appreciation and thanks are extended to Scott Aaronson, Gil Kalai, and Aram Harrow, for their sustained and admirably collegial commitment to asking good questions and seeking good answers. - -REPLY [5 votes]: I doubt that these theories can recreate the causal properties of nonrelativistic QM, i.e. the feature of local evolution/measurement not causing observable effects far away. I think they are also incompatible with the traditional Born rule (i.e. probability = absolute value squared of amplitude), because these theories plus the Born rule would mean that certain correlations are impossible for classical probability distributions, which would quick lead to undesirable behavior. -As written your models are incomplete, because they describe only Hamiltonian evolution, but not (a) how to combine several such systems, or (b) how measurement works, including local measurement. I really can't see how to complete your models in this way without breaking lots of other things. -(Also, I think Gil believes in the standard model of Hilbert space, and disagrees only about whether the noise assumptions of the threshold theorem are realistic. So I think his skepticism is distinct from this line of reasoning.)<|endoftext|> -TITLE: Is there a good way to understand the free loop space of a sphere? -QUESTION [22 upvotes]: I'd like to understand the structure of the free loop space of $S^n$ for small values of $n$. Here "understand" means roughly that I'd like to know a CW complex with the same homotopy type. -I already "understand" the pointed loop space of $S^n$ using the James reduced product. (See Hatcher, Algebraic Topology, Section 4.J.) Is there something analogous for the free loop space? Alternatively, is there some simple relationship between the free loop space and the pointed loop space? - -REPLY [6 votes]: I recently learned the following description of the topology of the loop space at the European Talbot on free loop spaces, and is probably what Ziller does in the paper in the answer of Lennart Meier. -The idea is to use Morse theory for the standard round sphere. The critical points of the energy functional $E:LS^n\rightarrow \mathbb R$ are the closed geodesics. -What are the closed geodesics? The trivial ones are the constant loops. The collection of all constant loops forms a family diffeomorphic to $S^n$. Other closed geodesics are great circles. The family of great circles can be described as $T_rS^n$ where $r$ is the energy of a prime great circle. But there are more closed geodesiscs. These are the iterates of the great circles. The iterates form families $T_{k^2r}S^n$, which are of course all diffeomorphic to the unit sphere bundle of $S^n$. -The wonderful luck is that $E$ is Morse-Bott and satisfies Palais-Smale. That means that you can build up $LS^n$ by attaching disc bundles along the critical sets. Thus we see that the homotopy type is -$$ -LS^n\simeq S^n\cup D_{\lambda(1)}T_{r}S^n\cup D_{\lambda(2)}T_{2^2r}S^n\cup\ldots -$$ -In the union we make some identifications on the boundary of the disc bundles which I do not want to specify here. The number $\lambda(k)$ is the Morse-Bott index of the critical set $T_{k^2r}S^n$ (If I remember correctly it is $(2k-1)(n-1)$). -One can show that the attachment of the discbundles is homological trivial in this situation. Thus the homology of the loop space the sum of the homology of the sphere and copies of the homology of the unit tangent bundle of the sphere shifted in dimension. -The unit tangent bundle of the sphere has two torsion if $n$ is even, but not when $n$ is odd. This explains the difference in the homology of the free loop space in the even and odd case. This is Lennart's comment on the Hairy Ball theorem. -I think that this tactic can be made to work for all manifolds that admit metrics all of whose geodesics are closed (and of the same length). -If one feels like it one can even compute string operations like the Chas-Sullivan product in this manner.<|endoftext|> -TITLE: Distinct manifolds with the same configuration spaces? -QUESTION [20 upvotes]: For a space $X$, let $C_k X$ denote the space of configurations of $k$ distinct unordered points in $X$. - -What is an example of a pair of smooth manifolds $M$ and $N$ that are not homeomorphic but such that for all $k \in \mathbb{N}$, the spaces $C_k M$ and $C_k N$ are homotopy equivalent? - -(This question is prompted by David Ayala's work on automorphisms of Ran spaces, and more generally the question of the strength of topological field theory invariants.) -Update: Qiaochu and Ricardo answered the question using examples of manifolds with boundary and non-compact manifolds, respectively, so let's now restrict attention to the (no doubt much harder) issue that is most directly relevant for TFT invariants: - -Is there an example where the manifolds are closed? - -REPLY [11 votes]: I will present an example involving only (non-compact) manifolds without boundary. As far as I know, the analogous problem for closed manifolds is wide open. Nevertheless, the article Configuration spaces are not homotopy invariant by Paolo Salvatore and Riccardo Longoni is very much worth looking at (it was published in Topology, volume 44, number 2, 2005, pages 375-380).$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\Conf}{\mathrm{Conf}}$$\newcommand{\To}{\longrightarrow}$ -Proposition: Assume $n \geq 3$. Let $W$ be a contractible topological $n$-manifold without boundary, and $O \subset W$ an open subspace homeomorphic to $\RR^n$. Then the inclusion of $O$ into $W$ induces homotopy equivalences on all ordered and unordered configuration spaces. -A few consequences of this are: - -An embedding $V \to W$ between contractible $n$-manifolds without boundary induces a homotopy equivalence on all ordered and unordered configuration spaces. -If $V$ and $W$ are contractible $n$-manifolds without boundary, then there are homotopy equivalences $\Conf(V,k) \simeq \Conf(W,k)$ between their ordered configuration spaces, and $C_k V \simeq C_k W$ between their unordered configuration spaces. -In particular, the configuration spaces, both ordered and unordered, of the Whitehead manifold are homotopy equivalent to the corresponding configuration spaces of $\RR^3$. - - -The remainder of this post discusses the proof of the proposition above. Here is a lemma which will be used. -Lemma: Let $S$ be any finite subset of $O$. Then the inclusion $O\setminus S \to W\setminus S$ is a weak equivalence. -Proof: -Apply the van Kampen theorem to the open cover of $W$ by $O$ and $W\setminus S$. Since $O\setminus S$ is simply connected (since it is equivalent to a wedge sum of $(n-1)$-dimensional spheres), it follows that $\Pi_1(W\setminus S) \simeq \Pi_1(W)$ is trivial. -Now apply the Mayer–Vietoris sequence for homology to the same open cover of $W$ by $O$ and $W\setminus S$. The contractibility of $W$ implies that the inclusion $O\setminus S \to W\setminus S$ is a homology equivalence. Since both spaces are simply connected, this inclusion is also a weak equivalence. ■ -Now we prove that the map on ordered configuration spaces $\Conf(O,k) \to \Conf(W,k)$ is a weak equivalence for any $k>0$. Note that we have a map of (horizontal) fibration sequences -$$ -\begin{matrix} -O\setminus S & \To & \Conf(O,k) & \To & \Conf(O,k-1) \\ -\big\downarrow & & \big\downarrow & & \big\downarrow \\ -W\setminus S & \To & \Conf(W,k) & \To & \Conf(W,k-1) -\end{matrix} -$$ -where $S$ is any subset of $O$ of size $k-1$. The previous lemma states that the map on the fibres is a weak equivalence. Thus, by induction on $k$, we conclude that the map $\Conf(O,k) \to \Conf(W,k)$ is a weak equivalence. -Now the result for unordered configuration spaces follows since we have a commutative diagram of fibration sequences -$$ -\begin{matrix} -\Sigma_k & \To & \Conf(O,k) & \To & C_k O \\ -\big\downarrow & & \big\downarrow & & \big\downarrow \\ -\Sigma_k & \To & \Conf(W,k) & \To & C_k W -\end{matrix} -$$ -which is an equivalence on the fibres and total spaces. We conclude that the map on the base spaces is also a weak equivalence.<|endoftext|> -TITLE: Reference for a conjecture on the first primes congruent to 1 modulo other primes -QUESTION [13 upvotes]: Given a prime $p$, define $f(p)$ to be the smallest prime congruent to $1$ modulo $p$. For example, $f(7)=29$. It has been conjectured that $f(p) -TITLE: Coherence between different ranking methods of a graph's vertices -QUESTION [9 upvotes]: Given a (connected) graph $G$ it is natural to want to rank its vertices, with the more "central" vertices ranked higher. -Two natural ways of doing it are: - -By the degrees. -By the entries in a Perron eigenvector of the adjacency matrix. - -These two methods coincide for regular graphs and for so-called harmonic graphs (defined as graphs in which the degree vector is an eigenvector) - which is all a tad trivial. -What is more interesting is that for many random graphs I've checked the two orderings coincide as well and I am able to show that they coincide for threshold graphs. - -Have such graphs been studied? - -I did find in the mathematical sociology literature some work on the question when the most central vertex w.r.t both rankings is the same but nothing for the whole vector. -(If such graphs haven't been named yet, I propose to call them tranquil, since tranquility is a lesser form of harmony). -P.S. -Method 2 is essentially what Google does in its PageRank algorithm. - -REPLY [2 votes]: I'm not sure how common this is. It may depend on how you are determining random graphs. -There may be ties so let me separate the issues of the equivalence relation "same rank" from those of the linear order among equivalence classes. In the case of ranking by degrees there will have to be at least one tie. In the case of the Perron ranking there might or might not be any ties. Vertices in the same orbit of the automorphism group will have the same rank in the Perron ranking (and equal degree). In the simplest graph with no automorphisms ( A seven vertex tree with leaves at distances $1,2,$ and $3$ from a central vertex) no two vertices get the same weight. It is true for this graph that the three lowest weights occur on the degree $1$ vertices and the highest weight on the unique degree $3$ vertex. Is this what you mean? Otherwise I'd expect "most" trees to fail. -Since vertices in the same orbit of the automorphism group will have the same rank in either system. If the automorphism group is degree transitive there is some chance that the two equivalence relations will be the same. This would partly explain threshold graphs. I say partly because I have not shown that the linear order on equivalence classes is according to degree. Perhaps this additional condition (which holds for threshold graphs) is enough: If there is a path $u,v,w$ with $v$ having lower degree than $u$ or $w$, then there is also an edge $u,w.$ That is just a guess, but I give an example below where this condition does not hold and the linear orders are not the same. -At the end I note a weaker condition than degree transitivity which suffices for same degree vertices to have equal Perron rank. It is useful for analyzing examples such as the following one: -Here is a rather symmetric graph where the linear order fails: Consider a graph with $26$ vertices belonging to classes $A,B,C,D$ of respective sizes $9,3,12,2$ and respective degrees $1,7,2,6.$ Each vertex in $B$ is connected to $3$ vertices from $A$ and $4$ from $C$ while each vertex in $D$ is connected to $6$ vertices from $C.$ Then the largest eigenvalue is roughly $3.38$ and a Perron eigenvector assigns weights of roughly $0.267,0.904,0.563,1$ to the classes. So the two degree $6$ vertices receive higher weight than the three degree $7$ vertices. In the similar example with degrees $1,5,2,4$ the vertices of degrees $4$ and $5$ both get equal weight. -Somewhat more general than degree transitivity is degree regularity: Suppose that the existing degrees are $d_1,d_2,\cdots,d_k$ and that there are $k^2$ integers $n_{ij}$ so that each vertex of degree $d_i$ has exactly $n_{ij}$ neighbors of degree $d_j$. Then the (right) Perron eigenvector of the $k \times k$ matrix with entries $n_{ij}$ lifts in an obvious way to the graph.<|endoftext|> -TITLE: Is the perimeter of an ellipse with integer axes irrational? -QUESTION [15 upvotes]: Let $Q$ be an ellipse with integer-length axes $a$ and $b$: -$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \;.$$ -The perimeter of $Q$ is given by the complete elliptic integral of the 2nd kind, $E(\;)$: -$4 a E(\epsilon)$ where the eccentricity $\epsilon = \sqrt{1-(b/a)^2}$. -Is the following known? - -Q. Is the perimeter of an ellipse with integer-length axes $a$ and $b$ always - irrational? If not, for which $a,b$ is it rational? - -            -           $a,b=2,1$. Perimeter is approximately $9.051752335744220113037500073330267105760842885887$. - -Answered by paul Monsky in the comments: The perimeter is transcendental! -Established by Theodor Schneider in 1937. - -REPLY [10 votes]: Answered by paul Monsky in the comments: The perimeter is transcendental! -Established by Theodor Schneider in 1937: - -  - -  - -Roughly (Those with better knowledge of Deutsch, please correct!): - -The value of an elliptic integral of the first or second kind with algebraic - coefficients and between algebraic limits is transcendental.<|endoftext|> -TITLE: Fourier transform of the unit sphere -QUESTION [22 upvotes]: The Fourier transform of the volume form of the (n-1)-sphere in $\mathbf R^n$ is given by the well-known formula -$$ -\int_{S^{n-1}}e^{i\langle\mathbf a,\mathbf u\rangle}d\sigma(\mathbf u) = (2\pi)^{\nu + 1}\|\mathbf a\|^{-\nu}J_\nu(\|\mathbf a\|), -\qquad\nu=\frac n2 -1, -\tag1 -$$ -found e.g. in [1, p. 198] or [2, p. 154]. - -Does anyone here know earlier references, and perhaps who first published this formula? - -According to Watson [3, p. 9] the case n=2, -$$ -\frac1{2\pi}\int_{S^1}e^{ia\cos\theta}d\theta=J_0(a) -\tag2 -$$ -goes back to Parseval [4], but I am mainly curious about the case n=3, -$$ -\frac1{4\pi}\int_{S^2}e^{i\langle\mathbf a,\mathbf u\rangle}d\sigma(\mathbf u)=\frac{\sin\|\mathbf a\|}{\|\mathbf a\|}. -\tag3 -$$ - - -I. M. Gel'fand & G. E. Shilov, Generalized functions, vol. 1, Academic Press (1964). -E. M. Stein & G. Weiss, Introduction to Fourier analysis on Euclidean spaces, Princeton UP (1971). -G. N. Watson, A treatise on the theory of Bessel functions, Cambridge UP (1922) -M. A. Parseval, Mémoire sur les séries et sur l'intégration complète (etc.) (1805) - -REPLY [2 votes]: Sonin computed (according to Fichtenholz, but no reference given) $$\int\limits_{\sum_{k=1}^n x_k^2\leq 1}\exp(\langle a,x\rangle) dx_1...dx_n.$$ Fichtenholz did research in multivariate integration, so he knew all these things from folklore, I guess.<|endoftext|> -TITLE: Primes $p$ for which $pk+1$ is prime for small $k$ (or approximating Sophie Germain) -QUESTION [9 upvotes]: The twin prime conjecture says there are infinitely many pairs $p,p+2$ that are both prime, and although we still don't know whether it's true there's been a lot of progress recently showing that there are infinitely many pairs $p,p+k$ that are both prime, for $k$ bounded by a constant. -It's also an old conjecture that there are infinitely many pairs $p,2p+1$ that are both prime (Sophie Germain primes and safe primes respectively). Is there any hope of adapting the recent twin prime progress to approximate the constant 2 in the Sophie Germain conjecture by something larger? That is, can it be proven that there are infinitely many pairs $p,kp+1$ that are both prime, with $k=O(1)$? -Of course Dirichlet and Linnik say that for every $p$ there is eventually a $k$ such that $pk+1$ is prime, but I'm more interested in the case of infinitely often rather than for all $p$, and I'd like $k$ to be significantly smaller than the proven bounds in Linnik's theorem. The best bound I know how to prove is $k=O(p)$, which follows from the existence of infinitely many "pseudo Sophie Germain primes", numbers $q$ that are prime or the product of two primes and for which $2q+1$ is prime (see arXiv:math/0603439). Letting $p$ be the largest prime factor of $q$ gives $k=2q/p\le 2p$. So I think any bound $k=o(p)$ would be interesting. - -REPLY [4 votes]: Consider a tuple of the form $(n,a_1 n +1, a_2 n+1, \ldots a_k n+1)$ where the $a_k$'s are chosen such that there is no local obstruction to the numbers in the tuple being simultaneous prime. Then the GYP/Zhang method allows one to prove that two of these are prime infinity often if $k$ is larger than some fixed $K_0$ (see: http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes for the most up-to-date value of $K_0$). The Maynard/Tao argument allows one to get more primes in the tuple at the expense of making $K_0$ larger. -The main issue with getting something of the form of the Sophie Germain conjecture is that one does not have any control over where in the tuple the primes occur. If one could somehow force the first term of the tuple to be one of the primes then we could deduce that the pair $(p, ap+1)$ were simultaneously prime for some a in any admissible tuple of sufficiently large size. However, there seems to be no way to deduce/force restrictions of this form with the current method. [Of course getting the Sophie Germain conjecture with precisely $2$ is probably even harder and of a similar difficulty to the twin prime problem.] -On the other hand, by the pigeonhole principle, one can deduce that there are infinity many prime pairs of the form (say) $(an+1, bn+1)$ which is of a similar flavor. -Of course, given the recent breakthroughs in this area, problems of this form do seem tantalizingly close.<|endoftext|> -TITLE: Difference of j-invariant values and the abc conjecture -QUESTION [35 upvotes]: I learned of the following example in a recent seminar: if $j(\tau)$ denotes the usual $j$-invariant, and $\alpha = (-1+i\sqrt{163})/2$, then -\begin{align*} -\frac{j(i)}{1728} &= 1 \\ -\frac{-j(\alpha)}{1728} &= 151931373056000 = 2^{12}5^323^329^3 \\ -\frac{j(i)-j(\alpha)}{1728} &= 151931373056001 = 3^37^211^219^2127^2163. -\end{align*} -A computation revealts that $(a,b,c) = (1, 151931373056000, 151931373056001)$ is a reasonably high-quality abc triple: if $R$ denotes the radical of $abc$ (the product of the distinct primes dividing $abc$), then -$$ -q(a,b,c) = \frac{\log c}{\log R} = 1.20362\dots. -$$ -My question is: is this a freak coincidence (including the fact that $163$ divides $c$), or is this example part of a larger theory? Are there families of high-quality abc triples that come from differences of $j$-invariants? - -REPLY [27 votes]: As T.Dokchitser explained, there is indeed good reason for this, -but $j(\alpha) = -640320^3$ is the last such example for the -classical $j$-invariant. (One of the motivations for my -"Shimura curve computations" -(LNCS 1423 [=ANTS-3 proceedings, 1998]) was to find exotic ABC -triples coming from CM points on Shimura curves, but no spectacular -example turned up.) Still the connection can be used in -the opposite direction. If we somehow knew the ABC conjecture -(with effective constants) but didn't yet know that ${\bf Q}(\alpha)$ -is the last quadratic imaginary field of class number one, then we could -use the the fact that $j(\alpha)$ is a cube and $j(\alpha)-12^3$ is almost -a square to solve the $h=1$ problem. In 2000 Granville and Stark -(Invent. Math. 139 #3, 509$-$523) used this idea to prove that -an ABC conjecture over arbitrary number fields (with effective constants) -would imply a strong enough lower bound on $h(-D)$ to banish the Siegel zero -for imaginary quadratic characters! Unfortunately this approach has yet -to yield an unconditional proof. -P.S. (In response to the OP's comment on T.Dokchitser's accepted answer) -Here's the list of $13$ discriminants $-D$ of imaginary quadratic orders -of class number 1 and the corresponding integers $j = j(\alpha_D)$ with -$\alpha_D = (D + \sqrt{-D})/2$: -$$ -\begin{array}{c|cccccccc} --D & -3 & -4 & -7 & -8 & -11 & -12 & -16 & -19 & -\\ \hline - j & 0 & 12^3 & -15^3 & 20^3 & -32^3 & 2\cdot 30^3 & 66^3 & -96^3 & -\end{array} -$$ $$ -\begin{array}{c|ccccc} --D & -27 & -28 & -43 & -67 & -163 \\ -\hline -j & -3 \cdot 160^3 & 255^3 & -960^3 & -5280^3 & -640320^3 -\end{array} -$$<|endoftext|> -TITLE: Square of primary ideals -QUESTION [12 upvotes]: Is there any example of a $P$-primary ideal $I$ in a noetherian domain $R$ such that $I^2=PI \not=P^2$? - -REPLY [7 votes]: What a fun problem for the holiday season! Yes, an example is $R=\mathbb C[a,b,c,d]/(a^4-bd, b^3-cd, c^2-ad, a^3b^2c-d^3) \cong \mathbb C[t^9,t^{13},t^{16},t^{23}]$. Let $I=(a,b,c)$ and $P=(a,b,c,d)$.<|endoftext|> -TITLE: The number of distinct prime factors of $n\in\mathbb N$ -QUESTION [6 upvotes]: Let $\omega(n)$ be the number of distinct prime factors of a natural number $n$. -Note that $\omega(n)=0\iff n=1$, and that $\omega(24)=\omega(2^3\cdot 3^1)=2\ (\not = 4)$. -(For more details, you can see "Distinct Prime Factors" on wolfram math world) -Then, for $m\in\mathbb N$, let $S_m=\{n|n\in\mathbb N, \omega(n)=m\}$. Also, Let $N_m(x)$ be the counting function that gives the number of the elements of $S_m$ less than or equal to $x$, for any real number $x$. See the line graph below. This graph shows $N_m(x)\ (m=1,2,3,4)$ in $1\lt x\le 2309=2\cdot3\cdot5\cdot7\cdot11-1.$ Note that $N_m(2309)=0$ for $m\ge 5$, and that $\omega(2310)=\omega(2\cdot 3\cdot 5\cdot 7\cdot 11)=5$. -$\ \ \ \ \ \ \ \ \ \ $ -In addition to this, let $P_m(x)=\frac{N_m(x)}{x}\times 100$. See the line graph below. This graph shows $P_m(x)\ (m=1,2,3,4)$ in $1\lt x\le 2309=2\cdot3\cdot5\cdot7\cdot11-1.$ -$\ \ \ \ \ \ \ \ \ \ $ -Then, here are my questions. - -Question : Are the followings true? -$(1)$ For every $m\in\mathbb N$, $N_m(x)=O(x)\ (x\to\infty)$. -$(2)$ For every $m\in\mathbb N$, $\lim_{x\to\infty} P_m(x)=0$. -$(3)$ If $x\ge 50$, then $P_2(x)\gt P_m(x)$ for every $m\ge 3$. - -The above graphs led me to the these conjectures, but I don't have any good ideas. I would like to know any relevant references. Can anyone help? - -REPLY [7 votes]: I don't quite understand question (1). Isn't $N_m(x) \leq x$ trivially? You seem to be interested in fixed $m$. Then an affirmative answer to (2) and a negative answer to (3) follows from a classical extension of the prime number theorem due to Landau. See, for example, -http://www2.imperial.ac.uk/~bin06/M34PM16-Analytic-Number-Theory/m3pm16div.pdf -In fact, the answer to (2) is "yes, uniformly in $m$." Indeed, a theorem of Hardy and Ramanujan implies that $N_m(x) \ll x/\sqrt{\log\log{x}}$ uniformly in $m$.<|endoftext|> -TITLE: What is the crucial difference the Maynard/Tao approach and Goldston-Pintz-Yildirim that extends to prime k-tuples with $k>2$ -QUESTION [29 upvotes]: Suppose $m$ is a positive integer. A quantity of interest is -$$ -H_m = \liminf_{n\to\infty} \left(p_{n+m} - p_n \right) -$$ -The twin prime conjecture, is, of course $H_1 = 2$, the the prime k-tuples conjecture of Hardy and Littlewood asserts that $H_2 = 6$, $H_3 = 8$ and so on. Goldston-Pintz-Yildirim showed that under the Elliot-Halberstam conjecture, $H_1 < \infty$, a result that Yitang Zhang has famously recently established unconditionally. -It is my understanding that the methods of GPY were essentially restricted when dealing with $m>1$. One of their non-trivial results was that $g_n = p_{n+1} - p_n$ satisfies $g_n = o(\log p_n)$, or in other words that $\liminf \frac{g_n}{\log p_n} = 0$, which obviously trivially follows from Zhang's result. -However, it is my understanding that even under the full Elliot-Halberstam conjecture, GPY were unable to prove even their weaker result for $m\geq 2$. Maynard and Tao (see http://arxiv.org/abs/1311.4600) have shown that in fact, that $H_m < \infty$. Apparently, this work only required the classical Bombieri-Vinogradov theorem, not even the variation of that Zhang's proof uses. -So my question is this: what shortcoming did the GPY approach have that did not allow it to extend to $m>2$, and how has Maynard's approach solved this problem? -PS: I apologize if this question is somehow inappropriate for Mathoverflow. I am an undergraduate trying to read up on these recent results by myself as much as I can, and I figured this would probably be the best place to ask my question. - -REPLY [38 votes]: The major difference is the choice of the Selberg Sieve weights. I strongly recommend reading Maynard's paper. It is well written, and the core ideas are nicely explained. - -In what follows, I'll give a brief explanation of what Selberg Sieve weights are, why they appear, and what was chosen differently in Maynard's paper compared to that of Goldston Pintz and Yildirim. -Let $\chi_{\mathcal{P}}(n)$ denote the indicator function for the primes. Given an admissible $k$-tuple $\mathcal{H}=\{h_11$. If this is the case, then by the non-negativity of $a(n)$, we have that for some $n$ between $x$ and $2x$ $$\left(\sum_{i=1}^k\chi_{\mathcal{P}}(n)(n+h_i)\right)\geq \lceil\rho \rceil,$$ and so there are at least $\lceil\rho \rceil$ primes in an interval of length at most $h_k-h_1$. This would imply Zhangs theorem on bounded gaps between primes, and if we can take $\rho$ arbitrarily large, it would yield the Maynard-Tao theorem. However, choosing a function $a(n)$ for which equation $(1)$ this holds is very difficult. Of course we would like to take $$a(n)=\begin{cases} -1 & \text{when each of }n+h_{i}\text{ are prime}\\ -0 & \text{otherwise} -\end{cases} ,$$ - as this maximizes the ratio of the two sides, but then we cannot evaluate $\sum_{xR$ where $R1$, with $k$ depending on $\rho$. - -Here are four great resources where you can read more: - -Maynard's paper. -Tao's blog posts. -Granville's article on the Maynard-Tao theorem and the work of Zhang. -Soundararajan's exposition of Goldston Pintz and Yildirim's argument.<|endoftext|> -TITLE: "Circular" domination in ${\mathbb R}^4$ -QUESTION [37 upvotes]: The following problem is related to (and motivated by) the first open case of this MO question. It is difficult to believe that this is a hard problem; and yet, I do not have a solution. -For two vectors $x,y\in {\mathbb R}^4$, let us say that $x$ is dominated by $y$ if there are (at least) two coordinates in which $x$ is strictly smaller than $y$. With this convention, the problem goes as follows: - -Does there exist a finite, non-empty set $S\subset {\mathbb R}^4$ with the property that for any two vectors $x,y\in S$, there exists yet another vector $z\in S$ such that the coordinate-wise maximum of $x$ and $y$ is dominated by $z$? - -Notice that it is easy to find a set of vectors with the property in question in ${\mathbb R}^5$ (see comments below), or to find a finite, non-empty set of vectors in ${\mathbb R}^4$ such that any vector is dominated by another one. On the other hand, there is no finite, non-empty set in ${\mathbb R}^4$ in which for any two vectors there is a third one exceeding their maximum in at least three coordinates. - -Post-factum: make sure you have not missed the extremely clever solution by zeb! It is not easy to follow, but I made an effort to trace it carefully and, as far as I can see, it is perfectly correct. Already the first step is very unusual: instead of looking at the smallest counterexample, zeb considers a counterexample which is by no means smallest, but instead has a comprehensible structure. Whether you have voted for the problem itself or not, consider voting for zeb's solution! - -REPLY [28 votes]: There is no such set $S$. Suppose for a contradiction that there was. By rescaling the coordinates, we can assume all coefficients of points in $S$ are positive integers. Now construct a set $S'$ as follows: for every point $(a,b,c,d)\in S$, put points $(a,b,0,0), (a,0,c,0), (a,0,0,d), (0,b,c,0), (0,b,0,d), (0,0,c,d)$ into $S'$. $S'$ will then be a solution to the problem if $S$ was, so replace $S$ by $S'$. -By adding finitely many additional points to $S$, we may further assume that if we decrease a coefficient of a point in $S$ by $1$, then the resulting point is still in $S$ as long as the coefficient was greater than $1$ to start with. $S$ now has the following properties: -Property 0: Every point in $S$ has exactly two positive coefficients, and the maximal $M$ such that $(M,1,0,0)\in S$ is the same as the maximal $M'$ such that $(M',0,1,0)\in S$. -Property 1: If $(a,b,0,0)\in S$ and $(0,0,c,d)\in S$, then at least one of $(a+1,0,c+1,0), (a+1,0,0,d+1), (0,b+1,c+1,0), (0,b+1,0,d+1)$ is in $S$. -Proof: Let $A$ be maximal such that $(A,b,0,0)\in S$ and let $D$ be maximal such that $(0,0,c,D)\in S$. Some element of $S$ dominates $(A,b,c,D)$, and by the choice of $A$ (respectively $D$) it can't have its last two (respectively first two) coefficients both equal to $0$. -Property 2: If $(x,y,0,0)\in S$, then at least one of $(x+1,0,0,2), (0,y+1,0,2)$ is in $S$. -Proof: Let $X$ be maximal such that $(X,y,0,0)\in S$, and let $M$ be maximal such that $(0,0,M,1)\in S$. Then some element $(a,b,c,d)$ of $S$ dominates $(X,y,M,1)$. By Property 0 we have $c\le M$, so $c = 0$. Since we can't have $(X+1,y+1,0,0)\in S$, $d$ must not be $0$, so $d \ge 2$ and either $a \ge X+1 \ge x+1$ or $b \ge y+1$. -Now consider the following property, depending on a parameter $k$: -Property $k$: If $(x,y,0,0)\in S$, then at least one of $(x+1,0,0,k), (0,y+1,0,k)$ is in $S$. -If $S$ has Property $k$ for every integer $k$, then clearly $S$ must be infinite. We'll now prove that in fact Property $k$ implies Property $k+1$ for $k \ge 2$, and this will give our desired contradiction. -Property $k$ implies Property $k+1$: Choose $A,B,C$ maximal such that $(A,0,0,k), (0,B,0,k), (0,0,C,k) \in S$. To see that such $A,B,C$ exist at all, we apply Property $k$ to the point $(1,1,0,0)$ to see that either $(2,0,0,k)$ or $(0,2,0,k)$ is in $S$, and then we apply Property $0$ to see that all of $(1,0,0,k), (0,1,0,k), (0,0,1,k)$ are in $S$. -Now we construct a sequence of points $s_i \in S$, each with fourth coordinate equal to zero, as follows. Start with $s_0 = (x,y,0,0)$. For each $i$, split into several cases: - -If $s_i = (0,u,v,0)$, apply Property 1 to $(0,u,v,0)$ and $(A,0,0,k)$, and call the resulting dominating point $s_{i+1}$. If the fourth coordinate of $s_{i+1}$ is nonzero, stop here. -If $s_i = (u,0,v,0)$, apply Property 1 to $(u,0,v,0)$ and $(0,B,0,k)$. Proceed similarly to the above. -If $s_i = (u,v,0,0)$, apply Property 1 to $(u,v,0,0)$ and $(0,0,C,k)$. Proceed similarly to the above. - -The claim is that this process must stop, and the final $s_{i+1}$ produced will be the point we are looking for. To see this, first note that we can never end up in the same case twice in a row during this process. Furthermore, we can never pass through all three cases in the course of this process: for instance, if we were to start in case 3 at step $i-2$, then go through case 2 at step $i-1$, and then end up in case 1 at step $i$, then if we write $s_i = (0,u,v,0)$ we find that $u = B+1$ and $v = C+2$. Applying Property $k$ to $s_i$, we see that one of $(0,B+2,0,k), (0,0,C+3,k)$ is in $S$, contradicting the choice of $A,B,C$. -Thus, the process alternates between case 3 and some other case, say case 2 for concreteness. At each step, the first coordinate of $s_i$ will then increase, and inductively we see that for $i\ge 0$, the first coordinate of $s_i$ is $x+i$. Since it can't increase without bound, the process must eventually end. If it ends after step $0$, then the dominating vector $s_1$ is either $(x+1,0,0,k+1)$ or $(0,y+1,0,k+1)$. If it ends after step $i$ for $i\ge 1$, then the dominating vector $s_{i+1}$ must be $(x+i+1,0,0,k+1)$, since otherwise it would be either $(0,B+2,0,k+1)$ or $(0,0,C+2,k+1)$ depending on whether $i$ was even or odd, contradicting the choice of $A,B,C$.<|endoftext|> -TITLE: Cobordism modulated by a cohomology operation -QUESTION [5 upvotes]: I've recently encountered the following cobordism theory modulated by a class $\sigma \in H^{d+1}(B^2\mathbb{Z}/2,U(1))$. -My objects are $d$-dimensional spin manifolds with chosen spin structure. Note that a spin structure, thought of as a null-homotopy of $w_2$, gives a trivialization of $w_2^*\sigma$. -My morphisms are given by cobordisms by manifolds with $w_2^*\sigma = 0$ for which the trivializations of $w_2^*\sigma$ on the boundary components given by the chosen spin structures there all come from restricting a single trivialization of $w_2^*\sigma$ on the cobordism. -Call the resulting group $\Omega^{{\rm spin},\sigma}_d$. There is a natural surjective map $\Omega^{\rm spin}_d \to \Omega^{{\rm spin},\sigma}_d$. I'd like to calculate the kernel of this map, at least for $d=1,2,3,4,5$. -There is another map $\Omega^{{\rm spin},\sigma}_d \to \Omega^{\rm SO}_d$ forgetting all the trivialization business. The composition $\Omega^{\rm spin}_d \to \Omega^{{\rm spin},\sigma}_d \to \Omega^{\rm SO}_d$ is known to have kernel in degrees $8k+1,8k+2$ with rather explicit generators. For $k=0$, these are the circle with antiperiodic spin structure and its Cartesian square, respectively. The kernel we're after is contained in this one, so to do the calculation in the range I'm interested in, it might be possible to do things case by case. -Let's do the circle case. Trivializations of $w_2^*\sigma$ on $S^1$ form a torsor over $H^1(S^1,U(1))$. Trivializations on a surface $\Sigma$ bounding $S^1$ are a torsor over $H^1(\Sigma,U(1))$. Since the class of $\partial \Sigma$ in $\Sigma$ is zero, by universal coefficients the map $H^1(\Sigma,U(1)) \to H^1(\partial\Sigma,U(1))$ is zero. Thus, any two trivializations of $w_2^*\sigma$ on $\Sigma$ restrict to the same trivialization on the boundary. Thus, the two spin structures of the circle are distinct in $\Omega^{{\rm spin},\sigma}_d$ when $\sigma \neq 0$ since in this case the spin structures actually give different trivializations of $w_2^*\sigma$ (as can be checked from the long exact sequence coming from $\sigma:\mathbb{Z}/2 \to U(1)$). -When $\sigma = 0$, then we get the full kernel. This is the case for $d=2$ since $H^3(B^2\mathbb{Z}/2,U(1)) = 0$, but this is the last degree where this cohomology vanishes. -This finishes the calculation in the range I originally considered, so how about the general case? -The next case of interest is $d=9$, for which $H^{10}(B^2\mathbb{Z}/2,U(1))$ is nontrivial but indescribable to me. -A first subproblem for the general attack might be ``when is a cohomology operation injective?" I don't know the answer to this either, except in the case $d=1$ when $\sigma:H^2(-,\mathbb{Z}/2) \to H^2(-,U(1))$ is given by a homomorphism $\mathbb{Z}/2 \to U(1)$ and we have a long exact sequence where the kernel of $\sigma$ is the image of $H^1(-,U(1))$. -Any help or references would be much appreciated. - -REPLY [5 votes]: To bring this into a classical setting, it seems to me that one can use the bordism groups $\Omega_d^{\sigma}$ where both d-manifolds and (d+1)-bordisms are not necessarily spin, but have only a trivialization of $w_2^*\sigma$. -Then there is a natural forgetful map $\Omega_d^{Spin}\to \Omega_d^{\sigma}$ with image -$\Omega_d^{spin,\sigma}$. This fits into a long exact sequence -$$\dots \to \Omega_{d+1}^{\sigma,spin}\to \Omega_d^{Spin}\to \Omega_d^{\sigma}\to \dots$$ where $\Omega_{d+1}^{\sigma,spin}$ is by definition the bordism group of $(d+1)$-manifolds with boundary equipped with a trivialization of $w_2^*\sigma$, and, on the boundary, a refinement to a spin structure. -For computations, this all translates into homotopy groups of Thom spectra, -in particular the Thom spectrum for $\Omega_*^{\sigma}$ is the one for $F\to BO$, where $F$ is the homotopy fiber for -$$ \sigma\circ w_2: BSO\to K(\mathbb Z/2,2)\to K(\mathbb Z,d+2).$$ -Stong's book is a reference for this. For computations the Adams spectral sequence might be helpful.<|endoftext|> -TITLE: What are the obstructions to showing that $\zeta$ doesn't vanish on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$ -QUESTION [18 upvotes]: Most (if not all) of the proofs of the Prime Number Theorem that I have seen in the -literature rely on the fact that the Riemann zeta function, $\zeta(s)$, does not vanish -on the line ${\rm Re}(s) = 1$. Since we have a lot of numerical evidence for the validity of RH, one would suspect that for small $\varepsilon$ (to start with) we have that $\zeta(s) \neq 0$ on the strip $1- \varepsilon < {\rm Re}(s) \leq 1$. Does anyone know what are the main difficulties for proving such a result? - -REPLY [2 votes]: The third approach looks at a known zero free region. The simplest such theorem is that there is some $A$ such that -$$ -\zeta(s)\ne0\quad\text{for}\quad \sigma>1-A/\log(t). -$$ -What prevents us from 'straightening it out' so the boundary is a vertical line? -The proof begins like the proof that $\zeta(1+it)\ne0$ (see previous answer), but considers instead of $\log(\zeta(s))$, rather $\log(\zeta(s))^\prime=\zeta^\prime(s)/\zeta(s)$, again expanded as a sum over primes. For $\sigma>1$, and any $t$, taking real parts and using the same trig identity shows that -$$ -0\le -3\frac{\zeta^\prime}{\zeta}(\sigma)-4\text{Re} \frac{\zeta^\prime}{\zeta}(\sigma+it)-\text{Re}\frac{\zeta^\prime}{\zeta}(\sigma+2it). -$$ -We will estimate each of the three terms on the left side. The first is easy: -$$ --3\frac{\zeta^\prime}{\zeta}(\sigma)=\frac{3}{\sigma-1}-3C+O(\sigma-1)<\frac{3}{\sigma-1}, -$$ -where $C$ is the Euler constant. -For the other two, we need the expression of $\zeta(s)$ as a product over nontrivial zeros $\rho=\beta+i\gamma$. Taking logarithmic derivatives gives -$$ --\frac{\zeta^\prime(s)}{\zeta(s)}=-\log(2\pi)+1+C+\frac{1}{s-1}+\frac12\frac{\Gamma^\prime(s/2+1)}{\Gamma(s/2+1)}-\sum_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right). -$$ -Via Stirling's formula, for $s\to\infty$ in a vertical strip -$$ -\frac{\Gamma^\prime(s)}{\Gamma(s)}\sim \log(s)\sim\log(t). -$$ -This is where the $\log(t)$ in our zero free region will come from, but so far -$$ --\frac{\zeta^\prime(s)}{\zeta(s)}+\sum_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)\sim\frac12\log(t/2). -$$ -Taking real parts, the asymptotic relation above gives that -$$ --\text{Re}\frac{\zeta^\prime(s)}{\zeta(s)}+\sum_\rho\left(\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2}+\frac{\beta}{\beta^2+\gamma^2}\right)<\log(t). -$$ -Recalling that $\sigma>1$, every term in the sum over $\rho$ is positive, so we estimate the last of our three terms as -$$ --\text{Re}\frac{\zeta^\prime(\sigma+2it)}{\zeta(\sigma+2it)}<\log(2t). -$$ -To get a zero free region, we now fix a zero $\rho_0=\beta_0+i\gamma_0$, and set $t=\gamma_0$ in the $s$ parameter. In estimating the middle of our three terms, we drop every term in the sum except that corresponding to $\rho=\rho_0$. This gives -$$ --4\text{Re}\frac{\zeta^\prime(\sigma+i\gamma_0)}{\zeta(\sigma+i\gamma_0)}<4\log(\gamma_0)-\frac{4}{\sigma-\beta_0}. -$$ -Combining the three estimates, we now have -$$ -0\le \frac{3}{\sigma-1}+4\log(\gamma_0)-\frac{4}{\sigma-\beta_0}+\log(2\gamma_0)\le\frac{3}{\sigma-1}-\frac{4}{\sigma-\beta_0}+5\log(\gamma_0). -$$ -Thus -$$ -\frac{3}{\sigma-1}+5\log(\gamma_0)\ge \frac{4}{\sigma-\beta_0} -$$ -$$ -\sigma-\beta_0\ge \frac{4(\sigma-1)}{3+5(\sigma-1)\log(\gamma_0)}, -$$ -and -$$ -1-\beta_0\ge \frac{4(\sigma-1)}{3+5(\sigma-1)\log(\gamma_0)}+1-\sigma. -$$ -Finally setting the parameter $\sigma=1+1/(10\log(\gamma_0))$, the right side of the inequality simplifies and we obtain -$$ -1-\beta_0\ge\frac{1}{70\log(\gamma_0)}. -$$ - -How might we do better? As mentioned above, the $\log(t)$ arises from the contribution of the logarithmic derivative of the $\Gamma$ function. Except for the factor of $1/2$ we dropped, this is the best estimate possible. But we lost a lot by first introducing all the zeros of $\zeta(s)$, and then throwing out all but one. A better zero free region may be obtained by subtracting off from the logarithmic derivative only the contribution of the nearby zeros. The best known bounds replace $1/\log(t)$ with $1/(\log(t)^{2/3}\log\log(t)^{1/3})$.<|endoftext|> -TITLE: "Derived" polyhedra and polytopes -QUESTION [18 upvotes]: The notion of derived polygon is natural and leads to remarkable convergence. -Start with a polygon, and replace it by locating a point on every edge -a fraction $\alpha$ between the two endpoints. For example, here is what happens for a random polygon of $8$ vertices and $\alpha=\frac{1}{4}$ (successive images rescaled): -           - - -MathWorld: "the derived polygons ... approach a shape with opposite sides parallel and equal in length, and all have the same centroid." - -Now imagine generalizing this to polyhedra, let's say, convex polyhedra in $\mathbb{R}^3$ with triangular faces. -Start with a polyhedron $P$ of $F$ faces and $V$ vertices. For a given triple of positive reals $(\alpha,\beta,\gamma)$ that sum to $1$, locate a point $p$ in each face by -weighting the face's three vertices by $(\alpha,\beta,\gamma)$. (So if $\alpha=\beta=\gamma=\frac{1}{3}$, $p$ is the centroid of the face.) Now replace the original polyhedron -$P$ by the convex hull of the new points. Here is an example with $(\frac{1}{4},\frac{1}{2},\frac{1}{4})$ (again, each image rescaled): -           - -You can see what happens. At first $V=8$ and $F=10$, but next $V=10$ (because every face -generates a point), and by Euler's formula, $F=16$ ($F=2 V-4$). Etc.: the number of vertices and faces grows. Here are enlargements of the first and 10th polyhedra, -the latter of which has $1540$ vertices: -  -It seems that, despite the combinatorial growth in $\mathbb{R}^3$ absent in $\mathbb{R}^2$, it is likely the shape is converging to a limit that may bear -similarities to what can be established in 2D. -My question is: - -Q. Has this process been studied before? If so, does the shape approach a limit largely independent of $(\alpha,\beta,\gamma)$, analogous to the situation in 2D? - If so, what characteristics does this limit shape bear to the initial polyhedron? - -REPLY [3 votes]: Just a bit more "data" in the form of images. I started with a "random polyhedron" -built from the convex hull of a small number of points. Then I iterated the process -of replacing each face by its centroid, i.e., $\alpha=\beta=\gamma=\frac{1}{3}$ -in my original description, and -again taking the convex hull. At least visually, the data support Gjergji's suggestion -that the process converges to ellipsoids (of different axes lengths). -    - -So, to make an explicit conjecture out of these observations: - -Conjecture. Given any convex polyhedron $P$ in $\mathbb{R}^3$, - let $c(P)$ be the convex hull of the - centroids of the faces of $P$. Then $c^k(P)$ converges to an ellipsoid - as $k \to \infty$. - -Secondarily, it makes sense to conjecture the same holds for any convex -polytope $P$ in $\mathbb{R}^d$.<|endoftext|> -TITLE: arctan of a square root as a rational multiple of pi -QUESTION [7 upvotes]: I know that if $x$ is a rational multiple of $\pi$, then $tan(x)$ is algebraic. -Is there a fairly simple way to express $x$ as $\pi\ m/n$, if $tan(x)$ is given as a square root of a rational? - -REPLY [5 votes]: If $x \in \mathbb{Q}$ and $tan^2(x\pi) \in \mathbb{Q}$, then $tan(x\pi) \in \{0, \pm\sqrt{3}, \pm\frac{1}{\sqrt{3}}, \pm 1 \}$. -Chapter 11 of the A Concrete Approach to Abstract Algebra has a simple proof of this fact. -Also, a similar question has proofs that can be extended to this case.<|endoftext|> -TITLE: Isolated critical points -QUESTION [5 upvotes]: Is the following statement true or false? -Let $f:U\subset{\bf R}^n\to{\bf R}$ be a $C^2$-function (or $C^k$, with $k>2$; or real analytic) defined in a neighborhood $U$ of $0$. Assume that $0$ is the unique critical -point of $f$ in $U$, and that it is totally degenerate (i.e., $\mathrm d^2f(0)=0$). -Let $V\subset{\bf R}^n$ be a subspace, denote by $\pi_V:{\bf R}^n\to V$ the orthogonal projection. -Then, there exists a positive definite quadratic -form $Q$ on $V$ such that the function $\widetilde f(x)=f(x)+Q\big(\pi_V(x)\big)$ has $0$ as an isolated critical point. - -REPLY [6 votes]: In the real analytic case the problem reduces to an algebraic one. Denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA$ the ring of germs at $\newcommand{\bR}{\mathbb{R}}$ $0\in\bR^n$ of real analytic functions defined in some neighborhood of the origin. To such a function $f$ we associate its Jacobian ideal $J_f\subset\eA$ generated by the germs at $0$ $\newcommand{\pa}{\partial}$ of the first order partial derivatives of $f$, $\pa_{x_1} f,\dotsc, \pa_{x_n} f$. Then -$$ \mbox{$\dim \eA/J_f <\infty $}\;\;\Rightarrow 0 \;\;\mbox{is an isolated critical point of $f$} . $$ -The dimension of the $\bR$-algebra $\eA/J_f$ is called the Milnor number of $f$ at the critical point $0$ and it is denoted by $\mu(f,0)$. (A similar result holds if $\eA$ is replaced with the ring of germs at $0$ of smooth functions.) The natural condition to impose is the finite multiplicity of the critical point, $\mu(f,0)<\infty$. -Thus, in the real analytic case, one can ask the closely related question: given that $\mu(f,0)<\infty$, is it true that for any homogeneous quadratic polynomial $q:\bR^n\to \bR$ we have $\mu(f+q,0)<\infty$. -Here is a general perturbation result: if $p:\bR^n\to\bR$ is a polynomial function then there exists $\newcommand{\ve}{\varepsilon}$ $\ve_0>0$ such that, for any $|s|<\ve_0$ we have -$$\mu(f+s p,0)<\infty . $$ -Much more refined results are available. I refer you two two sources I found useful. - -T. de Jong, G. Pfister: Local Analytic Geometry, Vieweg, 2000 -Arnold, Gusein-Zade, Varchenko: Singularities of Differentiable Maps. Vol.1 - -Theorem 6.4.5. in deJong-Pfister is particularly relevant.<|endoftext|> -TITLE: Shortest Paths in the "Cantor Graph" -QUESTION [10 upvotes]: First, let me explain, what I understand by a "Cantor Graph": -it is an infinite, directed graph with self loops and countably many vertices labelled with the natural numbers; every ordered pair of vertices is connected by an arc, whose weight equals the quotient of the labels of its tail and head vertex (the graph is inspired by Cantor's famous proof of the countability of the rationals). -Now, my questions are, given two natural numbers, $m$ and $n$, -1.) how long is the shortest path from $m$ to $n$? and, -2.) how many arcs are on the shortest path from $m$ to $n$? -3.) can the above questions be answered without partially constructing the "Cantor Graph"? -Going, for example, directly from $10$ to $2$ would result in the length of the arc from $10$ to $2$, whose length is 5; going via $5$ would result in a path length of 10/5 + 5/2 = 9/2 and is thus shorter than going via the direct connection. - -REPLY [9 votes]: Let me propose an answer to a slightly different question which in some sense is an approximation of the original. -Modified question: Let the vertices of the Cantor graph be the real numbers greater than $1$. And let the weight of the edge $x \rightarrow y$ be $\frac{x}{y}$. What is the shortest path from any $x$ to $1$? -The connection to your original question can be seen by dividing every number with $n$ (if we would like to get from $m$ to $n$). -Answer to the modified question: Let us examine the case where we are using exactly $k$ arcs. We are visiting exactly $k-1$ vertices from $x$ to $1$. Let $\alpha_i$ denote the ratio of the $i$-th and $(i+1)$-th vertex. Thus we are visiting the following vertices: $$ x\, ,\, x \alpha_1 \, , \, x \alpha_1 \alpha_2 \, , \, \dots \, , \, x \prod_{j=1}^{i}\alpha_j \, , \, \dots ,1 $$ -The value of this path is exactly: $$x\prod_{i=1}^{k-1} \alpha_i + \sum_{i=1}^{k-1}\frac{1}{\alpha_i} \geq k x^{\frac{1}{k}}$$ -Where the inequality is just the AM-GM. Where equality holds if and only if every summand is the same, in other words where every $\alpha_i=x^{-\frac{1}{k}}$. From this it follows that the length of the shortest path is $\min_{k\in \mathbb{Z}^+ }( kx^{\frac{1}{k}} )$. Easy calculations shows that $k$ arcs are optimal if $$\left(1+\frac{1}{k}\right)^{k(k+1)} \leq x \leq \left(1+\frac{1}{k+1}\right)^{(k+1)(k+2)} $$ which bounds are in some sense near $e^{k+1} \leq x \leq e^{k+2}$. -From this it follows that the length of the shortest path is $\Omega(\log{x})$.<|endoftext|> -TITLE: Paley graphs over $p^{2}$ vertices -QUESTION [5 upvotes]: I have proved that every Paley graph $P(p^{2})$ over $p^{2}$ vertices, where $p\geq 5$ is a prime number has a cospectral mate, i.e. for every prime number $p\geq 5$ there exists a graph $\Gamma_{p}$ such that $P(p^{2})$ and $\Gamma_{p}$ are cospectral but non-isomorphic. Is it well-known? If so, Could one please giving me the references? - -REPLY [4 votes]: Choose a projective plane of order $p$, where $p$ is odd. Choose a line and view it as a line at infinity. Choose a partition of the points on the line into two classes $C_0$ and $C_1$ of size $(p+1)/2$. Now construct a graph on the affine points, the $p^2$ points not on the line, where two affine points are adjacenct if the line through them meets the line at infinity at a point in $C_0$. The resulting graph is a conference graph and so is cospectral with the Paley graph on $p^2$ vertices. -There is a problem: we have to decide if the graph is isomorphic to a Paley graph. -For moderate values this can be decided by computer. The computational evidence is -that we get large families of non-isomorphic graphs using the above construction. -But there is a construction of conference graphs on $q^2$ vertices due to Peisert ("All Self-Complementary Symmetric Graphs"). Here $q$ is a power of a prime $p$, -and Peisert focusses on the graphs when $p\equiv3$ mod 4, because the graphs in -this case are arc-transitive and self-complementary. But his construction works when $p\cong1$ mod 4, as shown in Natalie Mullin's M.Math thesis: -https://uwspace.uwaterloo.ca/bitstream/handle/10012/4264/nm_thesis.pdf?sequence=1 -So we do know that if $p>3$ then a Paley graph on $p^2$ vertices is not determined by its spectrum. (There may well be earlier proofs of this.) -It is possible that the graphs you have constructed are new. -I believe that if $p\equiv1$ mod 4 and $p\ge29$, the Paley graph on $p$ vertices is not determined by its spectrum. But this is only proved when $p=29$ (by computer search); for larger $p$ I am not aware of any construction. This is a very interesting question.<|endoftext|> -TITLE: Well founded induction attributed to Noether -QUESTION [8 upvotes]: What I know as well founded induction, namely the rule -$$ \big(\forall y.(\forall z.z\lt y\Rightarrow\phi z)\Rightarrow\phi y\big)\Longrightarrow\big(\forall x.\phi x\big), $$ -whose validity is the definition of well-foundedness of the relation $\lt$, -I have recently heard called Noetherian induction. In fact Wikipedia includes this attribution. -With Excluded Middle and Dependent Choice, $\lt$ is well founded iff it has no infinite descending sequence -$$ x_0 \gt x_1 \gt x_2 \gt x_3 \gt \cdots $$ -This suggests at least an analogy with the ascending chain condition that -any sequence -$$ I_0 \subset I_1 \subset I_2 \subset \cdots $$ -of ideals in a Noetherian ring is eventually constant. -Does the attribution of this idea to Emmy Noether amount to more than this analogy? Did she identify the principle at the top as one of logic? Or is this simply a case of the common phenomenon that names of concepts in Mathematics get transferred to more general or simply analogous situations without very much sound historical basis? -Thanks in particular to Joel Hamkins and Antoine Chambert-Loir (ACL) for their comments. Whilst it would be nice to fill in the small gap between Noether and Bourbaki, we have a plausible explanation of the connection between Emmy Noether the algebraist and this logical idea. -Generally speaking, I am a great admirer of Emmy Noether as a conceptual mathematician and rarely have much to say in favour of set theory. However, on this occasion, it would appear that the latter has a much better historical claim to this idea than algebraists do. -The work of Dimitry Mirimanoff that Joel mentions is in the paper Les antinomies de Russell et de Burali-Forti: et le problème fondamental de la théorie des ensembles, which appeared in Enseignement mathématique in 1917 but seems not to be available online. As Joel says, it defines the rank of the $\in$ relation of a set and uses induction in the form of infinite descent. -However, infinite descent in algebra goes back to Euclid: Book 7, Proposition 31 proves that any composite number is measured by some prime number, saying in the proof that "if the prime number is not found, an infinite series of numbers will measure the number $A$, each of which is less than the other: which is impossible in numbers". -The well founded relation $x\lt y$ in Euclid is that $x$ is a proper divisor of $y$. Algebraists from Kummer to Noether generalised divisibility to inclusion of ideals, $(y)\subset(x)$, which accounts for the reversal of the order. Noether generalised Euclid's arguments about prime factorisation from numbers to ideals. -Another thing that bugs me, both as a conceptual and constructive mathematician, is that the principle of well founded induction that I stated at the top is not the same thing as infinite descent. Quite apart from priority, it seems to me that one should take account of how ideas are formulated when naming or attributing them. I don't know who first wrote down the principle above, but Gentzen seems plausible. -In fact, the place where I recently heard well founded induction attributed to Noether was neither in the context of algebra nor logic: it was at a workshop on induction in automatic theorem provers. -I am not sure which rule Joel is labelling "well founded induction" in his further comments below. The rule at the top that I gave that name is an appropriate style to use as a proof rule in a constructive setting, whereas using infinite descent as a formulation of induction depends on an ambient logic with Excluded Middle and Dependent Choice. -One can also generalise the principle at the top by restricting the class of $\phi$s to which it is applicable. For example if the domain has an order with directed joins and $\phi$ is required to be closed under them then we have Scott induction. This proves properties of fixed points, in particular in the semantics of recursion in programming languages. I imagine that this idiom might be adapted to finitary constructive algebra too. - -REPLY [10 votes]: It seems to me to be much more than merely an analogy, because (assuming DC) the ascending chain condition is exactly equivalent to asserting that the collection of ideals is well-founded under (reverse) inclusion. Thus, one can make arguments by induction on ideals, where each instance reduces to the instances in larger ideals, if any, and it seems that Noether did this quite well. For example, under the ACC, every nonempty set of ideals must contain a maximal element, and this is simply another way of stating the induction principle: to prove $\phi(I)$ for every ideal, simply prove that if $\phi(J)$ for all $J\supsetneq I$, then $\phi(I)$, since this will rule out a maximal element of the set $\{I\mid \neg\phi(I)\}$, which must therefore be empty. When the ACC holds, therefore, one may assign an ordinal rank to every ideal, in the manner that any well-founded relation supports such ranks, namely, the maximal ideals get rank 0, and penultimate ideals get rank 1 and so on, with the rank of an ideal equal to the supremum of the ranks+1 of the ideals properly containing it. -So my perspective is that the ACC is a quite robust and important instance of well-foundedness, rather than merely analogous to it. -Concerning the history of the terminology, I noticed that the Wikipedia entry on Noetherian induction redirects to the page on well-founded relations, and Wikipedia cites "Bourbaki, N. (1972) Elements of mathematics. Commutative algebra, Addison-Wesley" specifically in connection with this terminology. So perhaps Bourbaki is the origin of the terminology? (See ACL comments below.) Transfinite recursion itself certainly pre-dates Noether, tracing back to Cantor's use of it in the Cantor-Bendixson theorem, which is also the theorem that led Cantor to the ordinals. Meanwhile, the Wikipedia entry on the axiom of foundation asserts that "the concept of well-foundedness and rank of a set were both introduced by Dmitry Mirimanoff (1917)." -Lastly, let me add that one doesn't generally much see this Noetherian terminology used for well-foundedness or well-founded induction in the parts of logic or set theory with which I am familiar, where the use of well-foundedness is pervasive and often a central concern. But I suppose it wouldn't be surprising to find this terminology more commonly used in algebra, because of Noether's successful use of it there.<|endoftext|> -TITLE: Geodesics on the twisted pseudosphere (Dini's surface) -QUESTION [14 upvotes]: I wonder how difficult it is to compute geodesics on Dini's Surface, -a twisted pseudosphere? -Here is one parametrization, from -Alfred Gray's Modern Differential Geometry of Curves and Surfaces, p.495: -\begin{eqnarray} -x(u,v) &=& -a \cos (u) \sin (v)\\ -y(u,v) &=& -a \sin (u) \sin (v)\\ -z(u,v) &=& -a \left[\cos (v)+\log \left(\tan - \frac{v}{2}\right)\right]+b u -\end{eqnarray} -Dini's surface has constant curvature of $\frac{-1}{a^2+b^2}$. -And here is an image, for $a=1,\; b=\frac{1}{12}$, with $u \in [0,8\pi]$ -(The curve defined by $v=\pi/2$ is shown green): -        - -What I especially wonder is if there is a geodesic that spirals down through every turn, -which would be kinda cool. :-) - -The green curve below might be one of Robert Bryant's geodesics—the computation is -complicated enough that I am quite uncertain. Have to leave it there for the nonce... - -REPLY [5 votes]: Since Joseph asked for pictures, I will add to Robert's excellent response with the following images. -For the purposes of implementing Robert's change of variables in Mathematica, I simplified the composition as shown below: -(*twist parameter*) -t = Pi/6; - -(*auxiliary function*) - -g[x_, y_] := Sqrt[x^2 - y^2 + (x^2 + y^2)*Cos[2*t]]; - -(*change of variables for u*) - -u[x_, y_] := - Csc[t]*(Log[x^2 + y^2] - - Log[Sqrt[2]*x + - g[x, y]] + (Cos[t]* - Log[(-(x*Sqrt[1 + Cos[2*t]]) - - g[x, y])/(-(x*Sqrt[1 + Cos[2*t]]) + g[x, y])])/2); - -(*parameterization*) - -r[x_, y_] := {Sin[t]*Cos[u[x, y]]*y/x, Sin[t]*Sin[u[x, y]]*y/x, - Cos[t]*(Sqrt[1 - Tan[t]^2*y^2/x^2] + - Log[(Tan[t]*y/x)/(1 + Sqrt[1 - Tan[t]^2*y^2/x^2])]) + - u[x, y]*Sin[t]}; - -The domain of r[x,y] is the region where $y\geq0$, $x\geq y\tan t$. This parameterization is an isometry from its domain (endowed with the Poincare half-space model metric) to its image. -First, here is an image of the parameterized surface using Mathematica's standard mesh function -r1 = 1; r2 = 10; ParametricPlot3D[r[x, y], {x, 0, r2}, {y, 0, r2}, - RegionFunction -> - Function[{a, b, c, x, y}, - x >= Tan[t]*y && r1^2 <= x^2 + y^2 <= r2^2], PlotPoints -> 100] - - -Next, let's visualize some geodesics. Below we show some geodesics in the half-plane model and their images under the parameterization. -(*parameter for paths*) -rad = 1; - -(*Geodesics in half-plane model*) -path1[x_] := {x, Cos[t]*rad}; -path2[th_] := {rad*Cos[th], rad*Sin[th]}; -path3[th_] := rad*Sin[t]/Sin[2 t]*{Cos[th], Sin[th] + 1}; - -(*Show geodesics in half-plane model with domain*) -Show[ - RegionPlot[domain, {x, 0, 10}, {y, 0, 10}, ImageSize -> Medium, - PlotStyle -> GrayLevel[0.9], BoundaryStyle -> None], - ParametricPlot[path1[x], {x, Tan[t]*rad, 10}, PlotStyle -> Red], - ParametricPlot[path2[th], {th, 0, Pi/2 - t}, PlotStyle -> Green], - ParametricPlot[path3[th], {th, -Pi/2, Pi/2 - 2 t}, - PlotStyle -> Blue] - ] - - -Now, those same three geodesics mapped onto Dini's Surface. -Show[ - ParametricPlot3D[r[x, y], {x, 0, 10}, {y, 0, 10}, - RegionFunction -> Function[{a, b, c, x, y}, domain], - PlotPoints -> 100, Mesh -> False, PlotStyle -> Opacity[0.5], - PlotRange -> {{-1, 1}, {-1, 1}, {-4, 4}}], - ParametricPlot3D[r @@ path1[x], {x, Tan[t]*rad, 10}, - PlotStyle -> Red], - ParametricPlot3D[r @@ path2[th], {th, 0, Pi/2 - t}, - PlotStyle -> Green], - ParametricPlot3D[r @@ path3[th], {th, -Pi/2, Pi/2 - 2 t}, - PlotStyle -> Blue] - ] - - -The blue geodesic seems to be what Joseph was imagining in the first place, following the surface to $z=-\infty$ and getting ever closer to the $z$-axis. The green geodesic leaves the rim at a right angle and follows the shortest path to the $z$-axis. Finally, most surprising to me anyway, is the red geodesic, which winds up the surface to $z=\infty$.<|endoftext|> -TITLE: Weyl algebra and its nontriviality -QUESTION [9 upvotes]: The Weyl algebra (say, over $\mathbb{C}$) is an universal unital algebra with two generators $x,y$ subject to the relation $xy-yx=1$. This algebra can be constructed in the following way: take two dimensional vector space with basis $\{x,y\}$ and construct the tensor algebra $T(V)$. Then take an ideal $I=(xy-yx-1)$ generated by $xy-yx-1$ and form a quotient $T(V)/I$. -It is possible to find a norm on $T(V)$ in which $T(V)$ becomes a normed algebra, therefore we can take the closure $\overline{I}$ of $I$ and form a quotient $T(V)/\overline{I}$. This is again a normed algebra but generators of this algebra satisfy the relation $xy-yx=1$ but this is impossible in any normed unital algebra. This apparent contradiction actually shows that $I$ must be dense in $A$. This leads me to the question: it is really obvious and immediate that $I$ is not the whole $T(V)$? I'm pretty sure that this must be true but it seems to me that it would be hard to give a "two line" proof of this fact. Forgive me if this question is to elementary: but in the literature no one seems to see any problem with the existence (and nontriviality) of Weyl algebra. - -REPLY [6 votes]: Generally the strategy for showing that some syntactic construction is nontrivial is to find a semantic model of it. E.g. the strategy for showing that some relations don't force a group to be a trivial group is to find, say, some matrices satisfying the relations, and the strategy for showing that some axioms in some theory don't contradict each other is to find a model satisfying all the axioms. -In this case the strategy for showing that some relations don't force a ring to be nontrivial is to find a module. An alternative definition of the Weyl algebra, at least in characteristic zero, is that it is algebra of differential operators on $k[x]$, where one generator acts as multiplication by $x$ and the other acts as $\frac{\partial}{\partial x}$. In characteristic zero, this action can be used to prove the stronger statement that the Weyl algebra has a basis given by the monomials $x^i \frac{\partial}{\partial x^j}$. -In positive characteristic $k[x]$ actually fails to be a faithful module of the Weyl algebra because $\frac{\partial}{\partial x^p}$ acts by zero. A module which is faithful regardless of the characteristic can be constructed using the lattice of Young tableaux; see this math.SE answer for details. -The basis above should remind you of a PBW basis, and in fact the Weyl algebra is almost the universal enveloping algebra of the Heisenberg Lie algebra spanned by $1, x, \frac{\partial}{\partial x}$. It is precisely the quotient of this universal enveloping algebra by the relation that $1$ acts by the identity, and either the PBW theorem or a slight modification of it can be used to prove the stronger statement above. (A less coordinate-dependent way of saying this is that the Weyl algebra has a natural filtration and you can show that its associated graded is $k[x, y]$.) - -REPLY [6 votes]: A generic argument that can be used to show that an algebra is non-trivial is Bergman's Diamond Lemma. In this case, it immediately applies (there are no ambiguities) so for example $x$ is not in the ideal. -This method is useful even in situations in which you cannot find a representation to implement the idea in the other two answers, or when the characteristic gets in the way.<|endoftext|> -TITLE: Is there "nonorientable Heegaard Floer homology"? -QUESTION [6 upvotes]: I have a Heegaard diagram which produces a non-orientable 3-manifold. I want to know any 3-manifold invariant which can be calculated from Heegaard diagrams for non-orientable 3-manifold. (As far as I know, Roklin invariant or Heegaard Floer homologies are defined on oriented 3-manifolds, right?) - -REPLY [7 votes]: In general, any invariant of oriented manifolds is an invariant of non-orientable ones, since one can pass to the oriented double cover. Note that this cover admits an orientation-reversing diffeomorphism, so in some sense it shouldn't matter which orientation you choose. (This point should be treated with great care, especially if you are interested in maps between various groups defined on the double cover, or gradings that depend on the choice of orientation.) So in this sense the Heegaard Floer homology groups of the oriented double cover $\tilde{M}$ are invariants of your manifold $M$; in principle one could probably find a way of expressing those in terms of a Heegaard diagram for $M$. One thing to keep in mind is that Heegaard diagrams for non-orientable manifolds involve non-orientable handlebodies. -One should consider not just the Heegaard Floer groups of $\tilde{M}$, but also the action of the covering transformation. Since the covering transformation doesn't fix any points, it's not so clear how to define this, as the definition of HF involves a choice of a basepoint; see Juhász-Thurston. Presumably you have to work modulo automorphisms that result from moving the base point around. The point above about how you would define maps is also relevant here. -With regard to Rohlin invariants, you should note that the Rohlin invariant of an oriented manifold depends on a choice of spin structure. Presumably you have to choose something like a pin structure on $M$ and again try to define a Rohlin-type invariant of $\tilde{M}$. Rohlin invariants of oriented spin 3-manifolds live in $Q/16Z$; the resulting invariant would at best live in $Q/8Z$ or something like that.<|endoftext|> -TITLE: Yang-Mills flow, Ricci flow and the holonomy -QUESTION [5 upvotes]: Is the holonomy group (based at some point) preserved along the Yang-Mills flow/ Ricci flow? -(1) For Yang-Mills case, we know that the centralizer of the holonomy $H_x$ is the isotropy group of the connection $\Gamma_A$. Hence, by the uniqueness of the flow, $\Gamma_A$ can not get smaller along the flow. But can it get larger? -(2) Can we derive anything for $H_x$ given some results of $\Gamma_A$? -(3) What about the Ricci flow case? I am not very sure about this. It seems that special holonomy means (almost) Einstein, except being Kahler. In either case, they are preserved? - -REPLY [2 votes]: If your compact manifold (assume irreducible WLOG, i.e. not a product) has Einstein metric, then Ricci flow just scales, so we're good. But as a counterexample in general, consider $SO(n)$ with an almost-flat torus (i.e. perturb the flat metric). The holonomy is $SO(n)$, but Ricci flow can make it flat in the limit, so that holonomy drops to trivial. -Now for the Kahler case $U(n/2)$, Ricci flow preserves Kahlerness of the metric, so that the holonomy cannot jump above $U(n/2)$. But it could possibly drop as in the above counterexample. -(I remember learning about this from Robert Bryant)<|endoftext|> -TITLE: Asymptotic value of a multivariate integral -QUESTION [6 upvotes]: The following question is a simple case of a type of problem that occurs in combinatorial enumeration problems. -Define -$$F(x_1,\ldots,x_n) = \frac{1}{(2\pi)^{n/2}}\exp\biggl( -\frac12\sum_{j=1}^n x_j^2 + i\,n^{-1/4} \sum_{j=1}^n x_j^3 \biggr), $$ -where $i$ is the imaginary unit. -Let $\epsilon\gt 0$ be small enough. Let -$$I_0(n) = \int_{-n^\epsilon}^{n^\epsilon}\cdots \int_{-n^\epsilon}^{n^\epsilon} - F(x_1,\ldots,x_n) ~ dx_1\cdots dx_n. $$ -We can estimate $I_0(n)$ as $n\to\infty$ by factoring -$$F(x_1,\ldots,x_n) = \prod_{j=1}^n \frac{1}{\sqrt{2\pi}} - \exp\bigl(-x_j^2/2+in^{-1/4}x_j^3\bigr),$$ -which separates the integral into a product of $n$ 1-dimensional integrals. This very easily gives -$$I_0(n) = \exp\bigl( -n^{1/2}/2 + O(n^{-3})\bigr).$$ -So far so good. But now let $A=A(n)$ be a symmetric positive definite matrix. We can assume it is pretty nice, say all eigenvalues bounded between two positive constants independent of $n$. Now define -$$G(x_1,\ldots,x_n) = \frac{1}{(2\pi)^{n/2}}\exp\biggl( -\frac12 \mathbf{x}^TA\mathbf{x} + i\,n^{-1/4} \sum_{j=1}^n x_j^3 \biggr). $$ -How do we estimate -$$I_1(n) = \int_{-n^\epsilon}^{n^\epsilon}\cdots \int_{-n^\epsilon}^{n^\epsilon} - G(x_1,\ldots,x_n) ~ dx_1\cdots dx_n? $$ -The source of the difficulty is that $\int G$ is exponentially smaller than $\int |G|$, so as soon as you approximate the integrand the answer goes away. What to do? -Note that in this case (and very commonly in practice) symmetry implies that $I_1(n)$ is real. So we can discard the imaginary part of the integrand and integrate only the real part, which is -$$ \mathfrak{R}G(x_1,\ldots,x_n) = \frac{1}{(2\pi)^{n/2}}\exp\biggl( -\frac12 \mathbf{x}^TA\mathbf{x}\biggr)\cos\biggl(n^{-1/4} \sum_{j=1}^n x_j^3 \biggr). $$ -Not sure that helps. -ADDED 2 Jan 2018. Mikhail Isaev and I now have a method that can precisely estimate such integrals in many general cases. It isn't completely written up yet. - -REPLY [2 votes]: I'll consider the integrals extended from $-\infty$ to $\infty$ and show that the integral of $G$ is exponentially small compared to the -integral of $|G|$ -- precisely, the oscillating integral is smaller by a factor of $C_0\exp(-C \sqrt{n})$ for some positive constants $C_0$ and $C$ -that depend only on the eigenvalues of the quadratic form $A$. One can get the same estimate for the truncated integrals. It may be possible to refine this to get asymptotics but -I have not thought about that. -The problem asks for an estimate for -$$ -I = \frac{1}{(2\pi)^{n/2}} \int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \exp\Big( -\frac{1}{2} A(x_1,\ldots,x_n) +\frac{i}{n^{\frac 14}} \sum_j x_j^3\Big) dx_1\ldots dx_n. -$$ -where $A$ is a positive definite quadratic form. We wish to show that $I$ is small compared to the same integral without the oscillating term; this is -$$ -\frac{1}{(2\pi)^{n/2} } \int_{-\infty}^{\infty}\cdots \int_{-\infty}^{\infty} \exp\Big(- \frac 12 A(x_1,\ldots,x_n )\Big) dx_1 \ldots dx_n = (\text{det} A)^{-\frac 12}. -$$ -Think of the integrals in $I$ as contour integrals being integrated on the real axis. The idea is to replace the integrals on the real axis by -integrals along the line from $-\infty + i\alpha$ to $\infty +i\alpha$ for some real number $\alpha$ to be chosen carefully. Thus -$$ -I =\frac{1}{(2\pi)^{n/2}} \int_{-\infty+i\alpha}^{\infty+i\alpha} \cdots \int_{-\infty+i\alpha}^{\infty+i\alpha}\exp\Big(-\frac12 A(z_1,\ldots,z_n) + \frac{i}{n^{\frac 14}} -\sum z_j^3 \Big) dz_1 \ldots dz_n, -$$ -and writing now $z_j =x_j +i\alpha$ this equals -$$ -\frac{1}{(2\pi)^{n/2}} \int_{-\infty}^{\infty}\cdots \int_{-\infty}^{\infty} \exp\Big( -\frac 12 A(x_1+i\alpha,\ldots, x_n+i\alpha) + \frac{i}{n^{\frac 14}} -\sum_j (x_j+i\alpha)^3\Big) dx_1\ldots dx_n. -$$ -We now estimate the integral above by just taking the absolute value of the integrand (and choosing $\alpha$ carefully). The integrand is in modulus -$$ -\exp \Big( -\frac 12 A(x_1,\ldots, x_n) + \frac{\alpha^2 }{2} A(1,\ldots, 1) +\alpha^3 n^{\frac 34} -\frac{3\alpha}{n^{\frac 14} } \sum_j x_j^2\Big). -$$ -We will take $\alpha =\beta/n^{\frac 14}$ for a suitably small positive constant $\beta$. Since the eigenvalues of $A$ are bounded between two positive constants independent of $n$, we have that $A(1,\ldots,1) \le 2C_1 n$ for some positive constant $C_1$, and that $3\sum_j x_j^2 \ge \frac{C_2}{2} A(x_1,\ldots, x_n)$ for some positive constant $C_2$. -Thus the quantity above is -$$ -\le \exp\Big(C_1 \beta^2 \sqrt{n} + \beta^3 - \frac{1}{2} A(x_1,\ldots,x_n) \Big(1+\frac{C_2\beta}{\sqrt{n}}\Big) \Big). -$$ -Integrating this, we find that -$$ -I \le \frac{\exp(C_1 \beta^2\sqrt{n} +\beta^3)}{(2\pi)^{n/2}} \int_{\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \exp\Big(-\frac 12 A(x_1,\ldots,x_n) \Big(1+\frac{\beta C_2}{\sqrt{n}}\Big) \Big) dx_1 \ldots dx_n -$$ -which is readily seen to be -$$ - \exp(C_1 \beta^2\sqrt{n} +\beta^3) (\text{det} A)^{-\frac 12} \Big( 1 +\frac{\beta C_2}{\sqrt{n}}\Big)^{-n/2}. - $$ -In other words we have shown that $I$ is smaller than the trivial bound by a factor of - $$ - \exp(C_1 \beta^2\sqrt{n} +\beta^3) \Big( 1 +\frac{\beta C_2}{\sqrt{n}}\Big)^{-n/2} \le - \exp(C_1 \beta^2 \sqrt{n} +\beta^3 - C_3 \beta\sqrt{n} \Big) - $$ - for a suitable positive constant $C_3$. By choosing $\beta$ small enough, we find that this is $\le C_0 \exp(-C\sqrt{n})$ - for some positive constants $C_0$ and $C$. This finishes the proof.<|endoftext|> -TITLE: Distribution of polynomials mod 1 using co-prime integers -QUESTION [9 upvotes]: If $a$ is any real irrational, then the set of numbers of the form $ax+y$ with $x$ and $y$ co-prime integers is dense in $\mathbb{R}$. I managed to prove this, in what I suspect is an overly complicated way, in response to -this question. -I think there must be a more perspicuous proof of this fundamental fact. Furthermore, I expect that the following more general assertion is true: -If $f\in \mathbb{R}[x]$ is any polynomial at least one of whose coefficients other than the constant term is irrational, then the numbers of the form $f(x)+y$ with $x$ and $y$ co-prime integers are dense in $\mathbb{R}$. -Now it seems to me that this ought to be a very well known fact or open problem (or maybe it is false!) but I can't find a single reference to it. Can anyone provide a proof or a reference? - -REPLY [3 votes]: The continued fraction for $a$ gives rational approximations $$\frac{p_i}{q_i} \approx a$$ with $|q_ia-p_i| \lt \frac1{q_i}$. Furthermore, successive approximants satisfy $p_{i+1}q_q-p_iq_{i+1}=\pm1.$ -Fix an index $i$ and consider the sequence of expressions $$r_k=x_ka+y_k=(q_{i-1}+kq_{i})a-(p_{i-1}+kp_{i})$$ where $k \in \mathbb{Z}$. The values $r_k$ form a two way infinite arithmetic progression with common difference $q_ia-p_i \lt \frac{1}{q_{i}}$ and thus allow us to approximate any desired real to an accuracy of better than $\frac{1}{2q_i}$. It remains only to note that - -The denominators $q_i$ increase (exponentially) and thus allow us to get any accuracy we desire. -$x_{k+1}y_k-x_{k}y_{k+1}=p_{i+1}q_i-p_iq_{i+1}=\pm1$ so the pairs $(x_k,y_k)$ are relatively prime.<|endoftext|> -TITLE: On the vanishing of the generalized von Mangoldt function $\Lambda_k(n)$ when $n$ has more than $k$ prime factors -QUESTION [14 upvotes]: It is a well-known fact that the generalized von Mangoldt function, defined by -$$\displaystyle \Lambda_k(n) = \sum_{d | n} \mu(d) \left(\log \frac{n}{d}\right)^k$$ -vanishes whenever $n$ has more than $k$ distinct prime factors. I was able to prove this fact through a relatively lengthy and cumbersome combinatorial argument by proving first for the squarefree case, say when $n = p_1 \cdots p_s$ with $s > k$, and showed that the vanishing of $\Lambda_k(n)$ can be deduced from the following polynomial identity: -$$\displaystyle (x_1 + \cdots + x_s)^k - \sum_{\substack{S \subset \{1, \cdots, s\} \\ |S| = s-1}} \left(\sum_{i \in S} x_i\right)^k + \sum_{\substack{S \subset \{1, \cdots, s\} \\ |S| = s-2}} \left(\sum_{i \in S} x_i \right)^k - \cdots $$ -$$ + (-1)^{s-1}\sum_{i=1}^s x_i^k = 0.$$ -The general case is then done through a similar argument (which depends on the above identity) and strong induction. -However, I find the above argument to be somewhat insipid and not very 'number theoretic', as it is deduced from a general polynomial identity rather than using any properties of numbers. Is there any conceptually simpler, more number theoretic proof? It would be an added bonus if the proof is shorter than the above. - -REPLY [2 votes]: It can be seen as a consequence of the multinomial theorem. We would like to prove the following. -The function $\Lambda_k(n)$ is always non-negative and supported only on integers $n$ with $\omega(n)\leq k$, with $\omega(n)$ the distinct prime factors counting function. Moreover, it verifies the following recursive relation -$$ -\Lambda_{k+1}(n)=\Lambda_k(n)\log n+\sum_{d|n}\Lambda_k(d)\Lambda(n/d). -$$ -Proof: -If $\mu(n)$ indicates the Moebius function, we begin with the following series of identities -\begin{align*} -\sum_{\substack{b|n}}\mu(b)(\log b)^k &=\sum_{\substack{b\geq 1\\ b|n}}\frac{\mu(b)(\log b)^k}{b^{\sigma}}\bigg|_{\sigma=0}\\ -&=(-1)^{k}\frac{d^{k}}{d\sigma^{k}}\sum_{\substack{b\geq 1\\ b|n}}\frac{\mu(b)}{b^{\sigma}}\bigg|_{\sigma=0}\\ -&=(-1)^{k}\frac{d^{k}}{d\sigma^{k}}\bigg(\prod_{p|n}\bigg(1-\frac{1}{p^{\sigma}}\bigg)\bigg)\bigg|_{\sigma=0}\\ -&=(-1)^{k}\sum_{j_1+j_2+...+j_{\omega(n)}=k}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}\bigg(1-\frac{1}{p_i^{\sigma}}\bigg)^{(j_i)}\bigg|_{\sigma=0}, -\end{align*} -by the multinomial theorem. By differentiating each binomial above, we can rewrite the previous sum as -\begin{align*} -&=(-1)^{k+\omega(n)}\sum_{\substack{j_1+j_2+...+j_{\omega(n)}=k\\ j_i\neq 0,\ \forall i}}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}(-\log p_i)^{j_i}.\\ -&=(-1)^{\omega(n)}\sum_{\substack{j_1+j_2+...+j_{\omega(n)}=k\\ j_i\neq 0,\ \forall i}}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}(\log p_i)^{j_i}. -\end{align*} -Moreover, we have -$$\Lambda_k(n)=\sum_{b|n}\mu(n/b)(\log b)^k=\mu(n)\sum_{b|n}\mu(b)(\log b)^k=(-1)^{\omega(n)}\sum_{b|n}\mu(b)(\log b)^k.$$ -We thus deduce that -$$\Lambda_k(n)=\sum_{\substack{j_1+j_2+...+j_{\omega(n)}=k\\ j_i\neq 0,\ \forall i}}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}(\log p_i)^{j_i},$$ -from which it is immediate to obtain the first assertion. Regarding the second one, we first notice that the Dirichlet series of $\Lambda_k(n)$ is given by -$$(*)\ -\sum_{n\geq 1}\frac{\Lambda_k(n)}{n^s}=(-1)^k\frac{\zeta^{(k)}(s)}{\zeta(s)}\ \ \ (\Re(s)>1), -$$ -where $\zeta(s)$ is the Riemann zeta function. This follows immediately by looking at $\Lambda_k(n)$ as the Dirichlet convolution of the Moebius function and the logarithm function at $n$ and using some basic identities in the theory of Dirichlet series. -Moreover, we clearly have -$$(-1)^{k}\sum_{n\geq 1}\frac{\Lambda_{k-1}(n)\log n}{n^s}=\frac{d}{ds}\frac{\zeta^{(k-1)}(s)}{\zeta(s)}=\frac{\zeta^{(k)}(s)}{\zeta(s)}-\frac{\zeta^{'}(s)}{\zeta(s)}\frac{\zeta^{(k-1)}(s)}{\zeta(s)}.$$ -Plugging the relation (*), for $k$, $k-1$ and $1$, into the above, we get also the second assertion.<|endoftext|> -TITLE: Prevalent singular cardinals hypothesis -QUESTION [5 upvotes]: The following notion is introduced by Assaf Rinot: -Definition. A singular cardinal $\kappa$ is a prevalent singular -cardinal iff there exists a family $\mathbb{A}\subset P(\kappa)$ with $|\mathbb{A}| = \kappa$ -and $sup\{|A| : A\in \mathbb{A}\} < \kappa$ such that any $B\subset \kappa$ with -$|B| < cf(\kappa)$ is contained in some $A\in \mathbb{A}$. -Prevalent Singular Cardinals Hypothesis (PSCH) states that any singular cardinal is a prevalent singular cardinal. -Clearly any singular cardinal of countable cofinality is prevalent. -Question. What is known about the consistency of the failure of PSCH? -Remark 1. PSCH follows from $GCH, SCH, PFA$ and many other combinatorial principles. -Remark 2. As stated in Assaf Rinot's paper "On topological spaces of singular density and minimal weight, Topology and its Applications 155 (2007) 135–140", $PSCH$ is a very weak assertion and all currently known methods for violating statements -of similar flavor, will fail to violate the $PSCH$. -Question 2. Is there any relation between PSCH and some combinatorial principles introduced by Shelah in PCF theory like $Cov$ or ...? - -REPLY [5 votes]: A singular cardinal $\lambda$ is prevalent iff there exists some cardinal $\mu<\lambda$ such that $Cov(\lambda,\mu,cf(\lambda),2)=\lambda$. In his solution of the pcf conjecture, Gitik has constructed a model where there exists a singular cardinal $\lambda$ of cofinality $\aleph_1$ such that $pp(\mu)>\lambda$ for cofinally many $\mu<\lambda$ of countable cofinality. Given $\mu<\lambda$, pass to a bigger $\mu'<\lambda$ of countable cofinality with $pp(\mu')>\lambda$. Then, we have $pp(\mu')\le Cov(\mu',\mu',cf(\mu')^+,2)\le Cov(\lambda,\mu',cf(\lambda),2)\le Cov(\lambda,\mu,cf(\lambda),2)$. So $Cov(\lambda,\mu,cf(\lambda),2)>\lambda$ for every $\mu<\lambda$, meaning that $\lambda$ is not prevalent.<|endoftext|> -TITLE: Avoiding multiples of $p$ -QUESTION [21 upvotes]: Let $p$ be a prime number and $P=\{1,2,...,p-1\}$ -In how many ways we can sum all the elements of $P$ in such a way that we will reach a multiple of $p$ -only when we sum the last summand? -For example let $p=7$ . -Clearly, $1+2+3+4+5+6$ is such a sum (In fact there exist $408$ such sums) - but $2+3+5+4+1+6$ is not, because already $2+3+5+4=2\cdot7$. -We can see after a little investigation that if the total number of sums is $f(p)$,then -$\frac{(p-1)!}{p-2}\leq f(p)\leq (p-1)!$ (equality holds iff $p=3$) -Is it possible to improve this (to find asymptotic bounds or -even better- something precise)? -Thanks in advance! -EDIT : It is possible to prove in an elementary way that for every $n\in \mathbb{N}$, $\phi(n)\mid f(n)$ -($\phi(n)$ is Euler's function) - -REPLY [3 votes]: As I mentioned in the comments above, the number of permutations of elements $1,2,\dots,p$ (i.e., including $p$) is just by factor of $p-2$ larger than the amount in question (in fact, this is true for any odd $p$). Such permutations are now counted in http://oeis.org/A232663 -Here is an explicit formula for the number of such permutations for a prime $p$, which I got by playing with inclusion-exclusion: -$$\sum_{A\in M_p} (-1)^{m+n}\cdot \frac{1}{n!} \cdot p^{n-\mathrm{rank}(A)} \cdot \prod_{j=1}^n (c_j-1)! \cdot \prod_{i=1}^m \binom{r_i}{a_{i1}, \dots, a_{in}},$$ -where: - -$M_p$ is the set of matrices with nonnegative integer entries that sum up to $p$, with no zero rows or columns. Size of $M_p$ is given by http://oeis.org/A120733 -$A=(a_{ij})$ is a matrix of size $m\times n$ (i.e., $m$ and $n$ are respectively the number of rows and columns in $A$); -$c_j = \sum_{i=1}^m a_{ij}$ is the sum of $j$-th column of matrix $A$; -$r_i = \sum_{j=1}^n a_{ij}$ is the sum of $i$-th row of matrix $A$; -matrix rank is computed over $\mathrm{GF}(p)$. - -The formula may not be so useful due to its complexity but it's still a nice one. But the same reason I do not post here its derivation as it would take at least a couple of pages. -The formula has been numerically tested for $p=5$ and $p=7$.<|endoftext|> -TITLE: On non-split extensions of $\mathrm{SL}_d(q)$ -QUESTION [7 upvotes]: Throughout this question, the following notation holds: Let $q$ be a power of a prime $p$, and let $d>4$ be a positive integer. Let $G$ be a finite group with a normal subgroup $E$ which is an elementary-abelian $p$-group, and such that $G/E\cong\mathrm{SL}_d(q)$. -I am interested in the situation where this extension is non-split, i.e. when there is no subgroup $H\frac34n$, then the extension is split? - -Note: I chose the number $\frac34$ out of thin air. I'd be happy to hear of a `yes' answer, with that $\frac34$ replaced by your favourite positive real number. - -REPLY [2 votes]: Here are some further comments, to supplement Derek's focused answer to Q2 as well as my own somewhat cryptic comments. -1) Q2 is fairly narrow, so the kind of ad hoc treatment Derek suggests is quite appropriate. Note here that you do need to have some information about the $p$-modular representations, for which the dimension bound is crucial. In general, even for finite special linear groups not much is known about the details of their modular representations. -2) I share Derek's view that Q1 is too general to admit a sensible answer. However, there has been a lot of organized study of extensions and cohomology for finite groups of Lie type, including subtle connections with the corresponding ideas for ambient algebraic groups. Here the 1977 Invent. Math. paper on rational and generic cohomology for which I gave a link is basic. (This paper combines independent work by CPS and by van der Kallen, whom I had the privilege to introduce to each other soon afterward at an Oberwolfach meeting.) For a much more recent treatment of issues related to $H^2$ along with a useful reference list, see the paper (later published in J. Algebra) here. The earliest work tends to rely just on finite group techniques, but by now it's natural to think also about algebraic groups and to deal with general groups of Lie type. There are of course some special features of your groups which might make things easier, but it's a good idea to have the general picture in mind. (For the algebraic groups, the possible non-existence of Levi factors will be one related theme.)<|endoftext|> -TITLE: Smoothing of the distance function on a Riemannian manifold -QUESTION [11 upvotes]: Suppose $(M,g)$ is a complete Riemannian manifold. $p\in M$ is a fixed point. $d_{p}(X)$ is the distance function defined by $p$ on M (i.e., $d_p(x)$=the distance between $p$ and $x$). Let $\epsilon>0$ be an arbitrary positive number. Is there a smooth function $\tilde{d}_p(x)$ on $M$, such that -$$ | d_p(x)-\tilde{d}_p(x) | < \epsilon$$ -$$ |\textrm{grad}(\tilde{d}_p)(x)|<2$$ -for $\forall x \in M$ ? -Functions satisfying the first condition can be constructed easily by partition of unity and the standard technique of mollifiers. However, I can't see how to control the gradient of the approximate function. -I need this result when I'm trying to follow a proof. The existence of such a function is taken for granted in that proof, and is called the "regularization" of the distance function. This question also seems to be interesting by itself. -I'm not sure if this question is too easy for math overflow. I put it on stack exchange last week but nobody answered it. Could you please help me? Thanks a lot. - -REPLY [5 votes]: This is an additional comment about higher derivative bounds. Consider the -class of pointed complete noncompact Riemannian manifolds $(M^{n},g,O)$ with -$\left\vert \operatorname{sect}\left( g\right) \right\vert \leq K$. For the -distance-like function constructed from mollification by Greene and Wu, one -also has a uniform upper bound for the Hessian of $u$. To obtain a two-sided -bound for the Hessian, one can smooth $u$ by the heat equation. In a paper -titled "Construction of an exhaustion function on complete manifolds", -Luen-Fai Tam proves that there exists a $C^{\infty}$ function $f:M\rightarrow -\mathbb{R}$ with $d\left( \cdot,O\right) +1\leq f\leq d\left( -\cdot,O\right) +C$, $\left\vert \nabla f\right\vert \leq C$, and $\left\vert -\nabla\nabla f\right\vert \leq C$, where $C$ depends only on $n$ and $K$. -Techniques that Tam uses include heat kernel estimates and weighted $L^{2}$ -estimates. An exposition of Tam's result is given in Section 4 of Chapter 26 -in the book "The Ricci Flow: Techniques and Applications: Part III: -Geometric-Analytic Aspects"; see therein for references to previous related work.<|endoftext|> -TITLE: Is there a combinatorial version of PL ambient isotopy in dimension $>3$? -QUESTION [7 upvotes]: The classical PL Reidemeister Theorem reads: - -Reidemeister Theorem: Two knots in $S^3$ are PL ambient isotopic if and only if any diagram of one can be transformed into a diagram of the other by Reidemeister moves. - -Recall that PL ambient isotopy of $K_1\colon\, S^1\hookrightarrow S^3$ to $K_2\colon\, S^1\hookrightarrow S^3$ is an orientation-preserving PL map from $S^{3}\times I$ to $S^3$, whose restriction to a map from $S^{3}\times \{t\}$ to $S^3$ is a PL homeomorphism for every $t\in I$, whose restriction to $S^{3}\times \{0\}$ is the identity map, and whose restriction to $S^{3}\times \{1\}$ composed with $K_1$ is $K_2$. -However, every source that I know (see here) proves an a-priori weaker version of the theorem, in which "PL ambient isotopy" is replaced by "combinatorial isotopy" which means: "related by a finite sequence of triangle moves"- replacing 2 (or 1) edges of a triangle which does not intersect the knot in its interior with the other edge (edges). See e.g. these notes by J. Roberts. Indeed, this is the version of the Reidemeister Theorem which Reidemeister himself proved. -I think that the seldom-cited but often-used fact that PL ambient isotopy coincides with combinatorial isotopy of knots in $S^3$ is proved (in the more general case of links, which extends also to knotted graphs as pointed out by Kauffman) in: -Graeub, W. (1950). Die semilinearen abbildungen (pp. 3-70). Springer Berlin Heidelberg. -Question 1: Does there exist a proof of this fact in English? I have trouble with the German. - -Question 2: Is there a combinatorial version of PL ambient isotopy in higher dimensions? Specifically, I am interested in the case of a mixed 1- and 2-dimensional polyhedron embedded in a PL 4-ball, so that the only non-manifold points are at endpoints of broken lines, where they meet other broken lines or broken planes. The obvious candidate is triangle moves, analogous "tetrahedron moves", and a "mixed dimension move" to move a neighbourhood of the intersection of a line with a plane, which is a bit harder to write down. Is the proof of this to be found anywhere, and is the general version known and shown? (I'm fairly confident it's not in Graeub's book, but perhaps it follows trivially somehow). - -REPLY [3 votes]: Unfortunately, this is not quite an answer. So for knotted surfaces in 4-space, there is Roseman's Theorem. I think that the context of Dennis's proof is in the smooth category. Or certainly the proof that I know depends on a smooth structure. -I would imagine that the proof could be cranked up to PL-locally flat, but I have my hesitations about non-simple branch points. Roseman outlines the higher dimensional case as well, and this outline follows the classification of multi-germs, or actually real pictures of complex multi-germs. If you give me N, and a large enough amount of time, I think that I can work out all the moves by hand, but again in a smooth category. -So fundamentally the idea is to take a codimension 2 embedding, and project it into (n+1)-space. Do so generically. Now examine the critical points and intersections of an isotopy. You can sort of build the whole theory up inductively. Your R-moves and critical events in dimension N become the singularities in dimension N+1. My paper on RR moves for foams on the arxiv can be used to fill in the details in the smooth case.<|endoftext|> -TITLE: Representing de Rham cohomology by smooth maps -QUESTION [13 upvotes]: It is well-known that in the homotopy category of, say, CW-complexes the singular cohomology functor is represented by the Eilenberg-Maclane spaces: $H^n(M,\mathbb{Z})=[M,K(\mathbb{Z},n)]$. My question is, if $M$ is a smooth Riemannian manifold, can one choose the representing map to be smooth (in any sense)? -For example, every de Rham cohomology class contains a unique harmonic form and for $n=1$ we have the "Albanese" map from $M$ into a circle, given by the integration of this form. -Can we do something similar for $n>1$? - -REPLY [2 votes]: I think I understand the construction for $H^2(M)$ when $H_1(M, \mathbb{Z})=0$ and can make it more explicit than the other answers. As I noted in a comment above, when $H_1(M, \mathbb{Z})$ has torsion, we cannot hope for a canonical construction. When $H_1(M, \mathbb{Z})$ is torsion-free but nontrivial, there should be a construction, but I haven't found it. Also, I am terrible with functional analysis and infinite dimensional spaces so I will be vague about what "nice" means. -Let $\omega$ be a closed integral $2$-form on $M$. Let $\gamma$ be a closed path in $M$. Since $H_1(M)=0$, there is a surface $S$ with $\partial S = \gamma$. Define $\theta(\gamma) = \exp(2 \pi i \int_S \omega)$. We must check that this is independent of the choice of $S$: If $\partial S_1 = \partial S_2$ then $S_1 - S_2$ is a cycle in $H_2(M, \mathbb{Z})$ so $\int_{S_1-S_2} \omega \in \mathbb{Z}$ and we deduce that $\int_{S_1} \omega \equiv \int_{S_2} \omega \bmod \mathbb{Z}$. -Fix a base point $x_0$ in $M$. Let $T$ be the space of all nice paths in $M$ starting at $x_0$. If you fear infinite dimensional spaces more than you hate making choices, choose a metric on $M$, let $T$ be the tangent space to $M$ at $x_0$ and interpret "nice" as "geodesic"; by the Hopf-Rinow theorem, we can identify these with $M$. (I can't decide whether my hatred or my fear is greater.) -Let $V$ be the vector space of all nice functions $f:T \to \mathbb{C}$ with the property that, if $\gamma_1$ and $\gamma_2$ are two paths $x_0 \leadsto x_1$ then $f(\gamma_1) = \theta(\gamma_1-\gamma_2) f(\gamma_2)$. (If $T$ is the tangent space to $M$ at $x$, I think we can take "nice" to be "smooth".) -Any path $\gamma$ now gives a linear functional $V \to \mathbb{C}$ by $f \mapsto -f(\gamma)$. If $\gamma_1$ and $\gamma_2$ both have endpoint $x_1$, then the corresponding linear functionals are proportional. So each $x \in M$ defines a point in $\mathbb{P}(V^{\ast})$. Up to the question of how to put a smooth structure on $\mathbb{P}(V^{\ast})$, we have embedded $M$ into an infinite dimensional projective space. -Again, if you fear infinite dimensional spaces more than you hate choices, rather than working with all of $V$, choose a finite dimensional subspace such that, for any $x \in M$, there is some $f \in V$ which is nonzero at $x$. We could try doing something like restricting ourselves to harmonic functions $f$ in the hope of getting a canonical finite dimensional subspace, but then we would get into the question of what the analogue of "ample" is in this setting. I don't know, but maybe some differential geometer does. - -This was the result of composing two tricks. The second trick is familiar to algebraic geometers: If $L$ is a vector bundle on $X$ and $V$ is its space of global sections, then $X$ embeds, choice-free, into $\mathbb{P}(V^{\ast})$. -The first trick was that, given a line bundle $L$ with connection, I can think of sections of $L$ over $M$ as functions $T \to \mathbb{C}$ whose holonomy respects the holonomy of the connection. I then remembered enough about how curvature works to see that I could write down the holonomy without building the line bundle or the connection. -If $H_1(M)$ is nonzero, define $Z_1(M)$ to be the integer $1$-cycles and $B_1(M)$ the integer $1$-boundaries, so $H_1(M) = Z_1(M)/B_1(M)$. The map $\theta$ still makes sense as a map $B_1(M) \to U(1)$. I think what we are supposed to do is lift $\theta$ to a map $\tilde{\theta}: C_1(M) \to U(1)$. Once we have done that, there is no problem with repeating the above construction with $\tilde{\theta}$. Presumably, there is a theorem that it only matters what choice we make on torsion elements of $C_1(M)/B_1(M)$, but that isn't clear to me.<|endoftext|> -TITLE: Failure of Shoenfield's Absoluteness -QUESTION [7 upvotes]: Shoenfield's absoluteness states that if $M \subseteq N$ are models of $ZF$ and $M \supseteq \omega_1^N$, then every $\Sigma^1_2$ formula with parameters in $M$ is absolute between $M$ and $N$. In particular, $\Sigma^1_2$ properties are preserved under generic extensions of the universe. -What I'm looking for are examples of failures of $\Sigma^1_2$ absoluteness between $M \subseteq N$ models of $ZF$ when $M$ does not contain all countable ordinals of $N$. I'll be even happier if the examples are in $ZF$ and as natural as possible, although any help will be appreciated of course. -Thanks ! - -REPLY [10 votes]: I'm turning my comment into an answer. With $\Sigma^1_2$ statements we can discuss well-foundedness: A real codes a well-founded model of enough set theory iff it codes a model (which is an arithmetic statement) and the model is well-founded (this you can express by saying that no sequence through the ordinals of the model is strictly decreasing). So, to say that there is such a well-founded model is $\Sigma^1_2$. As for "enough set theory", pick $T$ a finite and sufficiently strong fragment of $\mathsf{ZF}$. -By the reflection theorem, there are transitive models of $T$, so there are countable transitive ones (by Lowenheim-Skolem and Mostowski). Pick an example $M$ of smallest height. We have that, in $M$, there are no transitive set models of $T$. That is, the $\Sigma^1_2$ statement discussed in the previous paragraph is true in $V$ (and in set models of certain stronger fragments of $\mathsf{ZF}$), but fails in $M$. -If we assume that there are transitive set models of $\mathsf{ZF}$, then we can run this argument without using reflection, of course. We can even pick $M$ to be an $L_\alpha$. But the point of picking $T$ finite is so that we can formalize this: If it were the case that Shoenfield's absoluteness result goes through without the requirement on $\omega_1$, then this would be provable in $\mathsf{ZF}$, and the proof of course only uses a finite set of axioms, and would apply to finite fragments of set theory, as long as they are strong enough. We can then let $T$ be so strong that, in particular, it contains all these axioms. This, naturally, leads to a contradiction, and the conclusion is that Shoenfield's theorem indeed needs the uncountability assumption.<|endoftext|> -TITLE: Double coset formulas for Orthogonal groups [Solved] -QUESTION [5 upvotes]: According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute -the composition -$BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ -where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. -The result is that it should be the sum with possibly signs of the identity and -the switch map. -Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ -where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach -"The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ. -On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( -\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition -$BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same. -[Added afther the comment by Oscar Randal-Williams] -The map $2(id-Switch)$ should be non-trivial for the following reason. -Let's compute the induced maps in homology of infinite loop space by $2id$ -and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the -generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image -in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) -$ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas -$\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$. -As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ. -[Added Nov.30] -Here is another argument to show that $2id$ -and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I_n$-adic completion (probably a simple -2-adic completion should suffice). Now, the identity and the switch map -induce different maps on $E^*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^*(BO(1)^2)$ as well. -[Added Dec. 07 after a private communication with Oscar Randall-Williams] -(edited Dec 09) -What goes wrong in the above was the assumption that the "coefficients" -are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate -Burnside ring c.f. Mitchell-Priddy "A double coset formula for levi subgroups and splitting $BGL _n$" This issue is more ore less adressed in -Remark (1.6), as well as in the remark following Theorem (2.3). - -REPLY [2 votes]: What goes wrong in the above was the assumption that the "coefficients" -are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate -Burnside ring c.f. Mitchell-Priddy "A double coset formula for levi subgroups and splitting $BGL _n$" This issue is more ore less adressed in -Remark (1.6), as well as in the remark following Theorem (2.3).<|endoftext|> -TITLE: Quasi-isometric rigidity of certain products of groups -QUESTION [7 upvotes]: Let $G_1,G_2$ be two Gromov-hyperbolic groups. If $G_1\times \Bbb{Z}$ is quasi-isometric to $G_2\times \Bbb{Z}$, must it be true that $G_1$ is quasi-isometric to $G_2$? -This is similar to the classic topology problem where crossing two non-homeomorphic spaces with $\Bbb{R}$ can create two homeomorphic spaces which are homeomorphic. - -REPLY [9 votes]: See Kapovich and Leeb On asymptotic cones and quasi-isometry classes of fundamental groups of 3-manifolds, or Kapovich, Kleiner and Leeb -Quasi-isometries and the de Rham decomposition. This result can be also proven using the arguments of Eleanor Rieffel in Groups quasi-isometric to $H^2\times R$ (Journal of the London Mathematical Society 64 Issue 1 (2001) 44–60 doi:10.1017/S0024610701002034) which avoid asymptotic cones (in the case you do not accept the Axiom of Choice).<|endoftext|> -TITLE: Full-rank rectangular matrices over GF(2) -QUESTION [5 upvotes]: Given positive integers $k$, $m$, $n$, let $A$ be an $m \times n$ matrix over $GF(2)$ constructed as follows. Let $X_1, \ldots, X_m$ be independent random subsets of $\{1,\ldots,n\}$ with cardinality $k$, and take $A_{i,j} = 1$ if $j \in X_i$, $0$ otherwise. What can be said about the probability $P(k,m,n)$ that $A$ has rank $m$ over $GF(2)$? -In particular, I suspect that there is some -$t(k) > 1$ such that as $m, n \to \infty$ with $n/m < t(k) - \epsilon$, $P(k,m,n) \to 0$, while for $n/m > t(k) + \epsilon$, $P(k,m,n)$ is bounded below. Is this true? - -REPLY [9 votes]: Actually, something a bit stronger is true: For fixed $k\geq 3$ the threshold is sharp in the sense that above $t(k)$ the probability that $A$ is of full rank tends to $1$. In computer science this question has received a fair amount of study under the name of "random XOR-SAT". The existence of a threshold was apparently first shown by Creignou and Daude in 2003 (I haven't seen the paper and don't have access to it). -Dubois and Mandler determined the precise value of the threshold for $k=3$, and a recent preprint of Pittel and Sorkin extends this to larger $k$. The rough idea is to start by doing a peeling process -- if there's a column with exactly one non-zero entry, we can delete the column and the row of that entry without affecting whether the matrix has full rank. After repeatedly doing this and deleting any columns that are entirely $0$, we'll be left with a "core" matrix where each row has $k$ non-zero entries and each column has at least two non-zero entries. -What ends up being true (but requires a long technical argument to prove) is that the size of the core pretty much determines everything -- if the core has more rows than columns, it can't be full (row) rank, while if it has significantly more columns than rows it almost surely has full row rank.<|endoftext|> -TITLE: What is the strength of chains of 1-extendibles? -QUESTION [12 upvotes]: Let $X$ be a collection of cardinals such that if $\kappa,\lambda\in X$ and $\kappa<\lambda$, then there is a non-trivial elementary embedding $j:V_{\kappa+1} \to V_{\lambda+1}$ with $crit(j) = \kappa$. -What is the consistency strength of the claim that there are such $X$ for various sizes of $X$? In particular, what is the consistency strength when $X$ is a proper class of cardinals? - -REPLY [8 votes]: I think for the Theorem one can get by with a subcompact cardinal, which is weaker than a $\kappa^+$-supercompact. -( $\kappa$ is subcompact if for any $A\subseteq H_{\kappa^+}$ there are $j,\mu < \kappa, B\subseteq H_{\mu^+}$ with $j:(H_{\mu^+},\in, B) \longrightarrow_e (H_{\kappa^+},\in, A)$. ( where $ \longrightarrow_e$ means ``is an elementary map''. ) ) -Let $\kappa$ be subcompact as above, and let $A = \mathop{Sat} = -{Sat}_{\kappa^+}$ Then $\mathop{Sat} \subseteq H_{\kappa^+}$, and -applying the definition of subcompactness there are $\mu < \kappa$, $j$, -and $\overline{\mathop{Sat}}$ with -$j : (H_{\mu^+}, \overline{\mathop{Sat}}) \longrightarrow_e (H_{\kappa^+}, -\mathop{Sat})$. -Then $\overline{\mathop{Sat}}= {Sat}_{\mu^+}$ can be checked easily. $j$ is easily in $H_{\kappa^+}$ but the point of having the Sat predicate is that we can express in a first order way (in the extended language) that there is a $j$ with $ran(j)$ an elementary substructure of $(H_{\kappa^+},\in)$. -Thus $(H_{\kappa^+}, {Sat}) \models$``${ran} (j) \prec_e (V, \in)$'' indeed: -$(H_{\kappa^+}, {Sat}) \models$``There is $k, \kappa_0,$ with $k -: (H_{\kappa_0^+}, \in) \longrightarrow (V, \in)$ and ${ran} (k) -\prec_e (V, \in) ,$ and thus $k$ is an elementary map.'' -This goes down to $(H_{\mu^+}, {Sat}_{\mu^+})$. This starts out links in the chain, using the restriction of the map(s) $k$ to the domain $V_{\kappa_0 +1}$ etc. and the rest of the argument is as before.<|endoftext|> -TITLE: Gini Coefficient and Renyi Entropy -QUESTION [18 upvotes]: Gini coefficient (aka Gini Index) is a quantity used in economics to describe income inequality. It is 0 for uniformly distributed income, and approaches 1 when all income is in hands of one individual. -For probabilities we have: -$$ -G(\{p_i\}_{i\in\{1...n\}}) = \sum_{i=1}^n \frac{2i-n-1}{n}p_i, -$$ -where $p_i$ is a non-decreasing sequence, -or in the continuous case: -$$ -G(p) = \int_0^1 \left(1-2\int_0^xp(y)dy\right)dx = \int_0^1 (2x-1)p(x)dx -$$ -where probability density function $p(x)$ is non-decreasing. -As it is a measure of uniformness of a probability distribution, I was curious if it is possible to relate Gini coefficient to some Rényi entropy $H_q$ (for example, collision entropy $H_2$)? (Or Tsallis entropy $T_q$ as for the continuous case formulas are shorter.) -I got an upper bound -$$ -G(p) \leq \sqrt{\tfrac{1}{3}(e^{-H_2(p)}-1)} = \sqrt{-\tfrac{1}{3}T_2(p)} -$$ -with Schwarz inequality for $(2x-1)$ and $(p(x)-1)$. -It works similarly for discrete probabilities: -$$ -G(\{p_i\})\leq\sqrt{\tfrac{n}{3}}\sqrt{1-\tfrac{1}{n^2}}\sqrt{e^{-H_2(\{p_i\})}-\tfrac{1}{n}}. -$$ -Is there any non-trivial lower bound? That is, except for -$$G(p)\geq 1 - e^{H_0(p)} = -T_0(p),$$ -$$G(\{p_i\})\geq 1 - \tfrac{1}{n}e^{H_0(\{p_i\})}.$$ -Is there an upper bound that is saturated for $p = \{0,0,\cdots,0,1\}$? - -REPLY [4 votes]: As of now, the best I got is: -$$ -1 - e^{H_{1/2}(p)} \leq G(p) \leq \sqrt{1-e^{2 H_{1/2}(p)}}, -$$ -where $e^{H_{1/2}(p)} = 2 \ln \int_0^1p^{1/2}(x)dx$. -Both inequalities are saturated for both $G(p)=0$ and $G(p)=1$. -The lower bound is better than the previous one ($1 - e^{H_{0}(p)}$), as Rényi entropy $H_q(p)$ is non-decreasing in $q$. -The higher bound is different from $\sqrt{\tfrac{1}{3}e^{-H_2(p)}-1}$ (there is no inequality between them). -And for practical purposes their quadratic mean seems to be a good approximation, -$$ -G(p) \approx \sqrt{1 - e^{H_{1/2}(p)}}. -$$ -To get some taste of it, here is a plot for the lower, the upper bound and the approximation: - -(Sampled for distributions of the form $p(x) \propto (1-x+r)^{-\alpha} + c$, where $r$, $\alpha$ and $c$ are some parameters.) -Proof -We have: -$$ -G(p) = \int_0^1 (2x-1) p(x) dx = \tfrac{1}{2} \int_0^1 \int_0^1 |p(x) - p(y)|dx dy. -$$ -(Actually, the right hand side is more general, as it does not require any ordering of p(x).) -For the lower bound we use -$$ -|p(x) - p(y)| = \left|\sqrt{p(x)} + \sqrt{p(y)}\right| \left| \sqrt{p(x)} - \sqrt{p(y)} \right| \geq \left( \sqrt{p(x)} - \sqrt{p(y)} \right)^2 -$$ -and perform the integration. -For the upper bound we use Schwartz inequality for $\left|\sqrt{p(x)} + \sqrt{p(y)}\right|$ and $\left| \sqrt{p(x)} - \sqrt{p(y)} \right|$.<|endoftext|> -TITLE: Is the injective structure on unbounded chain complexes simplicial? -QUESTION [5 upvotes]: In Mark Hovey's article Model category structures on chain complexes of sheaves (arXiv:math/9909024) a model structure on the category $Ch(A)$ of unbounded chain complexes for a Grothendieck abelian category $A$ is constructed. This is called the injective structure and has the derived category of $A$ as its homotopy category. -Can this injective model be given the structure of a simplicial model category? How is the simplicial enrichment given? -The Dold-Kan correspondence identifies non-negative chain complexes with with simplicial objects in $A$ and there is an evident simplicial enrichment but the question is for unbounded. - -REPLY [7 votes]: It is instructive to consider a simpler case, namely that of chain complexes over a ring. Hovey deals with this example in great detail in his book Model Categories. In particular, on page 114 Hovey states that $Ch(R)$ is not a simplicial model category. In algebraic situations like this, it's not really the right question to ask a model category to be simplicial. Instead, one should ask it to be a dg-model category, a.k.a. a $Ch(\mathbb{Z})$-model category. This does hold for $Ch(R)$ with the injective model structure, and should probably hold in the generality of your question as well, though I haven't checked it. A good reference is chapter 4 of Hovey's book, where he defines the notion of a $\mathscr{C}$-model category $\mathscr{M}$ where $\mathscr{C}$ is a model category. This is a generalization of the notion of a simplicial model category (when $\mathscr{C}$ is simplicial sets), i.e. the coherence between $\mathscr{M}$ and $\mathscr{C}$ is analogous to Quillen's SM7 axiom, though Hovey was the first to realize the necessity of the unit axiom for general $\mathscr{C}$. -Note that the model categories $Ch(A)$ in your question are certainly Quillen equivalent to simplicial model categories, e.g. by Dan Dugger's work (because they are combinatorial model categories). Dugger's paper is called "Combinatorial model categories have presentations" -EDIT: Because you're in an additive setting, you actually have more machinery at your disposal than I realized. If you wish to compare an enrichment over $\mathscr{C}$ with an enrichment over $\mathscr{D}$ (e.g. comparing dg-model categories with simplicial model categories), then a great resource is the paper Enriched Model Categories by Dugger and Shipley, especially their notion of Quillen adjoint module. I only just found this paper today. Their machinery can be used to say when two enrichments are quasi-equivalent, i.e. equivalent up to homotopy. Now, monoids in sSet and Ch(k) are not so different (thanks to Dold-Kan) so you may be able to use this machinery to beef up the answer I gave from "Quillen equivalent to a simplicial model category" to something stronger. For example, their Proposition 5.2 and their section 9 let you get your hands on what's going on a bit more than the fully general results of Dugger I mentioned previously. On the other hand, if what I wrote before is enough for whatever application you had in mind then feel free to ignore this.<|endoftext|> -TITLE: Shrink polygon to a specific area by offsetting -QUESTION [7 upvotes]: I have a 2D polygon that I want to shrink by a specific offset (A) to match a certain area ratio (R) of the original polygon. Is there a formula or algorithm for such a problem? I am interested in a simple solution for a triangle/quad but also a solution for complex polygons. -I attached an image for explanation. The original polygon is offset by A (equal-distant for each edge). A has to be chosen so that the new polygon has a specific area. In this example it should have half the area of the initial polygon. - -REPLY [8 votes]: I am afraid there is no truly simple solution, unless you ignore the complexities. -Just think of how you would shrink a sharp V-shape, like this: -       -In the above, the offset was larger than the V-gap, and so the shape naturally -fractured into two disconnected regions. This issue arises even with a quadrilateral. -To delve into the topic more thoroughly, -see the result of Ron Wein's 2007 definitive Ph.D. thesis, described in this paper: - -"Exact and approximate construction of offset polygons." - (ACM link). - -  -His work is implemented in CGAL (CGAL manual link), and now nicely described in this book: - -Fogel, Efi, Dan Halperin, and Ron Wein. CGAL Arrangements and Their Applications: A Step-by-Step Guide. Vol. 7. Springer, 2012. - -Finally, if you want to skim over the complexities, here is a decade-old survey that is nevertheless useful as a high-level viewpoint: survey link.<|endoftext|> -TITLE: Branch loci of Ramified covers -QUESTION [9 upvotes]: Let $X\to Y$ be a $d:1$ ramified covering map from a smooth complex projective variety $X$ to a smooth complex projective variety $Y$. - -Question 1. What does the smoothness imply on the branch locus in $X$ or its image in $Y$? - Does it imply that its a normal crossing? -Question 2. If not, assuming $Y=\mathbb{P}^n$, can we always deform $X$ to get a normal-crossing branch locus in $\mathbb{P}^n$ (or at least for its preimage in $X$)? - -Comment 1. Regarding the nice answer of Tony, you may assume $X$ is simply-connected. -Comment 2. By normal crossing, I mean the reduced branch locus to be normal crossing. - -REPLY [8 votes]: Even in the simply connected case there is no hope. -Let $X$ be a smooth cubic surface in $\mathbb{P}^3$ and take a general projection $f\colon X \to \mathbb{P}^2$. Then $f$ is a branched triple cover and it is well known that the branch locus $B \subset \mathbb{P}^2$ is a sextic curve with six ordinary cusps lying on a conic. -There is no way to deform this cover (mantaining $X$ smooth) in order to obtain a branch locus which is normal crossing. -Notice that $X$ is a rational surface, hence $\pi_1(X)=\{1\}$.<|endoftext|> -TITLE: Which quaternion algebras have class number one? -QUESTION [9 upvotes]: Over Q, the definite quaternion algebras with a unique conjugacy class of maximal orders, i.e. "with class number one", are those with discriminant 2,3,5,7, and 13. -Three questions: - -What is a reference for this result? -What are some examples of definite quaternions of class number one, over other totally real fields? -For a fixed real quadratic field, is there a known classification of definite quaternion algebras class number one? - -REPLY [14 votes]: For the first question, the statement goes back to Brzezinski (http://archive.numdam.org/ARCHIVE/JTNB/JTNB_1995__7_1/JTNB_1995__7_1_93_0/JTNB_1995__7_1_93_0.pdf), who treats all definite quaternion orders over $\mathbb{Z}$, not just maximal ones. For maximal orders, this is an almost-immediate consequence of Eichler's mass formula. The simplest proof I know of the mass formula follows in the same way as the proof of the analytic class number formula and can be found in his "Lectures on Modular Correspondences"; for prime discriminant this has many nice proofs (Is there a nice proof of the fact that there are (p-1)/24 supersingular elliptic curves in characteristic p?). -For the second and third questions, you can see my paper with Markus Kirschmer (http://www.math.dartmouth.edu/~jvoight/articles/quatideal-fixed-errata-052613.pdf) where we find all Eichler orders of class number one: you're asking for Table 8.2. Kirschmer and David Lorch are working on generalizing this in several directions. The simplest (and arguably most beautiful) example is to take a maximal order in $B=(-1, -1 \mid \mathbb{Q}(\sqrt{5}))$, the base change of the Hamilton quaternions to $\mathbb{Q}(\sqrt{5})$. Such a maximal order is unique up to conjugation in $B^\times$ and has unit group (modulo scalars) given by the rigid motions of the icosahedron (or dodecahedron); explicitly, it is generated as a $\mathbb{Z}[(\sqrt{5}+1)/2]$-algebra by $i$ and the element -$$ \frac{1}{2}\left(\frac{\sqrt{5}+1}{2}+\frac{\sqrt{5}-1}{2}i + j\right).$$<|endoftext|> -TITLE: capturing small sets in small factors -QUESTION [15 upvotes]: Suppose $\kappa$ is a regular cardinal and $P$ is a $\kappa$-c.c. partial order. I want to know when are small sets added by subforcings of size $<\kappa$. The following seems well-known: - -Fact: If $\kappa$ is weakly compact, and $P$ has size $\kappa$ and is $\kappa$-c.c., then for any $P$-name $\tau$ for a set of ordinals of size $<\kappa$, there is a $Q \lhd P$ and a $Q$-name $\sigma$ such that $|Q| < \kappa$, and $1 \Vdash_P \sigma = \tau$. - -I think the easiest way to see the fact is to use the extension property. Every such $P$ can be coded as a set of ordinals $A \subseteq \kappa$, and there is some transitive $X$ of rank > $\kappa$ and a set $B$ such that $(V_\kappa,\in,A) \prec (X,\in,B)$. Any $P$-name for a small set is seen by the larger structure as captured by $P$, so this reflects. -Question 1: Are there counterexamples for some large cardinals which are weaker than weakly compact? -If $\kappa$ is supercompact, then the same conclusion holds for all $\kappa$-c.c. $P$. This is easy to see by taking a supercompactness embedding $j$ with closure at least $|P|$ and noting that $j[P]$ is a regular suborder of $j(P)$. Supercompactness must be a total overkill hypothesis. -Question 2: For what cardinals $\kappa$ do we have that every $\kappa$-c.c. partial order captures small sets in small factors? - -Update: I think I have a partial answer to Question 2. Supercompactness is indeed overkill, and weak compactness is enough after all. Let $\kappa$ be weakly compact, and $P$ be $\kappa$-c.c. Let $\theta > \kappa$ be regular such that $P \in H_\theta$. Let $\tau$ be any $P$-name for a $<\kappa$ sized set of ordinals. Let $M \prec H_\theta$ be such that $P,\tau \in M$, $|M| \leq \kappa$, and $M^{<\kappa} \subseteq M$. This is possible just because $\kappa$ is inaccessible. Then $P \cap M$ is a regular suborder of $P$, since all antichains contained in $P \cap M$ are members of $M$, and $M$ knows which ones are maximal. By the Fact, there is $Q \lhd P \cap M \lhd P$ and a $Q$-name $\sigma$ such that $|Q| < \kappa$ and $\Vdash_{P \cap M} \tau = \sigma$. - -REPLY [7 votes]: We now have a full answer. The capturing property is equivalent to weak compactness. -Yair already covered the non-inaccessible case. Now we show that this property implies the tree property at $\kappa$. If there is a $\kappa$-Aronszajn tree $T$, then there is a $\kappa$-c.c. forcing due to Baumgartner that makes $\kappa = \aleph_1$ and makes $T$ special. Conditions consist of finite partial functions from $T$ into $\omega$ with the property that comparable nodes are assigned different colors. Call this forcing $\mathbb P$ and let $\dot f$ be the canonical name for the generic specializing function. -Assume there is a regular $\mathbb Q \subseteq \mathbb P$ that captures $\dot f \restriction T_\omega$. WLOG $\mathbb Q$ is upward-closed in $\mathbb P$ (we can weaken conditions). Let $S$ be the set of nodes in $T$ that appear in any condition in $\mathbb Q$. -First we claim that for all $n \in \omega$ and all $s \in S$, $\{ (s,n) \} \in \mathbb Q$. If not, then for some $n$, $\Vdash_\mathbb Q \dot f(\check s) \not= n$. But we can take a generic $G \subseteq \mathbb P$ with $\dot f(\check s) = n$ and $G \cap \mathbb Q$ is generic, contradiction. -Second, if $X \subseteq S$ is any infinite ground model set, then a density argument shows that $\Vdash_{\mathbb Q} \mathrm{ran}(\dot f \restriction \check X) = \omega$. -Therefore take any $t \in T \setminus S$, and let $X = \{ s \in S : s < t \}$, which is infinite. If $G \subseteq \mathbb Q$ is generic, then a further forcing adds the specializing function on all of $T$. But there is no possible value left for $f(t)$.<|endoftext|> -TITLE: Combinatorial identities -QUESTION [9 upvotes]: I have computational evidence that -$$\sum_{k=0}^n \binom{4n+1}{k} \cdot \binom{3n-k}{2n}= 2^{2n+1}\cdot \binom{2n-1}{n}$$ -but I cannot prove it. I tried by induction, but it seems hard. Does anyone have an idea how to prove it? -What about -$$\sum_{k=0}^n \binom{4n+M}{k} \cdot \binom{3n-k}{2n}$$ -where $M\in\mathbb{Z}$? Is there a similiar identity? - -REPLY [12 votes]: Here is a proof repeating the trick I used in this earlier answer. By negating the second binomial coefficient, the identity in the question is equivalent to -$$ \sum_{k=0}^n (-1)^k \binom{4n+1}{k} \binom{-(2n+1)}{n-k} = (-1)^n 2^{2n} \binom{2n}{n}. $$ -This is the special case $x= 4n+1$ of the following identity -$$ \sum_{k=0}^n (-1)^k \binom{x}{k}\binom{2n-x}{n-k} = (-1)^n 2^{2n} \binom{(x-1)/2}{n}. $$ -When $x=2r+1$ is odd the left-hand side is zero since the summands for $k=r-j$ and $k=r+1+j$ cancel. So the two sides agree when $x=1,3,\ldots,2n-1$. When $x = 2n+1$ the left-hand side is $\sum_{k=0}^n (-1)^n \binom{2n+1}{k} = (-1)^n 2^{2n+1}/2 = (-1)^n 2^{2n}\binom{n}{n}$. So the two sides agree at $n+1$ points. Since they are polynomials of degree $n$ in $x$, they must agree for all $x$.<|endoftext|> -TITLE: How large can abelian subgroups of class 2 nilpotent groups or simple groups be? -QUESTION [5 upvotes]: If $G$ is a finite simple group then is it true that an abelian subgroup $H$ of $G$ of maximal order has order $|H| < |G|^{\frac{1}{3}}$? If so, could you please point me to a reference for this, or where a proof is given? Secondly, if $G$ is now a finite nilpotent group of class $2$ and $H$ is again an abelian subgroup of $G$ of maximal order, then is it true that $|H| \leq \sqrt{|G|}$? If so, is there a reference or proof for this? -Thanks in advance for any help. -Sandeep Murthy - -REPLY [6 votes]: The result for simple groups is almost true. It's a result of Vdovin: - -Theorem: Let $G$ be a finite non-abelian simple group, with $G\not\cong \mathrm{PSL}_2(q)$ for any prime power $q$. Let $A$ be an abelian subgroup of $G$. Then $|G|>|A|^3$. - -The result appeared in Algebra and Logic, 38, no. 2 (1999). (Note that any simple group isomorphic to $\mathrm{PSL}_2(q)$ has an abelian subgroup $A$ of order greater than $|G|^{1/3}$: just take $A$ to be a Sylow $p$-subgroup where $p$ is the characteristic of the field.) -Vdovin has another result, which appeared in Algebra and Logic, 39, no. 5 (2000) and which may also be of interest: - -Theorem: Let $G$ be a finite non-abelian simple group. Let $N$ be a nilpotent subgroup of $G$. Then $|G|>|N|^2$.<|endoftext|> -TITLE: The solutions of a system of polynomials -QUESTION [12 upvotes]: Given positive integers $m_1,...,m_n$, is it possible to solve the following equation system over the field of complex numbers? -$$m_1x_1+\cdots+m_nx_n=0$$ -$$m_1x_1^2+\cdots+m_nx_n^2=0$$ -$$\cdots$$ -$$m_1x_1^{n-1}+\cdots+m_nx_n^{n-1}=0$$ -$$x_1x_2\cdots x_n=1.$$ -When $n=1,2,3,4$, I can find a formula for the solutions. But I can not do it for $n>4$. Also, when $m_1=m_2=...=m_n=1$, the solutions can be written as -$$(\zeta^{\sigma(1)},\cdots,\zeta^{\sigma(n)})$$ -where $\zeta$ is a $n$-th primitive root of unity and $\sigma$ is an element of the symmetric group $S_n$. -For other concrete examples, the Mathematica numerical computation shows that the number of solutions to this equation system would be $n!$. Does anybody have an idea to prove it? -In general, if we do not assume $m_1,...,m_n$ are positive integers, what is the condition on $m_1,...,m_n$ such that this equation system has a solution? - -REPLY [11 votes]: I believe this follows using Vandermonde matrices. First I will prove a proposition that gives a bit more than what the OP asks. -For every integer $n \geq 2$, for every integer $r=1,\dots,n$, define -$$ F_r(X_1,\dots,X_n) = F_{n,r}(m_1,\dots,m_n;X_1,\dots,X_n) = m_1X_1^r + \dots + m_nX_n^r.$$ Let $P\subset \mathbb{C}$ be a nonempty subset that is stable under addition and which does not contain $0$, e.g., the set of positive integers. -Proposition $Q_n$. For every integer $n\geq2$, for every choice of $m_1,\dots,m_n$ in $P$, every nonzero solution $(A_1,\dots,A_n)\neq (0,\dots,0)$ of $$ -F_{n,1}(m_1,\dots,m_n;X_1,\dots,X_n)=\dots=F_{n,n-1}(m_1,\dots,m_n;X_1,\dots,X_n)=0$$ is not a solution of $F_{n,n}(m_1,\dots,m_n;X_1,\dots,X_n) = 0$, it has no coordinate equal to $0$, and it has no repeated coordinates. -Proof This will be proved by induction on $n$. It is straightforward to prove $Q_2$, the base case of the induction. Basically it boils down to the fact that none of $m_1$, $m_2$ nor $m_1+m_2$ can equal $0$. -Now by way of induction, assume that $n>2$ and that $Q_m$ holds for all integers $2\leq m < n$. Let $(A_1,\dots,A_n)$ be a nonzero solution of the system $F_1=\dots=F_{n-1}=0$. If any $A_i$ is zero, then after permuting the coordinates and $(m_1,\dots,m_n)$, we may assume that $A_n=0$. -Then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system -$$ F_{n-1,1}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,\dots ,$$ -$$ F_{n-1,n-2}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,$$ -and which also solves the equation $F_{n-1,n-1}=0$. This contradicts the induction hypothesis. Thus, no $A_i$ equals $0$. -Similarly, if any coordinates are repeated, say $A_{n-1}=A_n$, then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system, -$$ F_{n-1,1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0, \dots, $$ -$$ F_{n-1,n-2}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0,$$ -and which also solves the equation $$F_{n-1,n-1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1})=0.$$ Again this contradicts the induction hypothesis. Thus, there are no repeated coordinates. -Therefore, let $(A_1,\dots,A_n)$ be an $n$-tuple with no zero coordinates and no repeated coordinates. Then, by the theory of the Vandermonde determinant, there is only the trivial solution $(Y_1,\dots,Y_n)$ of the following linear system, -$$ -\left\{ \begin{array}{rrrrrrr} -Y_1A_1 & + & \dots & + & Y_nA_n & = & 0, \\ -Y_1A_1^2 & + & \dots & + & Y_nA_n^2 & = & 0, \\ -\vdots & & \ddots & & \vdots & & \vdots \\ -Y_1A_i^r & + & \dots & + & Y_nA_n^r & = & 0, \\ -\vdots & & \ddots & & \vdots & & \vdots \\ -Y_1A_1^n & + & \dots & + & Y_nA_n^n & = & 0. -\end{array} \right.$$ Indeed, the determinant of the coefficient matrix is -$$ -\text{det}([A_s^r]_{1\leq r,s \leq n}) = \pm A_1\cdots A_n \prod_{k -TITLE: Realizing a subgroup of a Lie group as a stabilizer subgroup -QUESTION [11 upvotes]: Let $G$ a compact semisimple Lie group, $H$ a subgroup of $G$. Is it always possible to find an irreducible representation $R$ of $G$ such that the stabilizer of an $x\in R$ is "locally isomorphic" to $H$? I am only interested in the case when $H$ is a continuous subgroup, and "locally isomorphic" means has the same Lie algebra. -If the result is not true, what would be the simplest counterexample? -What about weaker results when $R$ is not required to be irreducible, or when stabilizers of a finite set and not just of a single element are allowed? -Let me give an example of what I have in mind. I am only interested in the case when both $G$ and $H$ are finite dimensional compact Lie groups. As a representative example let's take $G=SU(3)$ and two subgroups $H_1=SU(2)$, $H_2=SU(2)\times U(1)$ (all groups over $\mathbb{C}$). I can realize $H_1$ as the stabilizer of $x=(0,0,1)^t$ in the fundamental representation of $SU(3)$. I can also realize $H_2$ as the stabilizer of $x=diag(1,1,-2)$ in the adjoint representation of $SU(3)$. Can I always do this? What would be an algorithm to construct the representation given the set of generators of $G$ which generate $H$? -Update: the answers below show that the answer is yes, if you allow reducible representations (which I don't mind). There remains a problem of how to construct $R$ concretely and simply, given the list of generators for $H$ inside $G$. Such constructions can be also extracted from the proofs below, however they do not look elementary (for me). - -REPLY [8 votes]: Though it's not so readily available online, there is an elementary textbook treatment of the basic theory here in Representations of Compact Lie Groups by Brocker and tom Dieck (GTM 98, Springer, 1985). -1) As they note early in the book, the isotropy group $H$ of a point $v$ must be closed in an arbitrary topological group $G$ acting continuously: here $H$ is the inverse image of $v$ under the (continuous) orbit map $g \mapsto g \cdot v$. -2) As a corollary of the Peter-Weyl theorem (using representative functions), -they derive easily in III, (4.6) for any compact Lie group $G$ (not necessarily semisimple): Every closed subgroup $H$ of $G$ appears as the isotropy group of an element of some $G$-module. -3) Unfortunately, since all of this theory is somewhat abstract, it doesn't seem to shed light on your question about finding an irreducible representation. However, you do have complete reducibility here of all (necessariy finite-dimensional) representations, which are usually studied in the essentially equivalent (complexified) Lie algebra setting. So if there is a counterexample it would probably be best located there. It seems hard to compute directly with group elements and group representations, but for example Willem de Graaf has worked extensively with computer algorithms for both real and complex Lie algebras. -P.S. Though Dan Mostow (who just turned 90) was on my thesis committee, I don't think you need to get into his more general results involving group actions on manifolds.<|endoftext|> -TITLE: From elementary equivalence to isomorphism -QUESTION [5 upvotes]: A few years ago, when I took the basic course in Logic, I was very surprised to discover that given a signature $\sigma$ and two structures $M$ and $N$ of $\sigma$ which are elementarily equivalent (that is, they satisfy the same first-order $\sigma$-sentences), they are not necessarily isomorphic, even if they are of the same cardinality. -My question is, are there some "reasonable" conditions that one may pose on either the signature $\sigma$ or the structures $M$ and $N$ that guarantee the implication: "If $M$ and $N$ are elementarily equivalent, then $M \cong N$"? - -REPLY [2 votes]: First, you should note that the signature is determined by the structures and any two structures that are elementarily equivalent or isomorphic have to have the same signature. -For your question, I don't think much can be done in the realm of first order logic. Asaf's answer is correct that you need $Th(M)$ to be categorical for $M \equiv N$ to imply $M \cong N$. Fixing the language is even worse as, for every signature, there are non-categorical theories in it. -Things get a little more interesting when you move to infinitary languages. $L_{\omega_1, \omega}$ is the logic formed by allowing countably many conjunctions and disjunctions (as opposed to finitely many in first order), while still requiring that there are only finitely many free variables. In this logic, every countable structure has a Scott sentence which characterizes its countable isomorphism type. That is: if $M$ is a countable structure, then there is a sentence $\phi_M \in L_{\omega_1, \omega}$ (of the language of $M$) such that if $N$ is a countable structure so $N \vDash \phi_M$, then $M \cong N$. -This doesn't generalize as nicely to larger infinitary languages as one might like, but I believe that if $cf(\lambda)=\omega$ and $M$ is a $\lambda$ sized model, then there is a $L_{\lambda^+, \omega}$ sentence with the same property.<|endoftext|> -TITLE: bialgebra cohomology -QUESTION [5 upvotes]: It seems that the Gerstenhaber-Schack (bi)complex of an associative bialgebra carries a homotopy e_3-algebra structure and a degree 2 Lie algebra bracket, up to homotopy. Does anyone know about a reference where these structures have been explicitly shown? - -REPLY [2 votes]: It's done in Ginot-Yalin Deformation theory of bialgebras, higher Hochschild cohomology and formality (arXiv:1606.01504 - -More precisely, my undertsanding is that what's true by abstract non-sense is that under some (co)nilpotency condition there's an equivalence between $E_2$-algebras and (dg-)bialgebras, and that the deformation complex of an $E_2$-algebra is $E_3$. What is done in that paper apart from carefully writing this up, is identifying this with the standard GS complex explaining and precisely in what way t controls deformations of bialgebra in this setting.<|endoftext|> -TITLE: Lower bounds for $|A+A|$ if $A$ contains only perfect squares -QUESTION [11 upvotes]: Let $A$ a set with $|A|=n$ that contains only perfect squares of integers. -What lower bounds can we give for $|A+A|$? -I think the lower bound $\gg \frac{n^2}{\sqrt{log \,n}}$ holds (this would be the best bound possible, with "equality" for $A=\{1^2,2^2,...n^2\}$). However, this estimate seems really hard. -I couldn't even prove the bound $|A+A|>Cn$ for all constant $C$ (for big $n$). I think proving this bound would already be a good start. Any idea or technique related to the problem is appreciated. -Thanks in advance. - -REPLY [17 votes]: This is a well-known (and difficult) problem. -The current record is $|A+A| \geq \log(|A|)^{c\log\log(|A|)}|A|$ due to Schoen (in 2011), using his near optimal form of Freiman's theorem. Note this is just shy of a power gain. -See Chang's paper On problems of Erdos and Rudin for more discussion of this and related questions.<|endoftext|> -TITLE: is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles? -QUESTION [23 upvotes]: This might be obvious to experts, but I'm not sure where to look for the answer. On a reasonably nice, at least noetherian, scheme (or variety, algebraic space, stack), can the category of coherent sheaves be constructed categorically from the category of vector bundles? I am thinking of Coh being some kind of 'abelian envelope' of Vect. -(in the affine case this question can be rephrased as: can the category of finitely generated modules be defined via the category of projective modules of finite rank)?) - -REPLY [17 votes]: Here are a few comments that might be useful. I don't think there is a chance that this can work unless the scheme in question has the resolution property (meaning every coherent sheaf is a quotient of a locally free sheaf of finite rank). Otherwise the category of locally free sheaves does not even form a generator of the category of all quasi-coherent sheaves, so it clearly contains more information. -Secondly, Quiaochu Yuan's construction for the affine case does not work globally for most schemes. What he does is indeed freely adding cokernels (to get to coherent sheaves) or freely adding all colimits (to get to quasi-coherent sheaves). The free cocompletion under all colimits of an additive category is given by taking the category of additive presheaves on it. (The free cocompletion under cokerenels is simply the closure of the representables under cokernels.) So, if we do that to the category of vector bundles on a scheme, we obtain a category of presheaves. However, any category of presheaves has a projective generator, while the category of quasi-coherent sheaves rarely does. -Finally, something positive that can be said: If you do assume that your scheme satisfies the resolution property (and I'll assume it is quasi-compact, not sure if that's necessary), then the full subcategory of vector bundles is a dense subcategory of the category of quasi-coherent sheaves. This is actually a quite amazing result: in a Grothendieck abelian category, any strong generator is dense, see -Brian Day and Ross Street, Categories in which all strong generators are dense, J. Pure Appl. Algebra 43 (1986), no. 3, 235–242. MR 868984 -Thus we know that the category of quasi-coherent sheaves is a reflective subcategory of the free cocompletion of the category of vector bundles. Any reflective subcategory is the localization of the surrounding category at the morphisms that the reflector inverts (that is, it can be obtained by formally inverting a class of morphisms). Since we're dealing with a locally finitely presentable category, this can be further reduced to inverting a generating set of these morphisms. In some sense this says that the category of quasi-coherent sheaves can be obtained by first freely adding colimits, and then imposing some relations (formally turn a certain set of morphisms into isomorphisms). -It seems however rather difficult to get an explicit such set of morphisms in general. -Edit: I noticed that you're also interested in algebraic spaces and algebraic stacks. The above argument about the category of quasi-coherent sheaves also works at that level of generality as long as the resolution property holds. Specifically, if you have a quasi-compact stack $X$ on the fpqc-site of affine schemes which has the resolution property (in algebraic topology these are sometimes called Adams stacks, since they are precisely the stacks associated to Adams Hopf algebroids), then the category of quasi-coherent sheaves on $X$ is given by a localization of the free cocompletion of the category of dualizable quasi-coherent sheaves on $X$ at a set of morphisms. -Note that it is not clear from this argument wether or not this set of morphisms is entirely determined by the subcategory of dualizable quasi-coherent sheaves. If that is not the case, there could be Adams stacks in the above sense with equivalent categories of dualizable sheaves but inequivalent categories of quasi-coherent sheaves.<|endoftext|> -TITLE: How "accidental" are equalities between parts of Ehrhart quasi-polynomials? When do they persist to Euler-Maclaurin? -QUESTION [7 upvotes]: Background -What I think of Ehrhart theory (http://en.wikipedia.org/wiki/Ehrhart_polynomial) asserts that if we take a lattice polytope $P$, and count the number of lattice points in the $t$th dilation of $P$, the result is polynomial in $t$. -If, however, $P$ is only a rational polytope, then in general we get a quasi-polynomial; that is, there is some period $N$, so that if for any $k$, I only look at $t$ congruent to $k$ mod $N$, then again I have a polynomial behavior; however, different residue classes will in general result in different polynomials. -One might expect that any two distinct residue classes have distinct polynomials, but this need not be the case: two different residue classes can have the same polynomial. These are the "accidental" equalities of the question title. -A simple example -For a simple example, take the triangle bounded by $x=0,y=3x$ and $y=1$; scaling by $t$ just changes $y=1$ to $y=t$. The count is quasi-polynomial of period 3, but only has two distinct polynomials. The number of lattice points is -$$ 1+\frac{t^2+5t}{6}$$ -if $t$ is congruent to 0 or 1 mod 3, and -$$ 1+ \frac{t^2+5t-2}{6}$$ -if $t$ is congruent to 2 mod 3. -Questions, general and specific -I'm interested in what we know about when or why these accidental equalities occur. That is a rather broad and open-ended question; so here's something a bit more specific. -If instead of just counting the lattice points in the $t$th dilate, we sum a polynomial function over the lattice points, we again get (quasi-)polynomial functions; call this Euler-Maclaurin theory. I had the naive hope that "Accidental" equalities in Ehrhart quasi-polynomials might extend to equalities in Euler-Maclaurin theory, but this appears not be always the case: if we try to sum the function $x$ over the lattice points in the $t$th dilate of the polytope in our example, we get -$$\frac{t^3}{54}+\frac{t^2}{9}+\frac{t}{6}$$ -for $t$ congruent to 0 mod 3, but -$$\frac{t^3}{54}+\frac{t^2}{9}+\frac{t}{18}-\frac{5}{27}$$ -for $t$ congruent to 1 mod 3. -If I have some rational polytope that has accidental equalities in its Ehrhart polynomials, and a specific polynomial I want to sum over it; is there some conditions in which the same accidental equalities will hold for the Euler-Maclaurin problem? - -REPLY [4 votes]: This is an interesting and wide open question. The easiest case when considering Ehrhart quasipolynomials, namely, when all constituent polynomials are equal, goes by the name of period collapse. The most famous instance comes from representation theory (see this paper by J. De Loera & T. McAllister, which appeared in Discrete Comp. Geom. 32 (2004)). You might also find this paper by C. Haase & T. McAllister (which appeared in Cont. Math. 451 (2008)) illuminating.<|endoftext|> -TITLE: Orbits of group scheme action -QUESTION [5 upvotes]: I am interested in orbits of the action of a group scheme on a scheme and I'm particularly interested in the following special case: Let $k$ be an algebraically closed field, let $G$ be an affine algebraic group over $k$ and let $X$ be an affine $k$-variety. Then it's a basic fact that orbits $Gx$ are locally closed subsets of $X$. Let us now pass to schemes, i.e., $G$ is an affine group $k$-scheme acting on an affine $k$-scheme $X$. How are orbits defined in this setting? Are these simply the set-theoretic orbits? Is the orbit of $G$ in a closed point of $X$ the same as above? (I'm confused because I think in this case the orbit consists entirely of closed points which seems wrong). Is it a locally closed subset and thus canonically a subscheme of $X$? If so, what is the relation between $G(k)x$ and $(Gx)(k)$ for a closed point $x$? -All this should be well known but I couldn't find references. In Mumford's GIT there is a very general definition of orbits (as scheme-theoretic images of a morphism obtained by the action, see Definition 0.4) but I don't know what this gives in my special basic setting above. If someone can deduce it from there, this would be awesome. - -REPLY [10 votes]: Presumably you meant to assume the schemes are finite type over $k$. To work naturally with orbit questions for such schemes one just has to bring in appropriate use of flatness to adapt intuition and experience from the traditional smooth setting over algebraically closed fields (e.g., one uses the robust theory of quotients modulo possibly non-smooth closed subgroup schemes). -For the orbit of a $k$-point $x \in X$ everything is fine: one has the orbit morphism $G \rightarrow X$, and its constructible image is locally closed because this can be checked over $\overline{k}$ (where we can pass to underlying reduced schemes). Likewise, for such $x$ one has the (possibly not smooth) stabilizer scheme $G_x$ inside $G$ over $k$ and a well-defined monomorphism $j_x:G/G_x \rightarrow X$. This is a locally closed immersion when $G$ is smooth (so in such cases we denote its smooth locally closed image with reduced structure as $Gx$, onto which the orbit map from $G$ is faithfully flat). Indeed, by descent theory we can extend the ground field to be algebraically closed, and then if $Z$ is the locally closed orbit with reduced structure we see that the orbit map factors through $Z$ and as such $G \rightarrow Z$ must be generically flat and hence faithfully flat, whence $j_x$ is an isomorphism onto $Z$ by descent theory. -Mere surjectivity of an orbit map (without flatness) is not a good notion of "homogeneous space". -In general with $x \in X(k)$ there is an absolutely huge gulf between $(Gx)(k)$ and $G(k)x$ (e.g., think about a surjective homomorphism between smooth affine groups!). In effect, say assuming $G$ is smooth so that the orbit is $G/G_x$ as shown above, you're asking about the existence of $k$-points in fibers of $\pi:G \rightarrow G/G_x$ over $k$-points of the target. Since $\pi$ is a $G_x$-torsor for the fppf topology, the obstructions lie in the fppf cohomology set ${\rm{H}}^1(k, G_x)$ (almost by definition), and in favorable cases one can analyze that by using knowledge of the structure of $G_x$. -Note however that considering the "orbit" of a point that is not a $k$-point doesn't quite make sense and is asking for trouble unless one is absolutely clear about definitions (and usually it is not what you want to ever consider -- it is technically usually much better to simply extend the ground field to work with orbits through rational points, and to use descent theory to go back down if you wish). However, there is a kind of substitute. You could contemplate $G$-stable non-empty locally closed subschemes $Z$ of $X$ such that the $G(\overline{k})$-action on $Z(\overline{k})$ is transitive and the surjective orbit map $G_{\overline{k}} \rightarrow Z_{\overline{k}}$ through some (equivalently, any) $z \in Z(\overline{k})$ is flat or equivalently identifies $Z_{\overline{k}}$ with $G_{\overline{k}}/(G_{\overline{k}})_z$. (If $G$ is smooth and $Z$ is geometrically reduced then this is equivalent to transitivity of the action on $\overline{k}$-points, for reasons explained above.) In effect, such $Z$ could be regarded as (non-empty) homogeneous spaces for $G$ inside $X$, and as an "orbit" for $G$ through any of closed point of $Z$ (it is easily checked that such a $Z$ is uniquely determined by any single closed point $x$ that it contains, but using the notation $Gx$ for such a $Z$ is going to lead to mistakes if $x \not\in X(k)$, so don't do that). If $Z(k)$ is empty then one should tread carefully to call such $Z$ an "orbit"; better to just say "homogeneous space". -Of course, for this viewpoint of homogeneous spaces inside $X$ (possibly without $k$-points) to be useful, say assuming $G$ is smooth to avoid some confusion, you want to prove an existence property: for any closed point $x \in X$ is there such a (necessarily smooth) $Z$ through $x$, or in other words does the orbit of $G_{\overline{k}}$ through a $\overline{k}$-point over $x$ -descend to a locally closed subscheme of $X$? -The existence of such $Z$ fails most of the time when $G$ is not connected and $k$ is not algebraically closed. For example, let $G = \mu_n$ acting on $X = {\rm{GL}}_1$ by scaling with ${\rm{char}}(k) \nmid n$ and suppose $x \in X$ is a closed point such that $[k(x):k]$ does not divide $n$. In such cases, there is no way for $x$ to lie in a $k$-finite closed subscheme of $X$ of order dividing $n$, so there is no meaningful notion of $Gx$ over $k$. Even if $k$ is separably closed but not algebraically closed (say with characteristic $p > 0$) you're going to have problems when $G$ is finite etale since such a $Z$ would have to also be finite etale and hence cannot be found containing $x$ such that $k(x)$ is not separable over $k$ (so you get more counterexamples when $X = \mathbf{A}^1_k$ on which $G = \mathbf{Z}/(p)$ acts by translation, even when $[k(x):k] = p$ so one cannot use the more elementary degree obstruction). -In view of those basic counterexamples, it is an interesting exercise with Galois descent to prove that such a $Z$ does exist when $k(x)$ is separable over $k$ provided that the smooth $G$ is also connected. (The fun part is to figure out where you use connectedness.) If $k(x)$ is not separable over $k$ (and $G$ is smooth and connected) then this can fail. For instance, if $k$ is imperfect with characteristic $p$ and $t \in k - k^p$ then $G := \{v^p = u - tu^p\}$ is a smooth connected 1-dimensional $k$-subgroup of $\mathbf{G}_a^2$ and its Zariski closure $X$ in $\mathbf{P}^2_k$ has 1 point $x$ on the line at $\infty$, with $k(x) = k(t)$, so the $G$-action on $\mathbf{P}^2_k$ via $(u,v).[a,b,c] = [a + uc, b + vc, c]$ restricts to an action on $X$ in which $X - \{x\} = G$ with the translation action. There is no reasonable sense in which the $G$-stable complement $\{x\}$ of $X - \{x\}$ in $X$ can be regarded as a homogeneous space for $G$ since it is not $k$-smooth (and in particular, there is no $Z$ as above in this case).<|endoftext|> -TITLE: Who is attributed with the conjecture that every multiply-perfect number greater than $1$ is even? -QUESTION [5 upvotes]: I know that Descartes is considered to be the first to ask whether or not odd perfect numbers exist ($n$ such that $\sigma(n)=2n$, where $\sigma(n)$ is the sum of divisors of $n$), and he also discovered several multiply-perfect numbers ($n$ such that $\sigma(n)=kn$ for some integer $k\geq 2$). No odd multiply-perfect numbers greater than one are yet known, and I've heard it called a conjecture that they don't exist several times, but I haven't found an attribution to anyone in particular. Is there an agreed upon origin for the conjecture? -The context of my question (please, feel free to skip this): -I'm looking for someone to attribute in a paper I'm writing that makes a refinement of this form of conjecture for a generalization of the multiply-perfect numbers. Specifically, let $$\dfrac{\sigma(n)}n=\dfrac{k}m$$ where $\gcd(k,m)=1$ and $m$ is a practical number i.e. $m=1$, or $p_i^{a_i+1}\leq 1+\sigma(p_1^{a_1}p_2^{a_2}...p_i^{a_i})$ for every $i\in[1,\omega(m)]$, where $p_1^{a_1}p_2^{a_2}...p_{\omega(m)}^{a_{\omega(m)}}$ is the canonical prime factorization of $m$. I call these pad (practical abundancy denominator) numbers, and conjecture that every such number is practical (i.e. a practical number). $\dfrac{\sigma(n)}n$ is called the abundancy of $n$. -It is easy to see that every even perfect number is practical, and every practical number greater than $1$ is trivially even, so the conjecture that every perfect number is practical is equivalent to the conjecture that they're all even. From this perspective it's natural to ask whether every multiply-perfect number is practical. I've tested essentially every known multiply-perfect number using the factorizations from Achim Flammenkamp's database and all of them so far are practical, so already this seems like a promising lead. -Clearly the multiply-perfect numbers are the special case of the displayed equation where $m=1$, which is trivially practical, so every multiply-perfect number is a pad number. For an idea of their relative size, there are $6484$ pad numbers less than $10^6$, all of them practical, compared with only $10$ multiply-perfect numbers in the same range. An additional fact that makes this conjecture seem quite intuitive is that all small multiples of a practical number are also practical. Specifically, if $a$ is practical and $b\leq \sigma(a)+1$ then $ab$ is practical. Since, $k/m$ is just the abundancy of $n$ in lowest terms, there exists some positive integer $r$ such that $n/m=\sigma(n)/k=r$. -When $r=n$, the multiply perfect numbers, we have no idea why every term would be practical, and yet this is the special case I've tested the furthest, and no counterexamples have yet been found. On the other hand, if $r\leq 1+\sigma(m)$ then $n$ is obviously practical. While in retrospective it seems rather intuitive to me, this was a purely numerical observation. Please let me know if you have ideas about this problem, since I believe it has the potential to remain open for a long time, but someone to attribute is what I'm asking for, since right now I feel I have very few people to attribute in my paper. This is for my first paper, which I think I will try to have published in Experimental Mathematics, since computational experiments were the method by which I reached this generalization. If you have strong opinions about other papers or people who should be attributed then please mention them at least in the comments. I'm not looking to downplay anyone's contributions, and I realize the importance of showing how your work ties in with that of others. -Edit: Note that the same conjecture appears to hold with $\sigma(m)$ replaced by $\sigma^*(m)=\prod_{i=1}^{\omega(m)}(p_i^{a_i}+1)$, the sum of the unitary divisors of $n$, which might be simpler to work with. - -REPLY [5 votes]: The earliest reference seems to be from 1966: E.A. Bugulov, On the question of the existence of odd multiperfect numbers, Kabardino-Balkarskaya State University Učen. Zap. 30 (1966) 9-19. [I could not find this article online.] -Bugulov showed that an odd multiperfect number must have at least 11 distinct prime divisors, more precisely, odd $k$-perfect numbers contain at least $\omega$ distinct prime factors, where $(k,\omega) =$ (3, 11), (4,21), (5, 54),... This result is discussed and improved by Shigeru Nakamura, On k-perfect numbers, Journal of the Tokyo University of Mercantile Marine (Natural Sciences), 33 (1982) 43–50. [Listed here, but not available online.]<|endoftext|> -TITLE: Question about the Aganagic-Vafa A-brane -QUESTION [5 upvotes]: According to Aganagic-Vafa (hep-th/0012041) and Fang-Liu (arXiv:1103.0693), for a semi-projective toric Calabi-Yau 3-manifold $X$, the Aganagic-Vafa A-brane $L_{AV}\subset X$ is defined by the equations -$\sum_{i=1}^{k+3}l_i^1|X_i|^2=c_1$, $\sum_{i=1}^{k+3}l_i^2|X_i|^2=c_2$, $\sum_{i=1}^{k+3}\phi_i=c_3$ -In the above, $X_i=\rho_ie^{i\phi_i}$ denotes the coordinates on $\mathbb{C}^{k+3}$, and $X$ is defined as the quotient $X=\mu^{-1}(r_1,\cdot\cdot\cdot,r_k)/G_\mathbb{R}$. Here $G\cong(\mathbb{C}^\ast)^k$, and $\mu$ is induced by the Hamiltonian $G_\mathbb{R}$-action. Also $l_i^1,l_i^2\in\mathbb{Z}$, and we require that $\sum_{i=1}^{k+3}l_i^\alpha=0$ for $\alpha=1,2$. $c_i$ are fiexed constants. -In the special case when $X=\mathbb{C}^3$ and $G$ is trivial, clearly $k=0$ and we get the equations which characterize a Harvey-Lawson fiber. -I think generically ($c_3\neq0$ or $c_3=0$ but $c_1\cdot c_2\neq0$ and $c_1\neq c_2$), an Aganagic-Vafa A-brane is just a special Lagrangian fiber of the Harvey-Lawson fibration, so it should be diffeomorphic to $T^2\times\mathbb{R}$. But in both of the two papers mentioned above, the topology of $L_{AV}$ is taken to be $\mathbb{R}^2\times S^1$. Why? - -REPLY [6 votes]: I believe that the special Lagrangians that Aganagic-Vafa want to consider are contained in special fibres of the Harvey-Lawson fibration. First, we had better -take $k=0$ in your equations, since otherwise the three equations will give a subspace which is of too high dimension to be a Lagrangian. (But these equations do generalize to higher dimension if you take more equations of the form $\sum l_i^2|X_i|^2=c$.) Taking specifically -the Harvey-Lawson fibration $(X_1,X_2,X_3)\mapsto (|X_1|^2-|X_2|^2, |X_1|^2-|X_3|^2, -Im(X_1X_2X_3))$, it is not difficult to see that set of critical points of the fibration are given by the points where at least two of the coordinates $X_1,X_2,$ -and $X_3$ are zero. A fibre over a critical value (except for the fibre over zero) is actually a union of two manifolds, each homeomorphic to ${\mathbb R}^2\times S^1$. -To see this topologically, it is best to think of the fact that there is a $T^2$-action which acts fibrewise: this is given by $(\theta_1,\theta_2)$ -acts by $(X_1,X_2,X_3)\mapsto (\exp(i\theta_1)X_1,\exp(i\theta_2)X_2, -\exp(-i(\theta_1+\theta_2))X_3)$. A general fibre of the Harvey-Lawson fibration then has this $T^2$ acting freely, and the quotient is ${\mathbb R}$, with -coordinate given by $Re(X_1X_2X_3)$. However, the $T^2$-orbits whose dimension is -$<2$ are precisely the orbits of points where at least two of the coordinates -are $0$. From this one sees that the general singular fibre is obtained by taking -${\mathbb R}\times T^2$ and contracting $\{0\}\times T^2$ to $\{0\}\times S^1$. -This decomposes as a union of two copies of ${\mathbb R}^2\times S^1$. Aganagic-Vafa are using one of these two copies. There are a lot more details of this kind of construction in my paper "Examples of special Lagrangian fibrations," http://arxiv.org/abs/math/0012002.<|endoftext|> -TITLE: Spaces over which every vector bundle is a summand of the trivial bundle -QUESTION [6 upvotes]: Let X be a Hausdorff space such that every real vector bundle on X is summand of a trivial bundle. Does this imply that X is homotopy equivalent to a compact Hausdorf space? This question is a "compact version" of the following question; -Paracompactness and inner product on vector bundles - -REPLY [13 votes]: Let $X$ be the wedge of infinitely many circles (equipped with the CW topology). Every vector bundle $\xi$ over $X$ is a summand of a trivial bundle, namely it is $\xi\oplus\xi$ is trivial because any vector bundle over a circle has this property (alternatively, one could appeal to the fact that $X$ is homotopy equivalent to a smooth manifold, if the number of circles is countable, as explained in my comment above). -Suppose there is a compact space $K$ and a homotopy equivalence $f: K\to X$. The image $f(K)$ is compact, so $f(K)$ lies in a finite subcomplex $X_0$ of $X$, i.e. $X_0$ is a wedge of finitely many circles. Thus any loop on $X$ is freely homotopic to a loop in $X_0$, which is false (because a circle that forms the wedge -can be homotoped into $X_0$ only if it lies in $X_0$. -EDIT: A key feature of the above example is that there is a homotopic to the identity $X\to X$ whose image lies in a compact subset, and it allows for the following optimal generalization. -Let $X$ be a (say path-connected) space homotopy equivalent to a locally compact ANR (such as a manifold, or a locally finite CW complex) which we denote $\bar X$. Suppose $X$ is homotopy equivalent to a compact space. Then there is a homotopic to the identity map $\bar X\to \bar X$ whose image lies in a compact set. By a standard argument (see e.g. proposition 3.18 of http://arxiv.org/abs/1210.6741 -of Guilbault's survey), this is equivalent to assuming that $\bar X$ is finitely dominated (by a finite CW complex). Note that finitely dominated spaces have finitely generated homology groups, and finitely presented fundamental groups (see e.g proposition 3.16 in the above paper). Of course, the fundamental group of a wedge of infinitely many circles is not finitely presented. -Conversely, Ferry proved in "Homotopy, simple homotopy, and compacta" (Topology 9 (1980) pp 101-110) that any space that is dominated by a compact Hausdorff space is homotopy equivalent to a compact Hausdorff space. Thus we conclude that a locally compact ANR is finitely dominated if and only if it is homotopy equivalent to a compact Hausdorff space. -In particular, if $X$ is homotopy equivalent to a smooth manifold that is not finitely dominated, then $X$ is not homotopy equivalent to a compact Hausdorff space, even though any vector bundle over $X$ is a summand of the trivial bundle.<|endoftext|> -TITLE: What's wrong with the Courant nodal domain theorem? -QUESTION [9 upvotes]: The Courant nodal domain theorem (for Neumann boundary conditions) says that the $n$-th eigenfunction has at most $n$ nodal domains (connected components where the eigenfunction has the same sign. However, Chavel in Eigenvalues in Riemannian Geometry seems to point out (P23) that Divergence theorem is used, so the regularity of the nodal set matters, whose proof by Cheng in dimension $\ge 3$ is incomplete. My question is: does it really affect the proof of the nodal domain theorem? More generally, does it affect the application of divergence theorem to a nodal domain (say to functions that are restrictions of $C^{\infty}(\mathbb R^n)$. - -REPLY [8 votes]: Not an precise answer : -As far as I remember this is not really an issue. One has to replace by an other version of divergence theorem ( for less regular domain) using geometric measure theory . The key point I think is that the set where the nodal lines are not regular (often called the singular set $\{u=0\}\cap\{\nabla u=0\}$ is of dimension at most $n-2$ so it has measure 0 for the $H^{n-1}$ measure. -But this is quite specific to solutions of elliptic pde and I don't think that smoothness is enough to guarantee any kind of regularity of the nodal set. -A more precise answer : -Take a look (for example) at the proof of the nice Sogge-Zelditch formula which precisely uses this type of div theroem (formula (8) in the paper).<|endoftext|> -TITLE: Density of all n such that 2^n-1 is square free -QUESTION [13 upvotes]: Is it true that the set $$S:=\{n\in \mathbb N\ |\ 2^n-1 \ \mbox{ is square free}\}$$ has positive density? What can we say when we replace $2^n-1$ with $\frac{a^n-1}{a-1}$? - -REPLY [11 votes]: User43383's answer linking the convergence of the sum of the reciprocals of the order of $2 \pmod{p^2}$ to the existence of a positive density is very nice. Let me add a little to this by saying what the density should be heuristically (approximately $0.754$), and by obtaining upper bounds for this density unconditionally (approximately $0.761$). For an odd prime $p$ let $\omega(p)$ denote the order of $2 \pmod{p^2}$. If $k$ is a square-free odd integer -define $\omega(k)$ to be the l.c.m. of all the $\omega(p)$'s with $p|k$; thus $\omega(k)$ is the order of $2 \pmod{k^2}$. -Note that $k^2$ divides $2^n-1$ if and only if $\omega(k)$ divides $n$. Now note that $\omega(3)=6$, $\omega(5)=20$, $\omega(7)=21$, $\omega(11)=110$, $\omega(13)=156$ and so on. Thus if $n$ is divisible by $6$, or $20$ or $21$, or $110$, or $156$ etc then $2^n-1$ is not squarefree. The density of the set of multiples of $\omega(p)$ for $3\le p\le z$ say may be calculated by using inclusion-exclusion. Put $P(z) = \prod_{3\le p\le z} p$; then this density is -$$ -\sum_{d|P(z)} \frac{\mu(d)}{\omega(d)}. -$$ -It is not immediate that this density is at least -$$ -\prod_{3\le p\le z} \Big(1-\frac{1}{\omega(p)}\Big) -$$ -as indicated in User43383's answer, but this is true by an inequality of Heilbronn and Rohrbach (see Theorem 0.9 in Hall's book Sets of Multiples; Cambridge tract 118). In any case, computing this density for any $z$ would give an upper bound for the density of squarefree Mersenne numbers. Thus taking $z=31$ one gets an upper bound of $0.761645...$ for the density. -So far our argument is easy to make rigorous. Now for the heuristic part. The density computed above is a monotone decreasing function of $z$ clearly. Therefore it converges to some number, which will be strictly positive if $\sum 1/\omega(p)$ converges (thanks to the Heilbronn-Rohrbach inequality and the argument of User43383). One would expect that this limiting value is the right density. -Alternatively we can argue as follows: Using that $\sum_{k^2|n} \mu(k) = 1$ if $n$ is square-free and $0$ otherwise, we may -write -$$ -\sum_{n\le N} \mu(2^n-1)^2 = \sum_{k \text{ odd}} \mu(k) \sum_{n\le N, k^2|2^n-1} 1. -$$ -The inner sum over $n$ may be expected to be roughly $N/\omega(k)$. Omitting error terms, which are significant since the sum over $k$ is over a large range, we may expect an asymptotic of -$$ -N \sum_{k \text{ odd}} \frac{\mu(k)}{\omega(k)}. -$$ -In other words, the density should be -$$ -\sum_{k \text{ odd} } \frac{\mu(k)}{\omega(k)}, -$$ -which presumably converges, and presumably to the same answer obtained by considering $P(z)$ as before. -Numerics: There are only two known Wieferich primes $1093$ and $3511$ up to $4 \times 10^{12}$. I calculated the conjectured density above up to $10000$ and the probability seems to be around $0.754$ (the first two digits seem reliable and the third fluctuates a bit). I also found $7550$ square-free Mersenne numbers with $n$ up to $10000$ (to be precise the numbers I found here are highly likely to be squarefree -- they are not divisible by the squares of the known Wieferich primes, and also not by the squares of all non Wieferich primes -- but there is a possibility that some of them could be divisible by the square of some ginormous Wieferich prime); the fit seems very good. -Remarks: I also note that the convergence of the sum $\sum_p 1/\omega(p)$ has come up earlier in work of Granville and Soundararajan on an Erdos problem of writing odd numbers as a squarefree plus a power of $2$. See http://www.dms.umontreal.ca/~andrew/PDF/wieferich.pdf for a copy of that paper, especially page 9 there which shows under reasonable conjectures on Wieferich primes that $\sum_p 1/\omega(p)< 0.9091$. -Let me also refer to a somewhat related MO problem Probability of coprime polynomials -for which I discuss a heuristic similar to the one above. That problem may be viewed as a function field analog of asking what is the probability that $2^n-1$ and $3^m-1$ are coprime if $m$ and $n$ are chosen randomly. In problems of this type the expected densities do not appear to be products over primes as one may first think, since the divisibility conditions imposed by different primes are highly dependent.<|endoftext|> -TITLE: Atiyah-Singer for pseudodifferential operators via heat kernel? -QUESTION [13 upvotes]: The Atiyah-Singer theorem for Dirac-type operators can be proved using the heat kernel and this proof has an advantage over the proof via K-theory, because the first is local but the latter is not. But the K-theoretic proof has the advantage that it proves Atiyah-Singer not only for Dirac-type operators, but much more generally for elliptic, symmetric pseudodifferential operators (PDOs). - -I was wondering whether one could do the heat kernel proof also for pseudodifferential operators? - -I know that we can conclude Atiyah-Singer for PDOs from the theorem for Dirac operators by using Poincare duality (i.e., by using that on compact spin$^c$-manifolds every K-homology class can be represented by the class of a twisted Dirac operator). -But I'm asking if the proof via the heat kernel directly generalizes to PDOs? One problem that I can see is that it is often used that $e^{itD}$ has finite propagation speed, and I think that this does not hold anymore if D is merely pseudodifferential (right?). But maybe it is possible to work around that? Or what is the reason why I only encounter heat kernel proofs for Dirac type operators but not for PDOs? - -REPLY [16 votes]: The answer is no. Even for the differentil elliptic operator, the heat kernel method can not give the result. -In the heat method proof, we use McKean-Singer formula, -$$\mathrm{Ind} D= \mathrm{Tr} \left[e^{-tD^*D}-e^{-tDD^*}\right]=\int_M \mathrm{Tr} [p_t(x,x)-q_t(x,x)]dx,$$ -where $p_t(x,y), q_t(x,y)$ are the heat kernel. -This is always true, even for PDO. -McKean-Singer conjectured, and proved by a lot of people (Gilkey, Getzler, Bismut...) using different methods, for Dirac operator, when $t\to0$, -$$\mathrm{Tr} [p_t(x,x)-q_t(x,x)]\to a(x). \quad (1)$$ -Then we can deduce the Atiyah Singer index theorem -$$\mathrm{Ind} D=\int_X a(x)dx.$$ -When $t\to0$, the limit of $\mathrm{Tr} [p_t(x,x)],\mathrm{Tr} [q_t(x,x)]$ do not exist -but the limit of the difference exsit. This is a very deep result relating to the supersymmetry and the structure of Dirac operator. -Your question is to ask if $\mathrm{Tr} [p_t(x,x)-q_t(x,x)]$ has a limit when $t\to0$ for PDO. This is not true even for general elliptic oprator. -An example is that for the Kahler manifold $(M,\omega)$, $D=\sqrt{2}(\bar{\partial}+\bar{\partial}^*)$ is a Dirac operator. the limit in (1) exist. But when $M$ is not Kahler, the limit in (1) maybe does not exist. In Bismut's paper, A local index theorem for non Kähler manifolds http://link.springer.com/article/10.1007%2FBF01443359, -he find such limit exist if $\partial \overline{\partial}\omega=0 $, and this is almost a necessary condition. -Let's final remark that the finite propagation speed is not essential, just a trick.<|endoftext|> -TITLE: What properties do "large topoi" share with actual topoi? -QUESTION [14 upvotes]: Fix Grothendieck universes $\mathcal{U} \in \mathcal{V}$ and suppose that $C$ is a locally $\mathcal{U}$-small category which is $\mathcal{V}$-small. Denote by $Set$ the category of $\mathcal{U}$-small sets (which is also $\mathcal{V}$-small). Suppose that $\mathcal{E}$ is a category which can obtained as a left exact localization of $$Set^{C^{op}}.$$ $\mathcal{E}$ is not (usually) a Grothendieck topos in either universe, but in many ways, it behaves as if it is. It can fail to be locally presentable however. -My question: Exactly which characteristic properties of topoi (e.g. Giraud's axioms, being Cartesian closed...) hold for $\mathcal{E}$ and which do not? - -REPLY [12 votes]: If one is not comfortable with Grothendieck universes, then a good heuristic is to replace $\mathcal{U}$-small set by "finite set", and $\mathcal{V}$-small set by "set" (or even in some cases by "countable set"), try to answer analogical questions in this setting, and then transfer answers back to the original setting (this is possible because $\aleph_0$ behaves almost like a strongly inaccessible cardinal). -Here are some trivial observations for finite sets: - -$\mathbf{FinSet}^{\aleph_0}$ does not have finite generating family, -$\mathbf{FinSet}^{N}$, where $N$ is the free monoid $\langle 0, +1\rangle$ on a single generator $0$, is not cartesian closed, neither it has a subobject classifier -Every category of the form $\mathbf{FinSet}^{\mathbb{C}^{op}}$ has finite limits and colimits (constructed pointwise); therefore every left exact localization of $\mathbf{FinSet}^{\mathbb{C}^{op}}$ has finite colimits (the same is true for finite limits, but for limits this is not actually a useful heuristic).<|endoftext|> -TITLE: Does ZF have an initial model? -QUESTION [5 upvotes]: Definition 1: A class $\mathcal{K}$ of countable transitive models of $\text{ZF}$ has an "initial member" $M$ if each member of $\mathcal{K}$ is a forcing extension of $M$ for some partial order $\mathbb{P}\in M$ and some $\mathbb{P}$-generic $G$ over $M$. -Definition 2: An extension $T$ of $\text{ZF}$ has an "initial c.t.m" if the collection of all countable transitive models of $T$ has an "initial member". -Question 1: Assuming some consistency assumptions is it consistent that $\text{ZF}$ has an initial c.t.m? -Question 2: Assuming some consistency assumptions is it consistent that $\text{ZF}$ has a consistent extension like $T$ with an initial c.t.m? - -REPLY [7 votes]: Your question has a certain affinity with the concept of a solid bedrock model, which arises in the theory of set-theoretic geology. Namely, $W$ is bedrock for $V$ if $V$ is a forcing extension of $W$ and $W$ satisfies the ground axiom, meaning that it is not a forcing extension of any deeper ground, or in other words, that it is minimal among all the grounds of $V$. The model $W$ is a solid bedrock if $W$ is least among all the grounds of $V$. -Although assertions about whether there is a bedrock or whether there are grounds of a certain nature seem at first to be second-order assertions about $V$, because they quantify over the inner models $W$ that might be grounds, in fact these are all first- order expressible in the language of set theory. The reason is that the collection of grounds of $V$ is uniformly definable, in that there is a definable family of classes $W_r$ such that every $W_r$ is a ground of $V$; every ground of $V$ is $W_r$ for some $r$; and the relation $x\in W_r$ is definable in $x$ and $r$. Thus, one may quantify over the collection of grounds by quantifying over the parameter $r$ used in this definition. -In his dissertation, Jonas Reitz proved that there are bottomless models $V$, which have no bedrock models; that is, a bottomless model $V$ can be realized as a set-forcing extension $V=W[G]$ of a ground $W$, but one can always go deeper, and realize $W=W_0[G_0]$ as a forcing extension of a still deeper ground, with no bottom.<|endoftext|> -TITLE: (Fri)end(l)y way to express Kan liftings -QUESTION [8 upvotes]: I would like to have an explicit description of the left/right Kan lift of a functor $F$ through $G$, $\text{Lift}_GF$/$\text{Rift}_GF$ in terms of coends/ends (this can be done for Kan extensions, so I hardly believe there is no way to dualize the argument). -Google is of little help, a part from a couple of hints on how to define $\text{Rift}$ of a profunctor through another. -Can you help me? Feel free to suppose the completeness/cocompleteness you need to define the right end/coend. - -REPLY [10 votes]: Such a characterisation is not possible in $\mathbf{Cat}$ for trivial reason --- the dual of a functor generally cannot be thought of as a functor. In an analogical situation in $\mathbf{Set}$, for exactly the same reason, we do not expect that every colimit can be expressed in a canonical way as a limit. -For example, let us write $|2|$ for the discrete category consisting of two objects: $0$ and $1$, and let $\delta_0 \colon \mathbb{C} \rightarrow |2|$, $\delta_1 \colon \mathbb{D} \rightarrow |2|$ be the obvious constant functors (with disjoint images) from non-trivial categories $\mathbb{C}, \mathbb{D}$. Then there are no liftings of $\delta_0$ through $\delta_1$, no matter how complete and cocomplete $\mathbb{C}$ and $\mathbb{D}$ are. In fact, for any functor $F \colon \mathbb{C} \rightarrow \mathbb{D}$, there are no natural transformations $\delta_1 \circ F \rightarrow \delta_0$ nor $\delta_0 \rightarrow \delta_1 \circ F$. The same is true, if we substitute category $|2|$ with its free cocompletion $\mathbf{Set}^{|2|^{op}}$. -Notice also, that the formula for Kan liftings of profunctors, defines liftings internal to the bicategory of profunctors, and not internal to the 2-category of categories. - -Let me recall that a Kan lifting in $\mathbf{Cat}$ is defined as a Kan extension in $\mathbf{Cat}^{op}$. The deep reason, why we do not have nice formulae for Kan liftings, is that $\mathbf{Cat}^{op}$, from the perspective of our covariant world, is not as nice as $\mathbf{Cat}$. Specifically, there is no way to define the concept of (co)ends in $\mathbf{Cat}^{op}$, because the internal calculi of $\mathbf{Cat}^{op}$ is not expressible enough ($\mathbf{Cat}^{op}$ does not have the canonical Yoneda structure). -Saying that, it is good to know, that in most situations we care about absolute Kan liftings, which may be expressed in the usual hom-set manner (these formulae can be further generalised inside any 2-category): -$$\hom(F(-), G(=)) \approx \hom(-, \mathit{Rift}_F(G)(=))$$ -and: -$$\hom(F(-), G(=)) \approx \hom(\mathit{Lift}_G(F)(-), =)$$<|endoftext|> -TITLE: A generalization of van der Waerden's conjecture -QUESTION [7 upvotes]: I am wondering if the following generalization of van der Waerden's conjecture is true. -Suppose A is an n x n non-negative matrix with all column sums equal to 1, and the sum of row i equal to $T_i$. Then $per(A) \geq T_1\ldots T_n \frac{n!}{n^n}$. This obviously implies van der Waerden's conjecture. I can check it by hand for 2x2 matrices, and I did not have the patience to try larger examples (so it may be false for some easy example). I couldn't modify Gurvits's proof to work either. - -REPLY [13 votes]: The conjecture is false. Here is a counterexample. -\begin{equation*} - A = \begin{bmatrix} - \tfrac18 & \tfrac4{15} & \tfrac1{10}\\ - \tfrac18 & \tfrac4{15} & \tfrac1{10}\\ - \tfrac68 & \tfrac7{15} & \tfrac8{10} - \end{bmatrix}. -\end{equation*} -For this matrix, $\text{per}(A)=\frac{21}{200}=0.105$. The row sums $T_1,\ldots,T_3=\left\{\frac{59}{120},\frac{59}{120},\frac{121}{60}\right\}$, so that $T_1T_2T_3 n!/n^n = \frac{421201}{3888000} \approx 0.108334...$, which is greater than $\frac{21}{200} = 0.105$.<|endoftext|> -TITLE: When are all greater cardinals sharply greater? -QUESTION [6 upvotes]: Makkai and Paré introduced the following binary relation on regular cardinals: given $\kappa$ and $\lambda$, $\kappa \vartriangleleft \lambda$ (read, $\kappa$ is sharply less than $\lambda$) when $\kappa < \lambda$ and, for every set $X$ of cardinality $< \lambda$, the set $P_\kappa (X)$ of all subsets of $X$ of cardinality $< \kappa$ has a cofinal subset of cardinality $< \lambda$. -It is not hard to see that $\kappa \vartriangleleft \kappa^+$ for all regular cardinals $\kappa$. On the other hand, $\aleph_1$ is not sharply less than $\aleph_{\omega + 1}$, so $\vartriangleleft$ is not the same as $<$. Nonetheless, it is true that $\aleph_0 \vartriangleleft \lambda$ for every uncountable regular cardinal $\lambda$, simply because $P_{\aleph_0} (X)$ has the same cardinality as $X$ when $X$ is infinite. More generally, if for all (not necessarily regular) cardinals $\kappa' < \kappa$ and all cardinals $\lambda' < \lambda$, we have ${\lambda'}^{\kappa'} < \lambda$, then $\kappa \vartriangleleft \lambda$. In particular if $\lambda$ is an inaccessible cardinal then $\kappa \vartriangleleft \lambda$ for all regular cardinals $\kappa < \lambda$. -Question. Do there exist uncountable regular cardinals $\kappa$ such that $\kappa \vartriangleleft \lambda$ if and only if $\kappa < \lambda$? Is there a proper class of them? - -REPLY [4 votes]: No. -Given regular cardinals $\kappa$ and $\lambda$, $\kappa \vartriangleleft \lambda$ -iff $\kappa < \lambda$ and -$$ -\operatorname{cf} ([ \mu ]^{< \kappa} , \subseteq) = -\operatorname{cov} (\mu , \kappa , \kappa) < \lambda -$$ -for every cardinal $\mu$ with $\kappa \leq \mu < \lambda$. -Let $\kappa$ be any regular cardinal > $\aleph_0$. Then, let $\delta$ be the ordinal such that $\kappa = \aleph_{\delta}$. -Define $\lambda = \aleph_{\delta + \omega + 1}$. -Now, -$$ -\operatorname{cov} (\aleph_{\delta + \omega} , \kappa , \kappa) \geq -$$ -$$ -\operatorname{cov} (\aleph_{\delta + \omega} , \aleph_{\delta + \omega} , \aleph_1) = -$$ -$$ -\operatorname{cov} (\aleph_{\delta + \omega} , \aleph_{\delta + \omega} , {(\operatorname{cf} (\aleph_{\delta + \omega}))}^+) \geq -$$ -$$ -\operatorname{cov} (\aleph_{\delta + \omega} , \aleph_{\delta + \omega} , {(\operatorname{cf} (\aleph_{\delta + \omega}))}^+ , \operatorname{cf} (\aleph_{\delta + \omega})) \geq -$$ -$$ -\aleph_{\delta + \omega + 1} = \lambda , -$$ -the last inequality by Fact 1 in - -Andreas Liu, Bounds for covering numbers, - The Journal of Symbolic Logic 71 (2006), 1303-1310. - doi:10.2178/jsl/1164060456 - -Hence, $\kappa < \lambda$ but -$\neg (\kappa \vartriangleleft \lambda)$.<|endoftext|> -TITLE: How to show that the chromatic number > aleph_0 -QUESTION [7 upvotes]: Let $V =\{ f | \exists _{\alpha<\omega_1} ( f:\alpha \rightarrow \mathbb{N} \wedge f $ is $ 1-1) \}$. We define $E\subseteq [V]^2 $, such that $\forall_{f,g\in V } (\in E \longleftrightarrow ( f\subseteq g \vee g\subseteq f))$. We need to show that the chromatic number of $G=(V,E)$ is bigger than $\aleph_ 0$. It is obvious why it cannot be $<\aleph_0$, but I'm having trouble showing that it cant be $=\aleph_0$. Could anyone help? - -REPLY [6 votes]: There is a beautiful argument going back to, I think, Galvin. -Assume that $F:V\to\omega$ is a good coloring. By transfinite recursion define $f_\alpha:\alpha\to\omega$ as follows. $f_0=\emptyset$. If $\alpha$ is limit, set $f_\alpha=\bigcup\{f_\beta:\beta<\alpha\}$. If $f_\alpha$ is given, let $f_{\alpha+1}$ be that extension of it to $\alpha+1$ for which $f_{\alpha+1}(\alpha)=F(f_\alpha)$ holds. Now one can show that each $f_\alpha$ is in $V$ and $f_{\beta}\subseteq f_\alpha$ for $\beta<\alpha$. But then $\bigcup\{f_\alpha:\alpha<\omega_1\}$ would be an injection $\omega_1\to\omega$, an impossibility.<|endoftext|> -TITLE: Elementary proof for Hilbert's irreducibility theorem -QUESTION [14 upvotes]: I have tried to find a complete proof for Hilbert's irreducibility theorem, but everything I found was way beyond my level of understanding. -I am only interested in the simple case where the polynomial is in two variables over the rationals. Specifically, if $f\in \mathbb{Q}[T,X]$ be an irreducible polynomial, then there exist infinitely many $t_j\in\mathbb{Q}$ such that $f(t_j,X)\in\mathbb{Q}[X]$ is irreducible. -Is there a way to prove this using mostly elementary results? - -REPLY [3 votes]: Kaltofen's 1985 proof (Wayback machine) seems completely elementary and effective. -E. Kaltofen. Polynomial-time reductions from multivariate to bi- and univariate integral polynomial factorization. SIAM J. Comput., 14(2):469-489, 1985; DOI: 10.1137/0214035.<|endoftext|> -TITLE: Which spaces have the (weak) homotopy type of compact Hausdorff spaces? -QUESTION [23 upvotes]: Inspired by the discussion in the comments of this question, I'd like to ask the following question: is it possible to characterize the class of spaces that are homotopy equivalent (or weak equivalent) to compact Hausdorff spaces? As noted in the linked question's comments, no locally connected space with infinitely many components can be homotopy equivalent to a compact Hausdorff space. Are there any other restrictions? Is every path-connected space homotopy equivalent to a compact Hausdorff space? It seems plausible to me that every space might at least be weak equivalent to a compact Hausdorff space: perhaps the topology on an infinite CW-complex can be coarsened to be compact Hausdorff without changing the weak homotopy type. -Update: I've accepted Jeremy Rickard's answer, as it seems to more or less completely answer the case of weak equivalence (amazingly, every space is weak equivalent to a compact Hausdorff space iff there does not exist a measurable cardinal). The comments indicate that spaces having the (strong) homotopy type of compact Hausdorff spaces are much more restricted; I'd still welcome answers elaborating further on this. - -REPLY [15 votes]: Expanding on my comment, if there are measurable cardinals then it follows from the results of -A. Przeździecki, Measurable cardinals and fundamental groups of compact spaces. -Fund. Math. 192 (2006), no. 1, 87–92. -that there are spaces not weakly equivalent to any compact Hausdorff space, as Przeździecki proves that (if and only if there is a measurable cardinal) there are groups $G$ that are not the fundamental group of any compact Hausdorff space, and so the classifying space of such a group is a counterexample. -He also proves that every group of non-measurable cardinality is the fundamental group of a path-connected compact space, answering a question of Keesling and Rudyak who had earlier proved this with "connected" in place of "path-connected" in -J.E. Keesling and Y.B. Rudyak, On fundamental groups of compact Hausdorff spaces. -Proc. Amer. Math. Soc. 135 (2007), no. 8, 2629–2631.<|endoftext|> -TITLE: Choquet theory and Hilbert's fourth problem -QUESTION [13 upvotes]: The following text is an attempt to see Hilbert's fourth problem in a new light. -Definition. A pseudometric $d$ on $\mathbb{R}^n$ is called projective if whenever a point $z$ belongs to a line segment $xy$, it follows that $d(x,z)=d(x,y) + d(y,z)$. In other words, a pseudometric is projective if line segments are geodesic segments. -It is immediate (and important) that the set of all projective pseudometrics in - $\mathbb{R}^n$ is a convex cone. This suggest the following problem: -Problem. Construct all projective pseudometrics $d$ that are indecomposable in the sense that if $d$ is a positive linear combination of projective pseudometrics $d_1$ and $d_2$, then both of these metrics are multiples of $d$. -Examples of indecomposable projective pseudometrics. Given a closed half-space $H \subset \mathbb{R}^n$, the pseudometric $\delta_H(x,y)$ that takes the value $1$ when exactly one of the two points lies on $H$ and takes the value $0$ in any other case is an indecomposable projective pseudometric. -Question. Are there other examples when $n = 2$? Find other examples for -$n > 2$. -I suspect there are no other indecomposable projective pseudometrics in two dimensions, but I know there have to be some (and more than just some) other -such pseudometrics in dimension greater than two. To explain why, I have to motivate the question. -Motivation. Hilbert's fourth problem is somewhat vaguely posed as "find all geometries in which the straight line is the shortest connection between two points". In Wikipedia the problem appears with the note "too vague to be stated resolved or not." In this question I'm trying to get my thoughts straight on a way in which the problem can be stated not only clearly, but also insightfully through Choquet's (integral) representation theory. -Consider the vector space $V_n$ of locally bounded real-valued functions on $\mathbb{R}^n \times \mathbb{R}^n$ that are identically zero on the diagonal. The set of all pseudometrics in -$\mathbb{R}^n$ is a convex cone in $V_n$ as is the set of all projective pseudometrics in $\mathbb{R}^n$. I shall call this last cone ${\cal H}_n$ and think of it as the set of weak solutions of Hilbert's fourth problem. The space $V_n$ can be topologized in a number of ways to make it into a locally convex topologically vector space in which ${\cal H}_n$ is a closed convex cone. For example, If $k$ is a positive integer and $\epsilon$ is a positive real number, we can consider the following system of neighborhoods of the origin: -$$ -U_{k,\epsilon} := \{d \in V_n: d(x,y) < \epsilon \text{ if $x$ and $y$ belong to the Euclidean ball of radius k} \}. -$$ -However, I'm not convinced this will be the good topology for what I want at the end (namely, the application of Choquet's theory to Hilbert's fourth problem). -In any case, the problem that was stated at the beginning of this questions is simply: describe the set of extreme rays of the cone ${\cal H}_n$. -This question, while interesting by itself, becomes much more interesting if one could set things up so that the following vaguely-stated representation theorem holds: for every projective pseudometric $d$, there exists a probability measure $\mu$ supported on the set of extreme rays so that $d$ is the barycenter of $\mu$. -Why should something like this be true? Well, faith and optimism are unexplainable phenomena, but the Busemann-Pogorelov-Alexander-Ambartzumian solution of Hilbert's fourth problem in two dimensions can be re-interpreted as follows: -Theorem. If $d$ is a continuous projective metric on the plane, there exists a Borel measure $\mu$ defined on the space of closed half-planes such that -$$ -d(x,y) = \int \delta_H(x,y) \, d\mu(H) . -$$ -In other words, there is a Choquet representation for continuous projective metrics on the plane. -In higher dimensions, we can construct all sufficiently smooth projective metrics in the same way, but we need to use signed measures. I guess then that there must be other indecomposable projective pseudometrics besides the $\delta_H$. -Apology and conclusion. I'm sorry about this horribly long attempt to get my ideas straight the upshot of which is: if every projective pseudometric in $\mathbb{R}^n$ can be written as the barycenter of some probability measure supported on the set indecomposable projective pseudometrics, then possibly more precise interpretation of Hilbert's fourth problem would be to construct all indecomposable projective pseudometrics in $\mathbb{R}^n$. - -REPLY [3 votes]: It seems to me that the set of projective semi-distances that you describe in the example (let's call them of type I) can be slightly generalized the following way. In the definition of $d_H$, let $H$ be any open half-plane, plus an open half-line contained in its boundary. In other words, up to an affinity, these other distances (say of type II) are defined by $d(x,y)=0$ or $1$ according whether $x \mathbf{\ge} y$ or not, where $\ge$ denotes the lexicographic order in $\mathbb{R}^2$ (we may also parametrize all of them by the unit tangent bundle of $\mathbb{R}^2$). -In contrast, the distance of type I are analogously related to the order $x_1\ge y_1$. These two classes cover all $\{0,1\}$-valued projective semi-distances, and are all indecomposable (even in a slightly more general sense). -Indeed, given a semi-distance $d$ we may consider the equivalence relation $x\sim y\ \mathrm{iff}\ d(x,y)=0$, whose classes are the closures of points. (Let me anticipate a remark here: if $d$ is a semi-distance that is topologically stronger than another semi-distance $d'$, then the partition into closures of points induced by $d $ refines the partition induced by $d'$). -If $d$ is projective, the closure of any point is a convex set (a convex combination of a pair of points at zero distance, must be also at zero distance from them). -If $d$ is a $\{0,1\}$-valued projective semi-distance, then the set of points at distance $1$ from any given point $p$ is also a convex set: for, if $d(p,x)=d(p,y)=1$ and $z:=tx+(1-t)y$ for some $0\le t\le 1$, then since $d(x,z)+d(z,y)=d(x,y)\le1$, at least one among $d(x,z)$ and $d(y,z)$ is zero, so that in any case $ d(p,z)= 1$. -But a convex set of $\mathbb{R^2}$ whose complement is also convex is necessarily either an open half-plane, or an closed half plane, or an open hals-plane plus an (either open or closed) half-line contained in its boundary. As a consequence, the partition induced by a (non-zero) $\{0,1\}$-valued projective semi-distance $d$ necessarily consists of two classes, and therefore is of the above types I or II. -A $\{0,1\}$-valued projective semi-distance $d$ that is topologically stronger than a non-zero semi-distance $d'$ refines the partition of the $d'$-closures of points. But this easily implies that $d'$ induces the same partition, and that it can only take two values. Thus it is a positive multiple of $d$. In particular, if $d=d_1+d_2$, both $d_1$ and $d_2$ are multiples of $d$, (even if we do not assume they are projective).<|endoftext|> -TITLE: symplectic structure of tangent bundle of $\mathbb{S}^{n-1}$ -QUESTION [5 upvotes]: It is well known that $T\mathbb{S}^{n-1}$ is diffeomorphic to $M= f^{-1}(1)$ where -$f:\mathbb{C}^n\rightarrow \mathbb{C}$ is $f(z):=\sum_{i=1}^{n} z_{i}^{2}$. -Two questions: -1) Is $M$ a symplectic submanifold of $\mathbb{C}^n\sim \mathbb{R}^{2n}$ (with the standard symplectic structure)? -If the answer is affirmative, we can consider two symplectic structures for $T\mathbb{S}^{n-1}$. The first is the original structure of the tangent or cotangent bundle, the second one is the pull back structure of $M$. -2) Are these structures equivalent? - -REPLY [10 votes]: Here is a formula for an explicit symplectomorphism $F$ from $T^*S^{n-1}$ to the affine quadric $\{\sum z_{j}^{2}=1\}$ in $\mathbb{C}^n$: -$$ -F(p,q) = \left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{1/2} q - i\left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{-1/2}p -$$ -Here I view $T^*S^{n-1}$ as consisting of pairs $(p,q)\in \mathbb{R}^n\times\mathbb{R}^n$ with $p\cdot q=0$ and $|q|=1$. To check that $F$ is a symplectomorphism one can just check that $F$ pulls back the one-form $\alpha=-\sum y_jdx_j$ on $\mathbb{C}^n$ (which is a primitive for the standard symplectic form) to the canonical one-form on $T^*S^{n-1}$. The coefficients involving $\sqrt{1+4|p|^2}$ are designed to cause the map to have image in the quadric. -To indicate how I came up with this formula, the reasoning was that a suitably natural symplectomorphism ought to be equivariant with respect to the natural Hamiltonian $O(n)$-actions on both spaces. The easiest way to do this seems to be to take $F(p,q)=f(|p|)q-ig(|p|)p$ for some real functions $f$ and $g$. For the image to be in the quadric one needs $$f(|p|)^2- |p|^2g(|p|)^2=1.$$ Meanwhile if this is to be a symplectomorphism it should pull back the moment map for the $O(n)$-action on $\mathbb{C}^n$ to the moment map for the $O(n)$-action on $T^*S^{n-1}$. The norms of these moment maps are, respectively $(x+iy)\mapsto |x||y|$ and $(p,q)\mapsto |p|$, giving an equation $|f||g||p|=|p|$, i.e., $$|g|=\frac{1}{|f|}.$$ Solving these for $f$ and $g$ yields the formula at the top, which can then be directly confirmed to have the required properties.<|endoftext|> -TITLE: Weakly amenability and exactness for discrete groups -QUESTION [6 upvotes]: A countable discrete group $\Gamma$ is said to be weakly amenable with Cowling-Haagerup constant 1 if there exists a sequence of finitely supported functions $(\phi_n)$ on $\Gamma$ such that $\phi_n\rightarrow 1$ pointwise and $\sup_n ||\phi_n||_{cb}\leq 1$, where $||\phi||_{cb}$ denotes the (completely bounded) norm of the Schur multiplier on $B(l_2\Gamma)$ associated with $(x,y)\mapsto \phi(x^{-1}y)$. -A countable discrete group $\Gamma$ is exact if there exists a sequence of positive-definite kernels $\phi_n:\Gamma\times \Gamma\rightarrow \mathbb{C}$ with the following two properties: -1 For every finite set $F\subseteq \Gamma$ and every $\epsilon>0$ there is an $N$ such that $g_1^{-1}g_2\in F\Longrightarrow |\phi_n(g_1,g_2)-1|< \epsilon, \forall n>N$. -2 For every $n$ there is a finite set $F\subseteq \Gamma$ such that -$\phi_n(g_1,g_2)\neq 0\Longrightarrow g_1^{-1}g_2\in F$. -It is well-known that for discrete groups weakly amenability with Cowling-Haagerup constant 1 implies exactness. (More generally, AP implies exactness). -Is there a direct proof of this result without using the fact that the reduced group C*-algebra $C^*_r(\Gamma)$ is exact as a C*-algebra implies that $\Gamma$ is exact as a group? - -REPLY [2 votes]: See Theorem B/Theorem 3.4. in "Exactness of locally compact groups" http://arxiv.org/abs/1603.01829<|endoftext|> -TITLE: Tverberg's theorem in CAT(0) spaces -QUESTION [12 upvotes]: Does Tverberg's theorem hold for CAT(0) spaces of covering dimension $d<\infty$: -Is it true that for any $d$-dimensional $CAT(0)$-space $X$ and a subset $E\subset X$ of cardinality $(d + 1)(r - 1) + 1$, there exists a point $x\in X$ and a partition of $E$ into $r$ subsets $E_1,...,E_r$, such that $x$ belongs to the intersection of closed convex hulls of the subsets $E_i$? - -REPLY [14 votes]: No. Let $X$ be a tripod (three segments with one common endpoint), $d=1$, $r=2$ and $E$ the set of the 3 leaf points.<|endoftext|> -TITLE: What prefix and factors determine a ultimately periodic word uniquely -QUESTION [5 upvotes]: Let $\xi$ be an ultimately periodic sequence, i.e. there exists finite sequences $p, q \in X^*$ such that $\xi = pq^{\omega}$. Does there exists a $n > 0$ such that the prefix of length $n$ and all infixes of length $n$ determine $\xi$ uniquely, meaning it is the only word with this prefix and infixes. -Some context: I am currently searching for conditions under which a prefix and a set of infixes (or factors) determine a word uniquely. I guess this problem should be solvable for ultimately periodic words. If $\xi$ is just periodic, i.e. $\xi= q^{\omega}$ then this is easy, because choose $q$ minimal, meaning such that it is not a power of some other finite word (such words are called primitive), then the factors of length $n := |q|$ determine $\xi$ unique. For suppose $\eta$ is another word with the same prefix and factors of length $n$, then $\eta = q\tau$, now look at the word $\eta[2...n+1] = q[2...n] x$. Because $\eta$ has the same factors as $\xi$, and all factors of $\xi$ are conjugates (i.e. cyclic permutations) of $q$, there must be $i$ such that $q[2...n] x = q[i+1...n] q[1...i]$. As $q$ was choosen primitive it must have exactly $n$ different conjugate words (this is a well known fact about primitive words) so that $i = 1$ follows (otherwise there would be fewer then $n$ conjugates) which implies $x = q[1..1] = q_1$. Proceeding inductively in this way we see that $\eta = q^{\omega}$. But I am unable to extends this to ultimately periodic sequences, because the prefix $p$ could be anything. So any suggestions or help? -Some remarks on notation: If $w$ is a finite sequence, by $w[i...j]$ I denote the subsequence from the $i$-th up to the $j$ position $w[i...j] = w_i w_{i+1} \cdots w_{j}$. - -REPLY [2 votes]: This is a more detailed version of Alessandro's comment. By a classical theorem of Morse-Hedlund, one has that an infinite word $\eta$ is ultimately periodic iff there exists $m\geq 0$ so that $\eta$ has the same number of factors of length $m$ and $m+1$. Moreover, if $M$ is the number of subwords of length $m$, then the period of $\eta$ is bounded by $M$. -Given $\xi=pq^{\omega}$ with $q$ primitive, we can compute $m$. I guess $m=|p|+|q|+1$ works (probably I don't need the $+1$). Also one can prove $M\leq |p|+|q|$. (This can be found in Combinatorics, Automata and Number Theory handbook edited by Berthe and Rigo.) Now surely $n=2|p|+2|q|+1$ is more than safe enough. If $\eta$ has the same factors and prefixes of length $n$ as $\xi$, then it will have the same number of factors of length $m$ and $m+1$ so be ultimately periodic with period bounded by the same $M=|p|+|q|$. Hopefully you can work out the details from here.<|endoftext|> -TITLE: A subgroup intersects conjugacy class of every prime power order element -QUESTION [11 upvotes]: Let $G$ be a finite group and $H$ be a subgroup of $G$. Suppose that for any prime power order element $x$ of $G$, there exists some element $g$ in $G$ such that $x^g$ is contained in $H$. Does it follow that $H=G$? - -REPLY [10 votes]: As Geoff Robinson pointed out in his comment, the answer is yes by a deep and difficult theorem in finite groups. It was proven in Fein; Kantor; Schacher: Relative Brauer groups. II., J. Reine Angew. Math. 328 (1981), 39–57. As far as I know, no simpler proof has been found yet (maybe Theo Johnson-Freyd has a better one ...?).<|endoftext|> -TITLE: Which statements and arguments of Hovey's "Model categories" fail without functorial factorizations of morphisms? -QUESTION [10 upvotes]: I would like to study the homotopy theory of the category of pro-objects over a proper model category $M$. $Pro-M$ is endowed with the strict model structure; it seems that functorial functorizations of morphisms in $M$ do not extend to ones for $Pro-M$. - My question is: which arguments and statements of Hovey's book cannot be applied in the absense of functorial factorizations? In particular, can one apply the dual to the argument used in the proof of Theorem 7.3.1 in Hovey? Though $Pro-M$ is not fibrantly generated, it seems that one can replace Hovey's (co)small object argument here with the one provided by Theorem 6.1 of B. Chorny's paper "A generalization of Quillen’s small object argument"; yet this is not quite clear from www.math.tamu.edu/~plfilho/wk-seminar/Hovey_book.ps p. 186‎ -(probably this version of Hovey's argument requires certain correction). -Also, is there an easier way to prove that $Ho(M)$ cogenerates $Ho(Pro-M)$? -Upd. I am deeply grateful to Tim Porter; it seems that the preprint http://arxiv.org/abs/1305.4607 solves all my current problems with non-functorial factorizations. Yet comments concerning the general question could also be very interesting. - -REPLY [5 votes]: Have a look at: Functorial Factorizations in Pro Categories by Ilan Barnea Tomer M. Schlank (ArXiv: http://arxiv.org/abs/1305.4607). This will in part answer your question. It is worth stating that Pro - M only fails to be fibrantly generated because there are too many generating fibrations, and Chorny has worked with another form of fibrant generation see his paper with Rosicky: Source: Homology Homotopy Appl. Volume 14, Number 1 (2012), 263-280.<|endoftext|> -TITLE: Can any suspension spectrum be realized as Waldhausen K-theory? -QUESTION [7 upvotes]: If we consider the category of finite, pointed sets and declare cofibrations to be inclusions and weak equivalences to be bijections, we get a Waldhausen category whose $K$-theory spectrum is the sphere spectrum. -Given a pointed space $X$, is there a nice description (analogous to the one above) of a Waldhausen category whose $K$-theory spectrum is the suspension spectrum of $X$? -My first guess is something like finite, pointed sets equipped with a pointed map to $X$, but I don't really know enough about $K$-theory to prove that this guess is correct (if it even is....) - -REPLY [5 votes]: (1) Given a simplicial monoid $G$ let $R^0(*, G)$ be the Waldhausen category of pointed finite free $G$-simplicial sets weakly equivalent to $(\coprod^k G)_+$, for varying $k\ge0$. This is the special case $n=0$ of the notation from sections 2.1 and 2.2 of Waldhausen's ''Algebraic $K$-theory of spaces''. He obtains a homotopy equivalence -$$ -|hR^0_k(*, G)| \simeq BH^0_k(G) = B(\Sigma_k \ltimes |G|^k) , -$$ -so by Segal's theorem -$$ -\Omega |h N_\bullet R^0(*, G)| \simeq Z \times colim_k |hR^0_k(*, G)|^+ -$$ -and the Barratt-Priddy-Quillen-Segal theorem -$$ -Z \times colim_k B(\Sigma_k \ltimes |G|^k)^+ \simeq Q(B|G|_+) -$$ -you know that the ``direct sum'' $K$-theory $\Omega |h N_\bullet R^0(*, G)|$ of $R^0(*, G)$ is a model for $Q(B|G|)_+)$. I think the natural map -$$ -|hN_\bullet R^0(*, G)| \to |hS_\bullet R^0(*, G)| -$$ -is an equivalence, since all cofibrations are split in this case, but I don't have a reference at hand. -For a space $X$, let $R^0(X)$ be the Waldhausen category of finite retractive spaces over $X$, homotopy equivalent to $X$ disjoint union finitely many points (or $0$-cells, if you like). If $X \simeq B|G|$, there is a homotopy equivalence -$$ -|hS_\bullet R^0(X)| \simeq |hS_\bullet R^0(*, G)| , -$$ -so $R^0(X)$ should do the trick for you, also if $X$ is not connected. -(2) For simplicial sets $X \colon [q] \mapsto X_q$ there is a different construction, in an unpublished preprint of Igusa-Waldhausen (my copy is from 1991), of a simplicial Waldhausen category $C_0(X)$ that in degree $q$ is given by a non-obvious Waldhausen category of finite sets over $X_q$. Their Corollary 1.5 is an equivalence -$$ -\Omega |iS_\bullet C_0(X)| \simeq Q(|X|_+) . -$$<|endoftext|> -TITLE: Two questions from combinatorics on words -QUESTION [15 upvotes]: Question 1. Assume that an infinite word $u\in\{0,1\}^{\mathbb Z}$ is not balanced. Is it true that there exists a finite 0-1 word $w$ such that $0w01w1$ or $1w10w0$ is a factor of $u$? Is it true that both are? (Perhaps, with different $w$.) - answered in the negative -Question 1 (modified). Assume that an infinite word $u\in\{0,1\}^{\mathbb Z}$ is not balanced. Is it true that there exists a finite (possibly, empty) 0-1 word $w=w_1\dots w_k$ such that $0w_1\dots w_k01w_k\dots w_11$ or $1w_1\dots w_k10w_k\dots w_10$ is a factor of $u$? Is it true that both are? (Perhaps, with different $w$.) - ANSWERED by Wolfgang -Question 2. Assume now that $u\in\{0,1\}^{\mathbb Z}$ is balanced and aperiodic (i.e., Sturmian). Is it true that for any $n\ge1$ there exists a finite 0-1 word $w=w_1\dots w_k$ with $k\ge n$ such that $w_1\dots w_k01w_k\dots w_1$ or $w_1\dots w_k10w_k\dots w_1$ is a factor of $u$? Again, is it true that both are (with different $w$)? - ANSWERED by domotorp -I am familiar with Lothaire's famous monograph "Algebraic Combinatorics on Words" but couldn't find these results there. -EDIT. Here's what is meant by `balanced'. For a finite 0-1 word $w$ let $\delta(w)$ denote the number of 1s in $w$. A 0-1 word $w$ (finite or infinite) is called balanced if for any $n\ge1$ and any two factors $u$ and $v$ of $w$ of length $n$ we have $|\delta(u)-\delta(v)|\le 1$. -For instance, $0100101$ is balanced whilst $101000$ is not, since $\delta(101)=2$ and $\delta(000)=0$. - -REPLY [3 votes]: Too late, but here's my proof that the answer to Question 1 (modified) is yes. -If $u\in\{0,1\}^\mathbb{Z}$ is not balanced, there exists a palindrome $w$ such that $0w0$ and $1w1$ are factors of $u$ (see Prop. 2.1.3 in Lothaire 2); if $w$ is of minimal length $n$, then $u$ has at most $k+1$ factors of each length $k\leq n+1$ (Prop. 2.1.2). -Moreover $\{0w0,1w1\}$ is the only imbalance among factors of length $\leq n+2$ in $u$. It follows that the set of factors of length $n+2$ is contained in $\{0w0,1w1\}\cup [w01]$, where [x] denotes the conjugacy class of $x$ (all of its cyclic shifts). Also, every factor of length $n+1$ except $0w$ and $1w$ are not right special, i.e., they always occur followed by the same letter. -From this we can derive that there exists $m\geq 0$ such that every occurrence of $0w0$ is followed by $(1w0)^m$ and every occurrence of $1w1$ is followed by $(0w1)^m$. If such $m$ is chosen to be maximal, then either $$0w0(1w0)^m1w1(0w1)^m=0(w01)^mw01w(10w)^m1$$ or -$$1w1(0w1)^m0w0(1w0)^m=1(w10)^mw10w(01w)^m0$$ is a factor of $u$.<|endoftext|> -TITLE: Relation between hypercompleteness and the property that Cech cohomology calculates sheaf cohomology -QUESTION [5 upvotes]: Let $C$ be a small site with fibre products. The (injective) Čech model structure on simplicial presheaves $\operatorname{sPre}(C)$ on $C$ presents an $(\infty,1)$-topos and one may ask if this $(\infty,1)$-topos is hypercomplete or not. -If I understand correctly, hypercompleteness means in this setting precisely that a Bousfield localization of the Čech model structure on $\operatorname{sPre}(C)$ at hypercovers (which results in Jardine's local (injective) model structure) doesn't change anything. In other words, the Čech model structure is already Jardine's local model structure. -Let $X$ be a scheme and $\mathcal{F}$ a quasi-coherent sheaf on $X$ with values in abelian groups. Sometimes, the Zariski-sheaf cohomology $H^n(X,\mathcal{F})$ of $X$ with values in $\mathcal {F}$ can be calculated by Čech cohomology since a certain spectral sequence degenerates. This is the case for example when $X$ is separated. -I hope that I recall it correctly but there seem to be results (stacks project, TAG 01H0) that this is true for a general $X$, if one considers ''Hypercover-Čech cohomology'', i.e. Čech cohomology with respect to hypercovers instead of just ordinary covers. For example, if $X$ is not separated, the intersection of two open affines does not have to be affine - but it is covered by affines, and so on. -Let $X_{Zar}$ denote the small Zariski-site on a scheme $X$. - -What is the relation between the $(\infty,1)$-topos associated to $X_{Zar}$ being hypercomplete and the property that sheaf cohomology can be calculated by ordinary Čech cohomology? - -REPLY [6 votes]: There is no relation between hypercompleteness and the property that Čech cohomology agrees with genuine cohomology, i.e., there is no implication either way. For example, étale cohomology of nice schemes can be computed using Čech cohomology even though the small étale (∞,1)-topos is typically not hypercomplete, and, conversely, Zariski cohomology cannot always be computed as Čech cohomology, even though the small Zariski (∞,1)-topos is often hypercomplete (for any noetherian scheme of finite Krull dimension). -Verdier's hypercovering theorem, which says that hypercovers can be used to compute cohomology, follows from the existence of a category of fibrant objects on locally fibrant simplicial presheaves in which hypercovers are the acyclic fibrations. There is no such thing for Čech covers, or even, as far as I know, for bounded hypercovers. In fact, since an (∞,1)-topos and its hypercompletion have the same cohomology, hypercovers also compute the cohomology in the Čech localization of $sPre(C)$. -Update: The reason that $sPre(C)_{Cech}$ and $sPre(C)_{hyper}$ have the same cohomology is that the Eilenberg–Mac Lane object $K(A,n)∈ sPre(C)_{Cech}$, which is a fibrant replacement of the presheaf $U\mapsto K(A(U),n)$, is already local with respect to all hypercovers, so it’s already fibrant in $sPre(C)_{hyper}$. This is because $K(A,n)$ is $n$-truncated, i.e., its sheaves of homotopy groups vanish in degree $>n$. So the mapping space from a hypercover $U_\bullet$ into $K(A,n)$ is the same as the mapping space from the $n$-bounded hypercover $cosk_n U_\bullet$ into $K(A,n)$, but every fibrant object in $sPre(C)_{Cech}$ is already local with respect to bounded hypercovers (see Dugger–Hollander–Isaksen).<|endoftext|> -TITLE: What is known about maximal free subgroups of surface groups? -QUESTION [5 upvotes]: Let $\Gamma_g=< a_1,...,a_g,b_1,...,b_g | \prod_{i=1}^g [a_i,b_i]>$ (a surface group). What is known about maximal free subgroups of $\Gamma_g$ for $g>1.$ (I.e. free subgroups which are not properly embedded into any other free subgroup)? -For example, - -Is $$ free? Maximal? -What are the lower, upper bounds on ranks of the maximal free subgroups? -Can the free subgroups of the minimal rank be classified somehow? - -REPLY [11 votes]: The subgroup is free, and therefore not maximal (see 2.). To see this, the subgroup generated by all the generators except $b_g$ is the fundamental group of a subsurface obtained by cutting along a curve (dual to $b_g$). This surface lifts to a 2-fold cover dual to the curve, and the subgroup is obtained by adding another element corresponding to $b_g^2$, so it has infinite index in the 2-fold cover (since its rank is too small to generate). -Maximal rank free subgroups (i.e. subgroups of infinite index) must be infinitely generated. More generally, this holds for lattices in $PSL_2(\mathbb{C})$ (the argument they gives works for $PSL_2(\mathbb{R})$, and is even easier, since finitely-generated subgroups are geometrically finite). -Since maximal free subgroups have infinite rank, I don't think there will be a nice description of them. They can't be normal, since the only possible quotient groups could have no proper non-trivial subgroups, so must be finite cyclic groups of prime order, a contradiction.<|endoftext|> -TITLE: The derivative of the Cholesky factor -QUESTION [9 upvotes]: Let $A$ be a symmetric, positive definite $p\times p$ matrix, and let $f(A)$ be its Cholesky factor. That is, $f(A)$ is a lower triangular $p\times p$ matrix such that $A = f(A) f(A)^{\top}$. I am wondering if the derivative -$$ -\frac{\mathrm{d}\operatorname{vech}\left(f(A)\right)}{\mathrm{d}\operatorname{vech}\left(A\right)} -$$ -is known, where $\operatorname{vech}$ is the half-vectorization function. -(I am conjecturing that it is something like $L^{\top} \left(f(A)\otimes A^{-1}\right) L$, where $L$ is the elimination matrix, but I would need a reference or proof anyway.) - -REPLY [7 votes]: I've written a relevant note on arXiv: http://arxiv.org/abs/1602.07527 -I included the neat closed form solution pete gives in a comment, and also a messy expression (converted to the notation f = chol(A)): -$$ - \frac{\partial f_{ij}}{\partial A_{kl}} = - \bigg(\sum_{m>j} f_{im}f_{mk}^{-1} + \tfrac{1}{2}f_{ij}f_{jk}^{-1}\bigg)f_{jl}^{-1} - + - (1-\delta_{kl})\bigg(\sum_{m>j} f_{im}f_{ml}^{-1} + \tfrac{1}{2}f_{ij}f_{jl}^{-1}\bigg)f_{jk}^{-1}. -$$ -However, if you're interested in differentiating a larger expression, you can do that in $O(N^3)$, without computing all $O(N^4)$ derivatives in $\frac{\partial \mathrm{vech}(f)}{\partial \mathrm{vech}A}$. The note explains different ways to do that. -(pete: if you tell me who you are, I'll add a proper acknowledgement to my note in any future revision.)<|endoftext|> -TITLE: State of knowledge of $a^n+b^n=c^n+d^n$ vs. $a^n+b^n+c^n=d^n+e^n+f^n$ -QUESTION [21 upvotes]: As far as I understand, both of the Diophantine equations -$$a^5 + b^5 = c^5 + d^5$$ -and -$$a^6 + b^6 = c^6 + d^6$$ -have no known nontrivial solutions, but -$$24^5 + 28^5 + 67^5 = 3^5+64^5+62^5$$ -and -$$3^6+19^6+22^6 = 10^6+15^6+23^6$$ -among many other solutions are known, when the number of summands is increased -from $2$ to $3$. -My information here is at least a decade out of date, -and I am wondering if the resolution of Fermat's Last Theorem has clarified -this situation, -with respect to sums of an equal number of powers...? - -REPLY [10 votes]: Label the equation, -$$x_1^k+x_2^k+\dots+x_m^k = y_1^k+y_2^k+\dots+y_n^k$$ -as a $(k,m,n)$. Let type of primitive solutions be polynomial identity $P(n)$, or elliptic curve $E$. Then results (mostly) for the balanced case $m=n$ are, -I. Table 1 -$$\begin{array}{|c|c|c|} -(k,m,n)& \text{# of known solutions}& \text{Type}\\ -3,2,2& \infty&P(n)\\ -4,2,2& \infty&P(n),E\\ -5,3,3& \infty&P(n),E\\ -6,3,3& \infty&P(n),E\\ -7,4,4& \text{many} &E\,?\\ -7,4,5& \infty&P(n)+E\\ -8,4,4& 1&-\\ -9,5,5& \text{many}&-\\ -9,6,6& \infty & E\\ -10,5,5& 0&-\\ -\end{array}$$ -Note: For $(7,4,5)$, see this MSE answer. -II. Table 2. (For multi-grades) -$$\begin{array}{|c|c|c|} -(k,m,n)& \text{# of known solutions}&\text{Type}\\ -5,4,4& \infty&P(n),E\\ -6,4,4& \infty&P(n),E\\ -7,5,5& \infty&P(n),E\\ -8,5,5& \infty&E\\ -9,6,6& \text{many}&E\,?\\ -10,6,6& \infty&E\\ -11,7,7& 0&-\\ -12,7,7& 0&-\\ -\end{array}$$ -Note: A multi-grade is simultaneously valid for multiple $k$. For example, the $(9,6,6)$ is for $k=1,3,5,7,9$ while the $(10,6,6)$ is for $k=2,4,6,8,10$.<|endoftext|> -TITLE: Why Yau's theorem implies the existence of hyperkähler metric on complex symplectic manifolds? -QUESTION [12 upvotes]: Every expository article on hyperkähler manifolds that I have read states without detailed proof the following fact: -It follows from Yau's theorem (i.e. a compact Kähler manifold $M$ with $c_1(M)=0$ admits a Ricci-flat Kähler metric) that if a compact Kähler manifold $M$ has a complex symplectic form $\omega_\mathbb{C}$ then there exists a Kähler form $\omega$ on $M$ such that $\omega$ and $\omega_\mathbb{C}$ constitute a hyperkähler structrue. -I have trouble trying to see this. Indeed, for the metric to be hyperkähler we require the holonomy group to be in $Sp(n)$ (where $\mathrm{dim}_\mathbb{R}M=4n$), so at least the holonomy should be in $Sp(2n,\mathbb{C})$, which means that the Levi-Civita covariant derivative of $\omega_\mathbb{C}$ is $0$. This is a condition stronger than Ricci-flatness at first glance and I can't see how to realize this only knowing the existence of Ricci-flat metric. - -REPLY [18 votes]: The point is that a Bochner formula shows that, if the Kähler metric is Ricci-flat and the manifold is compact, then every global holomorphic $p$-form must be parallel. -In particular, if you have a complex, compact symplectic manifold that is Kähler, then you have a nonvanishing holomorphic volume form (the top power of the holomorphic symplectic $2$-form). You can then use Yau's theorem to find a Kähler metric whose volume form is a constant multiple of this holomorphic volume form wedged with its conjugate and this metric will have vanishing Ricci curvature. Then, by the Bochner formula, the symplectic form has to be parallel with respect to the Levi-Civita connection of the Kähler metric. In particular, this implies that the holonomy preserves this form and hence is a subgroup of $\mathrm{Sp}(n,\mathbb{C})\cap\mathrm{U}(2n)=\mathrm{Sp}(n)$. If you know, for some reason, that the manifold is simply-connected and not a product, then deRham's Theorem shows that the holonomy must act irreducibly. Berger's classification of the possible irreducibly acting holonomy groups then implies that the metric must have restricted holonomy equal to $\mathrm{Sp}(n)$, since no proper subgroup of this group can act irreducibly and be the holonomy.<|endoftext|> -TITLE: Can the Burgess-Hazen analysis of Predicative Arithmetic be extended to Transfinite Types? -QUESTION [5 upvotes]: Around page 300 of his book "Mathematical Thought and its Objects", Charles Parsons discusses the work of Edward Nelson, who believes that mathematical induction is impredicative, because it can be applied to formulas with quantifiers ranging over natural numbers, even though we conceive of natural numbers as objects belonging to all inductive formulas, including the formula we happen to be applying induction to. Nelson argues that if we reconstruct arithmetic along predicative lines, then we can only accept weak forms of induction that are interpretable in Robinson's Q, like induction on formulas with bounded quantifiers, and on this basis he accepts the totality of addition and multiplication, but not exponentiation. -Parsons agrees with Nelson that there's something impredicative about induction, but he believes that the totality of exponentiation is still predicative. This is based on a paper by Burgess and Hazen, "Predicative Logic and Formal Arithmetic": projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.ndjfl/1039293018 -This paper is concerned with predicative second-order logic, which is like regular second-order logic, except we have a ramified theory of types, which breaks the comprehension schema into levels. The comprehension schema for level 0 sets only allows formulas that have no quantification over sets. The schema for level 1 sets allows quantification only over level 0 sets. For any natural number n, the schema for level n+1 allows quantification over sets of level n and below. Burgess and Hansen prove that predicative second-order logic plus the axiom of infinity implies Robinson's Q + induction on formulas with bounded quantifiers + the totality of exponentiation. This is the basis on which Parsons concludes that exponentiation is total from a predicative point of view. -But as Parson points out, there's no particular reason to stop at finite levels. We can define a comprehension schema for level ω sets, for instance, allowing quantifies to range over sets of finite level. And so on, going to bigger and bigger transfinite ordinals. This is analogous to the Feferman-Schutte analysis of predicative second-order arithmetic (except that Feferman and Schutte rely on a different notion of predicativity, known as "predicativity given the natural numbers", which accepts the natural numbers as a completed totality, in contrast to Nelson and Paraons who think of it only as a potential infinity). We allow a comprehension schema for level $\alpha$ sets as long as $\alpha$ is a transfinite ordinal that is "predicatively acceptable" in a well-defined sense using lower-level comprehension schemes. For starters, we can have comprehension for levels up to $\omega^3$, since as discussed above we can establish the totality of exponentiation using finite levels, and exponentiatial function arithmetic has proof-theoretic ordinal $\omega^3$. This process would presumably converge on some ordinal, akin to the Feferman-Schutte ordinal. And it would presumably allow us to establish a larger subsystem of first-order arithmetic than if we just stuck to finite levels as Burgess and Hazen did. -Parsons, who wrote his book in 2008, said that it was still an open problem as to what exactly that larger subsystem was, although he guesses that it won't be bigger than PRA. Has any progress been made on this since 2008, or was Paraons even mistaken about it being unsolved? Has it at least been shown that, say, the totality of superexponentiation is provable in this larger subsystem? -Any help would be greatly appreciated. -Thank You in Advance. -EDIT: @UlrikBuccholtz's answer points to a paper by Leivant which states that "predicative stratification in the polymorphic lambda calculus using levels $<\omega^\ell$ leads to definability of functions in Grzegorczyk's $\mathscr E_{\ell+4}$". I'm not that familiar with the lambda calculus, so can someone confirm that this implies that $EFA$ with predicative second-order logic with comprehension schemes for levels up to $<\omega^3$ proves that all the functions in Grzegorczyk's $\mathscr E_{7}$ are total? If that were true then the proof-theoretic ordinal of this system would be $\omega^7$, and then by similar methods, I think we can go to $\omega^{11}$, $\omega^{15}$, etc, all the way up to $\omega^\omega$, the proof-theoretic ordinal of $PRA$. -EDIT 2: As I discuss in this question, the Feferman-Schutte approach to extending the ramified hierarchy to transfinite levels seems to rely on some form of the omega rule, either the infinitary omega rule or the formalized omega rule. I don't know what the philosophical justification for invoking the omega rule is, but whatever it is, does it depend on the fact that Feferman and Schutte are analyzing "predicativity given the natural numbers", which takes the set of natural numbers as a completed totality, thereby justifying the omega rule somehow. If that's the case, then presumably we wouldn't be justified in using the oeega rule here, since the stricter notion of predicativity (as opposed to predicativity given thr natural numbers) that Parsons and Nelson espouse treats the natural numbers as only a potential infinity, leading to a skepticism of induction itself, let alone the omega rule. -So can anyone confirm that the omega rule is essential to how Feferman and Schutte extend the ramified hierarchy, and if so whether there's any other way to extend it in the context of h Burgess-Hazen analysis? - -REPLY [5 votes]: I'm not aware of anyone doing the setup exactly as you describe, although it is very likely that it has been done, because it is very similar to Kreisel's proposed method of analyzing finitism in Ordinal logics and the characterization of informal concepts of proof (of course, by many accounts he overestimated the reach of finitism and predicativity given the natural numbers). -However, I would suggest you take a look at Feferman and Strahm (2010), Unfolding of finitist arithmetic, where it is shown that the unfolding (in the sense of Feferman's unfolding program) of finitism is proof-theoretically equivalent to PRA (Primitive Recursive Arithmetic) and hence has proof-theoretic ordinal $\omega^\omega$. -The unfolding is relevant here because it gives a kind of predicative closure given certain base principles. For instance, Feferman and Strahm (2000), The unfolding of non-finitist arithmetic, show that the unfolding of a basic system NFA (of Non-Finitist Arithmetic) is proof-theoretically equivalent to predicative analysis and has proof-theoretic ordinal $\Gamma_0$. -Update: You may also be interested in the work of Leivant, in particular his paper with Danner, Stratified polymorphism and primitive recursion, where it is shown that predicative stratification in the polymorphic lambda calculus using levels $<\omega^\ell$ leads to definability of functions in Grzegorczyk's $\mathscr E_{\ell+4}$. But they don't study an autonomous system.<|endoftext|> -TITLE: Number of subgroups of a given index of a free group -QUESTION [9 upvotes]: Given $n,d\in \mathbb{Z}^+$, how many subgroups of index $d$ does the free group of -rank $n$ have? -In case $n=1$ the question is trivial, and in case $n=2, d=2$ there are 3 such subgroups. -I think I have got a algorithm to solve case $d=2$ for arbitrary $n$. -But the general case seems very difficult. For example in case $n=2$ and $d$ a -prime number, I don't know how to proceed. -That's my question. Any suggestions will be appreciated. -Thanks - -REPLY [15 votes]: This calculation was performed by Marshall Hall Jr. Let $N(d,n)$ be the number of subgroups of index $d$ in the free group of rank $n$. Then -$N(d,n)=d(d!)^{n-1}-\sum_{i=1}^{d-1}((d-i)!)^{n-1}N(i,n)$. -One can prove this inductively by analysing permutation groups as in abx's answer, or alternatively by thinking about covering spaces of graphs---see the answers to this question on MSE for a few more details.<|endoftext|> -TITLE: Completely Metrizable Space and Baire Theorem -QUESTION [10 upvotes]: Is well know that completely metrizable spaces are Baire's spaces. Reciprocally, if $X$ is a Baire's metric space, then $X$ is completely metrizable? - -REPLY [13 votes]: No, in order for a subspace of a complete metric space to be completely metrizable it is necessary and sufficient for it to be $G_\delta$. There are only $2^{\aleph_0}$ many $G_\delta$ subsets of $\mathbb{R}^2$ but there are $2^{2^{\aleph_0}}$ sets in $$\{X \subseteq \mathbb{R}^2 : (0,1)\times(0,1) \subseteq X \subseteq [0,1]\times[0,1]\},$$ each of which is a metrizable Baire space.<|endoftext|> -TITLE: Sums of reciprocals of prime numbers: $p \equiv a \!\! \mod m$ vs. $p \equiv b \!\! \mod m$ -QUESTION [13 upvotes]: Given positive integers $a$, $m$ and $n$, let $s_{a(m)}(n)$ denote the -sum of the reciprocals of the prime numbers less than or equal to $n$ -which are congruent to $a$ modulo $m$. -Is there an integer $n$ such that $s_{1(3)}(n) > s_{2(3)}(n)$? -For small $n$, the function $s_{2(3)}$ is clearly ahead -- -for example it exceeds $1$ already at $n = 59$, while for -$s_{1(3)}$ this takes until $n = 3560503$. -Even if the answer is no, are there still other -residue classes $a(m)$ and $b(m)$ such that the function -$$ - f_{a(m),b(m)}: \mathbb{N} \rightarrow \mathbb{R}, \ \ - n \mapsto s_{a(m)}(n) - s_{b(m)}(n) -$$ -has infinitely many sign changes? -Does $f_{a(m),b(m)}(n)$ converge for $n \rightarrow \infty$, -and if so, is there anything known about the constants -$c_{a(m),b(m)} := \lim_{n \rightarrow \infty} f_{a(m),b(m)}(n)$? - -REPLY [4 votes]: There is a detailed survey -Comparative prime number theory: A survey -Greg Martin, Justin Scarfy. -It contains numerous results on this "comparative prime number theory", -which goes back to Chebyshev, Knapowski and Turan, and others. -In this area there are several results which indicate that primes in "quadratic residue classes" occur (in some precise technical sense) less often than primes in quadratic non-residue classes. -For example, for your question modulo 3, Theorem 5.6 is of interest, -which answers your question for a different weight. -$\lim_{x \rightarrow \infty} \sum_p \chi_3(p) \frac{\log p}{p^{1/2}} -\exp(-(\log^2 p)/x)=-\infty$. (Here $\chi_3(p)=1$, if $p=1 \bmod 3$, and $\chi_3(p)=-1$ if $p=2 \bmod 3$,<|endoftext|> -TITLE: Can we simplify $\int_{0}^{\infty}\frac{{\sin}^px}{x^q}dx$? -QUESTION [33 upvotes]: We know the followings : -$$\int_{0}^{\infty}\frac{{\sin}x}{x}dx=\int_{0}^{\infty}\frac{{\sin}^2x}{x^2}dx=\frac{\pi}{2},\int_{0}^{\infty}\frac{{\sin}^3x}{x^3}dx=\frac{3\pi}{8}.$$ -Also, we can get -$$\int_{0}^{\infty}\frac{{\sin}^3x}{x^2}dx=\frac{3\log 3}{4},\int_{0}^{\infty}\frac{{\sin}^4x}{x^3}dx=\log 2.$$ -Then, I got interested in their generalization. - -Question : Letting $p,q\in\mathbb N$, can we simplify the following? - $$\int_{0}^{\infty}\frac{{\sin}^px}{x^q}dx$$ - -I don't have any good idea. Could you show me how to simplify this? -Remark : This question has been asked previously on math.SE without receiving any answers. - -REPLY [10 votes]: Suppose that $f$ is a smooth function which satisfies in the following condition -$$f(\pi+x)=f(x)\space \space and \space f(\pi-x)=f(x)$$ then, if the following integral exists -$$\int_0^\infty\frac{\sin^2x}{x^2}f(x)$$ -We have the following nice equalities -$$\int_0^\infty\frac{\sin^2x}{x^2}f(x)=\int_0^{\infty}\frac{\sin x}{x}f(x)=\int_0^{\pi/2}f(x)dx$$ -I proved the following formula as a part of my Bachelor project which extend the Lobachevsky's work that if we have - $f(\pi+x)=f(x)\space \space and \space f(\pi-x)=f(x)$ then -$$\int_0^\infty\frac{\sin^4x}{x^4}f(x)=\int_0^{\pi/2}f(x)dx+\frac{2}{3}\int_0^{\pi/2}\sin^2 x\,f(x)dx$$ -and I gave a method for finding an explicit formula for higher degree -$$\int_0^\infty\frac{\sin^{2n}x}{x^{2n}}f(x)$$ -see http://arxiv.org/pdf/1004.2653.pdf<|endoftext|> -TITLE: Proving that a semigroup is regular -QUESTION [5 upvotes]: In a number of diverse situations of interest to me (mostly associated with something called the abelian sandpile model), one can define a nonabelian semigroup generated by commuting elements $a_1,\dots,a_n$ and commuting elements $b_1,\dots,b_n$ such that for all $i$, $a_i$ and $b_i$ do not commute but are semigroup inverses, aka quasi-inverses (i.e., $a_ib_ia_i=a_i$ and $b_ia_ib_i=b_i$). I'd like to know more about the structure of the semigroup. In particular, I'd like to know that for every sequence $i_1, \dots, i_r$ with terms in ${1,\dots,n}$, the products $a_{i_1} \dots a_{i_r}$ and $b_{i_1} \dots b_{i_r}$ are semigroup inverses. Are there standard theorems or methods in semigroup theory that would help me? I should mention that for my applications, the elements $a_i$ and $b_i$ are not inverses in the group sense, even though the semigroup is unital; that is, there is an identity element $e$, but we do not have $a_i b_i = e$ or $b_i a_i = e$. -[In the original version of the question I wrote "$b_i=Ua_iU$ ($1 \leq i \leq n$) for some fixed involution $U$" but then Boris Novikov's question made me realize that for the applications of interest to me $U$ belongs to a larger semigroup, so it seemed best to omit $U$ from the statement of the problem for the time being. I also did not state explicitly that the semigroup is nonabelian.] -There is an extant notion of "sandpile semigroups", but I'm pretty sure that the semigroup I'm interested (introduced by Andrea Sportiello and his coworkers) is something different. -See http://jamespropp.org/pseudo.pdf for a (slightly out-of-date) one-page blurb about the questions that motivated the post, concerning the sandpile model, rotor-router model, and divisible sandpile model. I'm hoping that basic theorems from the theory of semigroups will provide a uniform approach to proving regularity of semigroups in all three contexts. - -REPLY [2 votes]: If your monoid is defined in terms of generators and relations, a simple tool you can use to check whether elements have quasi-inverses of the specified form is Bergman's diamond lemma. For instance, if your monoid is given the presentation $\langle a,b\ :\ aba=a,bab=b \rangle$ you can quickly check that the two relations give a "reduction system". The element $a^2$ does not have a quasi-inverse because if $m$ is any monomial on the letters $a,b$, then just compute the reduced form of $a^2ma^2$ (which must contain at least three instances of $a$). It is even easier to check that $a^2b^2a^2$ is already in reduced form, so $b^2$ is not a quasi-inverse to $a^2$. -Thus, taking $a_1=a_2=a$ and $b_1=b_2=b$, you have your desired counter-example. With more complicated relations, it may take a little more work but you should be able to rule out (von Neumann) regularity in many cases.<|endoftext|> -TITLE: Are there any fast algorithms for factoring integers that don't work by searching for smooth numbers? -QUESTION [7 upvotes]: All of the fast algorithms that I have seen which factor integers work by searching for smooth numbers. Are there any fast algorithms for factoring integers that don't work by searching for smooth numbers? Is there any reason to believe that it is possible to construct fast algorithms for factoring integers that don't work by searching for smooth numbers? - -REPLY [2 votes]: Depends a bit on what you call a "fast algorithm", but Pollard rho doesn't search for smooth numbers.<|endoftext|> -TITLE: Generalizations of Birkhoff's HSP Theorem -QUESTION [6 upvotes]: Let $\mathbf{C}$ be the class of algebraic structures of some fixed type satisfying some sentence $\phi$. Birkhoff's HSP theorem says that $\mathbf{C}$ is closed under homomorphisms, subalgebras and products if and only if $\phi$ is equivalent to some universally quantified conjuction of equations. In the following article - -Properties preserved under algebraic construction. -Author: R. C. Lyndon. -Journal: Bull. Amer. Math. Soc. 65 (1959), 287-299. -http://www.ams.org/journals/bull/1959-65-05/S0002-9904-1959-10321-9/ -Lyndon refers to an "obvious" HD theorem at the top of p293 (as opposed to Birkhoff's UHD(=HSP) theorem) which I think means that the same result holds if $\mathbf{C}$ is only closed under homomorphisms and products, except that we now allow existential as well as universal quantifiers. Let us say we weaken the assumption on $\mathbf{C}$ just a little more, and assume that the homomorphisms considered only come from coordinate functions on products, i.e. what if we only assume that $\mathbf{C}$ satisfies -$X\times Y\in\mathbf{C}\quad\Leftrightarrow\quad X\in\mathbf{C}\textrm{ and }Y\in\mathbf{C}$ -(and the same for infinite products, although I think this already follows from the above finite product assumption). What kind of sentence must $\phi$ now be equivalent to? - -REPLY [7 votes]: It is true that a first-order sentence, which is preserved by finite direct products, is also preserved by infinite direct products; see Corollary 6.7 of S. Feferman and R. L. Vaught, The first order properties of products of algebraic systems, Fund. Math. 47 (1959), 57-103. -In H. J. Keisler's terminology, a first-order sentence is called a product sentence if it holds in a product $X\times Y$ whenever it holds in $X$ and $Y$, a factor sentence if it holds in $X$ and $Y$ whenever it holds in $X\times Y$. If memory serves, rather complicated characterizations of product sentences and factor sentences were found by Keisler in the 1960s; the work of Keisler's student J. M. Weinstein may also be relevant. I'm pretty sure the results or at least the references are in the book Model Theory by C. C. Chang and H. J. Keisler. (I happen to have at hand a reference to Weinstein's dissertation: Joseph M. Weinstein, First order properties preserved by direct product, University of Wisconsin, Madison, 1965.) -Now, if you wanted to know which first-order sentences are preserved by reduced products (direct products reduced modulo a filter on the index set, like ultraproducts but with any old filter instead of an ultrafilter), the answer (also due to Keisler) is very nice: a first-order sentence is preserved by proper reduced products if and only if it's logically equivalent to a Horn sentence. ("Proper" here means that the index set is nonempty and the filter is a filter of nonempty sets; if you want to include the improper reduced product, insert "strict" before "Horn sentence".) I'm sure this is discussed in the Chang-Keisler book.<|endoftext|> -TITLE: probability of non-existence of a sum subset -QUESTION [5 upvotes]: Given a list of distinct integers $a_1,\dots,a_n$ chosen uniformly from the interval $(-2^n,2^n)$, what is the probability that the list contains $\underline{NO\mbox{ }SUBSETS}$ that sum to another integer $b:|b|<2^n$? (what is the probability that the np complete problem SUBSET SUM does not have a subset sum) -Is the odds that there is no subset that sums to b (when b is fixed) is $\frac{1}{2}\pm\epsilon$ or $\epsilon$ or $1-\epsilon$? - -REPLY [2 votes]: The probability that a particular subset of size $k$ has sum $b$ is bounded by $\frac{C}{2^n \sqrt{k}}$, (maximized by $b=0$). Thus the probability that some subset has sum $b$ is at most the sum over $k\leq n$, which is roughly $C/\sqrt{n}$. -[EDIT:] The above is for numbers chosen uniformly with repetition. However, this makes little difference, as the probability of a repetition is much smaller, of order $n^2 2^{-n}$.<|endoftext|> -TITLE: An inequality related to the number of binary strings with no fixed substring -QUESTION [8 upvotes]: This is basically a repost of this math.se question. At the time I was writing this I thought it has to have a straightforward solution so I posted it there. Now I am not so sure about it being so easy. The problem is as follows. -Let $f \in \{0,1\}^k$ and let $S_n(f)$ be the number of strings of $\{0,1\}^n$ that do not contain $f$ as a substring. Is it true that $$|f| > |f'| \implies S_n(f) > S_n(f')$$ -I am aware that the transfer-matrix method could be used to compute $S_n(f)$ given a concrete $f.$ I don't know though if it offers a solution to this problem or perhaps if there's any other obvious reason why this is true. - -REPLY [9 votes]: Your conjecture is true. Here is a proof. -Define $\operatorname{Av}_n(w)$ to be the number of binary words of length $n$ which avoid the pattern $w$. -Let $u$ and $v$ be binary words with $|u| = k$ and $|v|=m$ with $k < m$. We will show that $\operatorname{Av}_n(u) < \operatorname{Av}_n(v)$ for all $n$. -Define the special words - $$M_n = \overbrace{00\cdots00}^n\qquad\text{and}\qquad L_n = \overbrace{00\cdots01}^n.$$ -We can show fairly easily using the cluster method of Goulden and Jackson (and other ways as well, though the cluster method works easily for any pattern) that words avoiding $M_n$ and words avoiding $L_n$ have the generating functions - $$m_n(x) = \sum_{r \geq 0}\operatorname{Av}_r(M_n)x^r = \frac{1-x^n}{1-2x+x^{n+1}}$$ -and - $$\ell_n(x) = \sum_{r\geq 0} \operatorname{Av}_r(L_n)x^r = \frac{1}{1-2x+x^n}.$$ -Moreover, of all words $w$ with $|w|=n$, $M_n$ is the most avoided word and $L_n$ is the least avoided word. Formally, for $w$ with $|w|=n$ and all $r$ - $$\operatorname{Av}_r(L_n) \leq \operatorname{Av}_r(w) \leq \operatorname{Av}_r(M_n).$$ -This can be seen probabilistically by observing that the number of occurrences of a pattern of length $n$ in all words of length $r$ is independent of what the pattern is. Since $M_n$ "packs" the most easily (i.e., has a lot of overlaps) and $L_n$ does not "pack" at all (i.e., cannot overlap itself), it follows that $M_n$ appears as a pattern in less words overall than any other pattern and $L_n$ appears as a pattern in more words overall than any other pattern. -It should also be obvious that $\operatorname{Av}_r(L_n) \leq \operatorname{Av}_r(L_{n+1})$ for all $r$. -We need to prove one more fact: $\operatorname{Av}_r(M_{s-1}) < \operatorname{Av}_r(L_s)$ for all $r \geq s-1$. -Edit: As @DavidSpeyer pointed out in a comment, this is easily proved by observing that $M_{s-1}$ is a subword of $L_s$. I've removed my lengthier argument, but left the generating functions $m_n(x)$ and $\ell_n(x)$ defined above. -We now combine all of our results: for $r \geq k$ - $$\operatorname{Av}_r(u) \leq \operatorname{Av}_r(M_k) < \operatorname{Av}_r(L_{k+1}) \leq \operatorname{Av}_r(L_m) \leq \operatorname{Av}_r(v).\;\;\square$$<|endoftext|> -TITLE: Was lattice theory central to mid-20th century mathematics? -QUESTION [24 upvotes]: Four years ago, I read a book on the history of mathematics up to 1970 or so. It was very interesting up until the end. The last few chapters, though, were on lattices. The author claimed that lattices were taking math by storm, and that soon lattices would be one of the central objects in mathematics, and many other such claims similar to those now made in reference to category theory. -I couldn't find the book online, but I found references to similar ideas in this post:Good lattice theory books? -Question 1: Was lattice theory central to mid-twentieth century mathematics? -Question 2: Is it central now? -Edit: A simpler question with more definite answers: did any mathematicians in the mid-1900s claim that lattice theory was vital or central to mathematics, and what arguments did they give? - -REPLY [7 votes]: Answer to Question 1. In "The Many Lives of Lattice Theory," Gian-Carlo Rota wrote of "Professor [I. M.] Gelfand's oft-repeated prediction that lattice theory will play a leading role in the mathematics of the twenty-first century." So you'd better get on board now! -Von Neumann spent a few years working in lattice theory (writing "Continuous Geometry"), creating with Professor Birkhoff the logic of quantum mechanics. Nowadays there is the work of people like Fotini Kalamara Markopoulou of the Perimeter Institute, whom Scientic American "hailed as one of the world's most promising young physicists." She uses certain types of distributive lattices called Heyting algebras. -Going back to the middle of the last century, Horn also did work on Heyting algebras. (In fact, he had many papers dealing with lattices.) -I was introduced to lattice theory through "Algebra" by Birkhoff and Mac Lane, and I imagine many people who read that book over the decades were also somehow influenced, by that and Cohn's books, which took a universal algebraic approach. Tarski did work with people like C. C. Chang dealing with general algebras which necessarily involved lattice theory and order theory. They also worked with Bjarni Jonsson; when I mentioned his name to an Icelandic mathematician I knew in graduate school, the latter got very excited... -You'll find a couple of papers dealing with lattice theory by people like Shannon; Turing Award winner Hartmanis's first papers were in lattice theory. 2012 Nobel laureate Al Roth's first paper was in lattice theory. -Lurking behind what I heard was the most downloaded article on JSTOR, "College Admissions and the Stability of Marriage," by Gale and Shapley, is lattice theory; Conway proved that that the set of stable marriages forms a distributive lattice. -Dana Scott, another Turing Award winner, advanced the idea of "data types as lattices." -You find lots of applications of lattice theory---for instance, in imaging, I was told by a prominent figure in mathematical morphology that the lattice theory is essential. Last summer I even found out that the police in Amsterdam are using lattice theory for terrorist profiling. A group at Los Alamos National Laboratory was also using lattice theory in a terrorism context. Lattice theory (or order theory) has also been used in sociology, chemistry, epidemiology, linguistics.... Just to leave you with one example, you can look up a $749,226 Environmental Protection Agency grant from 2013 dealing with "Galois lattices."<|endoftext|> -TITLE: How is a descent datum the same as a comodule structure? -QUESTION [17 upvotes]: For a homomorphism of commutative rings $f:R\to S$, there are at least two notions of a descent datum for this map. One of these is to be an $S$-module $M$, with an isomorphism $M\otimes_R S\cong S\otimes_R M$ satisfying the cocycle condition. This can be thought of as saying something about "agreeing on intersections" since it's demanding that the two ways of tensoring $M$ up to an $S\otimes_R S$ module, i.e. either along the left unit or the right unit $S\to S\otimes_R S$ are equivalent. This is basically, I think, the dual of saying that it agrees on projections. -Another common way of phrasing descent data is to say that $M$ is an $S$-module which is also an $S\otimes_R S$-comodule, where $S\otimes_R S$ is a coring with structure map $\Delta:S\otimes_R S\to S\otimes_R S\otimes_R S\cong S\otimes_R S\otimes_S S \otimes_R S$ using the unit map of $S$. -I have written down some vague things about how these two are the same, but is there a functorial equivalence between the two categories of descent data? How does one obtain a comodule structure from the isomorphism $M\otimes_R S\cong S\otimes_R M$? And vice versa? And if you're feeling pedagogical, you might even mention how these two notions are equivalent to the notion of being a coalgebra for a certain comonad! -Thanks! - -REPLY [18 votes]: I find it easier to use geometric notation, so let $X = \mathrm{Spec}(S)$, $Y=\mathrm{Spec}(R)$, and $\phi: X \to Y$ be the morphism corresponding to $f$. We have adjoint functors -$$ -\phi_\ast : S\text{-}\mathrm{mod} \to R\text{-}\mathrm{mod}: \phi^\ast. -$$ -Consider the (beginning of the) Cech simplicial set correponding to the map $\phi$. -$$ \ldots X \times_Y X \rightrightarrows ^{\widehat{\phi}_1} _{\widehat{\phi}_2} X \stackrel{\phi}{\rightarrow} Y $$ -(sorry for the bad latex). It might be helpful to have in mind the example where $X$ is an open cover of $Y$, so that the space $X\times _Y X$ is consists of double intersections. -First of all, note that the structure of being a comodule for the coring $S \otimes_R S$ is a map -$$M \stackrel{a}{\to} M \otimes _S (S \otimes _R S) \simeq M \otimes _R S \simeq \phi ^\ast \phi _\ast M$$ -satisfying the usual conditions. You may notice that the functor $\phi^\ast \phi_\ast$ has the structure of a comonad, and indeed, being a comodule for $S \otimes_R S$ in the category of $S$-modules is the same thing as being a comodule (aka coalgebra) for the comonad $\phi^\ast \phi_\ast$. -By base change, we have $\phi ^\ast \phi_\ast M \simeq \widehat{\phi}_{1\ast} \widehat{\phi}_2 ^\ast M$, so by adjunction, a map $a$ as above corresponds to a gluing map -$$\widetilde{a}: \widehat{\phi}^\ast _1 M \to \widehat{\phi}^\ast_2 M$$ -i.e. a map -$$ M\otimes_R S \to S \otimes_R M$$. -This is the basic mechanism of the correspondence between comodules for $S \otimes _R S$ and the more down-to-earth notion of descent data. One can check that the conditions for $a$ to define a comodule correspond to the cocycle condition for the gluing map $\widetilde{a}$. -If the map $\phi$ is actually nice enough for descent (e.g. faithfully flat), then descent is a concequence of the Barr-Beck theorem: an $S$-module $M$ which is a comodule for $\phi^\ast \phi_\ast$ descends to an $R$-module $N$. -As you are a homotopy theorist, I should note that if you want to do this with complexes of modules (or more fancy things), then you will need the entire Cech simplicial set $\mathcal C(\phi)_\bullet$. Then, the data of being a comodule for $\phi^\ast \phi_\ast$ will be equivalent to being a simplicial module on $\mathcal C(\phi)_\bullet$.<|endoftext|> -TITLE: How to draw a Zoll surface? -QUESTION [16 upvotes]: I take into account that lots of questions on Zoll surfaces have already been asked on the forum. But I will stubbornly continue asking. Are there any chances to draw explicitely at least one Zoll surface which is not of revolution? Is there some addition to the not at all explicit argument of V.Guillemin who proves that there are many Zoll metrics on the sphere (functional parameter family)? Guillemin uses inverse function theorem to prove this fact so no hope for explicit calculations... -This is so strange, people do work a lot in the domain of Zoll surfaces although I do not find almost any pictures for this on Internet. -Additional questions that perturb me: - -Has somebody proven an existence of Zoll metric not close to the canonical one? (not by the "perturbation" argument?) -For the Zoll metrics in higher dimensions, even on the sphere $S^d$, do we have any examples except for the analogues of "revolution metrics"? That is, do we have any examples of metrics without symmetries? -What is known for Zoll metrics on the manifolds which are not $C^{\infty}$ smooth? -Is there a book on Zoll surfaces that is more digestible that A.Besse's Manifolds all of whose geodesics are closed? - -PS. I hope that this post could become a collection of the facts "known-by-now" on Zoll surfaces. I'm just getting lost in the abundance of information on the subject which is badly gathered. And I'm sorry that I do not have any picture in the post with the word "draw" in the subject. - -REPLY [2 votes]: If you want examples of Zoll (and also Tannery) metrics you will find some defined both on manifolds ($\Bbb S^2$) in fact and on some orbifolds (imagine a ``sphere" with a smooth south pole and a north pole having a conical singularity) -in the following reference: -G. Valent, Letters in Mathematical physics, ${\bf 104}$ ($2014$) $1121-1135$ (MR3229169) -where you will find a mechanical approach (a physicist view) to the problem of determining the geodesics equations.<|endoftext|> -TITLE: Does the 4-sphere have a nonzero Poisson structure as a Poisson homogeneous space? -QUESTION [5 upvotes]: It is known that the 4-sphere does not have a symplectic structure. However, it does admit Poisson structures, for example the zero Poisson structure, which is quite boring. Does it have other, more interesting Poisson structures? For example, are there Poisson structures on $\operatorname{Spin}(5)$ (or $SO(5)$) such that $SO(4)$ is a Poisson subgroup, such that $S^4$ is a Poisson homogeneous space? -edit: A Poisson homogeneous space $G/H$ is a quotient space of a Poisson Lie group $G$ by a Poisson Lie subgroup $H$. In this situation, $G/H$ inherits a Poisson bracket from $G$ by projection. I do not explicitly look for constant rank Poisson structure, as the below answer suggests. A definition is given for example in Chari & Pressley, "Quantum Groups", section 1.2.B. - -REPLY [7 votes]: Yes: -imho geometrically the most interesting one is obtained as a quotient $SO(5)/SO(4)$ where the Poisson stucture on $SO(5)$ is not the so-called standard one, but one determined by an element in the maximal torus (sometimes they are called twisted). This is the Poisson analogue of what is mentioned in Quantum symmetry groups of noncommutative spheres - Varilly, Joseph C. Commun.Math.Phys. 221 (2001) 511-523 (where only the quantum counterpart is developed), ie. the Poisson version of the Connes--Landi noncommutative 4--sphere. As I mentioned in the comment above $SO(4)$ is not a Poisson-Lie subgroup of a $SO(5)$ but only a coisotropic subgroup. -The symplectic foliation is very interesting. In fact you have a level function which is a Casimir, so that leaves are contained in the 3-dimensional spheres $t= const.$ (0-dim leaves when $t=0,1$). Inside such spheres 2-dimensional leaves correspond to the usual description of the 3-sphere as two solid tori glued together. Of course the rank drops down to zero in two copies of $S^1$. -I sort of got the impression this is the only way you obtain a Poisson homogeneous structure on $S^4$ starting from a compact Poisson-Lie group. -There is another interesting Poisson structure on $S^4$ coming from Poisson-Lie groups, not homogeneous one but as a double coset. There the symplectic foliation consists only of two leaves: a 0-dimensional one and a $4$--dimensional symplectomorphic to the standard one on $C^2$ (so a sort of "Poisson compactification"). But since I contributed to this maybe it is interesting only for me... -ADDED -About the notion of Poisson homogeneous spaces $G/H$ of a Poisson-Lie group $G$ one may consider: -1) $H$ is a Poisson-Lie subgroup of $G$ (def. of Chari-Pressley, indeed); -2) the projection $G:\to G/H$ is a Poisson map; -3) the action of $G$ on $G/H$ is both a homogeneous action and a Poisson action. -We have $1)\Rightarrow 2) \Rightarrow 3)$ but none of the arrows is reversible. $2)$ is equivalent to $H$ being a coisotropic subgroup of $G$ and is characterized by the fact that there exists a point in $G/H$ in which the Poisson bivector vanishes. Remark that in general the trivial $\pi=0$ Poisson structure on $G/H$ is not necessarily Poisson homogeneous, as you seem to assume, unless the Poisson-Lie structure on the whole $G$ is trivial.<|endoftext|> -TITLE: Strongly connected directed graphs with large directed diameter and small undirected diameter? -QUESTION [13 upvotes]: This question is an attempt to make progress on domotorp's interesting challenge. This question was originally asked in two parts; the former of which was answered by Ilya Bogdanov, and the latter of which is still stumping me. I'll keep both parts of the question for the record, but the interesting part is after the second dividing line. - -Let $G$ be a directed graph on $n$ vertices. For any vertices $u$ and $v$, let $\delta(u,v)$ be the length of the shortest directed path from $u$ to $v$ and let $d(u,v)$ be the length of the shortest path from $u$ to $v$ ignoring edge orientations. We will assume that $\delta(u,v)$ (and hence $d(u,v)$) if finite; the term for this is that the graph is strongly connected. I'll write $d=\max_{(u,v)} d(u,v)$ and $\delta = \max_{(u,v)} \delta(u,v)$. - -Is there a bound for $\delta$ in terms of $d$, independent of $n$? - -This question was answered in the negative by Ilya Bogdanov, below. - -The question I can't answer: -Define a directed graph to have the pairwise domination property if, for any two distinct vertices $u$ and $v$ of $G$, there is a vertex $x$ with $u \rightarrow x \leftarrow v$. (In particular, this implies $d \leq 2$.) What I really need is: - -Is there an integer $k$ such that every graph on $\geq 2$ vertices with the pairwise domination property contains an oriented cycle of length $\leq k$? - -Note that it is enough to study strongly connected graphs here: If $G$ has the pairwise domination property, and $H$ is a strongly connected component of $G$ with no edges coming out of it, then $H$ also has the pairwise domination property. -In fact, I can't even prove or disprove the following (hence the bounty): - -Does every graph on $\geq 2$ vertices with the pairwise domination property contain an oriented triangle? - -REPLY [8 votes]: Here's how to construct a counterexample to the last question. -It's quite easy to introduce a direction to the complete bipartite graph $K_{6,6}$ such that (a) every vertex has outdegree $3$ and (b) the graph has the dominance property on pairs of non-adjacent vertices. One possible orientation of the $36$ edges is as follows: -$$\begin{bmatrix} -+ & + & + & - & - & -\\ -+ & - & - & + & + & -\\ -+ & - & + & - & - & +\\ -- & + & - & + & - & +\\ -- & + & - & - & + & +\\ -- & - & + & + & + & -\\ -\end{bmatrix}$$ -(I left out the drawing, because it was too much of a mess.) This directed graph has no oriented triangles since it is bipartite. -Now replace every vertex in $K_{6,6}$ with an oriented $4$-cycle, so the total number of vertices becomes $12 \times 4 =48$ (and the number of edges is $36 \times 16+4 \times 12=624$). The newly constructed graph has the dominance property on all pairs of vertices and also contains no oriented triangles.<|endoftext|> -TITLE: Is there an integer a such that f(X)+a is irreducible in Z[X]? -QUESTION [5 upvotes]: Let $f\in\mathbb{Z}[X]$ be an irreducible polynomial. Is there an integer $a\neq 0$ such that $f(X)+a$ is also irreducible in $\mathbb{Z}[X]$? -Can this be also extended to $\mathbb{Q}[X]$? - -REPLY [23 votes]: Yes, and you don't need $f$ irreducible. The following irreducibility criterion suffices and shows that infinitely many $a$ work. -Lemma: Let $g(x) = a_n x^n + ... + a_0 \in \mathbb{Z}[x]$ be such that $a_0$ is prime and -$$|a_0| > |a_1| + ... + |a_n|.$$ -Then $g(x)$ is irreducible. -Proof. The condition on the coefficients ensures that all complex roots of $g$ have absolute value greater than $1$. But if $g$ is reducible then at least one of its irreducible factors has constant term $\pm 1$, hence at least one of its roots has absolute value less than or equal to $1$; contradiction. $\Box$<|endoftext|> -TITLE: Adelic methods for classical modular forms -QUESTION [8 upvotes]: Many conjectures about properties of automorphic forms on $\mathrm{GL}(2)$ can be formulated in the basic language of classical modular forms (i.e. Hecke forms that are holomorphic on $\mathbb{H}$ or nonholomorphic Maass forms). However, sometimes the proofs of these conjectures are best understood (or indeed only understood!) by first translating the problem to the adelic setting of automorphic representations of $\mathrm{GL}_2\left(\mathbb{A}_{\mathbb{Q}}\right)$. -An example is a recent paper of Templier disproving a folklore conjecture that $\|f\|_{\infty} \ll_{\varepsilon} N^{\varepsilon}$ for $f \in S_k(N,\chi)$, by (essentially) showing that the local newvector $W_p$ in the Whittaker model of the automorphic representation associated to a newform $f \in S_k^{*}(N,\chi)$ takes large values when $p^2 | N$. -My question is: what are other examples of adelic methods being used to solve problems in the classical theory of modular forms? - -REPLY [2 votes]: This answer is partly inspired by Joël's, but I think it's worth mentioning that the method of newforms is much coarser then theory of types. Modular Hecke cusp forms for fixed weight but different level often share the "same" L-function. The L-function does depend significantly on the irreducible cupsidal automorphic representation, not so much upon the precise vector given. If you are willing to use the theory of types and the corresponding congruence representation-valued forms, you are able to see where redundant things are happening. E.g. given a supercuspidal representation of $SL_2(F_p)$, you can inflate it to $ \tau_p$ of $SL_2(Z_p)$ and $\tau$ of $SL_2(Z)$. If you study the corresponding $\tau$ vector-valued forms, the corresponding local factor of the supercuspidal representations are precisely the supercuspidal given by the induced of $\tau_p$ up to $SL_2(Q_p)$. These also turn up as newvectors somewhere, but they can't be separated from the other stuff happening there anymore. Note that the corresponding $L$-factor is constant.<|endoftext|> -TITLE: Is there a 0-1 law for the theory of groups? -QUESTION [67 upvotes]: Several months ago, Dominik asked the question Is there a 0-1 law for the theory of groups? on mathstackexchange, but although his question received attention there is still no answer. By asking the question here, I hope to find some result solving partially the problem or motivating a possible answer. -For convenience, I paste the complete question below. - -For each first order sentence $\phi$ in the language of groups, define : -$$p_N(\phi)=\frac{\text{number of nonisomorphic groups $G$ of order} \le N\text{ such that } \phi \text{ is valid in } G}{\text{number of nonisomorphic groups of order} \le N}$$ -Thus, $p_N(\phi)$ can be regarded as the probability that $\phi$ is valid in a randomly chosen group of order $\le N$. -Now define $$p(\phi)=\lim_{N \to \infty}p_N(\phi)$$ -if this limit exists. -We say that the theory of groups fulfills a first order zero-one law if for every sentence $\phi$, $p(\phi)$ exists and equals either $0$ or $1$. I'm asking myself whether this 0-1 law holds indeed in group theory. -Since it is conjectured that "almost every group is a 2-group", statements like $\exists x: x\ne 1 \wedge x^2=1 \wedge \forall y:xy=yx$ (meaning $2|Z(G)$) or $\forall x: x^3=1 \to x=1$ (no element has order 3) should have probability $1$ and I don't see any possibility to construct any sentence with $p\not \in \{0,1\}$. Am I missing an obvious counterexample, or can you show (under the condition that almost every group is indeed a 2-group) that the theory of finite groups fulfills this 0-1 law? - -REPLY [8 votes]: Here is a proposal for a possible sentence with probability that doesn't converge. Actually proving it should be hard, and I'm not sure how confident I should be in it but I thought I'd put it out there: -$$\exists_{w_1,w_2} \forall_{x_1, y_1, x_2, y_2} \exists_{z} \ w_1 z w_1^{-1} z^{-1} = x_1 y_1 x_1^{-1} y_1^{-1} \ \mbox{and} \ w_2 z w_2^{-1} z^{-1} = x_2 y_2 x_2^{-1} y_2^{-1} \quad (\ast)$$ -As discussed in comments, it is thought that almost all groups are $2$-groups. The number of groups of order $p^n$ is $p^{(2/27) n^3 + O(n^{8/3})}$. (If we write $N = p^n$, this is $\exp ( (2/27) (\log N)^3/(\log p)^2 + \cdots)$, so $2$-groups overwhelm $p$ groups for other $p$.) This is a theorem of Sims. -Let's understand where the $(2/27) n^3$ comes from. Look at central extensions -$$0 \to C_p^{n-r} \to G \to C_p^r \to 0.$$ -If we look at isomorphism classes of extensions, this is classified by an $H^2$ group of dimension $f(r):= \binom{r}{2} (n-r) + r(n-r)$; Sims writes this down explicitly near the start of his paper. If we maximize $f(r)$ as a function of $r$, it is optimized at -$$r = \begin{cases} 2m & n=3m \\ 2m \ \mbox{and} \ 2m+1 & n=3m+1 \\2m+1 & n=3m+2 \\ \end{cases}$$ -and, at those values, it is $\approx (2/27) n^3$. Moreover, this maxima are sharply peaked: The value of $f(r)$ for any other $r$ is something like $n$ lower. So, if we were to choose $r$ in proportion to $\left| H^2(C_p^r, C_p^{n-r}) \right| = p^{f(r)}$, we would be choosing the values above with probability $1$. -In particular, if we were choosing $r$ in proportion to $|H^2|$, the probability that $r \geq 2(n-r)$ would approach $1$ for $n \equiv 0 \bmod 3$, would approach $1/2$ for $n \equiv 1 \bmod 3$ and would approach $0$ for $n \equiv 2 \bmod 3$. -For fixed $w_1$ and $w_2 \in C_p^r$, the map $z \mapsto (w_1 z w_1^{-1} z^{-1}, w_2 z w_2^{-1} z^{-1})$ gives a linear map $C_p^r \to C_p^{2(n-r)}$. So, if $r<2(n-r)$, then this map can't possibly be surjective and $(\ast)$ must fail. (Actually, it only fails if commutators generate $C_p^{n-r}$. That feels like a probability $1$ statement, but the issue should be checked.) On the other hand, if $r \geq 2(n-r)$, I see no reason that $(\ast)$ shouldn't be true. -This leaves two questions - -In the model where we select $r$ proportional to $\left| H^2(C_p^r, C_p^{n-r}) \right|$ and then select a random extension, how likely is condition $(\ast)$ in the cases where $r \geq 2(n-r)$? -A much more difficult question: How close is the random $H^2$ model to the original question? Higman and Sims prove some results along those lines, but they are very far from as strong as we'd want. Can we say heuristically whether we should expect the real situation to be as strongly peaked at a few values of $r$ as the toy model is?<|endoftext|> -TITLE: Cohomology ring of a flag variety and representation theory -QUESTION [12 upvotes]: I'm interested in the cohomology ring $H^*(G/B)$ of a flag variety $G/B$, where $G$ is a complex semi-simple Lie group and $B$ the Borel subgroup. Borel (1953) showed that this ring is isomorphic to the coinvariants algebra of the associated root system. $H^*(G/B)$ also has a distinguished basis given by Schubert cells. Demazure (1974) and Bernstein, Gelfand, Gelfand (1973) identified elements of the coinvariants algebra corresponding to the Schubert cells. -I know that flag varieties play an important role in representation theory (Borel-Weil-Bott theorem, for instance). I'd like to know whether the cohomology ring $H^*(G/B)$ carries any useful representation-theoretic information. References to literature would also be very appreciated. - -REPLY [6 votes]: Both the study of ordinary Schubert calculus (the ordinary, e.g. Borel-Moore, cohomology of Schubert varieties of the flag variety) and the study of Kazhdan-Lusztig theory (the intersection cohomology of Schubert varieties of the flag variety) are directly related to the representation theory of the Hecke algebra of the associated Weyl group and its specializations. -The literature is vast, but a couple of places to start would be Stephen Griffeth and Arun Ram's article Affine Hecke Algebras and the Schubert Calculus and the article by David Kazhdan and George Lusztig Schubert Varieties and Poincare Duality (the second link is the AMS review on mathscinet, available by subscription only).<|endoftext|> -TITLE: Can the Brun-Titchmarsh theorem be improved when the modulus is smooth? -QUESTION [9 upvotes]: For $q,a$ relatively prime, let $\pi(x,q,a)$ denote the number of primes less than $x$ which are congruent to $a$ modulo $q$. The Brun-Titchmarsh theorem states that $$\pi(x,q,a)\leq \frac{(2+o(1))x}{\phi(q)\log(x/q)}$$ for all $q0$ and you take $q$ to be $x^\theta$-smooth, then you do get that -$$ -\pi(x;q,a)\le (2+\delta) \frac{x}{\phi(q)\log x}, -$$ -with $\delta$ tending to 0 as $\theta\to0$ and $x\to\infty$. This follows by Theorem 10 in this paper of Chang http://math.ucr.edu/~mcc/paper/143%20Char060113.pdf and the zero-density estimates of Jutila (On Linnik's constant, Math. Scand. 41 (1977), no. 1, 45–62) and Huxley (Large values of Dirichlet polynomials. III. Acta Arith. 26 (1974/75), no. 4, 435–444). See also the derivation of Theorem 2 in Iwaniec's paper "On zeros of Dirichlet's L series", -Invent. Math. 23 (1974), 97–104. The constant 2 here corresponds directly to a potential exceptional character, whose contribution is bounded trivially. As Lucia also mentions, we cannot answer this question if $x$ is very close to $q$.<|endoftext|> -TITLE: Conductor of a representation of a $p$-adic group -QUESTION [8 upvotes]: Let $G$ be a connected split reductive group over $\mathbb{Z}$. Let $F$ be a local non-Archimedean field. Let $\rho$ be an irreducible smooth representation of $G(F)$. How does one define the conductor of $\rho$? -The conductor should be an integer canonically associated with the representation $\rho$ so that it matches up with the conductor of the corresponding Galois representation on the Langlands' dual side. -An example of an answer that I'm looking for is given in http://www2.imperial.ac.uk/~buzzard/maths/research/notes/old_introductory_notes_on_local_langlands.pdf -Here, Kevin Buzzard gives a nice simple algebraic definition of the conductor for representations $\mathrm{GL}_2$ using the mirabolic subgroup (see middle of page six of his notes). Does his definition generalize to $\mathrm{GL}_n$? Is there a simple algebraic definition for the conductor of representations of general $G$? What is a good reference for this subject? - -REPLY [7 votes]: for GL_n you could look at: -H.Jacquet,I.I.Piatetski-Shapiro,J.Shalika:Conducteur des representations du groupe lineaire, Math. Annalen 256, 199–214 (1981).<|endoftext|> -TITLE: Quantum Cartan matrices and Coxeter elements -QUESTION [7 upvotes]: Let $\Gamma$ be a bipartite graph, with the vertices partitioned into disjoint sets $\Pi_1$ and $\Pi_2$. Let $W$ be the associated Weyl group, with Coxeter generators $\{s_i\}_{i\in \Gamma}$. Let $\{\alpha_i\}$ be the simple roots of the associated root system. Here are two ways to associate a polynomial to $\Gamma$: -1) Let $c = (\prod_{i\in\Pi_1} s_i)(\prod_{j\in \Pi_2} s_j)$ be a bicolored Coxeter element. -Let $charCox_\Gamma(t)$ denote the characteristic polynomial of c (acting in the reflection representation). -2) Let $C_\Gamma$ denote the Cartan matrix, whose entries are $(\langle \alpha_i,\alpha_j \rangle)_{ij}$ (so in this case the entries are all 2,-1,or 0). Let $QC_\Gamma$ denote the quantum Cartan matrix, where all $2$'s have been changed to $1+q^2$, and all $-1$'s to $-q$'s. Then we have a polynomial $detQC_\Gamma(q)$, the determinant of the quantum Cartan matrix. (In fact this is a polynomial in $q^2$). -It appears to be the case that -$$ charCox_\Gamma(q^2) = detQC_\Gamma(q). $$ -I suspect this, or something very close, is already known, in which case it would be great to have a reference. One straightforward way to prove it, at least when $\Gamma$ is a tree, is by showing that both polynomials satisfy the same recursion for joining. (Probably something similar will prove the statement for more general bipartite graphs.) But that proof is not particularly enlightening. Is there a conceptual explanation for this fact? - -REPLY [3 votes]: Notice that -$$\det(c - q^2) = \pm \det\left( \prod_{s \in \Pi_1} s - q^2 \prod_{s \in \Pi_2} s \right).$$ -We write the reflection representation in the basis of simple roots, and we order our roots so that $\Pi_1$ comes before $\Pi_2$. The product $\prod_{s \in \Pi_1} s$ has block form -$$\begin{pmatrix} -\mathrm{Id} & A \\ 0 & \mathrm{Id} \end{pmatrix}$$ -where $A$ is the adjacency matrix between $\Pi_1$ and $\Pi_2$. Computing the other product similarly gives -$$\prod_{s \in \Pi_1} s - q^2 \prod_{s \in \Pi_2} s = \begin{pmatrix} -\mathrm{Id} & A \\ 0 & \mathrm{Id} \end{pmatrix} - q^2 \begin{pmatrix} \mathrm{Id} & 0 \\ A^T & -\mathrm{Id} \end{pmatrix} = \begin{pmatrix} -1-q^2 & A \\ -q^2 A^T & 1+q^2 \end{pmatrix}.$$ -Left multiplying by the block matrix $\left( \begin{smallmatrix} -q & 0\\0 & 1 \end{smallmatrix} \right)$ and right multiplying by $\left( \begin{smallmatrix} q^{-1} & 0\\0 & 1 \end{smallmatrix} \right)$ turns this into the quantum Cartan matrix you propose. -No insight into any higher context, I'm afraid.<|endoftext|> -TITLE: Stratifications and Filtrations of the Affine Grassmannian -QUESTION [6 upvotes]: Let $G$ be a connected, simply-connected complex semisimple group. Let $$\mathcal{G}r=G(\mathcal{\mathbb{C}((t))})/G(\mathcal{\mathbb{C}[[t]]})$$ be the affine Grassmannian of $G$. We know that $\mathcal{G}r$ has a natural increasing filtration $$\mathcal{G}r_0\subseteq\mathcal{G}r_1\subseteq\ldots\subseteq\mathcal{G}r_n\subseteq\ldots\subseteq\mathcal{G}r,$$ indexed by pole order considerations. Now, fix a maximal torus $T\subseteq G$. We also know that $\mathcal{G}r$ has a stratification into the subvarieties $$\mathcal{G}r^{\lambda}:=G(\mathbb{C}[[t]])t^{\lambda},$$ where $\lambda$ ranges over the dominant coweights of $T$. -Given a fixed $n$, I am seeking a description of those dominant coweights $\lambda$ for which $\mathcal{G}r^{\lambda}\subseteq\mathcal{G}r_n$. I would appreciate any references that might provide some details concerning this description. - -REPLY [3 votes]: I do not really answer you question but maybe this helps: -Let $\mathcal{K} =\mathbb{C}((t))$ and $\mathcal{O}:=\mathbb{C}[[t]]$. For $n\geq 0$ denote the $\mathcal{K}_n$ the $\mathcal{O}$ ideal in $\mathcal{K}$ generated by $t^{-n}$. Choose an embedding $G\hookrightarrow GL_m$. Let -$$G(\mathcal{K}_n):=\{A\in G(\mathcal{K})\vert \text{ all entries of } A \text{ and } A^{-1} \text{ are in }\mathcal{K}_n\}$$ -Then $$Gr_n=G(\mathcal{K}_n)/G(\mathcal{O})$$ -Via this you can in principle calculate all $\lambda$ which are in $Gr_n$. (Note that you can write every $\lambda$ in the form $\lambda: \mathbb{C}^\times \to T$, $t\mapsto \begin{pmatrix}t^{\lambda_1} & & \\ & \ddots \\ && t^{\lambda_m} \end{pmatrix}$ for some $\lambda_i\in \mathbb{Z}$ and all unspecified entries are zero.)<|endoftext|> -TITLE: Hurwitz integers represented as sums of two squares of Hurwitz integers -QUESTION [6 upvotes]: I wonder if there exists a characterisation of Hurwitz integers which are represented as sums of two squares of Hurwitz integers, up to multiplication by a unit. And if so, could you please point to a reference? -By a Hurwitz integer I mean an integer in the ring of quaternions, that is, a quaternion whose components are either all integers or all half-integers. -It can be proved that we only need to concentrate on Hurwitz integers with integer components. -Thanks, and regards, -Guillermo - -REPLY [2 votes]: This is not an answer, but gives some (more) evidences. -Let's say that an algebra $A$ satisfies property $(P)$ if for any element $a$ of $A$ there exists an element $x$ in $A^\times$ and elements $y,z$ in $A$ such that the following equality hold $$a=x(y^2+z^2) \ \ .$$ -$\bullet$ Let $R$ be a principal ideal domain, and assume $2$ is invertible in $R$. Then $A:=\mathrm{M}_2(R)$ satisfies property $(P)$. -Indeed, we can always write -$$M=\left(\begin{array}{cc} \alpha &\beta\\ 0&\gamma\end{array}\right)= -\left(\begin{array}{cc} u &\beta\\ 0&1-u\end{array}\right)^2 -+ -\left(\begin{array}{cc} 0&1\\ \alpha-u^2&0\end{array}\right)^2$$ -with -$u=\frac{\alpha-\gamma+1}{2}$. -$\bullet$ The algebra $\mathrm{M}_2(\mathbf Z)$ satisfies proprty $(P)$. Let $M$ be an upper triangular matrix as above. -a) The above equality shows that if $\alpha-\beta$ is odd, we have a solution. -b) If $M$ is an even matrix, $M=2M'$, then the result follows from the result for $M'$ and the remark $$(\star)\ \ \ (x-y)^2+(x+y)^2=2x^2+2y^2\ \ .$$ -c) If $\alpha$ and $\gamma$ are even, but $\beta$ is odd, then by rows manipulations, we can obtain a lower triangular matrix from $M$, with the first second diagonal entry even and the second odd. Thus we are in the same situation as in a). -d) If $\alpha$ and $\gamma$ are odd then up to row manipulations we can assume $\beta$ is even, $\beta=2\beta'$, and we can also assume that $\alpha-\gamma$ is a multiple of $4$. We write -$$M=\left(\begin{array}{cc} \alpha &\beta\\ 0&\gamma\end{array}\right)= -\left(\begin{array}{cc} u &\beta'\\ 0&2-u\end{array}\right)^2 -+ -\left(\begin{array}{cc} 0&1\\ \alpha-u^2&0\end{array}\right)^2$$ -with -$u=\frac{\alpha-\gamma+4}{4}$. -$\bullet$ Note that it's not true that any matrix in $\mathrm{M}_2(\mathbf Z)$ is a sum of two squares. A counter-example is $M=\left(\begin{array}{cc} 1 &0\\ 0&3\end{array}\right)$. But any matrix in $\mathrm{M}_2(\mathbf Z)$ is a sum of three squares. This can be found in Morris Newman, Sums of squares of matrices, PJM 118, 1985 (thanks to Will Jagy for the reference). -$\bullet$ Let $\mathbf H$ denote the ring of Hurewitz integers. The above remarks show that the property is true for $\mathbf H\otimes\mathbf Z_p$ for odd primes $p$ (edit : since there is an isomorphism of algebras $\mathbf H\otimes\mathbf Z_p\simeq \mathrm{M}_2(\mathbf Z_p)$). It would be interesting to know what happens for $\mathbf H\otimes\mathbf Z_2$. -Edit : computations indicate at least that $A :=\mathbf H\otimes\mathbf Z/32$ has property $(P)$, but that even more is true : any quaternion $q\in A$ can be written as the product of the image in $A$ of a unit of $\mathbf H$ and a sum of two squares. The problem, for now, is that the map $x\mapsto x^2$ is far from being smooth, and no Hensel lemma can be brutally invoqued. -$\bullet$ New Edit : So it seems $A:=\mathbf H\otimes\mathbf Z_2$ satisfies a strong form of property $(P)$ as explained above : any quaternion $q\in A$ can be written as the product of the image in $A$ of a unit of $\mathbf H$ and a sum of two squares. -Here is a proof : -$\to$ By $(\star)$, we note that it suffices to prove the result for quaternions whose class is non-trivial in $A\otimes\mathbf Z/2$. -$\to$ Let $f:A\to A$ be the map $x\mapsto x^2$. We compute $\mathrm{det}(d_x(f))=4\mathrm{Tr}(x)^2N(x)$. Let's say that a unit of $A$ is nice if its trace is a unit. For $x$ such a nice unit, the image of d_x(f)) contains $2.\mathbf H_2$. Thus a quaternion that has the same class in $A\otimes\mathbf Z/8$ than the square of a nice unit is the square of a nice unit. -$\to$ It hapens that all quaternions in $A\otimes\mathbf Z/8$ that are not trivial in $A\otimes\mathbf Z/2$ are the product of - -the image in $A\otimes\mathbf Z/8$ of a unit of $\mathbf H$ - -by - -the sum of two images in $A\otimes\mathbf Z/8$ of squares of nice units (direct computation). - -$\to$ The result follows. -(Remark : $2$ is not a prime in $\mathbf H$. It is ramified. Thus these methods might well be refined by working mod $(1-i)^s$. In fact experiments show that the liftability of squares of nice units is already available from $A\otimes\mathbf Z/4$ on.)<|endoftext|> -TITLE: Prime numbers and limit ordinals -QUESTION [5 upvotes]: As a set, i.e. as a von Neumann ordinal, the $\omega$-th limit ordinal $\omega^2$ is fairly complex and not so easy to visualize (for the novice). But as an explicit well-ordering of $\mathbb{N}$, there is a chance, and even more: all limit ordinals less than $\omega^2$ come as good old natural numbers. -Let $\pi:\mathbb{N}\rightarrow \mathbb{N}$ be the function that maps each natural number to its smallest prime factor. Consider the following well-ordering of $\mathbb{N}$: -$$n \preceq m :\equiv \begin{cases} n &\leq m &\mbox{if } &\pi(n) = \pi(m) \\ -\pi(n) &< \pi(m) &\mbox{if } &\pi(n) \neq \pi(m) \end{cases} -$$ -This ordering is of type $\omega^2$ and it well-orders $\mathbb{N}$ like this: -$$2,4,6,8,\dots,3,9,15,\dots,5,25,35,\dots,7,49,\dots,11,121,\dots$$ -Note, that the limit ordinals less than $\omega^2$, i.e. $\omega,\omega\cdot 2,\omega \cdot 3,\dots$ correspond exactly to the (odd) prime numbers. - -I wonder if this specific well-ordering of order-type $\omega^2$ is by - any means distinguished - as the most simple or the most natural one - - like the natural ordering of $\mathbb{N}$ is the most natural - well-ordering of type $\omega$. - - -[Addendum:] For example this well-ordering looks like the limit of a sequence of "natural" well-orderings of type $\omega\cdot k$: -$\omega\cdot 2 = 2,4,6,8,\dots,3,5,7,9,\dots$ -the multiples of $2$, followed by the rest -$\omega\cdot 3 = 2,4,6,8,\dots,3,9,15,\dots,5,7,11,13,\dots$ -the multiples of $2$, followed by the multiples of $3$ (that are not multiples of $2$), followed by the rest -In comparsion, orderings of type $\omega^2$ based on arbitrary pairing functions (see Joel's answer) come somehow out of the blue. - -And I am looking for comparably easy to understand explicit well-orderings of $\mathbb{N}$ of types significantly larger than $\omega^2$, e.g. $\omega^\omega$ or $\epsilon_0$. - -REPLY [5 votes]: I find ordinal notations based on diagrams to be more natural than these numeric encodings, and they can go pretty high, like up to the Bachmann-Howard ordinal, while remaining intuitively understandable. Herman Ruge Jervell had some very nice presentations about how to do these diagrams, but they seem to be offline now. It's possible they are in his new book, which I haven't seen a copy of yet.<|endoftext|> -TITLE: Derivative of Riemann Zeta at nontrivial zeros -QUESTION [5 upvotes]: I would like to know whether the real part of the first derivative of the Zeta function at the non trivial zeros of Zeta is stricly positive and if so, is there a proof for it. -Also, are there tables or studies of first derivative values at zeros of the Zeta function? -Thank you all for your help. - -REPLY [11 votes]: No, it is not. -Using sage, format is -n-th zero, zero, Re(zeta'(s)): -127 (0.5 + 282.46511476505209623j) -0.051786600288820422906 -136 (0.5 + 295.57325487895829239j) -0.031941907138887597722 -196 (0.5 + 391.45608356363804577j) -0.047584869480501458447 -213 (0.5 + 415.45521499629459886j) -0.57814342438306688421 -233 (0.5 + 446.86062269642952253j) -0.27263428500279642628 -256 (0.5 + 478.94218153463482654j) -0.12771720375134527005 -289 (0.5 + 527.90364160127234523j) -0.96196701578032407318 -368 (0.5 + 637.39719315983730717j) -0.31309354934078881436 - -Code: -import mpmath -for n in [ 1 .. 1000]: - z=mpmath.zetazero(n) - d=mpmath.zeta(z,derivative=1).real - if d -TITLE: Is the conjugacy problem solvable in $Out(F_n)$? -QUESTION [6 upvotes]: There is a paper of Martin Lustig on his webpage giving a positive answer to the conjugacy problem for the outer automorphism group of the free group $F_n$. On the other hand, there seems not to be a publication in a journal about this, but several related publications. Is it possible to say which are the "best" results in this direction, with a complete proof ? (For example, a result by Sela if the outer automorphism group is finite, etc.). - -REPLY [3 votes]: The solution for "general Dehn twist" automorphisms of $Out(F_n)$, namely those which are products of powers of a commuting set of individual Dehn twists, is given in -MR1691946 (2000c:20058) -Cohen, Marshall M.; Lustig, Martin. -The conjugacy problem for Dehn twist automorphisms of free groups. -Comment. Math. Helv. 74 (1999), no. 2, 179–200. -For fully irreducible outer automorphisms, namely those for which no proper nontrivial free factor conjugacy class is periodic, although there is no publication that contains a complete description of the conjugacy problem, nonetheless it is a "folk theorem" [CORRECTION: see the post of User40911 for an attribution of this to J. Los, as I ought to have remembered since I wrote the MathSciNet review :-/ ] that this case is completely covered by the train track technology given in -MR1147956 (92m:20017 -Bestvina, Mladen; Handel, Michael. -Train tracks and automorphisms of free groups. -Ann. of Math. (2) 135 (1992), no. 1, 1–51. -That paper is not written with algorithmic issues in mind, so there are plenty of algorithmic details to fill in. But in outline, given a fully irreducible $\phi \in Out(F_n)$: it is understood that a train track representative of $\phi$ may be computed by following the procedure described in this paper (and there exist several computer implementations); and it is also understood that all of the finitely many train track representatives of $\phi$ (up to topological conjugacy) can similarly be computed; and, finally, this finite set of data is a complete conjugacy invariant for $\phi$. - -REPLY [2 votes]: Theorem 1 in http://www.zbmath.org/?q=an:00912201 (J. Loss On the conjugacy problem for automorphisms of free groups. Topology 35, n.3 p779-806, 1996) -states -"The conjugacy problem for the irreducible outer automorphisms of a free group admits a solution."<|endoftext|> -TITLE: Can we foliate the punctured space by tori? -QUESTION [9 upvotes]: Is it possible to have a 2 dimensional foliation of $\mathbb{R}^{3}-\{0\}$ such that each leaf is homeomorphic to the torus? what algebraic topological obstruction exist? -Another question: is there a foliation as above with the following additional property : -The foliation is stable at the origin. that is for every neighborhood V of origin there is a smaller neighborhood W such that the saturation of W is contained in V? -the motivation for this question is the concept of "Blow up" of singularities of vector field. when we blow up a singularity in R^3, we replace the singularity by a S^2. The blow up processes is based on the fact that R^3-{0} is foliated by a familly of S^2. Now if the answer of the above question is positive we can obtain a new type of torus- blow up. -However the blowing up was my main motivation for this question, but ,by this question, I never mean "is R^3-{0} homeomorphic to R x tori?" - -REPLY [3 votes]: I would like to offer another explanation of the impossibility of foliating $R^3-0$ by tori (or by higher genus closed surfaces), at least in the $C^\infty$ case. -Previously I commented that "foliations are rather far from fibrations". Closer to the truth, foliations are `submersions' onto their (potentially very weird!) leaf spaces. So the distance between foliations and fibrations is, in some respects, comparable with the distance between submersions and fibrations. In the case of smoothly foliating open manifolds by compact submanifolds, the distinction however is very small -- in fact, I want to claim below that foliations in this setting are exactly fibrations. -The starting point is Ehresmann's fibration theorem: if $f:V\to M$ is a proper submersion of smooth manifolds, then $f$ is a locally trivial fibration. A proof can be found in Brocker & Janich's "Introduction to differential topology", section 8.12. -Hence if we have a proper smooth function $f$ on $R^3-0$ having no critical points, then the fibres $f^{-1}(pt)$ foliate $R^3-0$ by compact embedded submanifolds. Ehresmann's theorem tells us $f$ is a locally trivial fibration, and of course as $R$ is contractible, $f$ actually defines a globally trivial fibration. In otherwords, all the fibres are diffeomorphic and we have $R^3-0 \simeq f^{-1}(pt) \times R$. From here we can determine that any fibre must be $\simeq S^2$. -An important point which needs some further justification is this: given a smooth foliation $\mathscr{F}$ of $R^3-0$ by, say, compact tori, how do I know that the quotient map from $R^3-0$ to the leaf space is a smooth submersion of $R^3-0$ onto a smooth $1$-manifold? I see how compactness ensures distinct leafs remain separated (and so the quotient is hausdorff), however i am not clear on pinning down the smooth structure of the quotient map. -Granting the above point, we would know that the quotient is necessarily a noncompact smooth 1-manifold, i.e. the real line $R$, and hence the total space must be a product $T\times R$ -- which we know it isn't. -These arguments, however, suffer the shortcoming of not being able to establish whether or not $R^3-0$ can be foliated by punctured surfaces, a question which seems interesting itself.<|endoftext|> -TITLE: Is Langlands reciprocity somehow analogous to the wave-particle duality of quantum mechanics? -QUESTION [36 upvotes]: Apologies for the vague question, and for the many inaccuracies (I am not a physicist and barely a number theorist). -In physics, there is the notion of gauge group of a field theory. The gauge group is the symmetry group of the Lagrangian describing the theory. It is the "symmetry group of reality" in a reality dictated by the Lagrangian. -On the other hand, the Galois group $G_{\mathbf Q}$ also occurs as the symmetry group of a theory, namely of the theory of algebraic varieties over ${\mathbf Q}$. We might like to think of it as the symmetry group of a hidden "arithmetic field theory" attached to $\mathbf Q$. -Irreducible representations of the gauge group of a field theory can be interpreted as elementary particles of a certain kind (the gauge bosons of the field theory); for this reason, people have said that adequate Galois representations of $G_{\mathbf Q}$ could be thought of as elementary particles (I believe I read this from Brian Conrad). -Elementary particles, however, can also be viewed as incarnations of their wave functions $f \in \Psi$. Moreover, suppose that we have a complete set of commuting observables $\{T_n :\Psi \to \Psi\}$. Let $\mathbf T$ be the $\mathbf Z$-algebra of $End(\Psi)$ generated by the $T_i$'s. Then it is natural to think that the simultaneous eigenfunctions of $\mathbf T$ should be precisely the wave functions of the gauge bosons: if two wave functions look the same under every element of our complete set of observables $\mathbf T$, then they should actually be indistinguishible. -An analogous situation on the arithmetic side is the following: start with the Tate motive $M$ of an elliptic curve $E/\mathbf Q$. Then, according to the modularity theorem, we can attach a "wave function" to the "elementary particle" $M$, namely the Hecke eigenform $f$ associated to $E/\mathbf Q$. Here of course $\mathbf T$ is the Hecke algebra. We can reconstruct the elementary particle $M$ from its wave function, using the algebra of observables $\mathbf T$ (Eichler-Shimura theory). -Thus, it appears to me that the passage from Galois representations to automorphic forms is analogous to the passage from "particle" to "wave" in quantum mechanics. -Is this right? -If so, what more can be said about this? - -REPLY [7 votes]: I believe that this is an interesting question that deserves attention even though it is difficult and perhaps too vague, as noted in the comments. Needless to say that the following will not be a detailed answer but my goal is to at least give some pointers of how to view the question in a slightly different way. -The first question is whether the structural outline given in the question is the appropriate one. In my view the picture that emerges is slightly different. While on the arithmetic side the absolute Galois group $G_{\mathbb Q}$ is usually emphasized as the primary object and their dual motives (via Grothendieck and Langlands) both types are also associated to reductive groups $G$. When thinking about motives in specific geometric situations a reductive group $G$ is chosen, much like a specific Lie group $G_L$ is associated to a specific physical theory. In the context of classical modular forms of some weight $w$ this would be SL$_2({\mathbb R}$) and its congruence subgroups $\Gamma_0(N)\subset {\rm SL}_2({\mathbb Z})$, and for more general automorphic forms higher rank groups enter. In this view the focus is more on the individual reductive groups, for example GL$(n)$ or GSp$(n)$, which therefore play the role of the physical Lie groups, such as SU$(n)$ or the exceptional groups $E_n$. The analogy then is between automorphic forms on the arithmetic side and sections of vector bundles associated to the principal bundle $P(G_L)$ determined by the Lie group $G_L$. While the absolute Galois group is of course still the unifying fundamental object, in physics this would shift the focus away from individual theories to considering all theories at the same time, something that could in principle be done at the level of effective field theories. The particle-wave duality is encoded in the interpretation of this section, either quantum mechanically or quantum field theoretically, where the quantization is implemented in a different way. -The second question raised is concerned with what else can be said about this potential analogy. In physics the key question is the dynamics of the theory. Historically, more often than not fundamental theories in particle physics were constructed by first extracting the symmetry group $G_L$ from observed patterns (an example would be Gell-Mann's discovery of quarks) and after the group had been extracted the question became what the dynamics is. This is a nontrivial task because while symmetry principles put constraints on the dynamics, they usually do not fix it. At the end of the day nature tells us what the right dynamics is, which can be encoded either in some system of hyperbolic differential equations or the Lagrangian with its associated action. If one views the Lie groups $G_L$ in physics as the analog of the reductive groups of the arithmetic theory, and the physical fields defined by sections in bundles as analogs of automorphic forms associated to motives, then the next question would be "what is the "dynamics" of automorphic forms".<|endoftext|> -TITLE: The line graphs of complete graphs and Cayley graphs -QUESTION [9 upvotes]: Let $n>3$ be an odd integer and let $K_n$ denote the complete graph on $n$ vertices. -For which integers $n$ the line graph $L(K_n)$ is a Cayley graph? -For even $n$, it follows from a result of Watkins that $L(K_n)$ is not a Cayley graph. For $n=5$, $L(K_n)$ is the complement of the Petersen graph and so it is not a Cayley graph. - -REPLY [13 votes]: If $L(K_n)$ is a Cayley graph for the group $G$, then $G$ is 2-homogeneous on $V(K_n)$, that is, it acts transitively on the set of unordered pairs of vertices -of $K_n$. However it is not 2-transitive. Kantor "Automorphism groups of designs" determines the 2-homogeneous groups that are not 2-transitive. He finds that such groups exist if and only if $n$ is a prime power congruent to 3 mod 4, and therefore these are the only values of $n$ for which $L(K_n)$ is a Cayley graph.<|endoftext|> -TITLE: Why is the dividing set nonempty when a convex surface has Legendrian boundary? -QUESTION [7 upvotes]: I am an undergrad and curious about the following question. Let $(Y,\xi)$ be a contact manifold, and $L\subset (Y,\xi)$ be a Legendrian knot which is the boundary of a convex surface $\Sigma$ embedded properly in Y. - -Why is the dividing set on $\Sigma$ nonempty? - -I know you can use Stokes' theorem to prove the statement for the closed case, but I don't see how it helps for the Legendrian boundary case. -Here is a related question, showing that the answer to the question above is trivial if ${\rm tb}(L)\neq 0$. -Thanks for your help! - -REPLY [5 votes]: When I first read the question, I found it really odd, but now I'm convinced that this is really true. -I'll use the setup of Etnyre's lecture notes on convex surfaces, around pages 5-6. Just to recap: take a 1-form $\alpha_1$ defining $\xi$. If $\Sigma$ is convex, we can find coordinates near $\Sigma$ such that $\alpha_1 = \beta_1 + u_1dt$, where $\beta_1 = \iota^*\alpha_1$ is the pull-back of $\alpha_1$ to $\Sigma$ and $u_1$ is a function on $\Sigma$ (and $t$ parametrises the direction transverse to $\Sigma$). The function $u_1$ is such that $u_1d\beta_1 + \beta_1\wedge du_1 > 0$ (this is half of Lemma 2.10 in the notes). -The dividing set $\Gamma$ is cut by $u_1$, in the sense that $\Gamma = \{u_1 = 0\}$. Suppose this is empty: then $u_1$ is everywhere positive or everywhere negative. I will suppose that $u_1>0$ everywhere (the other option being symmetric, as seen below). -Then $\alpha = \alpha_1/u_1$ is still a contact form for $\xi$, and $\iota^*\alpha = \beta_1/u_1 =: \beta$. It follows that $\alpha = \beta + dt$, and in particular the corresponding function $u \equiv 1$. Since convexity doesn't depend on the contact form, but only on the contact planes, it follows from Lemma 2.10 that $d\beta$ is a volume form on $\Sigma$. -Now you can apply Stokes' theorem to $\beta$ on $\Sigma$: since the boundary of $\Sigma$ is Legendrian, $\beta$ vanishes identically on the boundary, and you'd have -$$ 0 = \int_{\partial\Sigma} \beta = \int_\Sigma d\beta > 0.$$ -Had I found a negative $u_1$, I'd have probably found a < instead of a > in the last step, but the arguments runs in this case, too.<|endoftext|> -TITLE: Is the Milnor construction contractible -QUESTION [9 upvotes]: Let $G$ be any topological group. Then we can form the infinite join $E_G$ of $G$, i.e. the colimit $G*G*G\cdots$. - -Is $E_G$ contractible? - -I mean it is clear that $E_G$ is weakly contractible, but I dont see why it should be contractible if $G$ is not a CW-space. - -REPLY [4 votes]: Oliver Straser is correct in that Milnor himself in 1956 [1] only showed his model for $EG$ is weakly contractible (that is, all its homotopy groups vanish) with his coarse topology on the join. (For any compact Hausdorff spaces $X$ and $Y$, by the tube lemma, note that Milnor's $X \circ Y$ equals the quotient-topology join $X \ast Y$.) -Historically, for any topological group $G$, the first proof in the literature of the contractibility of Milnor's $EG := G^{\circ \aleph_0}$ is contained within the two-page proof of Theorem 8.1 in Dold's 1963 article [2]. -A few years later in 1966, tomDieck in Section 7 of [3] offers a more conceptual argument. -Contemporarily, I agree that the comment of Oldřich Spáčil is an acceptable answer, based on the simplified proof that tomDieck later gives in Proposition 14.4.6 of his 2008 book [4], which rehashes his earlier works. -[1] : John Milnor, Construction of universal bundles II, Annals Math 63(3):430–436, 1956. -[2] : Albrecht Dold, Partitions of unity in the theory of fibrations, Annals Math 78(2):223–225, 1963. -[3] : Tammo tomDieck, Klassifikation numerierbarer Bündel, Arch Math (Basel) 17:395–399, 1966. -[4] : Tammo tomDieck, Algebraic Topology, 2008.<|endoftext|> -TITLE: If $G \times G \cong H \times H$, then is $G \cong H$? -QUESTION [6 upvotes]: Let $G,H$ groups. Suppose that $G \times G \cong H \times H$. Then is necessarily $G \cong H$? -I know that if $G,H$ are finite groups then it is true (you show that the map $FinGrp \rightarrow \mathbb{N}$, $A \mapsto \left|Hom(A,G)\right| $ determines $G$, and then $\left|Hom(A,G)\right|=\sqrt{\left|Hom(A,G\times G)\right|}=\sqrt{\left|Hom(A,H\times H)\right|}=\left|Hom(A,H)\right|$), so I am interested in the infinite case. -Note that the proof for the finite case also works for finite topological spaces, finite graphs, and maybe other categories I didn't consider. So I also interested on the analogous questions about infinite topological spaces, infinite graphs, or any other interesting category. - -REPLY [8 votes]: In a word, no, even if $G$ and $H$ are assumed to be abelian. A. L. S. Corner has various results showing how strangely infinitely generated abelian groups behave. In particular, `given a positive integer $q$, there exist standard abelian $p$-groups $G$ and $H$ with no elements of infinite height such that $G^n≅H^n$ if and only if $q$ divides $n$'. (I'm quoting from the review on mathscinet.) So taking $q=2$ gives a counterexample to the question. This result is in - -Corner, A. L. S. - On endomorphism rings of primary abelian groups. - Quart. J. Math. Oxford Ser. (2) 20 1969 277–296. - -REPLY [6 votes]: Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981. -Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981): - -However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact. - -The first counterexample for part (c) was given by B. Jónsson, On direct decomposition and torsion-free abelian groups, Math. Scand. 5 (1957), 230-235. -P.S. The positive result in the finite case, proved by counting homomorphisms, is due to László Lovász; it applies to arbitrary finite structures, and it solved a problem of Tarski called the Unique Square Root Problem.<|endoftext|> -TITLE: Maximal Submodule of a Verma Module -QUESTION [7 upvotes]: Let $\mathfrak{h}$ be a Cartan subalgebra of a $\mathbb{C}$-semi simple Lie algebra $\mathfrak{g}$. Given $\lambda \in \mathfrak{h}^*$, $M(\lambda)$ the Verma module of highest weight $\lambda$ and $N(\lambda)$ its maximal submodule. - -Do we have examples of $\lambda$ integral and regular for which $N(\lambda)$ is not equal to the sum of the Verma modules it contains? -Is there any conditions on $\lambda$ to get 1.? - -I know BGG gave an example for $\mathfrak{sl}_4(\mathbb{C})$ but $\lambda=-\omega_1-\omega_3$ is singular. -Thanks. - -REPLY [9 votes]: For your first question the answer is yes, but probably you need to look at specific examples worked out using Kazhdan-Lusztig theory. It should be enough to look at type $A_3$ (as BGG did), where values $>1$ start to appear for some K-L polynomials evaluated at 1. I believe the hand computations by BGG were quicker for an irregular weight, but that's not essential. [As you've recognized, their notation for highest weights incorporates the $\rho$-shift in a sometimes confusing way. Similarly, I've tended to use "regular" in the BGG category as shorthand for "dot-regular" relative to the shifted action of the Weyl group. But that doesn't affect your questions.] -Concerning your second question, it seems (from K-L theory) that a simple description of $N(\lambda)$ occurs mainly when $\lambda$ is dominant integral (regular doesn't matter then). See for instance section 2.6 in my book on the BGG category (AMS, 2008), for the case of a dominant integral weight. This relies mainly on the early work of Harish-Chandra but was re-emphasized by Verma in his 1966 thesis and came to play a major role in the development of the BGG resolution. -[ADDED] If you want concrete examples showing that $N(\lambda)$ need not be the sum of Verma modules, you have to introduce more notation and do some careful bookkeeping with weights and Weyl group elements. This is probably not rewarding, as a quick review of the history shows. I'll refer to some sections of my book, which provide a unified exposition based on older papers. -By about 1970, BGG had refined Verma's ideas enough to see that $N(\lambda)$ is a sum of naturally embedded Verma submodules (neessarily having multiplicity 1) if $M(\lambda)$ has no repeated composition factors. (See my section 5.1, especially the remark and exercise.) This follows indirectly from their study of strongly linked subweights. A converse statement is less direct: If $M(\lambda)$ has a repeated composition factor $L(\nu)$, then for some strongly linked subweight $\mu$ of $\lambda$, $M(\mu)$ has a repeated composition factor $L(\nu)$ and $N(\mu)$ fails to be a sum of Verma submodules. -After another decade, the proof of the 1979 Kazhdan-Lusztig conjecture on composition factor multiplicities made it clear that (in principle) these multiplicities could be computed recursively in terms of values of certain (inverse) Kazhdan-Lusztig polynomials at 1. The full computation relies also on Jantzen's translation functors. Study of K-L polynomials then shows that multiplicities $>1$ only begin to occur in rank 3, as the $A_3$ example of BGG suggested. This depends on the Weyl group, which for the Lie algebra of type $A_3$ is the symmetric group $S_4$. Here there are two pairs of permutations which lead to multiplicities $>1$. For all integral weights in suitable Weyl chambers (and some singular weights on walls), one then gets examples of Verma modules which you ask about. Labelling relevant weights by Weyl group elements requires some care here. (See my sections 8.3-8.4.) -P.S. I should have mentioned a pre-KL paper by Deodhar and Lepowsky, which uses various methods to deal with the regularity question in most types; this is mostly superseded by the stronger information given by Kazhdan-Lusztig theory combined with the interpretation of coefficients of their polynomials in terms of the Jantzen filtration. Elsevier seems to have opened up their archives for online access, so the paper is freely available -here.<|endoftext|> -TITLE: The Theory of Transfinite Diophantine Equations -QUESTION [8 upvotes]: The theory of Diophantine equations is one of the main stream research areas in number theory. There are many known results and unknown conjectures about the existence of non-trivial solutions for equations of the form $a_1 x^{n_{1}}_1+\cdots+a_k x^{n_{k}}_k=b y^{m}$ which all of the variables are positive integers. -In this direction if we let variables to vary on transfinite ordinals as generalization of natural numbers, we will enter the realm of transfinite Diophantine equations like $\alpha_1 x^{\beta_{1}}_1+\cdots+\alpha_k x^{\beta_{k}}_k=\gamma y^{\delta}$ which $k$ is a positive integer and all other variables vary on non-zero ordinals. -It seems there are few known results in the literature about the theory of transfinite Diophantine equations. A possible reason could be the impossibility of calculation on transfinite numbers using super computers which gives an intuition about existence of a solution on integers. Also arithmetic operators lose some of their good properties on transfinite numbers. My questions are about existence of non-trivial solutions for such equations: -Question 1: For which non-trivial transfinite Diophantine equations like $\alpha_1 x^{\beta_{1}}_1+\cdots+\alpha_k x^{\beta_{k}}_k=\gamma y^{\delta}$ do we know that there is no non-trivial solution on both finite and infinite ordinals when $k$ is a positive integer and $\alpha_1,\cdots, \alpha_k,\beta_{1},\cdots, \beta_{k}, \gamma, \delta$ are positive (finite or transfinite) ordinals? -Question 2: Is there any non-trivial Diophantine equation like $\alpha_1 x^{\beta_{1}}_1+\cdots+\alpha_k x^{\beta_{k}}_k=\gamma y^{\delta}$ with positive integers $k, \alpha_1,\cdots, \alpha_k,\beta_{1},\cdots, \beta_{k}, \gamma, \delta$ which has no non-trivial solution on integers but has a solution on transfinite ordinal numbers? -Question 3: What is known about non-trivial transfinite numbers which can be solutions of the equations like $\alpha_1 x^{\beta_{1}}_1+\cdots+\alpha_k x^{\beta_{k}}_k=\gamma y^{\delta}$ when $k$ is a positive integer and $\alpha_1,\cdots, \alpha_k,\beta_{1},\cdots, \beta_{k}, \gamma, \delta$ are positive (finite or transfinite) ordinals? -Remark: The above question is similar to searching for Pythagorian triples in the finite case. -Question 4: Is there any geometric intuition about the theory of transfinite Diophantine equations as same as what developed for proving Fermat's conjecture? - -REPLY [5 votes]: I note that the equations arising in Fermat's last theorem $$x^n+y^n=z^n$$ all have nonzero solutions in the ordinals, since for any positive natural number $n$ we have $$17^n+\omega^n=\omega^n, $$ where $17$ is any finite natural number. This solution uses the usual ordinal arithmetic, but it may be interesting to consider the question with the alternative natural ordinal arithmetic, which is commutative, and these solutions no longer work. Is FLT provable in the ordinals for the commutative ordinal addition?<|endoftext|> -TITLE: Relation between the eigenvalue density and the resolvent? -QUESTION [5 upvotes]: Disclaimer: This is a cross-post from math.stackexchange. Given that there is little activity on the subject (random-matrice) on the aformentioned site, and given that many interesting discussion on this exact subject have taken place here, I feel that this might be a suitable place for this question, althought it is not exactly 'research level'. - -Many texts (e.g. 1-2-3) on random matrices start with some variation of the identity: -$$\rho_1(\lambda) = \frac{1}{\pi} \text{Im}\{\langle\text{Tr}(\mathbf{X}-\lambda\mathbf{I})^{-1}\rangle\}$$ -where the brakets denote the average over the ensemble of eigvenvalues and 'Tr' is the trace of the resolvant matrix $(\mathbf{X}-\lambda\mathbf{I})^{-1}$, where $\lambda\in\mathbb{C}\backslash\mathbb{R}$. The left side is the 1 point correlation function, i.e. the density, which is also defined as -$$\rho_1(\lambda)= \big\langle\sum_i \delta(\lambda-\lambda_i)\big\rangle $$ -(same ensemble). I would like to understand this identity better, so I re-did the complete derivation, as I could not find it anywhere, but I'm off by some $2N$ factor. And I cannot seem to identify a mistake. So here goes nothing. (disclaimer II: I'm a physicist, so some arguments might not be as sound as they could be, forgive me) - -We first express the matrices $\mathbf{X},\mathbf{I}_{N\times N}$ and $R_{\mathbf{X}}^{-1}(\lambda):=\mathbf{X}-\lambda\mathbf{I}_{N\times N}$ in term of their spectra -\begin{align} - &x_{mn} = \sum_{j=1}^N\lambda_j v_m^{(j)} v_{n}^{(j)} & &\Longleftrightarrow& &\mathbf{X} = \sum_{j=1}^N\lambda_j \mathbf{v}^T_j\mathbf{v}_j &\\ - &I_{mn} = \sum_{j=1}^Nv_m^{(j)} v_{n}^{(j)} & &\Longleftrightarrow& &\mathbf{I} = \sum_{j=1}^N\mathbf{v}^T_j\mathbf{v}_j&\\ - &R_{\mathbf{X}}^{-1}(\lambda)_{mn} = \sum_{j=1}^Nv_m^{(j)} v_{n}^{(j)} (\lambda_j -\lambda)& &\Longleftrightarrow& &R_{\mathbf{X}}^{-1}(\lambda)= \sum_{j=1}^N \mathbf{v}^T_j\mathbf{v}_j(\lambda_j-\lambda) -\end{align} -Notationally, I use $\mathbf{v}_j^T$ to express the $j^{th}$ eigenvetor as a column matrice, and $v_m^{(j)}$ to denote the m$^{th}$ element of the $j^{th}$ eigenvector. Moreoever, we assume that the norm of the basis is chosen to be equal to one such that all the above identities hold. -By inspection, it is evident that the spectral representation of the resolvent must be -\begin{align} - R_{\mathbf{X}}(\lambda)_{mn}= \sum_{j=1}^N\frac{v_m^{(j)} v_{n}^{(j)}}{(\lambda_j-\lambda)} \quad\Longleftrightarrow\quad R_{\mathbf{X}}(\lambda)= \sum_{j=1}^N\frac{\mathbf{v}^T_j\mathbf{v}_j}{ (\lambda_j-\lambda)}, -\end{align} -which is a straightforward thing to verify -\begin{align*} - R_{\mathbf{X}}^{-1}(\lambda) &R_{\mathbf{X}}(\lambda) \\ - &=\left(\sum_{j=1}^N \mathbf{v}^T_j\mathbf{v}_j(\lambda_j-\lambda)\right)\left(\sum_{k=1}^N\frac{\mathbf{v}^T_k\mathbf{v}_k}{ (\lambda_k-\lambda)}\right)\\ - &=\sum_{j,k} \frac{(\lambda_j-\lambda)}{(\lambda_k-\lambda)} \bigl(\mathbf{v}_j^T\mathbf{v}_j\bigr)\mathbf{v}_k^T\mathbf{v}_k\\ - &=\sum_{j,k} \frac{(\lambda_j-\lambda)}{(\lambda_k-\lambda)} \bigl( \delta_{jk}\bigr)\mathbf{v}_j^T\mathbf{v}_k\\ - &=\sum_{j} \mathbf{v}_j^T\mathbf{v}_j \equiv\mathbf{I}.\\ -\end{align*} -The trace of the resolvent is directly obtained from the above expression for its element $(m,n)$ -\begin{align} - \text{Tr}R_{\mathbf{X}}(\lambda+i\varepsilon) \equiv \sum_{k=1}^N R_{\mathbf{X}}(\lambda+i\varepsilon)_{kk}=\sum_{j,k=1}^N \frac{v^{(j)}_kv^{(j)}_k}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)}. -\end{align} -where the constraint that the argument of the resolvent must be complex and not on the real line is now explicit ($\in \mathbb{C}\backslash\mathbb{R}$, where $\lambda$ is the real part, and $i\varepsilon$ the imaginary part). -The average of this trace for some properly normalized and arbitrary pdf $\rho(\lambda_1,\lambda_2,...,\lambda_N)$ is thus -\begin{align} - \langle \text{Tr}(R_{\mathbf{X}}(\lambda+i\varepsilon))\rangle &= \int_{-\infty}^\infty d\lambda_1 ...\int_{-\infty}^\infty d\lambda_N\ \rho(\lambda_1,...,\lambda_N)\sum_{j,k=1}^N\frac{v^{(j)}_k v^{(j)}_k}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)}\notag\\ - &=\sum_{j,k=1}^Nv^{(j)}_kv^{(j)}_k\int_{-\infty}^\infty d\lambda_1... \int_{-\infty}^\infty d\lambda_j\ ...\int_{-\infty}^\infty d\lambda_N\ \frac{\rho(\lambda_1,...,\lambda_j,...,\lambda_N)}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)} -\end{align} -The integrals over $d\lambda_\ell$ for $\ell\neq j$ are trivial since $\{\lambda_\ell\}$ only appear in the probability density. Introducting the marginal probability $\rho_{(1)}(x)$, we thus have -\begin{align} - \langle \text{Tr}(R_{\mathbf{X}}(\lambda+i\varepsilon))\rangle &=\sum_{j,k=1}^Nv^{(j)}_kv^{(j)}_k \int_{-\infty}^\infty d\lambda_j\ \frac{\rho_{(1)}(\lambda_j)}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)}, -\end{align} -where we supposed that the variables $\lambda_1,...,\lambda_N$ are indistinguishable (i.e. same marginal probability density $\forall j$). - -We now consider the complex path integral over $C$ -- a closed semi-circle that goes to infinity in the upper half of the complex plane. The marginal pdf is normalizable by definition and there must therefore exist a region at $\infty$ where the integral over the open semi-circle $C_R$ is null, leaving us with an integral over the real line only. Mathematically: - \begin{align} - &\oint_C d z \frac{\rho_{(1)}(z)}{\bigl(z-(\lambda+i\varepsilon)\bigr)} = \\& \int_{-\infty}^\infty d\lambda_j\ \frac{\rho_{(1)}(\lambda_j)}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)} + \int_{C_R} d z \frac{\rho_{(1)}(z)}{\bigl(z-(\lambda+i\varepsilon)\bigr)} =\int_{-\infty}^\infty d\lambda_j\ \frac{\rho_{(1)}(\lambda_j)}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)} - \end{align} -But, this integral is also proportionnal to the sum of the residue in $C$. Here, the pdf cannot feature poles--yet again, by definition--which leaves us with a single pole of order 1 at $z=\lambda+i\varepsilon$, with corresponding residue -\begin{align} - \text{Res}\left(\frac{\rho_{(1)}(z)}{z-\lambda-i\varepsilon},\lambda+i\varepsilon\right)&=\lim_{z\to\lambda+i\varepsilon } \bigl(z-(\lambda+i\varepsilon)\bigr)\frac{\rho_{(1)}(z)}{z-(\lambda+i\varepsilon)}\notag\\ - &=\rho_{(1)}(\lambda+i\varepsilon) -\end{align} -i.e. -\begin{align} - \oint_C d z \frac{\rho_{(1)}(z)}{\bigl(z-(\lambda+i\varepsilon)\bigr)} = 2\pi i\text{Res}\left(\frac{\rho_{(1)}(z)}{z-\lambda-i\varepsilon},\lambda+i\varepsilon\right) -\end{align} -Putting all these results together yields -\begin{align} - \langle \text{Tr}(R_{\mathbf{X}}(\lambda+i\varepsilon))\rangle =2\pi i \rho_{(1)}(\lambda+i\varepsilon) \sum_{j,k=1}^Nv^{(j)}_kv^{(j)}_k = 2N\pi i \rho_{(1)}(\lambda+i\varepsilon) -\end{align} -Or in the limit $\varepsilon\to0$ -\begin{align} - \lim_{\varepsilon\to0}\langle \text{Tr}(R_{\mathbf{X}}(\lambda+i\varepsilon))\rangle = 2N\pi i \rho_{(1)}(\lambda) -\end{align} -Taking the imaginary part finally leads to -\begin{align} -\rho_{(1)}(\lambda) = \frac{1}{2N\pi} \lim_{\varepsilon\to0}\ \mathrm{Im} \bigg\langle \text{Tr} \big(R_{\mathbf{X}}(\lambda+i\varepsilon)\big)\bigg\rangle -\end{align} -which is the desired result divided by an extra 2N factor! - -REPLY [5 votes]: you're missing a factor $N$ because $\rho(\lambda_1,...,\lambda_j,...,\lambda_N)$ is normalized to unity, while $\rho_{(1)}(\lambda)$ is normalized to $N$: -\begin{align} - \int_{-\infty}^\infty d\lambda_1... \int_{-\infty}^\infty d\lambda_j\ ...\int_{-\infty}^\infty d\lambda_N\ \frac{\rho(\lambda_1,...,\lambda_j,...,\lambda_N)}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)}=\mathbf{\frac{1}{N}} \int_{-\infty}^\infty d\lambda_j\ \frac{\rho_{(1)}(\lambda_j)}{\bigl(\lambda_j-(\lambda+i\varepsilon)\bigr)}, -\end{align} -the factor of two appears because you first have to take the imaginary part before closing the contour (so that the integrand falls off as $1/\lambda^2$ rather than just as $1/\lambda$); so instead of $1/(\lambda_j-\lambda-i\varepsilon)$ you take the imaginary part $\varepsilon/[(\lambda_j-\lambda)^2+\varepsilon^2]=\varepsilon(\lambda_j-\lambda-i\varepsilon)^{-1}(\lambda_j-\lambda+i\varepsilon)^{-1}$ and then do the contour integration; the residue from the pole at $\lambda+i\varepsilon$ is now $1/2$, and you have found your missing factor of two!<|endoftext|> -TITLE: Irreducible polynomials with a root modulo almost all primes -QUESTION [10 upvotes]: Let $f \in \mathbb{Z}[x]$ be a non-zero polynomial which is irreducible over $\mathbb{Q}$. Suppose that $f$ has a root in $\mathbb{F}_p$ for almost all primes $p$. Must $f$ be linear? - - -Here by almost all, I mean all but finitely many. -Here is an equivalent way to state this question in terms of number fields. - - -Let $\mathbb{Q} \subset k$ be a number field. Suppose that for almost all primes $p$, the ring of integers of $k$ contains a prime ideal of norm $p$. Must we have $k = \mathbb{Q}$? - - -If $k$ is as above and is also Galois, then the Chebotarev density theorem implies that we indeed have $k = \mathbb{Q}$. The Galois case is quite special however, as here if a prime is split then it is also completely split. It is the non-Galois case that I am really interested in. - -REPLY [8 votes]: It seems this can be done "by hand", using much less than what goes into the full proof of Chebotarev. (But it's the same circle of ideas.) It is known that for any number field $K/\mathbf{Q}$, we have $\zeta_K(s) \sim \frac{\kappa}{s-1}$ as $s\downarrow 1$, for some $\kappa > 0$. Hence, $\log \zeta_K(s) = \log \frac{1}{s-1} + O(1)$, as $s\downarrow 1$. On the other hand, starting from the Euler product representation $\zeta_K(s) = \prod_{\mathfrak{p}} (1-|\mathfrak{p}|^{-s})^{-1}$, we also see that $\log\zeta_K(s) = \sum_{\mathfrak{p}} \frac{1}{|\mathfrak{p}|^s} + O(1) = \sum_{p} \frac{n(p)}{p^s} + O(1)$, where $n(p)$ is the number of degree one prime ideals $\mathfrak{p}$ of $\mathcal{O}_K$ that lie above the rational prime $p$. (Note that I'm using the somewhat nonstandard notation $|\cdot|$ for the norm map on ideals.) -In particular, when $K=\mathbf{Q}$, we get that $\sum_{p} \frac{1}{p^s} = \log\frac{1}{s-1} + O(1)$ as $s\downarrow 1$. -Now suppose that $k$ is a number field with the property that all but finitely many rational primes $p$ have a degree $1$ prime $\mathfrak{p}$ of $\mathcal{O}_k$ lying above them. Applying the result of the first paragraph to this $k$, we see that -$$ \log \frac{1}{s-1} \sim \log \zeta_k(s) = \sum_{p} \frac{n(p)}{p^s} + O(1)\geq \sum_{p} \frac{1}{p^s} + O(1) \sim \log \frac{1}{s-1}. $$ -Hence, -$$ \sum_{p:~n(p)>1}\frac{n(p)-1}{p^s} = o\left(\log\frac{1}{s-1}\right), $$ -as $s\downarrow 1$. Consequently, $$ \sum_{p:~n(p)>1}\frac{1}{p^s} = o\left(\log\frac{1}{s-1}\right). $$ -That is, the (Dirichlet) density of primes $p$ having more than one degree one prime ideal above them must be zero. But unless $[k:\mathbf{Q}] = 1$, this contradicts that a positive proportion of rational primes split completely in $k$. (It is in this final step that $n > 1$ is important.) -It might seem that I smuggled in Chebotarev to say that a positive proportion of rational primes split in $k$. But no, this follows again from the result in the first paragraph, now with $K$ taken as the Galois closure of $k/\mathbf{Q}$.<|endoftext|> -TITLE: Is the universal elliptic curve $\overline M_{1,2}$ a toric stack? -QUESTION [11 upvotes]: It is well-known that the compactification $\overline M_{1,1}$ of the moduli space of elliptic curves over $\mathbb C$ is a weighted projective line with -weights $4$ and $6$. As far as I can tell, this is more or less directly linked to the fact that the ring of holomorphic modular forms for $SL_2(\mathbb Z)$ -is a polynomial ring with generators $E_4$ and $E_6$. As a consequence, $\overline M_{1,1}$ is a smooth toric Deligne-Mumford stack. -It has been shown in Eichler-Zagier's book -http://carlossicoli.free.fr/E/Eichler_M.,_Zagier_D.-The_theory_of_Jacobi_forms.pdf -that the ring of weak Jacobi forms of even weight is a polynomial ring with a double grading by weight and index (see Theorem 9.3 on page 108). The bigradings of the variables are $(4,0),(6,0),(-2,1),(0,1)$. It is thus tempting to consider a GIT quotient of $\mathbb C^4$ with respect to the corresponding action of $\mathbb (C^*)^2$ by looking at the Gale dual of the set of weights. This would be a smooth toric DM stack of dimension two and Picard number 2. Dimension two makes a choice of triangulation automatic. -Does one get the universal elliptic curve $\overline M_{1,2}$ this way? If so, is there a reference for this? My guess is that it is a little bit off, but I am not positive. (Note: I believe the fibers of the projection map to $\overline M_{1,1}$ are elliptic curves modulo Kummer involution, which is why they are rational). - -REPLY [3 votes]: The fibers aren't rational curves when viewed as stacks, because they have four points with extra automorphisms. I think this will be problematic. -The universal family of elliptic curves is the quotient of the scheme with projective variablesz $x,y,z$, affine variables $g_2,g_3$, and equation $y^2z=x^3-g_2xz^2-g_3z^3$ by the torus action that takes fixez $z$ and takes $x \to t^2x, y \to t^3 y, g_2 \to t^4g_2, g_3 \to t^6 g_3$. So if $\overline{M}_{1,2}$ were a toric variety then this should be as well, but the equation has too many terms to be toric. -Here the extra automorphism is the case $t=-1$. Taking the even weight forms should basically by equivalent to taking the quotient by this action, where you replace $y$ by $y^2$ and then eliminating that variable, which I think gives your ring $(4,0)=g_2$, $(6,0)=g_3$, $(-2,1)=z$, $(0,1)=x$. I'm not completely sure of this because I don't think you actually can eliminate $y$ and get the Jacobi ring because of the extra factor of $z$, but you certainly get a toric variety by doing this (you just need to add the variable $x^3/z$.) However this is not going to be the same stack as you get from the full ring.<|endoftext|> -TITLE: Does Fermat's last theorem hold in the ordinals? -QUESTION [32 upvotes]: My question is whether there are no nontrivial solutions in the ordinals of the equations arising in Fermat's last theorem $$x^n+y^n=z^n$$ -where $n\gt 2$, and where we use the natural ordinal arithmetic, which is commutative. -(Note: If we had used the usual ordinal arithmetic, there are some easy counterexamples, such as $1^3+\omega^3=\omega^3$.) -The question spins off of my answer to Saint Georg's recent question, The Theory of Transfinite Diophantine Equations. -Feldmann Denis pointed out in the comments there that for very small ordinals (below $\omega^\omega$), the question reduces to the corresponding question in polynomials, where it has an affirmative answer. Can we extend this observation to work for all ordinals? - -REPLY [23 votes]: There are no nontrivial solutions. This follows from Wiles’s proof, and the following observation. -Proposition: If a set of Diophantine equations has a solution in (positive) ordinals using natural sum and product, then it has a solution in (positive) natural numbers. -Proof: Every ordinal can be uniquely written in Cantor normal form -$$\tag{$*$}\alpha=\sum_{i\alpha_{i+1}$ and $0 -TITLE: Thurston-Cannon $S^2$-filling curves -QUESTION [10 upvotes]: I have been looking into equivariant space-filling curves $S^1\to S^2$ as discussed by Cannon & Thurston in these two papers: - -Three-Dimensional Manifolds, Kleinian Groups and Hyperbolic Geometry -Group-Invariant Peano Curves - -The title of the second paper suggests a resemblance to the Peano Curve - -The Cannon-Thurston construction seems a little more delicate. In the first paper, near Theorem 5.7 we get - -Let $M_\phi$ be the mapping torus of a (pseudo-Anosov map) $\phi: S \to S$. This is $S \times [0,1]$ with the ends identified by $\phi$: $(S,0)\sim_\phi (S,1)$. -We can extend this map to the universal cover and get a map from the hyperbolic plane to hyperbolic space $\tilde{\phi}: \mathbb{H}^2 \to \mathbb{H}^3$. -For reasons I do not fully undestand, they ask if this extends to a map from $S^2 \to S^3$ which involves checking of the map exists $S^1_\infty \to S^2_\infty$ "near infinity". - -Thus we get a sphere-filling curve. The images supplied in Thurston's paper do not convince me they fill the sphere. - -Nor does this picture look very sphere-filling. - -In what sense does the figure about fill the sphere and how is this related to Peano space-filling curve of $[0,1]^2$ ? - -REPLY [7 votes]: There are multiple reasons that the Cannon-Thurston curves are related to the Peano curve. For instance, these are all continuous surjective paths onto 2-dimensional objects: for the original Peano curve the target is the square; for the Cannon-Thurston curves the target is the 2-sphere. -Also, the original Peano curve is constructed as a limit of a sequence of piecewise linear paths. If one thinks about the original Peano construction, one sees that the paths in the sequence can all be obtained from the final Peano curve by "connecting dots", that is to say by taking a finite sequence of points along the Peano curve and interpolating by straight line segments. The pictures of Cannon-Thurston curves you have exhibited were (I am pretty sure), also constructed by "connecting the dots", yielding some single term of an approximating sequence of paths. -As with any sequence, even say a sequence of rational numbers converging to a real number, there are important issues of convergence rate. "Connect-the-dots" approximations of Cannon-Thurston Peano curves converge rather rapidly for noncusped fibered 3-manifolds. But for cusped fibered manifolds they converge quite slowly. In Figure 12, right in the center is a cusp point, and the curve is actually differentiable exactly at the cusp point. The algorithm for drawing the connect-the-dots approximation is wasting lots of time near that cusp point (and others). Hence the lack of time to spend drawing any part of the curve that fills in the white spaces.<|endoftext|> -TITLE: Derived categories of singular varieties -QUESTION [10 upvotes]: Given my limited knowledge on derived categories, all the results on derived categories of complex of bounded sheaves are build upon smooth varieties, and people literally avoid singular case (as in the case of Bridgeland and Maciocia's proof of Fourier-Mukai transforms on abelian fiberation etc.) -I want to know -(1) What are the main difficulties working on singular and noncomplete varieties? (the resolution is not finite? the dimension of cohomology group is not finite? Or no spectral sequences? Then what's the exact trouble?) -(2) What is the current knowledge of derived categories on arbitrary varieties? -My question might be too big to be answered precisely -- any references are extremely welcome!!! - -REPLY [10 votes]: Probably the main reason people avoid singular varieties is because of boundedness. Arguments using induction and, as you say, spectral sequences need boundedness of the complexes in play in one direction or the other (and sometimes both). -Related to this, is the failure of Serre duality, which lies at the heart of many derived arguments. (the fact that the Serre functor commutes with any equivalence is a very nice tool to throw around) -In the smooth and proper case the three main functors to build integral transforms (pushforward, tensor, pullback) all preserve boundedness. -In the just proper case pushforward is still fine. Pullback from $X \times Y \to X$ is of course not the issue (unless you're doing some crazy relative situation) as the projection map is flat. So the real problem is in taking tensor products. -If the kernel is not perfect then tensoring with it will escape to the bounded-above derived category. -So if you planned to run an argument by induction on the length of the complex which started at the bottom you're done for. -For example, having nice formulae for the adjoints of an integral transform is no longer automatic as one cannot blindly apply Serre duality. -The derived category of singularities Fernando mentions is a bit of a silly name and refers to work of Orlov (originally, later extended by many many people). The point is that the category of perfect complexes sits inside the bounded derived category and one can take the (Verdier) quotient. You call this the derived category of singuarities. (for example, the structure sheaf of a singular point is not perfect, as it does not admit a finite resolution by vector bundles, so it will "survive" this quotienting procedure) In turn this category of singularities is often related to categories of matrix factorisations. -Perhaps I should mention that in the direction of Serre duality for singular varieties there is a whole school that thinks one should replace the unbounded derived category of quasi-coherent sheaves with the (bigger!) category of Ind-coherent sheaves. -Vaguely (ie at the level of my understanding) perfect complexes form the compact objects of $D(QCoh(X))$. As on a singular varieties there are more objects we care about (eg coherent sheaves!) one would like a category where the whole $D^b(Coh(X))$ is the subcategory of compact objects. This wild Ind-Coh is tailored to do just that. I'm still not sure what formal obstacles this improves upon. I think they were first introduced by Krause, with the very concrete description of being the homotopy category of unbounded complexes of injective quasi-coherent sheaves. I vaguely remember there being some work of Murfet relating this to matrix facotrisations, but I might be mistaken. -EDIT: Another example which is very relevant. For a blow up $Y \to X$, there is a nice formula for $D(Y)$ in terms of $D(X)$ and $D$ of the exceptional locus. This holds, of course, if everything (including the centre) is smooth. As soon as you blow something singular this is no longer true. It was mentioned somewhere on this website by the user Sasha (Kuznetsov?) that there are examples where $D(Y)$ is actually indecomposable over $D(X)$, which is very odd and disappointing.<|endoftext|> -TITLE: Semi-free resolutions -QUESTION [5 upvotes]: Let $\mathscr{C}$ be a DG category (not much will be lost if you assume that $\mathscr{C}$ has one object, i.e. is a DG algebra). One way to construct the unbounded derived category of $\mathscr{C}$-modules is by using semi-free resolutions. Recall that a DG module $F$ is free if it is a sum of shifts of corepresentable modules, or semi-free if it has an exhaustive filtration $0 = F_0 \subset F_1 \subset \cdots \subset F$ such that the subquotients $F_{n+1}/F_n$ are free. -I've seen the existence of semi-free resolutions stated as follows: for any DG module $M$, there is a semi-free module $F$ and a surjective quasi-isomorphism $F \to M$. But how does one actually construct $F$? I'm especially confused about how to make $F \to M$ surjective, since generators of free modules are cycles. -Edit: I found a construction of $F \to M$ in Drinfeld's paper "DG quotients of DG categories." Start by choosing a free module $F_1$ and a morphism $F_1 \to M$ which is surjective on cohomology. Now induct: given $F_n \to M$, let $C_n = \text{Cone}(F_n \to M)$ then find a free module $P_n$ and a morphism $P_n \to C_n$ which is surjective on cohomology. Using the map $C_n \to F_n[1]$ we get $P_n \to F_n[1]$, and Drinfeld takes $F_{n+1} = \text{Cone}(P_n[-1] \to F_n)$. Of course $F$ is the direct limit of the $F_n$. -This is nice, but I still don't understand how to make $F \to M$ surjective. Does this happen automatically for the construction given above, or do we have to do something extra? - -REPLY [6 votes]: There is a recent paper of Tobi Barthel, Emily Riehl, and myself -that answers this question in a model categorical framework. We -were lazy and only considered the one object case, although we -believe our work generalizes to DG categories. In another respect we -were not lazy: we work over a general commutative ring $R$, not just -a field, and this introduces interesting subtleties. The reference is -http://front.math.ucdavis.edu/1310.1159 -The standard model structure on (unbounded) DG modules over a DG -$R$-algebra $A$ takes quasi-isomorphisms as weak equivalences. -Cofibrant approximations are more general than semi-free resolutions, -but they are retracts of semi-free resolutions given by model theoretic -cellular DG modules. There is an early construction by my adviser, John Moore, -in the 1959-60 Cartan seminar, which we modernize. This uses bicomplexes, -as usual in differential homological algebra. There is another construction, -originally due to Gugenheim and myself, dating from 1974, which we also -modernize. Qiaochu, you will be interested that we drop surjectivity in that -construction, meaning that we do not have model theoretic fibrations. We use -multicomplexes, which are bigraded but have differentials that are sums of -pieces that mimic differentials in spectral sequences. The drop of -surjectivity and the use of multicomplexes gives a great gain of -computability, as we illustrate by modernizing 1974 applications to the -computation of the cohomology of many homogeneous spaces. -We also explain the role of the bar construction. When R is a field it -constructs cofibrant approximations in the standard model structure. In -general, it constructs cofibrant approximations in a relative model -structure for which the weak equivalences are the maps of DG $A$-modules -which are chain homotopy equivalences of underlying $R$-modules, and in -that generality it is not semi-free.<|endoftext|> -TITLE: C^2 submanifolds contained in a hypersurface -QUESTION [5 upvotes]: Suppose I have a smooth manifold $M$, and an embedded $C^2$ submanifold $N \subset M$, of codimension at least 3. Does there exist, for every point $x \in N$, a smooth ($C^\infty$) hypersurface in a ball in $M$, which contains the portion of $N$ in that ball? - -REPLY [9 votes]: The answer is negative.$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\abs}[1]{\lvert #1 \rvert}$$\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace}$$\newcommand{\ceiling}[1]{\left\lceil #1 \right\rceil}$ -More precisely, $f:\RR\to\RR^n$ defined by -$$ f(t) = \bigl( t,\abs{t}^{3+\frac{1}{2}},\abs{t}^{3+\frac{1}{3}},\ldots,\abs{t}^{3+\frac{1}{n}} \bigr) $$ -gives an embedded $C^2$ curve in $\RR^n$ such that no neighbourhood of zero in the curve is contained in a smooth (or even $C^4$) hypersurface in $\RR^n$. Taking products of $f$ with any $\RR^l$ produces examples of embedded $C^2$ submanifolds of $\RR^{n+l}$ of dimension $l+1$ for which there exists no neighbourhood of zero in the submanifold which is contained in a smooth hypersurface in $\RR^n$. -The argument below is divided into two parts: the first involves only the local description of submanifolds of $\RR^n$; the second part involves an estimate using Taylor expansions. The proofs are elementary and will be provided in some detail. -Smoothly independent functions -For convenience, let us introduce a little bit of terminology. -Definitions: - -Let $f:\RR\to\RR$, $g:\RR\to\RR^n$ be functions $\RR\to\RR$. Say that $f$ depends smoothly on $g$ at $t\in\RR$ if there exists a $C^\infty$ function $h:\RR^n\to\RR$ such that $f = h\circ g$ on some neighbourhood of $t$. -Call a function $f:\RR\to\RR^n$ smoothly dependent at $t\in\RR$ if $f_i$ depends smoothly on $(f_1,\ldots,f_{i-1},f_{i+1},\ldots,f_n)$ at $t$, for some $1\leq i\leq n$. Otherwise, say $f$ is smoothly independent at $t$. - -Let $f:\RR\to\RR^n$ be continuous, and denote by $L$ the image of $f$ as a subspace of $\RR^n$. -Claim 1: -Let $S$ be a $C^\infty$ submanifold of $\RR^n$ of codimension one (i.e. a smooth hypersurface) which contains some neighbourhood of $f(0)$ in $L$. Then $f$ is smoothly dependent at zero. -Proof: - -Since $S$ is a smooth hypersurface in $\RR^n$, it is locally the graph of a smooth function. More precisely, there exists $i\in\set{1,\ldots,n}$ such that some neighbourhood $U$ of $f(0)$ in $S$ is parametrized as -$$ -U = \set{ (x_1,\ldots,x_{i-1},g(x),x_i,\ldots,x_{n-1}) \mid x \in V } -$$ -where $V$ is an open in $\RR^{n-1}$ and $g:\RR^{n-1}\to\RR$ is a smooth function. Since $S$ contains some neighbourhood of $f(0)$ in $L$, we conclude that -$$ -f_i(t) = g\bigl(f_1(t),\ldots,f_{i-1}(t),f_{i+1}(t),\ldots,f_{n-1}(t)\bigr) -$$ -for $t$ sufficiently close to zero. ■ -Examples of smoothly independent functions -In view of claim 1, it remains to construct a $C^2$ embedding $f:\RR\to\RR^{n+1}$ which is smoothly independent at zero, for each $n>0$. -Claim 2: -Take any finite sequence $a_0=1,a_1,\ldots,a_n$ of positive real numbers such that none of them can be written as a linear combination of the others with non-negative integer coefficients. Then the function -$$ -f(t) = (t,\abs{t}^{a_1},\ldots,\abs{t}^{a_n}) -$$ -is smoothly independent at zero. -Here is a fairly explicit instance of claim 2 which recovers the example given at the beginning of the answer. -Example: -Pick some positive integer $l$ and a sequence of pairwise distinct real numbers $b_1,\ldots,b_n \in (0,1)$. Take $a_i = l+b_i$ for $i>0$ (and $a_0 = 1$). Then the condition in claim 2 above is satisfied, hence the corresponding $f$ is smoothly independent. Moreover, this choice of $f$ is a $C^{l-1}$ embedding. -Claim 2 is a direct consequence of the next lemma. -Lemma: Let $a_0,\ldots,a_n$ be positive real numbers such that $a_0$ is not a linear combination of $a_1,\ldots,a_n$ with non-negative integer coefficients. There exists no smooth function $h:\RR^n \to \RR$ such that $t^{a_0} = h(t^{a_1},\ldots,t^{a_n})$ for all $t>0$ in some neighbourhood of zero. -Proof: -Assume that such a smooth function $h$ does exist. Take the Taylor polynomial for $h$ of a sufficiently large degree $k$ (it actually suffices to take $k = \ceiling{ \frac{a_0}{ \min\set{a_1,a_2,\ldots,a_n} } }$), together with the corresponding Peano remainder term. Since $h$ is $C^k$, we have (in little-o notation) -$$ -h(x) = \sum_{\abs{I}\leq k} a_I x^I + o\left(\sum_{\abs{I} = k} \abs{x^I}\right) -$$ -where $I=(i_1,\ldots,i_n)$ ranges over $n$-tuples of non-negative integers, $\abs{I}=i_1+\ldots+i_n$, and $x^I = (x_1)^{i_1} \cdots (x_n)^{i_n}$. The little-o symbol above is considered in the limit $x\to 0$. -Replace the preceding equality into $t^{a_0} = h(t^{a_1},\ldots,t^{a_n})$ to produce: -$$ -t^{a_0} = \sum_{\abs{I}\leq k} a_I t^{i_1 a_1 + \cdots + i_n a_n} + o(t^{a_0}) -$$ -Given that $a_0$ is not a linear combination of $a_1,\ldots,a_n$ with non-negative integer coefficients, all the terms $a_I t^{i_1 a_1 + \cdots + i_n a_n}$ appearing above have degree different from $a_0$. Therefore, after grouping together monomials of the same degree, and moving the monomials of degree greater than $a_0$ into the little-o symbol, we get -$$ -t^{a_0} = \sum_{0\leq r -TITLE: Is there a differentiable but nonsmooth version of the continuous Implicit Function Theorem? -QUESTION [9 upvotes]: From the result discussed in Does the inverse function theorem hold for everywhere differentiable maps? (which I'll call the differentiable nonsmooth Inverse Function Theorem) one can obtain a differentiable but nonsmooth version of the Implicit Function Theorem by the usual argument. -There are two interesting closely related questions apparently still remaining: - -Is there a corresponding version of the continuous Implicit Function Theorem not requiring continuous differentiability in the variables being solved for? -In the original inductive proof of the Implicit Function Theorem, continuous differentiability was needed to insure a decreasing chain of locally nonvanishing minors for the Jacobian determinant. Does there always exist such a chain if simple differentiability with nonzero Jacobian is assumed? - -A positive answer to 2 gives a positive answer to 1 and an inductive proof of the differentiable nonsmooth Inverse Function Theorem. -For a more careful description of these problems, see the exposition of the Implicit Function Theorem in my real analysis manuscript on my website http://wolfweb.unr.edu/homepage/bruceb/ . - -REPLY [2 votes]: The answer to the first question is YES. There were two proofs of this result, one by Jean Saint Raymond (Mathematika, 2002), the other by L. Hurwicz and M. K. Richter (Discussion paper no. 279, Dept. of Economics, University of Minnesota, 1994).<|endoftext|> -TITLE: A Question About Pure States, Support Projections and Central Covers -QUESTION [7 upvotes]: I am trying to study the paper Consistency of a Counterexample to Naimark’s Problem by Charles Akemann and Nik Weaver, and there is a claim in Lemma 1 of the paper that I am stuck at, which is as follows. - -Claim: Let $ \mathscr{A} $ be a $ C^{*} $-algebra and $ g: \mathscr{A} \to \mathbb{C} $ a pure state on $ \mathscr{A} $. Let $ q \in \mathscr{A}^{**} $ be the support projection of $ g $. Also, let $ y $ be the central cover of $ q $ in $ \mathscr{A}^{**} $; in other words, $ y $ is the smallest projection $ p \in \mathcal{Z}(\mathscr{A}^{**}) $ such that $ q \leq p $. Then $ y \mathscr{A}^{**} $ is isomorphic to $ B(\mathcal{H}) $ for some Hilbert space $ \mathcal{H} $. - -The claim may seem obvious to the many operator theorists more capable than myself, but I am just not getting it. I believe that the GNS construction is needed here (as the claim is about a pure state, which corresponds to an irreducible $ * $-representation of $ \mathscr{A} $ on some Hilbert space $ \mathcal{H} $), but this is the furthest that I have gotten. - -REPLY [7 votes]: For each state $\phi$ on $A$ let $\pi_\phi: A \to B(H_\phi)$ be the corresponding GNS representation. Let $\pi: A \to B(H)$ be the direct sum of all of these representations. The von Neumann algebra $\pi(A)'' \subseteq B(H)$ is called the enveloping von Neumann algebra of $A$ and it is naturally isomorphic to $A^{**}$. (See C${}^*$-algebras and Their Automorphism Groups by Pedersen.) -For each $\phi$, since $H_\phi \subseteq H$ we have a natural projection from $\pi(A)'' \subseteq B(H)$ onto $\pi_\phi(A)'' \subseteq B(H_\phi)$. The kernel of this projection is a weak* closed ideal of $\pi(A)'' \cong A^{**}$ and hence it has the form $zA^{**}$ for some central projection $z$ in $A^{**}$. Then $y = 1-z$ is the central cover of $\phi$ and $A^{**} = yA^{**} \oplus zA^{**}$, so that $\pi_\phi(A)'' \cong A^{**}/zA^{**} \cong yA^{**}$. -So the general fact is that $\pi_\phi(A)'' \cong yA^{**}$. If $\phi$ is pure then $\pi_\phi$ is irreducible and $\pi_\phi(A)'' = B(H_\phi)$.<|endoftext|> -TITLE: Properness of the category of modules over a spectrum (that represents algebraic cobordism or motivic cohomology) -QUESTION [7 upvotes]: The abstract form of the question: let $C$ be a closed proper stable model category, $R$ is a ring object in it. Which conditions ensure that the category $R-mod$ is also proper? -Since weak equivalences and fibrations are detected via the forgetful functor $R-mod\to C$, right properness seems to be obvious. On the other hand, I do not understand how to check the left properness. Looking at the case of complexes of modules over a ring, I suspect that the forgetful functor respects pushouts; possibly $R-mod$ is proper if $R$ is cofibrant in $C$. Is this true? Does it make sense to pass to the category of cofibrant objects in $R-mod$? -Now a very concrete motivic question. I would like to prove that the category of modules over the Voevodsky's algebraic cobordism spectrum is proper. This assertion should probably very similar to its analogue for the case of modules over the (motivic Eilenber-MacLane spectrum) $MZ$. Yet in the paper Oliver R¨ondigs, Paul Arne Østvær, Modules over Motivic Cohomology, http://www.math.uni-bielefeld.de/~oroendig/MZfinal.pdf, I was not able to find the discussion of properness for the latter case (see Proposition 2.36). It is not stated that $MZ$ is cofibrant; one does not pass to cofibrant objects in $MZ-mod$. -Upd. It seems that Proposition 2.9 of Hovey's http://arxiv.org/abs/math/9803002 confirms my idea: everything is ok if MGl is cofibrant. Still I wonder whether the latter is known (for some model of $MGl$), and whether this result of Hovey was published somewhere. - -REPLY [2 votes]: Warning! This post may contain self-promotion. -If you only want to apply the left properness property to push-out squares where the underlying $R$-modules are cofibrant in $C$, then you can use Theorem 1.13 in this paper. This situation is actually rather common. The result is for operads rather than rings, but rings are operads concentrated in arity 1, so it applies. It assumes that $R$ is cofibrant in $C$ and that the tensor unit $I$ is cofibrant. You also have a version not assuming that the tensor unit is cofibrant, Theorem D.13. In that case it is enough that the unit $I\rightarrow R$ be a cofibration in $C$.<|endoftext|> -TITLE: Definability in HOD -QUESTION [6 upvotes]: Let $\mathfrak{M}$ be a countable transitive model of set theory, and consider HOD (the hereditarily ordinal definable elements of $\mathfrak{M}$). -Let $x$ be an object $x \in HOD$. So $x$ is hereditarily definable from ordinals. -Can we find a defining formula $\psi(x)$ of $x$ such that $\psi(x)$ contain only quantifiers ranging over the ordinals? -PS: by "ranging over ordinals" I mean a quantifier $Qy$ where either we have the restriction $Qy$($y$ is an ordinal ...) or that $Qy \in \alpha$ where $\alpha$ is an ordinal. -Edit: After Emil's comment, I am going to allow quantification over elements $y$ such that $y \in x$. The reason for this change is that in the application I have in mind we can allow this. And, yes, we can allow also arbitrary bounded quantifiers. - -REPLY [12 votes]: The answer is no, not necessarily. The main reason is that if $\varphi(x,\vec\alpha)$ is a formula all of whose quantifiers range only over ordinals or are bounded, with ordinal parameters $\vec\alpha$, then the truth of $\varphi(x,\vec\alpha)$ is invariant under extensions $V\subset W$ with the same ordinals and no new elements of sets in $V$, since the interpretation of the meaning of the formula does not change. In particular, the interpretation of $\varphi$ does not change by forcing. -Consider the following example. Start in $L$, and add a Cohen real $L[c]$, and the force to make that real $c$ definable in a forcing extension $L[c][G]$, by forcing to code its digits into the GCH pattern on the $\aleph_n$'s, for example. So $c$ is in $\text{HOD}^{L[c][G]}$, but by further forcing, we can collapse cardinals to $L[c][G][H]$, where $\text{HOD}=L$ again. Suppose that there were a nice formula $\varphi$ defining $c$ in $L[c][G]$. It follows that $\varphi(c)$ also holds in $L[c][G][H]$. But in this case, there is some condition forcing it, and this condition specifies only finitely much of $c$. Since the forcing overall is weakly homogeneous, we may find an automorphic image $d\in L[c]$ of $c$ that also contains that finite part of $c$, and so $\varphi(d)$ will also hold in $L[c][G][H]$, and therefore also in $L[c][G]$. This contradicts that $c$ was defined by $\varphi$ in $L[c][G]$.<|endoftext|> -TITLE: Institutional response to "Esquisse d'un programme" -QUESTION [33 upvotes]: It is well-known that Grothendieck's Esquisse d'un programme was submitted in 1984 as part as the author's application for the permanent position, Directeur de Recherche at CNRS (the main public research institution in France, employing thousands of full-time researchers.) This came -after Grothendieck had resigned his professorship at IHES and spent a -few years as a professor at the University of Montpellier. All of this is -for instance written in Esquisse d'un programme itself, or in Recoltes et Semailles. - -What I would like to know (with certainty): what happened to his application to CNRS? Did he receive a permanent position? Another kind of position? Nothing? If he got - some offer, did he accept it or refuse it? If he accepted an offer, did he effectively hold it, and when, how, and why did he eventually give up the position - (I assume he did, because before long he disappeared almost completely.) - -Now, Wikipedia (on the Esquisse d'un programme page) says something about it, namely that the application was successful and that he held the position held from 1984 to 1988. -It refers to an article in Science News, which, I seem to recall, did not say more. However, I have asked this of various people in France, receiving as many different answers as persons I asked. -So I would like confirmation of this point by knowledgeable people. Moreover, this Wikipedia sentence just answers part of the question: it doesn't say what kind of position -Grothendieck received, whether he effectively held it, and why and how he resigned from it. -On the other hand, this is a public event for which all or most -participants are still alive, so it should be relatively easy to obtain a knowledgeable answer. -I apologize that my question has absolutely no mathematical content, but there is a long sequence of well-accepted precedent questions about aspects of the lives of famous mathematicians and also about mathematical institutions. - -REPLY [37 votes]: I would think this is from an authoritative source, since apparently the author consulted with Bourguignon (chair of the hiring committee at CNRS). - -When Grothendieck reapplied to the CNRS in 1984, his application was - once again controversial. Jean-Pierre Bourguignon, now director of the - IHÉS, chaired the committee in charge of reviewing applications in - mathematics, among which was Grothendieck’s. According to Bourguignon, - in the handwritten letter required for the application, Grothendieck - listed several tasks he would not perform, such as supervising - research students. Because CNRS contracts obligate researchers to - perform some of these tasks, this letter was viewed by the CNRS - administration as proof of Grothendieck’s ineligibility. Bourguignon - said he tried to get Grothendieck to amend his application so that it - did not state explicitly all the tasks he refused to carry out, but - Grothendieck would not budge. After considerable effort on the part of - several people, Grothendieck was eventually put on a special kind of - position, called a position asterisquée, that was acceptable to him - and to the CNRS. The CNRS did not actually hire him but was in charge - only of paying his salary, and he retained his university - affiliation. So for his last few years at Montpellier before his - retirement in 1988, Grothendieck did not teach and spent less and less - time at the university.<|endoftext|> -TITLE: Regular lattice polygons -QUESTION [5 upvotes]: Suppose I want to construct an $N$-gon in the plane whose vertices are integer lattice points, and which is close to a regular $N$-gon (which means, the ratio of longest to the shortest side is within $\epsilon_1$ of $1,$ and the ratio of the largest to the smallest angle is within $\epsilon_2$ of $1$) The question is: is there any sort of reasonable upper/lower bound to the size (area or diameter) of such a thing? This is clearly related to diophantine approximation in the following way: take the "standard" regular $N$-gon (one whose vertices are $N$-th roots of unity,) and blow it up by a factor of $t$ until wall the vertices ($t$ times the roots of unity) are close to lattice points. How big does $t$ need to be in terms of some "closeness" bound? - -REPLY [3 votes]: Probably Dirichlet's approximation theorem gives best possible answer for large $N$. We can only reduce the number of simultaneous approximations using symmetries of $N$-gon. For example for $8$-gon it is necessary to approximate only one number $1/\sqrt2$ among $8$ numbers $e^{2\pi ik/8}$ $(k=0,\ldots 7)$. -But starting from arbitrary $N$-gon inscribed into the unit circle we can find $t\le R$ such that -$$|tx_1-a_1|\le\frac1{R^{\frac1{2N}}},|ty_1-b_1|\le\frac1{R^{\frac1{2N}}},\ldots |tx_N-a_N|\le\frac1{R^{\frac1{2N}}},|ty_N-b_N|\le\frac1{R^{\frac1{2N}}},$$ -where $(x_1,y_1), \ldots, (x_N,y_N)$ are coordinates of initial $N$-gon. It means that we can achieve $\epsilon_1=\epsilon_2\asymp t^{-1-\frac1{2N}}$ in the circle of radius $t$ for infinitely many $t$. -If we want a better exponent, we can do the following. For arbitrary $N$ we can take $x_1=1$, $y_1=0$ (no approximation needed). The rest part of vertices is symmetrical. So for odd $N$ we need to approximate $(N-1)/2$ points ($N-1$ coordinates), for $N\equiv2\pmod4$ $(N-2)/4$ points ($(N-2)/2$ coordinates), for $N\equiv4\pmod8$ $(N-4)/8$ points, for $N\equiv0\pmod8$ $(N-2)/4$ points $1+(N-8)/8$ coordinates. -Also there is an algorithm (based on discrete Fourier transform) of complexity $O(t^2)$ which allows to find all best possible solutions in the circle of radius $t$. This algorithm was published in Algebra and number theory for mathematical schools. Collected problems (Russian, see page 138). -It is very simple. We need so specify only one vertex to define our regular $N$-gon. So we take all points $A_1=(x_1,y_1)$ such that $0\le y_1\le x_1\le t$. For each $A_1$ we define $A_2',\ldots,A_N'$ rotating $A_1$ around the origin, and $A_2,\ldots,A_N$ as nearest integer points to $A_2',\ldots,A_N'$. After that it necessary to check the quality of resulting $N$-gon. We can say that $A_1,\ldots,A_N$ are $N$ complex numbers and calculate discrete Fourier series. First Fourier coefficient gives main cycle ($|C_1|\approx |OA_1|$) and the rest part of Fourier series forms an error term.<|endoftext|> -TITLE: In any Lie group with finitely many connected components, does there exist a finite subgroup which meets every component? -QUESTION [30 upvotes]: This question concerns a statement in a short paper by S. P. Wang titled “A note on free subgroups in linear groups" from 1981. The main result of this paper is the following theorem. -Theorem (Wang, 1981): For every field $k$ of characteristic 0 and subgroup $\Gamma$ of $GL(n, k)$, the group $\Gamma$ either has a nonabelian free subgroup or possesses a normal solvable subgroup of index < $\lambda(n)$. -To prove this theorem, Wang needs the following statement: -“It is easily shown that in any Lie group with finitely many connected components, there is a finite subgroup which meets every component.” -Please give me a proof of this statement. - -REPLY [5 votes]: I'm not at all sure what S.P. Wang had in mind as an "easy" proof, but it's worth adding some explicit references to published work concerning the well-studied structure of the normalizer of a maximal torus in a compact Lie group. (Some of this has also come up in an earlier MO question here.) -First, the reference for Wang's short paper is: "A note on free subgroups in linear groups". J. Algebra 71 (1981), no. 1, 232–234. This builds on an influential 1972 paper by Tits in the same journal (unfortunately not freely accessible online). -The paper "Normalisateurs de tores. I" by Tits (also in J. Algebra) is referenced in the compact Lie group setting by Dwyer and Wilkerson in their paper which can be freely downloaded here. Their reference list also includes a helpful older paper by M. Curtis et al.<|endoftext|> -TITLE: Spectrum of Laplacian in non-compact manifolds -QUESTION [13 upvotes]: What can be said about the spectrum of the Laplace-Beltrami operator on a non-compact, complete Riemannian manifold of finite volume? For example, is the point spectrum non-empty? -What would be a good reference for this? - -REPLY [14 votes]: As far as I know, this is a very delicate question. That is, already in two dimensions, there can be only finite discrete spectrum. (See Phillips-Sarnak and Wolpert.) -Even on the modular curve $SL(2,\mathbb Z)\backslash \mathfrak H$, it was highly non-trivial to prove existence of infinitely-many $L^2$ eigenvalues, apparently requiring Selberg's invention and application of his "trace formula" in the 1950's. -Similarly, it has only been after 2000 or so that various people (S. Miller, Lapid-Finis, and others) have proven "Weyl's Law" for higher-rank quotients of semi-simple Lie groups, that is, that the asymptotics for discrete (even cuspidal) spectrum are what one would expect for a compact Riemannian manifold. -That is, based on this perspective, I'd be amazed if much could be said in general... maybe some negative results for small deformations, analogous to Phillips-Sarnak and Wolpert?<|endoftext|> -TITLE: Is it worthwhile to give off-topic talks? -QUESTION [33 upvotes]: I am a graduate student. Occasionally for some reason I am asked to give a talk on my research at a conference whose stated purpose is almost completely unrelated to my research. To preserve my anonymity, I won't say what I actually do, but as an example: Imagine I do research in some particular aspect of exterior differential systems, and I am asked to speak at a special session on some particular aspect of numerical analysis. -In the past when I have done this, it has not been a pleasant experience. I sit through a bunch of talks I don't understand, try to socialize with a group of people who already know each other and who I'll likely never see again, and then give my talk, which I feel is seen as irrelevant. -But I am still encouraged (by my advisor and others) to give such talks, because it is good for my career. In particular, I've been told that it's good to advertise my work to a broad audience, and that people in my audience may someday referee my papers. This seems very unlikely to me, but I am uncertain. -Is giving such a talk really a worthwhile endeavor? - -REPLY [4 votes]: I support a nice Joël’s answer. -I think that success depends on communication and social activity, but personally I prefer to do more explicit carrier steps than those giving you only a hope that “maybe somebody somewhere will remember me”. I consider that if after a few months after the conference your listener will remember from where you are and that you are dealing with exterior differential systems, it will be very good. We may check this if we ask ourselves what we remember about conference speakers which were far from our mathematical interests, and I think that most of the mathematicians are narrow specialists. -I think that Internet is a good way to general (and, in particular to working and carrier) acquaintances and communications. In particular, we can use LinkedIn or MathOverflow for this purposes. Personally I gathered a material for two joint articles after few months at Mathematics Stack Exchange. When I was a graduate student and my English was more weak, I even had a sample letter to ask other mathematicians for papers. And I solved some open problems. -I think that success depends on many factors. For instance, I rarely visited conferences outside my city but I have 17 coauthors and big mathematical correspondence. Also my friend told me that I must attend conferences and directly contact with people to be invited, but I was invited to China when I had reviewed an article written by a Chinese professor. - -Concerning the mentioned differences between American and Russian scientific styles of work, I think that the style is strongly depending on a personal choice. -Andrew Wiles had rejected his partisipation in conferences and was proving Fermat’s Last Theorem on his own (see for the details Chapter 6 of Simon Singh’s “Fermat's Last Theorem”, which fragments are here and here). -Paul Erdős “spent most of his life as a vagabond, traveling between scientific conferences and the homes of colleagues all over the world. He would typically show up at a colleague's doorstep and announce “my brain is open”, staying long enough to collaborate on a few papers before moving on a few days later. In many cases, he would ask the current collaborator about whom to visit next”. He was generoulsy sharing his mathematicas ideas with community and was easily responding to others’s ideas. He died from a heart attack attending a conference in Warsaw and he had in his pocket a ticked to Vilnius, where should be his next conference. -Personally I was bred in L’viv topological school which is a branch of a topological school from Moscow State University. I was very disliking conferences and preferring not to leave my city. For instance, when I was in secondary school, I even skipped a participation in Soviet Union mathematical olympiad, because I was not interested in it, althouh I had a gold medal in Ukraine. :-) But now, living in a real world, which seems to be similar and to have similar problems both in America and Russia, I am forced to do scientific carrier in order to make a possibility to be a scientist. :-( -As a resume I recommend you to choose (or form your own) a scientific style that works better for you, gives you growth and is good for you.<|endoftext|> -TITLE: dual space of the subspace of the space of probability measures -QUESTION [5 upvotes]: I have a question which maybe so naive but I want to know the result about it. -Let $\mathcal{M}=\mathcal{M}(\mathbb{R})$ be the space of bounded measures. Then by some materiau such as Multidimensional diffusion processes and Large deviations, we know that the dual space of $\mathcal{M}$ is $\mathcal{C}_b^0(\mathbb{R})$ which is the space of continuous bounded functions defined on $\mathbb{R}$. Here the topology of $\mathcal{M}$ is induced by weak convergence. -Now we consider a subspace $\mathcal{M}_p$ of $\mathcal{M}$ such that: -$$\mathcal{M}_p=\{\mu\in\mathcal{M}: \int_{\mathbb{R}} x^2\mu(dx)<\infty\}$$ -I would like to know the dual space of $\mathcal{M}_p$, I guess it is the space of continuous functions $f$ satisfying -$$|f(x)|\leq C(1+|x|^2)$$ -for some constant $C$. But I don't know how to prove it. If someone knows it please let me know. Thanks a lot! - -REPLY [3 votes]: If $X$ is any locally convex space and $L$ is a subspace (endowed with the relative topology) then the (continuous) dual $L'$ of $L$ is a quotient of $X'$ by the subspace $L^\perp=\lbrace f\in X': f|_L=0\rbrace$ -(this follows from Hahn-Banach: the restriction map $X' \to L'$ is surjective). -Even if you modify the definition of $\mathcal M_p$ as suggested by Gerald Edgar -there is thus no reason to expect $\mathcal M_p'$ to be a subspace of $\mathcal M'$.<|endoftext|> -TITLE: Is every symplectic manifold a Hamiltonian reduction of a cotangent bundle? -QUESTION [14 upvotes]: Today I heard the claim that in practice, all symplectic manifolds that people care about arise as the Hamiltonian reduction of a cotangent bundle $T^{\ast}(M)$ under the action of a Lie group $G$ ($M$ and $G$ may both be infinite-dimensional in general, I think). For example many moduli spaces of interest arise in this way. Is it literally true that every symplectic manifold arises in this way? Can we moreover arrange for $G$ and $M$ to be finite-dimensional? - -REPLY [6 votes]: That reminds me of a paper that I believe should answer your question: $\mathbf R^{2n}$ is a universal symplectic manifold for reduction (available here): - -The authors show that if a manifold $Q$ is of finite type, that is, $H^k(Q,\mathbf Z)$ is finite-dimensional, then the noncanonical cotangent bundle $(T^∗Q,dθ_Q+τ^∗_QΩ)$ can be obtained by a Marsden-Weinstein reduction of $T^∗\mathbf R^n=\mathbf R^{2n}$ relative to a torus action. Here $Ω$ is a 2-form on $Q$, $τ_Q:T^∗Q→Q$ is the bundle map and $θ_Q$ is the canonical 1-form. Since any symplectic manifold $(Q,Ω)$ is a reduction of $(T^∗Q,dθ_Q+τ^∗_QΩ)$ relative to the zero section, it follows that any symplectic manifold can be obtained by a reduction of $\mathbf R^{2n}$ with the standard symplectic structure. (etc.)<|endoftext|> -TITLE: Jacobson ring = a ring whose nilradical and Jacobson radical coincide? -QUESTION [8 upvotes]: In Wikipedia it is claimed that "A ring is called a Jacobson ring if the nilradical coincides with the Jacobson radical." Here the word "ring" means a commutative ring. -However, I remember that Jacobson ring is defined to be a ring with the property "every prime ideal is equal to the intersection of some maximal ideals". -Are these two definitions really equivalent? - -REPLY [7 votes]: $\newcommand{\mm}{\mathfrak{m}}$The counterexamples given so far seem perhaps more elaborate than necessary. Here's another: -Let $A$ be a domain which is not a field and which has finitely many maximal ideals $\mm_1,\ldots,\mm_n$. (So e.g. take $\mathbb{Z}_p$ or $\mathbb{C}[[x_1,\ldots,x_n]]$ or $\mathbb{F}_{17}[[t^2,t^3]]$ or...you have some leeway here.) Let $R = A[t]$. Then: -$\bullet$ Like every domain, $R$ has zero nilradical. -$\bullet$ Like every polynomial ring over a domain, $R$ has zero Jacobson radical: if $0 \neq x \in R$, then $1+tx \notin A^{\times} = R^{\times}$. -$\bullet$: The ideal $\mathfrak{p} = \langle t \rangle$ of $R$ is prime. The maximal ideals of $R$ containing $\mathfrak{p}$ correspond to the maximal ideals $\mm_1,\ldots,\mm_n$ of $R/\mathfrak{p} = A$, so $\bigcap_{i=1}^n \mm_i \supset \mm_1 \cdots \mm_n \supsetneq (0)$. Thus $\mathfrak{p}$ is not the intersection of the maximal ideals containing it. -It may be helpful to mention the (easy) fact that a ring $R$ is Jacobson iff for all ideals $I$ of $R$, the nilradical of $R/I$ equals the Jacobson radical of $R/I$. This simplifies the last bullet point and is really the point of the construction. In fact for any ring $A$, the Jacobson radical of $A[t]$ is equal to the nilradical of $A[t]$ (Atiyah-Macdonald, Exercise 1.4), so taking any $A$ with nilradical smaller than its Jacobson radical works to give a counterexample: e.g. a domain which not a field and which has only finitely many maximal ideals.<|endoftext|> -TITLE: Inequalities for averaging over partially ordered sets -QUESTION [10 upvotes]: Let's start from a classical inequality: -If $0\le a_1\le\cdots\le a_k$ and $0\le b_1\le\cdots\le b_k$ then -$(a_1+\cdots+a_k)(b_1+\cdots+b_k)\le k(a_1b_1+\cdots+a_k b_k)$. -It can be written also in the form of averages: $Av(\{a_i\})Av(\{b_i\})\le Av(\{a_ib_i\})$ -(expressing convexity and many other properties) -I need the following generalization for finite partially ordered sets. Let $S$ be such a set, consider non-decreasing functions, i.e. $f:S\rightarrow\Bbb{R}_{\ge0}$, satisfying: if $a\ge b$ then $f(a)\ge f(b)$. -I need: $\underline{\text{ if $f,g$ are non-decreasing functions then } Av_S(f)Av_S(g)\le Av_S(fg).}$ -If $S$ is totally ordered, then one gets the classical version. -The inequality does not hold for arbitrary partial ordered sets (with obvious counterexamples). I guess a necessary condition is that $S$ has minimal and maximal elements. Even this is not enough (with obvious counterexamples). -In my particular case $S$ is the set of lattice points on a simplex, i.e.: $S_{n,r}:=\{(k_1,\dots,k_r)|\ k_1+\cdots+k_r=n,\ k_1,\dots,k_r\ge0\}$. (One can think about this as the set of monomials in r variables of total degree n.) The order is induced by recursive application of the rule $x^2_i\ge x_ix_j$. (So, e.g. $x^n_i\ge x^{n-1}_ix_j\ge x^{n-2}_ix_jx_k\ge\cdots$.) And the considered functions are symmetric (i.e. invariant w.r.t. to the permutation group $\Xi_r$, that acts on $S_{n,r}$.) -Alternatively, one can consider the quotient $S_{n,r}/\Xi_r$. (This set is partially ordered, with minimal and maximal elements.) -Probably in this particular case the inequality is well known? -I guess, a necessary condition on a partially ordered set to satisfy such an inequality (for any non-decreasing functions) is that $S$ is "ordered enough". Can this be made precise? Are there some sufficient conditions known? -For bookkeeping: in a very particular case (here) we proved this bound by terrible brute force. -upd: we have proved this inequality (for $S_{n,r}$) arXiv:1412.8200. - -REPLY [13 votes]: The natural generalization for your inequality is the setting of distributive lattices. The inequality is then known as the Fortuin–Kasteleyn–Ginibre (FKG) inequality, and has a long history. See for example Graham's article "Applications of the FKG inequality and its relatives". A generalization of the FKG inequality is the Ahlswede-Daykin inequality. For the appropriate specialisations, the FKG inequality implies -$$|A|\cdot |B|\le |A\cap B|\cdot |P|$$ -for all up sets $A,B$ as mentioned in domotorp's answer, and I don't know how this extends to lattices that are not distributive. The AD inequality, however specializes to -$$|A|\cdot|B|\le |A\wedge B|\cdot|A\vee B|$$ -which is always false when the lattice is not distributive. -Your Poset $S_{n,r}/\Xi_r$ is the lattice of partitions of $n$ with at most $r$ parts, with the dominance (majorization) ordering. This is always a lattice, however it is not distributive in general. For example $S_{n,n}/\Xi_n$ is not distributive for $n\geq 7$. It might still be worth investigating whether the FKG inequality holds for this lattice, even though we don't have distributivity, since it has other nice properties (for example the Mobius function is nice).<|endoftext|> -TITLE: Locally compact space that is not topologically complete -QUESTION [6 upvotes]: It is know that for a metric space, it is locally compact and separable iff exist an equivalent metric where a set is compact iff it is closed and limited. So, locally compact and seperable metric spaces are topologically complete. The question is: there is any metric space that is locally compact and not topologically complete? -Another motivation for this question is: Hausdorff locally compact spaces are Baire spaces, as well topologicaly complete space. But there are metric spaces that are Baire, but not topologicaly complete. So, the natural question is that above. -Obs: A metric space $(X,d)$ is topologically complete if there is a metric $d′$ on $X$ such that $Id:(X,d) \rightarrow (X,d′)$ is a homeomorphism and $(X,d′)$ is complete. - -REPLY [7 votes]: Every locally compact metric space can be given a compatible complete metric. -Suppose that $X$ is a locally compact metric space. Then $X$ is paracompact, so $X$ is a disjoint union of $\sigma$-compact locally compact spaces (There is a theorem proved in the book Topology by Dugundji that proves that every paracompact locally compact space is the disjoint union of $\sigma$-compact spaces. I prove this fact below.). Therefore we may without loss of generality assume that $X$ is $\sigma$-compact. Since we assume $X$ is $\sigma$-compact, there is a sequence of open sets $U_{n}$ such that $\overline{U_{n}}$ is compact and $\overline{U_{n}}\subseteq U_{n+1}$ for all $n$. From this sequence of open sets and Urysohn's lemma, there is a function $f:X\rightarrow[0,\infty)$ such that $\overline{U_{n}}\subseteq f^{-1}[0,n)\subseteq U_{n+1}$. Let $d$ be a metric on $X$, and define a new metric $d'$ on $X$ by letting $d'(x,y)=d(x,y)+|f(x)-f(y)|$. Clearly $(X,d')$ induces the original topology on $X$. I claim that $(X,d')$ is complete. Assume that $(x_{n})_{n}$ is a Cauchy sequence in $(X,d')$. Then the sequence $(x_{n})_{n}$ is bounded in $(X,d')$, so clearly the sequence $(f(x_{n}))_{n}$ is bounded as well. Therefore, there is some $N$ where $f(x_{n}) -TITLE: Fermat-quotient of "order" 3: I found $68^{112} \equiv 1 \pmod {113^3}$ - are there bigger examples known? -QUESTION [6 upvotes]: (I've taken this from MSE, it seems to be more appropriate here) -I'm rereading an older text on fermat-quotients (see wikipedia) from which I have now the -Question for -$$ b^{p-1} \equiv 1 \pmod{ p^m} \qquad \text{ with $p \in \mathbb P $, $1 \lt b \lt p$ and $m \gt 2$} $$ -(This is a generalization of the question for Wieferich primes). -Note that I ask here for examples, where the bases $b$ are smaller than the prime $p$, so a very well known weaker case $3^{10} \equiv 1 \pmod {11^2 } $ were an example, but only if the exponent at $11$ where one more; however frequent and well known cases like $18^6 \equiv 1 \pmod {7^3} $ were not because the base is bigger than the prime. -The only example that I've found so far is -$$ 68^{112} \equiv 1 \pmod {113^3 } $$ -but I've scanned only the first 2000 primes $p \in (3 \ldots 17389)$ and my primitive brute force algorithm has more than quadratic time-characteristic, so checking 10 000 or 100 000 primes were no fun - the quadratic regression prognoses 1 hour for testing 10 000 primes and 101 hours for testing 100 000 primes... -I'm aware of a couple of webpages containing lists of fermat quotients up to much higher primes, but either there is no explicite mention of the cases of $b \lt p$ and quotient $m \gt 2$ or I've been too dense when scanning through the listings (Richard Fischer, Wilfrid Keller, Michael Mossinghoff) - -For reference: my Pari/GP-code is -for(j=2,2000,p=prime(j);p3=p^3; - for(k=2,p-1, - r = lift(Mod(k,p3)^(p-1)); - if(r==1,print(p," ",k," ",r))); - ); - -(One correspondent took this up to the 10 000'th prime which is $p=104729$ ) - -[update] For the later casual reader I've included a more involved explanation and a table of data. See here (filesize 2Mb, inconvenient for modem-transfer) - -REPLY [6 votes]: Cases with $b -TITLE: Best known bounds on tensor rank of matrix multiplication of 3×3 matrices -QUESTION [16 upvotes]: Years ago I attended a conference where they taught us that matrix multiplication can be represented by a tensor. The rank of the tensor is important, because putting it into minimal rank form minimizes multiplications, so this seems like an important computational problem. -At the time (2007) the rank of 3×3 matrix multiplication was not known. Is this currently known? If not, what are the best kniwn bounds, and why is it difficult? - -REPLY [23 votes]: Federico Poloni's answer is up to date as far as I know, and his explanation of the difficulty is good, but here's another way to think about it. It's less concrete than his explanation, but perhaps useful. -The thing that confuses people is that matrix rank is not hard to compute. This is what we're used to, and it seems strange that computing tensor rank should be so much harder (for example, Håstad proved it is NP-hard). However, this is thinking about it backwards: tensor rank is the sort of thing we should expect to be hard, and the interesting part is that matrix rank is so much easier. The fundamental reason is symmetry. An $n \times n$ matrix involves $n^2$ parameters, and the general linear group is also $n^2$-dimensional, so it's reasonable to hope we can reduce every matrix to a simple canonical form by linear transformations. Indeed, we can reach row-echelon form, at which point we can simply read off the rank. By contrast, an $n \times n \times n$ tensor is $n^3$-dimensional, but there just isn't an equally large group acting on it. Instead, you still have quadratic-dimensional groups acting in several ways, which isn't nearly enough to reach any simple canonical form. Even after you mod out by all available symmetry, the dimension of the space of tensors is nearly as high as it was to start with, and you end up stuck. -For comparison, imagine trying to find the rank of a $9 \times 9$ matrix by a brute force search for rank-one summands, without using any simplifying transformations. That would be awful, and a $9 \times 9 \times 9$ tensor is far worse than that. Symmetry is our best tool by far, and when it is insufficient we don't have much to fall back on. - -REPLY [12 votes]: I think the best known result for the complexity of an exact algorithm for multiplying 3 by 3 matrices is 23 multiplications. Here is some recent work to find more algorithms since Laderman's 1976 result: -http://dl.acm.org/citation.cfm?id=2500627 -They find inequivalent algorithms, but still require 23 multiplications. -One thing to clarify from Federico's post. To specify any 9x9x9 tensor, you need to specify at most 9*9*9 = 729 parameters. To specify a rank 1 9x9x9 tensor, you could specify 9+9+9 = 27 parameters, but because of scaling you actually only need to specify 8 + 8 + 8 + 1 = 25 parameters (specify three 9-dimensional vectors up to scale, take their tensor product and then rescale globally). -To specify a rank 19 tensor (as a sum of 19 rank one tensors) you need to specify 19*25 = 425 parameters. This is much less than Federico estimated, but still very large, and still not feasible to do a brute force search. -About approximate rank, Schönhage found an approximate algorithm for 3x3 matrix multiplication that only uses 21 multiplications. -http://epubs.siam.org/doi/abs/10.1137/0210032 -The state of the art on lower bounds for rank and border rank for matrix multiplication (I think) is contained in these: -arxiv:1112.6007 -arxiv:1206.1530 - -REPLY [11 votes]: Stothers' thesis, from 2010, p. 18, lists 19 and 23 as the current lower and upper bounds. These are the same numbers that I have heard from some researchers in the field, and I don't think there has been any improvement since then. -Why is it a difficult problem? Simply because it is high-dimensional, non-convex optimization problem, with lots of local minima. To specify a rank-19 9x9x9 tensor, for instance, you need about 19x9x9x9 19x(9+9+9) entries (slightly less since you can normalize some things, actually), so that's the kind of dimensionality you are dealing with. Computers still can't deal reliably with this number of degrees of freedom for a problem like this. Even testing all possible decompositions with entries in $\{-1,0,1\}$ is not feasible. -Another difficulty for killing the problem numerically is that solutions can be "at infinity": for instance, let $T_\epsilon$ be the tensor such that -$$ -T(:,:,1)=\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}, T(:,:,2)=\begin{bmatrix}0 & 1 \\ 0 & \epsilon \end{bmatrix}. -$$ -$T_0$ has rank $3$, but any $T_\epsilon$ with $\epsilon \neq 0$ has rank 2. The decomposition can be parametrized explicitly and contains some $\epsilon^{-1}$ entries. So if you run an optimization algorithm to find a rank-2 decomposition of $T_0$, the residual will converge to 0, but at the same time the single entries of the decomposition will diverge. -In geometrical terms, the sets $S_\ell = \{T: \operatorname{rank} T \leq \ell \}$ are not closed, unlike in the matrix case. -EDIT: fixed error in number of degrees of freedom; see Luke Oeding's answer.<|endoftext|> -TITLE: Equivariant Stratifications of a Variety -QUESTION [7 upvotes]: Let $X$ be a complex variety acted upon algebraically by a complex torus $T$. Suppose that $\{X_{\beta}\}_{\beta\in S}$ is a finite $T$-equivariant stratification of $X$, so that the $X_{\beta}$ are smooth locally closed subvarieties and $$\overline{X_{\beta}}\subseteq\bigcup_{\gamma\leq\beta}X_{\gamma}.$$ For fixed $\beta\in S$, we have the equivariant Thom-Gysin sequences $$\ldots\rightarrow H_T^{i-2d(\beta)}(X_{\beta})\rightarrow H_T^i(\bigcup_{\gamma\geq\beta}X_{\gamma})\rightarrow H_T^i(\bigcup_{\gamma>\beta}X_{\gamma})\rightarrow\ldots,$$ where $d(\beta)$ is the complex codimension of $X_{\beta}$ in $X$. The idea is to inductively compute $H_T^*(X)$ from a knowledge of $H_T^*(X_{\beta})$ for each $\beta\in S$. This seems eminently possible if the partial order on $S$ is a total order. In this case, $X=\bigcup_{\gamma\leq\beta}X_{\gamma}$, where $\beta\in S$ is the maximal element. -However, there are some interesting examples in which $S$ is not totally ordered. Consider the nilpotent cone $\mathcal{N}$ of a finite-dimensional complex semisimple Lie algebra $\mathfrak{g}$. The nilpotent cone has a stratification into the nilpotent $G$-orbits, where $G$ is the simply-connected group with Lie algebra $\mathfrak{g}$. -$\textbf{Question}:$ Are there some general ways which to inductively compute $H_T^*(X)$ from the $H_T^*(X_{\beta})$ assuming only that $S$ is partially ordered? Are there some nice examples of this in the literature? - -REPLY [2 votes]: One possibility would be to follow the treatment of this subject given in Yang-Mills Equations over Riemann Surfaces by Atiyah and Bott. Call a subset $I\subseteq S$ open if whenever $\beta\in I$ and $\gamma\in S$ satisfy $\beta\leq\gamma$, we have $\gamma\in I$. For open $I$, we have the desirable property that $$X_I:=\bigcup_{\beta\in I}X_{\beta}$$ is open in $X$. Also, if $\beta$ is maximal in the complement of $I$, then $I\cup\{\beta\}$ is also open. So, we have the equivariant Thom-Gysin sequence, $$\cdots\rightarrow H_T^{i-2d(\beta)}(X_{\beta})\rightarrow H_T^*(X_{I\cup\{\beta\}})\rightarrow H_T^*(X_I)\rightarrow\cdots.$$ Assume that these sequences split into short-exact sequences. -We begin by selecting a maximal element $\beta_1\in S$. Set $I_1:=\{\beta_1\}$. Take a maximal element $\beta_2\in S\setminus I_1$ and set $I_2:=I_1\cup\{\beta_2\}$. Write the associated Thom-Gysin sequence to conclude that $$H^*_T(X_{I_2})\cong H_T^{*-2d(\beta_2)}(X_{\beta_2})\oplus H_T^*(X_{I_1})$$ as graded vector spaces. Next, choose a maximal element $\beta_3\in S\setminus I_2$ and set $I_3:=I_2\cup\{\beta_3\}$. Form the associated Thom-Gysin sequence to obtain $H_T^*(X_{I_3})$ as a graded vector space. Continue with this inductive procedure to obtain $H_T^*(X)$.<|endoftext|> -TITLE: Cotangent bundle lift theorem -QUESTION [8 upvotes]: Let $M$ be a smooth manifold and $T^\ast M$ be its cotangent bundle. Consider the tautological 1-form $\theta$ on $T^\ast M$ ($\theta=\sum y_i dx^i$ in local canonical coordinate systems). -A diffeomorphism $f:M\to M$ induces a pull-back lift $F=f^\ast:T^\ast M\to T^\ast M$. -It seems that we always have $F^\ast\theta=\theta$. But I don't know how to verify this. -I just heard the following cotangent bundle lift theorem: -If a diffeomorphism $F:T^\ast M\to T^\ast M$ preserves $\theta$, then $F=f^\ast$ for some diffeomorphism $f:M\to M$. -This looks too strong to me. Do you know how to prove this? -Thank you! - -REPLY [4 votes]: Just as a side-note. The proposition does not imply that every symplectomorphism of $T^*M$ is a cotangent lift of a diffeomorphism on $M$. As a counter example, let $\sigma$ be a non-zero closed one-form on $M$ and consider the map $F(x,p) = (x,p + \sigma(x))$. Notes that $F$ is not the cotangent lift of any diffeomorphism on $M$. Also $F^*d \theta = d(F^*\theta) = d( \sigma^H + \theta) = d\sigma^H + d\theta = d\theta$ where $\sigma^H$ is the one-form on $T^*M$ obtained by horizontal lift. Needless to say, if $\sigma$ is closed, then so is $\sigma^H$. Thus $F$ preserves the canonical symplectic form $d\theta$.<|endoftext|> -TITLE: Polyhedra from a tropical variety -QUESTION [5 upvotes]: it is known that tropicalization of a variety(irreducible and subvariety of some torus.) is a support of a polyhedral complex. I wonder which kinds of polyhedra can occur in this polyhedral complex. In other words, if P is any polyhedron of dimension d, then does some variety X always exist such that P is in a polyhedral complex of Trop(X)? - -REPLY [4 votes]: Provided that the facets of $P$ have integral normal vectors, this is always the case. -To construct such a tropical variety, first let $\Sigma$ be the normal fan to $P$. -This comes along with a piecewise linear function $\psi$ defined by -$\psi(n)=-inf\{ \langle n, m \rangle\,|\, m\in P\}$. Now choose any lattice polytope -$Q$ containing the origin as an interior point, and restrict $\psi$ to $Q$. -Take the tropical polynomial $$f=\sum_{m\in Q\cap {\bf Z}^n} \psi(m) z^m.$$ -It is then a simple exercise in convex geometry to see that this tropical polynomial -defines a tropical variety, which, when viewed as a polyhedral complex, contains -the boundary of the polytope $P$. In particular, the PL function defined by this tropical polynomial is the Legendre transform of the function $\psi$. You can see some -details of this in, say, Mikhalkin's paper http://arxiv.org/pdf/math/0312530.pdf, -in section 3. If you want to have the polytope $P$ itself appearing, then you -can take the tropical variety thought of as the graph of $f$, obtained by taking -the tropical polynomial $y+f$, where $y$ is an extra variable: see Proposition 3.5 -in the above cited paper.<|endoftext|> -TITLE: Maximizing ratio volume/diameter^n by an affinity -QUESTION [6 upvotes]: Suppose we have a convex compact body $D\subset \mathbb R^n$. We can try to apply affine transformation keeping the volume and decreasing the diameter of $D$. -It is clear that there is a constant $\lambda_n$ such that for any $D$ there is an affinity $F$ such that $diamater(F(D))^n\leq\lambda_n Volume(F(D))$. I'm interesting in the optimal value for $\lambda_n$. -The article "On the thinnest non-separable lattice of convex bodies" (E. Makai, p.23) gives an estimate $\lambda_n\leq (_n^{2n})n^{n/2}/k_n$ where $k_n$ is the volume of the convex hull of the unit sphere and $(\pm\sqrt{n},0,0,..)$, $\lambda_2$ is also computed. -I can not find in literature any better estimate for $\lambda_3$, but it seems that I can prove a few better estimate by school-like methods: -if there is no affinity decreasing diameter, that means that in $D$ there are a lot of diameters which will increase if we apply an infinitesimal affinity, so we get an estimate. I'm sure that I'm not the first one who apply such a simple idea to this problem. -Do you know any others references concerning this problem? - -REPLY [2 votes]: A closely related problem, considering the $(n-1)$-dimensional surface area instead of the diameter of the body, has been solved by Keith Ball, Volume ratios and a reverse isoperimetric inequality. J. London Math. Soc. (2) 44 (1991), no. 2, 351–359 [MR1136445]. The extreme case, as expected, is the $n$-dimensional simplex. Ball's proof may also work for the diameter. The article is available on the arXiv, see http://arxiv.org/abs/math/9201205 . The 2-dimensional case was solved long before, with a very short, elementary proof. -A quick estimate of the diameter can be obtained by taking the minimum volume ellipsoid containing the body $K$ and using the known (best possible, by the way) estimate of the ellipsoid's volume. An affine transformation that turns the ellipsoid into a ball yields a bound on the diameter - perhaps not the best possible, but a fairly good one.<|endoftext|> -TITLE: Name or references for minimal $N$ such that $\left(\frac{a}{b}\right)_n = \left(\frac{a}{b'}\right)_n$ whenever $b \equiv b' \bmod (N)$ -QUESTION [6 upvotes]: Let $\left( \dfrac{a}{b} \right)_n$ denote the nth power residue symbol, a generalization of the Legendre symbol. I have recently seen it quoted that there is a minimal ideal $N$ (minimal by ideal inclusion) such that if $b \equiv b' \bmod (N)$, then $\left( \dfrac{a}{b} \right)_n = \left( \dfrac{a}{b'} \right)_n$. This comes up when computing with automorphic forms on metaplectic covers of $GL(2)$. -Further, within the field $\mathbb{Q}(i)$ with ring of integers $\mathbb{Z}[i]$, I've heard it quoted that $N = \left( (1 + i)^3 \right)$. -I was hoping for references concerning a name for this $N$, a proof of its existence in general, and a method for finding $N$ given a reasonable number field. Can you help me out? - -REPLY [6 votes]: This integer is called the conductor of the power residue symbol. It coincides with the conductor of the Kummer extension $K(\sqrt[n]{a})/K$, where $K = {\mathbb Q}(\zeta_n)$ is the field of $n$-th roots of unity. This can be found in all decent books on class field theory, e.g. in Artin-Tate. For methods of computing conductors see the books by Cohen and Gras.<|endoftext|> -TITLE: Can the sum of two non-zero coprime fifth powers be powerful? -QUESTION [11 upvotes]: I am wondering if the sum of two non-zero coprime fifth powers can -be powerful. There are no small solutions. - -Q1 Can the sum of two non-zero coprime fifth powers be powerful? - -Got a partial result, possibly wrong. -Consider the surface: -$$ S: x^5+y^5-z^2 t^3=0 $$ -According to Magma it is a rational surface with complete parametrization: -x,y,z,t=u^8*s^2 + 10*u^7*v*s^2 + 43*u^6*v^2*s^2 + 104*u^5*v^3*s^2 + 155*u^4*v^4*s^2 + 147*u^3*v^5*s^2 + 90*u^2*v^6*s^2 + 36*u*v^7*s^2 + 8*v^8*s^2, u^7*v*s^2 + 9*u^6*v^2*s^2 + 34*u^5*v^3*s^2 + 70*u^4*v^4*s^2 + 85*u^3*v^5*s^2 + 62*u^2*v^6*s^2 + 28*u*v^7*s^2 + 8*v^8*s^2, u^5*s^5 + 10*u^4*v*s^5 + 40*u^3*v^2*s^5 + 80*u^2*v^3*s^5 + 80*u*v^4*s^5 + 32*v^5*s^5, u^10 + 10*u^9*v + 45*u^8*v^2 + 120*u^7*v^3 + 210*u^6*v^4 + 254*u^5*v^5 + 220*u^4*v^6 + 140*u^3*v^7 + 65*u^2*v^8 + 20*u*v^9 + 4*v^10 - -$$ -\begin{aligned} -x =& (u + v) \cdot s^{2} \cdot (u + 2 v)^{3} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}) \\ -y =& v \cdot s^{2} \cdot (u + 2 v)^{3} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}) \\ -z =& s^{5} \cdot (u + 2 v)^{5} \\ -t =& (u + 2 v)^{2} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4})^{2} -\end{aligned} -$$ -Which means $x^5+y^5 = z'^{10} t'^6$. -The inverse map is: -$$ -u=x^2 t - y^2 t, \; -v=x y t + y^2 t, \; -s=z t^2 -$$ -The inverse map implies for integer solution $(x,y,z,t)$, -$u,v,s$ must be integers. -From the parametrization -$$ -\gcd(x,y)=s^{2} \cdot (u + 2 v)^{3} \cdot (u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}) -$$ -And for coprimality, $|s|=1 ,\; | u + 2 v | =1 ,\; |u^{4} + 3 u^{3} v + 4 u^{2} v^{2} + 2 u v^{3} + v^{4}|=1$ which leads to solution of an -univariate polynomial, giving no solution. - -Q2 Is the above argument correct? - -It would be a proof to FLT for exponent $5$. -The related surface -$$ -S_7: x^7+y^7-z^3 t^4 = 0 -$$ -Is rational too. - -Q3 Is it true that for a prime $p \ge 5$, the surface - $ S_p: x^p + y^p - z^{\lfloor p / 2\rfloor} t^{p - \lfloor p / 2\rfloor} =0 $ is rational? - -Magma timeouts. According to a Macaulay2 program even $S$ is not rational. - -REPLY [10 votes]: Your formulas are only inverse up to rescaling. For example, $(u,v,s) = (1,1,1)$ maps to $(x,y,z,t) = (594, 297, 243, 1089)$ maps to $(u,v,s) = (288178803, 288178803, 288178803)$. More generally, -$$ (u,v,s) \mapsto (x,y,z,t) \mapsto (\Delta u,\Delta v,\Delta s)$$ -where -$$ \Delta = s^4 (u + 2 v)^9 (u^4 + 3 u^3 v + 4 u^2 v^2 + 2 u v^3 + v^4)^4. $$ -So $(x,y,z,t)$ integral doesn't imply $(u,v,s)$ integral, it only implies that $(\Delta u,\Delta v,\Delta s)$ is integral. - -Yes, $x^p+y^p = z^{\lfloor p/2 \rfloor} t^{\lceil p/2 \rceil}$ is rational. Put $p=2m+1$. I assume, since you call it a surface, you are considering this as a projective equation. Dehomogenize with respect to $t$ to get $x^{2m+1}+y^{2m+1} = z^m$. -In the open chart where $xyz \neq 0$, we can write this as $(x/y)^{2m+1} + 1 = z^m/y^{2m+1}$. Make the change of coordinates -$$a=xy^{-1},\ b=y^{-(2m+1)} z^m,\ c= y^2 z^{-1}$$ -with inverse -$$x=ab^{-1} c^{-m},\ y=b^{-1} c^{-m},\ z=b^{-2} c^{-(2m+1)}.$$ -In the new coordinates, we have $a^{2m+1}+1 = b$, with the obvious parametrization $(a,b,c) = (s,s^{2m+1}+1, t)$. -This is a standard trick for finding good rational forms of high degree equations with few variables: Find the lattice generated by the differences of the exponents -- in this case, the $\mathbb{Z}$-span of $(2m+1,-(2m+1),0)$ and $(0,-(2m+1), m)$. Saturate it -- we get the $\mathbb{Z}$-span of $(1,-1,0)$ and $(0,-(2m+1),m)$ and complete to a basis $\begin{pmatrix} 1 & -1 & 0 \\ 0 & - (2m+1) & m \\ 0 & 2 & -1 \end{pmatrix}$ of $\mathbb{Z}^3$. Inverting this determinant $1$ matrix gave me the formulas for $(x,y,z)$ in terms of $(a,b,c)$, and the equation become much lower degree in the new variables.<|endoftext|> -TITLE: How complicated is the formula expressing that a set is non-measurable? -QUESTION [6 upvotes]: The question is exactly stated by the title, i.e. how complicated is the formula $\psi(x)$ (in the language of set theory) expressing that a given set of reals $x$ is non-measurable? -A second question would be: can this formula be deduced from a formula (possibly with ordinal parameters or a countable sequence of ordinals) which involves only quantification over reals and natural numbers? -PS: By "deduced" I mean, given a formula $\varphi(x)$ defining a set $x$, then can we have $ZFC \vdash \forall x (\varphi(x) \rightarrow \psi(x))$. - -REPLY [10 votes]: Let me first consider the case where we use the usual notion of projectively definable set of reals, where $\varphi(x)$ is a property of the reals that might become a member of the set $X$ we are defining. Suppose that $X=\{ x\in\mathbb{R}\mid \varphi(x,z)\}$ is a projective set of reals, defined with a projective formula $\varphi$, which means that all quantifiers range only over reals and natural numbers, in a suitable language, with real parameter $z$. (One doesn't ordinarily allow ordinal parameters in a projective definition, and it isn't clear how one could use them anyway if all the quantifiers only range over reals and natural numbers.) -To say that $X$ is measurable is equivalent to the assertion that we may cover $X$ and its complement with open sets, whose intersection has as small a measure as we like: for every positive $\epsilon$, there are open sets $U\supset X$ and $V\supset\mathbb{R}\setminus X$ with $\mu(U\triangle V)\lt\epsilon$. Quantifying over open sets amounts to a real quantifier, since every open set is a countable union of rational intervals; and we may take $\epsilon=\frac1n$. -Thus, if $X$ has complexity $\Delta^1_n$ in the projective hierarchy, then the assertion "$X$ is measurable" has complexity at most $\Sigma^1_{n+1}$, and the assertion "$X$ is not measurable" has complexity at most $\Pi^1_{n+1}$, and these assertions are only a little bit more complicated than $X$ itself. In particular, these assertions are also projective. -For your second question, you seem to be asking whether there is a particular projective definition of a set of reals that we can prove (from ZFC?) that it is definitely not measurable. The answer is that we should not expect to do this, because the existence of certain large cardinals implies that every projective sets of reals is Lebesgue measurable, and so succeeding in your task would refute the consistency of those large cardinals, which would be an unexpected and amazing development. Meanwhile, of course, we cannot say for certain that we cannot find an example of what you seek, since the nonexistence of such a formula implies at the very least that ZFC is consistent, which is also something that, if true, we cannot prove. -Now let us consider your rather more idiosyncractic notion of definablity, where $\varphi(X)$ is a property of the set $X$ of reals, perhaps defined using only quantification over natural numbers and reals, where we allow $X$ to appear as a predicate in atomic subformulas of $\varphi$. The point to make now is that "$X$ is nonmeasurable" is expressible using quantifiers only over the natural numbers and the reals. So the family of all non-measurable sets is definable by a definition of your favored type. In this case, we can say, yes, indeed, there is a formula $\varphi(X)$ which ZFC proves holds of some sets, such that ZFC proves that $\forall X\ \varphi(X)\to X$ is not measurable, since in fact ZFC proves $\varphi(X)\leftrightarrow X$ is not measurable. - -REPLY [5 votes]: Let me add to Joel's excellent answer the following remarks: it follows from the classical work of Foreman, Magidor and Shelah that if there is a supercompact cardinal then all sets of reals in $L(\mathbb{R})$ are measurable. This means that starting with a supercompact cardinal, one can't contruct a wellordering of $\mathbb{R}$ in $L(\mathbb{R})$. In particular if there are $\omega$ Woodin cardinals, there is no projective wellordering of $\mathbb{R}$. So any formula expressing the non measurability of some set of reals would imply some anti-large cardinal statement (This is what Joel's is mentionning in his last paragraph). In other words under large cardinals, there are no wellorderings of a large initial segment of the Wadge Hierarchy (The Wadge Hierarchy is the hierarchy of complexity of the sets of reals, it goes all the way to $\Theta$ which is $\mathfrak{c}^+$ under $AC$ but it is very very large under large cardinals.)<|endoftext|> -TITLE: Status of the $x^2 + 1$ problem -QUESTION [18 upvotes]: It is a long-standing conjecture (probably just as old as the twin prime conjecture, which has gotten a lot of attention as of late since Zhang and Maynard's breakthrough results this year) that there exist infinitely many primes of the form $x^2 + 1$. The first few primes of this shape are $5, 17, 37, 101, \cdots$. Of course, no one has been able to prove this theorem, but perhaps if one relaxed the problem one would be able to obtain positive results. -So my question is: what is the least value of $n$ such that one can prove that infinitely many elements of $P_n$ (where $P_n$ is the set of numbers with at most $n$ prime divisors) are of the form $x^2 + 1$? -A similar result to this is a result of Selberg who proved that the polynomial $x(x+2)$ assumes values which have at most 7 prime factors, in particular there exist infinitely many elements of $P_7$ which are of the shape $x(x+2)$. -The parity problem in sieve theory would likely prevent any progress on obtain a lower bound for primes instead of almost primes. -Any insight would be appreciated. - -REPLY [30 votes]: Yes, Iwaniec proved in "Almost primes represented by quadratic polynomials" -(Invent. math. 47(1978) 171–188), that if $f$ is a quadratic polynomial with $f(0)$ an odd integer, then $f$ contains infinitely many elements of $P_2$. -For an arbitrary polynomial $f$ that is irreducible and doesn't have a fixed prime divisor, one can say that $f$ represents infinitely many elements of $P_n$, where $n=1+\deg f$. This was proven by Buhstab in "A combinatorial strengthening of the Eratosthenian sieve method", Usp. Mat. Nauk 22, no. 3, 199–226. - -REPLY [25 votes]: It is known that $n^2+1$ is infinitely often a number with at most $2$ prime factors, this is a result of Iwaniec, Almost-primes represented by quadratic polynomials, Invent. Math. 47 (1978), no. 2, 171–188. It is possible to get numbers with at most $3$ prime factors by a routine application of the weighted sieve, reducing this to $2$ requires a bilinear form of the error term and a bilinear level of distribution for the values of $n^2+1$. You can find a proof of the at most $2$ prime factors result in Friedlander and Iwaniec's book "Opera de Cribro".<|endoftext|> -TITLE: How to construct a group with specified growth function -QUESTION [10 upvotes]: Are there any procedures which given a nonnegative nondecreasing function on the integers will construct a finitely generated group with the same growth up to the usual equivalence of growth functions? Can this at least be done for nice functions? For example Bergman gives an explicit construction of semigroups of any growth between quadratic and cubic. -Also, is there an algorithmic way to do this if your function is recursive and given as input by a Turing machine? - -REPLY [15 votes]: Up to equivalence there is only one exponential type of growth in which case the answer is trivial. In polynomially bounded growth the answer follows from various old theorems (notably of Wolf and Gromov): the possible growths are $n^d$ for $d$ non-negative integer and things are classified. What's remaining is intermediate growth, and the first examples with growth determined up to equivalence were obtained quite recently by Bartholdi and Erschler (Inventiones, arxiv 1011.5266, 2012, see also their more recent arxiv 1110.3650). Although the growth of the first Grigorchuk group $\Gamma$ is not yet known up to equivalence, $\Gamma$ is used (in a permutational wreath product construction) in the construction of the Bartholdi-Erschler groups.<|endoftext|> -TITLE: A function composed with itself produces the identity -QUESTION [20 upvotes]: Let $B$ be the closed unit ball in $\mathbb R^3$ and $f: B\to B$ continuous, such that $f\circ f$ is the identity (i.e., $f\circ f=\mathbb 1_B$) and $f$ restricted on $\partial B$ is also the identity (i.e., $f|_{\partial B}=\mathbb 1_{\partial B}$). Does it imply that $f$ is the identity on $B$? -EDIT. If $B$ is instead the closed unit ball in $\mathbb R^2$, then the answer is also positive. (See Action on a Disk is Controlled by the Boundary, problem 10442, by -R. Bielawski & O. P. Lossers. Am. Math. Monthly, Vol. 105, No. 9 (Nov., 1998), pp. 860-861.) - -REPLY [27 votes]: Yes. Observe first that $f$ can be first extended to an involution of $\mathbb{R}^3$ and then to an involution $F : S^3 \rightarrow S^3$ of the one-point compactification of $\mathbb{R}^3$. A classical theorem of P. A. Smith then says that the fixed-point set of $F$ is homeomorphic to either $S^0$ or $S^1$ or $S^2$ or $S^3$; see Theorem 4 of -MR0000177 (1,30c) -Smith, P. A. -Transformations of finite period. II. -Ann. of Math. (2) 40, (1939). 690–711. -By the way, the proof shows that if $F$ is orientation-preserving, then the fixed-point set of $F$ must either be $S^1$ or $S^3$, while if $F$ is orientation-reversing, then the fixed-point set of $F$ must either be $S^0$ or $S^2$. In any case, from our construction it is clear that the fixed-point set of $F$ must be $S^3$, i.e. $F = \text{id}$. -I should remark that Smith's theorem is the beginning of a long story. See, in particular, the book -MR0758459 (86i:57002) -The Smith conjecture. -Papers presented at the symposium held at Columbia University, New York, 1979. Edited by John W. Morgan and Hyman Bass. Pure and Applied Mathematics, 112. Academic Press, Inc., Orlando, FL, 1984. xv+243 pp. ISBN: 0-12-506980-4 -The main theorem discussed in this book says that if $F$ is a nontrivial periodic orientation-preserving homeomorphism of $S^3$, then the fixed-point set of $F$ is an unknotted circle; this implies that $F$ is topologically conjugate to an element of the orthogonal group. This result was one of the first triumphs of Thurston's work on 3-manifolds. - -EDIT : As Mathieu Baillif points out in the comments, an easier way to finish this off would be to appeal to a theorem of M. H. A. Newman which asserts that if $M$ is any connected manifold and if $F:M \rightarrow M$ is a uniformly continuous periodic homeomorphism that fixes a nonempty open set in $M$, then $F$ is the identity. The reference for this result is -M. H. A. Newman, A theorem on periodic transformations of spaces, Q J Math (1931) (1): 1-8. -I remark that despite its age, this paper is very readable. - -EDIT 2 : I intended (but forgot) to mention in the above answer that the tricky point in this question is that $f$ is only assumed to be continuous. If $f$ is assumed to be smooth, then the question is much easier. Indeed, we have the following easy lemma. -LEMMA : Let $M$ be a smooth manifold with nonempty boundary and let $f : M \rightarrow M$ be a smooth map such that $f^k = \text{id}$ for some $k \geq 1$ (here the exponent means composition) and $f|_{\partial M} = \text{id}$. Then $f = \text{id}$. -To prove the lemma, we first prove that there is an open set $U \subset M$ such that $f|_U = \text{id}$. Choose a Riemannian metric $\mu'$ on $M$. Defining $\mu = \sum_{i=0}^{k-1} (f^i)^{\ast}(\mu')$, the Riemannian metric $\mu$ is $f$-invariant. Fix a point $p_0 \in \partial M$. Since $f|_{\partial M} = \text{id}$ we have $f(p_0)=p_0$. Even better, $D_{p_0} f : T_{p_0} M \rightarrow T_{p_0} M$ is the identity on a codimension $1$ hyperplane. Since $D_{p_0} f$ preserves the metric and orientation at $p_0$, we conclude that in fact $D_{p_0} f = \text{id}$. Using the exponential map, we deduce that $f$ is the identity on a neighborhood of $p_0$. -In particular, there exists a point $q_0$ in the interior of $M$ such that $f(q_0)=q_0$ and $D_{q_0}f = \text{id}$. Using an averaging trick as in the previous paragraph, we can choose a complete $f$-invariant Riemannian metric $\nu$ on $\text{Int}(M)$, which is a manifold without boundary. Since $\nu$ is complete, any two points in $\text{Int}(M)$ are connected by a $\nu$-geodesic. Using the exponential map at $q_0$, we thus deduce that $f|_{\text{Int}(M)} = \text{id}$, which implies that $f = \text{id}$.<|endoftext|> -TITLE: Mersenne almost primes -QUESTION [8 upvotes]: I asked earlier whether it can be proved that infinitely many elements of $P_n$ for some positive value of $n$ (here $P_n$ refers to the set of numbers with at most $n$ prime divisors). There I received a positive answer in the form of a 1978 paper of Iwaniec which confirmed that suitable quadratic polynomials do in fact hit $P_2$ infinitely often. -My question now concerns the situation with Mersenne primes. What is the least value of $n$, if it is known to be finite, such that it is known that infinitely many Mersenne numbers $2^p - 1$ (where $p$ is a prime) hits $P_n$ infinitely often. In other words, there exist infinitely many primes $p$ such that $2^p - 1$ has at most $n$ prime factors. -Edit: in view of the relatively clear answer given by http://www.math.ucsd.edu/~asalehig/SG_AffineSieveExpandersOverviewMSRI.pdf, it seems that what I asked is an open problem without much chance of being resolved in the near future, so I ask this modified question: -Let $M(x) = \{2^p - 1 : p \leq x\}$ where $p$ as usual denotes a prime. Is it possible to find two functions $f(x), g(x)$ that tend to infinity with $f$ taking on values in the positive integers such that -$$\displaystyle |M(x) \cap P_{f(x))}| \gg g(x)?$$ -In other words, can it be shown that the set of Mersenne numbers with $p \leq x$ having relatively few prime factors (here $f$ is understood to tend to infinity slowly) tends to infinity (perhaps slowly) as $x$ tends to infinity? - -REPLY [7 votes]: This is a classical problem, and it remains open to show that there exist arbitrarily large Mersenne numbers with a bounded number of prime factors. The work of Bourgain, Gamburd, and Sarnak on the affine sieve may be seen as a generalization of this kind of question, but their work does not say anything for Mersenne numbers. An exposition of this work may be found in Kowalski's Bourbaki article http://arxiv.org/abs/1012.2793 ; also see these lecture notes of Golsefidy http://www.math.ucsd.edu/~asalehig/SG_AffineSieveExpandersOverviewMSRI.pdf which explicitly mentiones the connection with Mersenne numbers.<|endoftext|> -TITLE: Did Gödel prove that the Ramified Theory of Types collapses at $\omega_1$? -QUESTION [5 upvotes]: Second-Order Arithmetic is considered impredicative, because the comprehension scheme allows formulas with bound second-order variables that range over all sets of natural numbers, including the set being defined. The standard resolution to this impredicativity is known as the Ramified Theory of Types, which divides the comprehension schema into levels. The comprehension schema for level $0$ sets does not allow formulas with any second-order quantification. The schema for level $1$ sets only allows quantification over level $0$ sets. For any natural number n, the schema for level $n+1$ sets allows quantification over sets of level $n$ and below. -But there's no obvious reason why we need to stop at finite levels. For instance, the schema for level $\omega$ sets allows quantification over sets of all finite levels. And so on, for higher and higher transfinite ordinals. The question is which ordinals to use. One viewpoint is that we should be predicative about our choice of ordinals as well. This means that we should only allow a comprehension schema for level $\alpha$ sets if we have already shown using lower-level comprehension schemata that $\alpha$ is a well-founded ordinal. Feferman and Schütte showed that if we proceed in this manner, then we'll ultimately get levels corresponding all ordinals up to a certain ordinal known as $\Gamma_0$, the Feferman-Schütte ordinal. -But it seems that before Feferman and Schütte, Gödel pursued an alternate approach in which we adopt a Platonistic as opposed to a predicativist view concerning what ordinals to use to index our comprehension schemes. And according to this Stanford Encylopedia of Philosophy article, he was apparently able to show that if you kept going to higher and higher ordinals, the Ramified hierarchy "collapses" at the ordinal $\omega_1$ (the least uncountable ordinal), in the sense that if you went to higher levels than that you can't prove any more statements. -First of all, is it true that Gödel proved such a result? If so, why is it that start with an arithmetical theory $T$ and keep adding consistency principles like $Con(T)$ and $Con(T+Con(T))$ and so on, this procedure collapses at the ordinal $\omega_1^{CK}$ (the least non-recursive ordinal), yet here you have to go all the way up to $\omega_1$? -More importantly, if you do make the levels go up to $\omega_1$, how much arithmetic can you prove in the resultant theory? Can you prove as much as second-order arithmetic can? Or can you only prove as much as some weaker subsystem of second-order arithmetic, and if so, how does that subsystem fit into the framework of reverse mathematics? What is the proof-theoretic ordinal of this theory? -Any help would be greatly appreciated. -Thank You in Advance. - -REPLY [3 votes]: I was reading Feferman's original 1964 paper "Systems of Predicative Analysis", and I think I've found an answer to my question about how much second-order arithmetic you can do if you let the ramified hierarchy go high enough. Feferman defines $M_\alpha$ to be the collection of all sets defined by comprehension schemes for level $\alpha$ and below. And on page 10 of the paper, Feferman says that Kleene proved in this paper that "$M_{\omega_1}$ consists exactly of the hyperarithmetic sets." Now one concern I have is that Feferman seems to be using the notation $\omega_1$ to refer to the Church-Kleene ordinal $\omega_1^{CK}$ rather than the first uncountable ordinal, which he denotes by $\Omega$. -But the work that Feferman does on page 11 somewhat alleviates this concern. For any collection of sets $D$, a formula $\phi$ is said to be "definite relative to $D$" if $\phi(x)$ if and only if $\phi_D(x)$, where $\phi_D$ is the formula obtained by restricting all the second-order quantifiers in $\phi$ to the sets in $D$. Let us define $D_0$ to be the collection of all sets defined by formulas with no second-order quantifiers. For any natural number $n$, let $D_{n+1}$ consist of all the sets in $D_n$ together with all the sets defined by formulas which are definite relative to $D_n$. And then $D_\omega$ is equal to the union of $D_n$ for all $n$. And so on, for higher and higher ordinals. Now Feferman shows that $D_{\Omega+1} = D_\Omega = D_{\omega_1}$, all of which are equal to the collection of hyperarithmetic sets. -Now Feferman doesn't explicitly say this, but I assume that analogously, $M_\Omega = M_{\omega_1}$. Can anyone back me up on that? If I'm right about that, and if the Stanford Encylopedia of Philosophy article is right that the ramified hierarchy collapses at $\Omega$, then that would mean that letting the ramified hierarchy go arbitrarily high would only get you to the hyperarithmetic sets, which is considerably weaker than what full second-order comprehension buys you.<|endoftext|> -TITLE: What kind of probability distribution maximizes the average distance between two points? -QUESTION [17 upvotes]: If $f$ is a probability distribution on the unit disk in $\mathbb{R}^2$, and $X_1$ and $X_2$ are two independent samples from $f$, then what is the distribution $f^*$ that maximizes the average distance between these two samples, $E\|X_1-X_2\|$? Should all of the probability mass be distributed along the perimeter? - -REPLY [19 votes]: The uniform distribution on the circle is optimal. -Every probability measure on the disc can be approximated by the sum of atomic measures with equal wieghts, that is, by measures of the form $\frac1n\sum_{i=1}^n \delta_{p_i}$ where $p_1,\dots,p_n$ are points in the disc. For such a measure, the average distance equals $\frac1{n^2}\sum_{i,j} |p_i-p_j|$. It suffices to show that for every $n$ and every collection of points $p_i$, this quantity is no greater than the average distance w.r.t. the uniform distribution on the circle. -Let us show that, for every fixed $n$, the maximum of the sum of pairwise distances $\sum_{i,j} |p_i-p_j|$ is attained when $p_i$'s are vertices of a regular $n$-gon insribed in the boundary circle. This implies the desired inequality. -First of all, the maximum exists by compactness. Then, since the distance is a convex function, the maximum is attained when all points are on the boundary. Now re-enumerate the points according to the cyclic order on the circle and split the sum into $\sum_i |p_i-p_{i+1}|$ (the sides), $\sum_i |p_i-p_{i+2}|$, etc. -For each fixed $k$, the sum $\sum_i |p_i-p_{i+k}|$ is maximized by a regular $n$-gon. Indeed, we have $p_i-p_{i+k}=2\sin(\alpha_i/2)$ where $\alpha_i$ is the angular measure of the arc between $p_i$ and $p_{i+k}$. So our sum equals $2\sum_i\sin(\alpha_i/2)$ where $0\le\alpha_i\le 2\pi$ and $\sum_i\alpha_i=2\pi k$. Since the function $t\mapsto\sin(t/2)$ is concave on $[0,2\pi]$, by Jensen's inequality the maximum is attained when all $\alpha_i$'s are equal, q.e.d.<|endoftext|> -TITLE: Has the Ramified Theory of Types been applied to NBG? -QUESTION [11 upvotes]: Questions of predicativity are well-studied in the context of arithmetic. We have a base theory, first-order Peano arithmetic. Some people, like Edward Nelson (in chapter 1 of his book) and Charles Parsons (in this paper), consider it to be impredicative, because induction allows a formula $\phi$ even if it contains quantifiers which range over the natural numbers, even though natural numbers are conceived as objects which satisfy all inductive formulas, including the formula $\phi$ itself. So Nelson has constructed a subsystem of first-order arithmetic which doesn't allow induction for much more than formulas with bounded quantifiers. -But some people like Weyl and Poincare are willing to accept the natural numbers as given in advance as a completed totality, so they're willing to accept first-order PA, but they're not willing to accept arbitrary sets of natural numbers on a Platonic basis. This leads them to consider second-order PA as impredicative, because its comprehension schema allows a formula $\phi$ even if it contain quantifiers that range over all sets of natural numbers, including the set currently being defined by $\phi$. This leads people to constructing subsystems of second-order PA such as $ACA_0$ whose predicative comprehension schema only allows formulas with no quantification over sets of real numbers. -Yet arithmetic, even though it gets the most focus from predicativist critics, is not the only subject where we run into these issues. They also arise in set theory. Like first-order PA, we have a base theory, ZF (or ZFC if you prefer). Some people consider ZF impredicative, because the seperation schema allows formulas that have quantifiers that range over all sets, including the one being defined. So people have made subsystems of ZF, like Kripke-Platek set theory and Myhill's constructivist set theory, which have a predicative seperation schema which only allows formulas with bounded quantifiers. (This is analogous to Nelson's predicative first-order arithmetic.) -But just like in the case of arithmetic, some people are willing to accept the universe of sets on a Platonic basis, so they accept ZF, but they're not willing to do the same for proper classes. This leads them to reject Morse-Kelley set theory as impredicative, because its class comprehension schema allows quantification over classes, including the one being defined. So just like $ACA_0$, people have constructed a subsystem of MK known as NBG with a predicative class comprehension schema that does not allow quantification over classes. -My question is, has anyone tried to apply the techniques developed to deal with predicativity for arithmetical theories, specifically the ramified theory of types, to predicative set theories? The ramified theory of types is a way to expand the strength of a predicative theory, by breaking an axiom schema into levels. Here is an illustration in the context of second-order arithmetic: we have a comprehension schema for level $0$ sets, which allows no quantification over sets of natural numbers. That's the comprehension schema for $ACA_0$, but we don't stop there. We have a comprehension schema for level $1$ sets, which only allows quantification over level $0$ sets. And in general, for any natural number $n$, the comprehension schema for level $n+1$ sets allows quantification over sets of level $n$ and below. And don't need to stop at finite levels. We can continue to a comprehension schema for level $\omega$ sets which allows quantification over sets of any finite level, and so on, for higher and higher transfinite ordinals. -The ramified theory of types allowed Feferman and Schutte to start from a relatively weak predicative theory $ACA_0$ which was conservative over first-order $PA$, and move to a much stronger predicative theory like $ATR_0$, as I discuss in this question. It similarly allowed Burgess and Hazen to start from Edward Nelson's theory, which was not much stronger than bounded arithmetic, and move to exponential function arithmetic, as I discuss in this question. -So has anyone tried to do similar things to predicative set theories? For instance, we can replace the predicative class comprehension schema of NBG with comprehension schemata for classes of various levels, just like was done for predicative second-order arithmetic. And we can take Kripke-Platek set theory and replace its predicative seperation schema with seperation schemata for sets of various levels. This may allow for the construction of much stronger predicative set theories. -I think this might have some relation to the constructible hierarchy. Can anyone confirm that? -Any help would be greatly appreciated. -Thank You in Advance. -EDIT: I haven't read it, but this chapter from Skolem's book "Abstract Set Theory" may be useful. - -REPLY [4 votes]: You might be interested in looking up some of the work Kentaro Fujimoto has been doing on predicativism about classes. See this abstract for a talk from last year, though I'm not finding anywhere that the ideas have been published yet. Looking at my notes from a similar talk by him in Warsaw, he defends what he calls "liberal predicativism", which on the formal side leads to an adoption of ETR—Elementary Transfinite Recursion, a set theoretic analogue of ATR. See also his paper "Classes and truths in set theory", which contains a thorough mathematical investigation of some of these ideas. -A second topic of potential interest to you is the construction of Gödel's L in the classes, which has been independently carried out by many set theorists. To my knowledge, the first published account of this is a 1973 paper by Wiktor Marek: "On the metamathematics of impredicative set theory". He uses what is essentially MK as his base theory, but the construction of L in the classes can be done from a much weaker base theory. In fact NBG + ETR suffices, as I show in section 2.3 of my dissertation. So if we accept NBG + ETR as being predicative, then we can predicatively carry out the construction of L beyond Ord. And ETR is the weakest principle extending NBG which suffices for this; cf. theorem 3.14 of my dissertation.<|endoftext|> -TITLE: runs of consecutive non squarefree integers -QUESTION [9 upvotes]: This question gained no attention at Math SE. -Call a sequence of $k$ consecutive naturals squary if each one of them is divided by a square > 1. The Chinese Remainder theorem trivially guarantees us squary $k$-sequences for each $k$ (even with pre-defined divisors in a given order). -Denote by $f(k,n)$ the number of squary sequences $(x+1,x+2,...,x+k)$ contained in $[1,n]$. For $k$ fixed and $n\to\infty$, what is known about the asymptotics of $f(k,n)$? -It is easy to show that $f(1,n)\sim (1-\frac6{\pi^2})n $. - -REPLY [3 votes]: The original question is answered by Lucia. I was thinking on, what if $k$ depends on $n$? I'm interested on question whether $k = \frac{\log n}{\log\log n}$. -In related question (Consecutive non squarefree integers) we have seen that $f\left(\frac{\log n}{\log\log n},n\right) > 0$ infinitely often. -Now I prove that, for any $\varepsilon$ -$$f\left(\frac{\log n}{\log\log n}, n\right) \geq n^{\frac{1}{2}- \varepsilon}$$ -infinitely often. -For some $k$ and for some $\pi$ permutation of $\{k+1, k+2, \ldots ,p_k^2\}$ let the following congruence system: -$$ -\begin{array}{ccll} - x_{\pi}& \equiv & k& \mod{p_k^2} \\ - x_{\pi}& \equiv & k+1& \mod{p_{\pi(k+1)}^2} \\ - & \colon & & \\ - x_{\pi}& \equiv & p_k^2& \mod{p_{\pi(p_k^2)}^2} - \end{array} -$$ -The system for any $\pi$ has a solution $x_{\pi}$ smaller than $\Pi_{i=k}^{p_k^2} p_i^2$, and $\{x_{\pi}-k, \ldots, x_{\pi}-p_k^2\}$ can not contain square-free numbers. -If $\pi_1 \neq \pi_2$, then the difference between $x_{\pi_1}$ and $x_{\pi_2}$ is at least $p_k^2-k$, because in $\{x_{\pi}-k, \ldots, x_{\pi}-p_k^2\}$ exactly one integer is divisible by $p_i^2$ for $k\leq i \leq p_k^2$, and only $x_{\pi}-k$ is divisible by $p_k^2$, so the intervals $[x_{\pi_1}-p_k^2, x_{\pi_1}-k]$ and $[x_{\pi_2}-p_k^2, x_{\pi_2}-k]$ are not overlapping. From that, there are at least $(p_k^2-k-1)!$ disjoint intervals of size $p_k^2-k$. -So let $m = p_k^2 \sim k^2\log^2k$, there are $(m-o(m^{1/2}))!$ disjoint intervals of size $m-o(m^{1/2})$ which doesn't contain square-free integers until $\Pi_{i=1}^{m} p_i^2 \leq e^{2(1+\varepsilon)m\log m}$. By letting $m = \frac{\log n}{\log\log n}$, the Stirling formula gives the solution. -Remarks: With more carefully written the congruence system, the same holds for $\frac{\pi^2}{6}\frac{\log n}{\log\log n}$. I think this also motivate that, $\frac{\pi^2}{6}\frac{\log n}{\log\log n}$ is not the best lower bound, but the fact $f\left((1+\varepsilon)\frac{\pi^2}{6}\frac{\log n}{\log\log n},n\right) > 0$ happens infinitely often, is not known.<|endoftext|> -TITLE: Counting 2m X 2m 0-1 matrices with m ones in each row and each column. -QUESTION [14 upvotes]: Given $m>1$, what is the number of $2m\times 2m$ matrices, made of $0$ and $1$, such that each row has exactly $m$ ones, and each column has exactly $m$ zeros. -I am not sure if this is a well-known problem. - -REPLY [25 votes]: An explicit formula for this was published about 30 years ago, but it was wrong. As the matter stands, there is no explicit formula. The values up to m=15 are here. The value for m=16 is known too, let me know if you'd like me to track it down. The asymptotic value appears in this paper. If I'm not mis-translating it is -$$ e^{-1/2+o(1)}\frac{\binom{2m}{m}^{4m}}{\binom{4m^2}{2m^2}}, $$ -which you could apply Stirling's formula to.<|endoftext|> -TITLE: Can there be only one (uncountable transitive model of ZFC)? -QUESTION [12 upvotes]: It is an immediate consequence of Cohen's forcing that if there is one countable transitive model of $\sf ZFC$, then there are many of them. Even if all of these models are of the same height, there are still many. -But the proof of existence of generic filters don't carry over for uncountable models of set theory. Not without further assumptions (like Martin's axiom) anyway. -Is it possible to have an uncountable transitive model which is the only uncountable transitive model? Since that model is going to have to be $L_\alpha$ for some $\alpha>\omega_1$, is it also consistent with $V\neq L^1$, that a unique uncountable model exists? -Can this be pushed arbitrarily high (perhaps with, or without the assistance of $\sf MA$ and the Tennenbaum-Solovay theorem) so there are many transitive models of size $<\kappa$, but only one of size $\kappa$? - -Footnotes: - -Clearly one can just add a Cohen real to the universe without damaging any uncountable sets in a mind-shattering way. And on the other hand, if $V\neq L$ in a strong enough way, i.e. $0^\#$ existing in the universe, then there are arbitrarily large uncountable transitive models of $\sf ZFC$. Simply $L_\kappa$ for $\kappa$ a cardinal, or a Silver indiscernible. -So my purposely vague question about $V\neq L$ is some middle-ground between the very uninteresting $L[r]$ for some "simple" real number $r$, and the existence of large cardinals (which automatically implies there are many uncountable transitive models). - -REPLY [12 votes]: No, this is not possible. -As Mohammad argued in the comments, such a model $M$ must contain all countable ordinals. -We may assume that $M = L^M$, otherwise $L^M$ is another transitive model of ZFC which is uncountable since $\omega_1 \subseteq L^M$. If $M = L^M$, then $M = L_\alpha$ for some $\alpha \geq \omega_1$. The cardinal $\omega_1$ must be an uncountable regular cardinal in $L$ and I will now write $\kappa$ instead of $\omega_1$ to avoid confusion with $\omega_1^L$. We may assume that $\alpha \lt (\kappa^+)^L$ (i.e. $L \vDash |M| = \kappa$). Otherwise we could apply Löwenheim-Skolem in $L$ to get $N \prec M$ with $\kappa \subseteq N$ and $|N| = \kappa$; the transitive collapse of $N$ would then be an uncountable model of ZFC different from $M$. -There are now two cases: - -If $\alpha \gt \kappa$, then the poset $(2^{\lt\kappa})^L$ is in $M$. Working in $L$, using the fact that $2^{\lt\kappa}$ is $(\lt\kappa)$-closed and that $|M| \leq \kappa$, we can easily construct a $M$-generic $G$ for $2^{\lt\kappa}$. Then, the generic extension $M[G]$ is an uncountable transitive model of ZFC different from $M$. -If $\alpha = \kappa$ then every set in $M$ is countable in $V$. Therefore, every set forcing in $M$ has an $M$-generic set in $V$, which leads to a plethora of different uncountable transitive models of ZFC.<|endoftext|> -TITLE: Why is there no product type in simply typed lambda-calculus? -QUESTION [12 upvotes]: Consider simply typed $\lambda$-calculus that has only the unit type as primitive. We would like to encode the product and the sum types. An encoding of the product type in the untyped $\lambda$-calculus is this: -Pair = $\lambda a.\lambda b.\lambda f. f\ a\ b$ -First = $\lambda p. p (\lambda x. \lambda y.x)$ -Second = $\lambda p. p (\lambda x. \lambda y.y)$ -Then p = Pair a b is a fully constructed pair, of which we can compute (First p) to recover a and (Second p) to recover b. -However, this construction cannot be given a consistent type in simply typed $\lambda$-calculus. The immediate problem is the type of $f$ that cannot be consistently assigned, because First and Second will not, in general, have equal types if $a$ and $b$ have different types. If $a$ has type $\alpha$ and $b$ has type $\beta$ then $f$ needs to have type $\alpha\to\beta\to\alpha$ and at the same time $\alpha\to\beta\to\beta$. This forces us to have $\alpha=\beta$, or else the types are inconsistent. -Well, if this construction does not work then maybe another one will work. But it seems to me that nothing can work here. It is not possible to define some $\lambda$-terms with consistent types to implement the product type. -How can one prove that the simply typed $\lambda$-calculus does not support the product type? Equivalently, to show that the category of types in this calculus does not contain product objects for unequal types. Equivalently, to show some appropriate statement about propositions that will follow from the Curry-Howard isomorphism. Equivalently, to show that some formula is not deducible in the intuitionistic logic... What is the method that can derive this kind of statement rigorously? Where can I find such a proof if it is readily found? -It is interesting to note that the sum type can be encoded in simply typed $\lambda$-calculus without such problems. -Left = $\lambda x.\lambda f. \lambda g.f\ x$ -Right = $\lambda y.\lambda f. \lambda g.g\ y$ -Choice = $\lambda c.\lambda f. \lambda g.c\ f\ g$ -If we assume that $x$ has type $\alpha$ and $y$ has type $\beta$, while (Choice c f g) returns a result of type $\rho$, it is straightforward to assign types consistently to Left, Right, and Choice. -Why is it that the product type is impossible (if this is true) while the sum type can be encoded? -PS -Here is an OCaml code that illustrates the problem with the product type. - # let pair (a:int) (b:bool) = fun f -> f a b;; - val pair : int -> bool -> (int -> bool -> ’a) -> ’a = - # let first p = p (fun (x:int)(y:bool) -> x);; - val first : ((int -> bool -> int) -> ’a) -> ’a = - # let p1 = pair 1 true;; - val p1 : (int -> bool -> ’_a) -> ’_a = - # first p1;; - - : int = 1 - # let second p = p (fun (x:int)(y:bool) -> y);; - val second : ((int -> bool -> bool) -> ’a) -> ’a = - # second p1;; - Error: This expression has type (int -> bool -> int) -> int - but an expression was expected of type (int -> bool -> bool) -> ’a - # p1;; - - : (int -> bool -> int) -> int = - -After applying first to a fully constructed pair p1, the type of the pair becomes monomorphic, and we cannot use second on p1 anymore. (Here I am telling OCaml that x and y are monomorphic and have particular types, in order to simulate the simply typed $\lambda$-calculus in Ocaml. If OCaml had a monomorphic, i.e. simply typed, implementation of $\lambda$-calculus, we would not be able to define pair, first, second at all.) -Here is OCaml code for implementing the sum type. - # let left (x:int) (f:int->bool)(g:unit->bool) = f x;; - val left : int -> (int -> bool) -> (unit -> bool) -> bool = - # let right (x:unit) (f:int->bool)(g:unit->bool) = g x;; - val right : unit -> (int -> bool) -> (unit -> bool) -> bool = - # let case (c:(int -> bool) -> (unit -> bool) -> bool) f g = c f g;; - val case : - ((int -> bool) -> (unit -> bool) -> bool) -> - (int -> bool) -> (unit -> bool) -> bool = - # let la = left 1;; - val la : (int -> bool) -> (unit -> bool) -> bool = - # let rb = right ();; - val rb : (int -> bool) -> (unit -> bool) -> bool = - # case la (fun x->x=1) (fun y->false);; - - : bool = true - # case rb (fun x->x=1) (fun y->false);; - - : bool = false - -Note that case, right, left have been fully specified and are never polymorphic. - -REPLY [7 votes]: You have not really encoded the sum type $\alpha \sqcup \beta$, but made a "virtual embedding" of the sum type into $(\rho^{\rho^\beta})^{\rho^\alpha}$. You could not encode sum types in such a calculus, simply because sum types are not expressible in it (you may know this result in a different form --- disjunctions are not definable in an implicational fragment of intuitionistic propositional logic). To see how your embedding works, let me assume for a moment that our calculus has both sum and product types. Then: -$$(\rho^{\rho^\beta})^{\rho^\alpha} \approx \rho^{\rho^\alpha \times \rho^\beta} \approx \rho^{\rho^{\alpha \sqcup \beta}}$$ -Therefore, your embedding may be thought of as canonical morphism: -$$\alpha \sqcup \beta \rightarrow \rho^{\rho^{\alpha \sqcup \beta}}$$ -Actually, $\rho^{\rho^{(-)}}$ is a continuation monad, and the embedding $\tau \overset{\mathit{ret}}\rightarrow \rho^{\rho^{\tau}}$ lifts types to the continuation semantics, where as you have figured out, coproducts "virtually exist". By the way, I have said "embedding", but $\mathit{ret}$ generally is not an embedding --- it is a monomorphism if and only if $\rho$ is an internal cogenerator --- which means that the internal contravariant hom-functor $\rho^{(-)}$ is faithfull (see also this answer). -As pointed out by Peter in the comments, you may similarly "embed" product types: -$$\alpha \times \beta \rightarrow \rho^{\rho^{\alpha \times \beta}} \approx \rho^{(\rho^{\beta})^\alpha}$$ -Then instead of working with $\alpha \times \beta$, work with continuation type $\rho^{(\rho^{\beta})^\alpha}$. For example you may define the first projection $\rho^{(\rho^{\beta})^\alpha} \rightarrow \rho^{\rho^\alpha}$ as: -$$\lambda p \colon \rho^{(\rho^{\beta})^\alpha} . \lambda f \colon \rho^\alpha . p (\lambda a \colon \alpha . \lambda b \colon \beta . f a \colon \rho)$$ -and in the same manner the second projection $\rho^{(\rho^{\beta})^\alpha} \rightarrow \rho^{\rho^\beta}$ as: -$$\lambda p \colon \rho^{(\rho^{\beta})^\alpha} . \lambda g \colon \rho^\beta . p (\lambda a \colon \alpha . \lambda b \colon \beta . g b \colon \rho)$$ -However, it is worth saying that to emulate product types you do not need the above construction --- functions from a product type $\alpha \times \beta \rightarrow \tau$ are tantamount to functions $\alpha \rightarrow \tau^\beta$, and function to a product type $\tau \rightarrow \alpha \times \beta$ are tantamount to pairs of functions $\tau \rightarrow \alpha$ and $\tau \rightarrow \beta$, therefore in both cases are (in some sense) "redundant". - -To not be worse, here is my GHC session: -pair :: Integer -> Bool -> (Integer -> Bool -> Bool) -> Bool -pair a b = \c -> c a b - -first :: ((Integer -> Bool -> Bool) -> Bool) -> (Integer -> Bool) -> Bool -first p = \f -> p (\a b -> f a) - -second :: ((Integer -> Bool -> Bool) -> Bool) -> (Bool -> Bool) -> Bool -second p = \g -> p (\a b -> g b) - -*Main> let p = pair 2 True in first p ((<)1) -True -*Main> let p = pair 2 True in first p ((<)3) -False -*Main> let p = pair 2 True in second p ((||) False) -True -*Main> let p = pair 2 False in second p ((||) False) -False<|endoftext|> -TITLE: Intermediate submodels and the continuum hypothesis -QUESTION [5 upvotes]: Let $V$ be a model of $ZFC+GCH$ and let $V[G]$ be a generic extension of $V$ in which $CH$ fails. -Question 1. Is there a model $W$ such that: -1) $V \subseteq W \subseteq V[G],$ -2) $W\models CH,$ -3) $W$ and $V[G]$ have the same cardinals. -Question 2. The same question as above, this time assuming $V=L$ and there are no large cardinals? - -REPLY [5 votes]: The answer is no in general. If $V[G]$ is a model of Martin's maximum and $W$ is an inner model of $V[G]$ with the same $\omega_2$, then $\mathcal P(\omega_1)^W=\mathcal P(\omega_1)^{V[G]}$, so $\mathsf{CH}$ fails in $W$. In fact, many strong reflection principles can be used here instead of $\mathsf{MM}$, for instance we could have $V[G]$ a model of Rado's conjecture. Also, (if there is a $\Sigma_1$-reflecting cardinal in $L$) there is a forcing extension $L[G]$ of $L$ where $\mathsf{BPFA}$ holds, and if $W$ is an inner model of $L[G]$ with the same $\omega_2$, then $W=L[G]$.<|endoftext|> -TITLE: Representation-theoretic operations on modular forms -QUESTION [7 upvotes]: Let $A$ and $B$ be Hecke eigenforms of some weight $k$ and level $N$. We know that there are irreducible representations $\rho_a$, $\rho_b$ of the absolute Galois group of $\mathbb{Q}$ whose trace of Frobenius of $p$ is given by $a_p$ or $b_p$, the Fourier coefficients of $A$ or $B$. -$AB$ is a modular form of weight $2k$. It isn't clearly a Hecke eigenform, but its Fourier coefficients are still algebraic numbers, and it can be expressed as a sum of Hecke eigenforms and possibly an Eisenstein series of weight $2k$. -We also have $\rho_a \otimes \rho_b$ a representation. It isn't irreducible, but can be expressed as a sum of irreducibles. The obvious hope is that this would correspond to the product, but that cannot be true because the characters don't match up. But convolution if appropriately defined would produce a form related to $\rho_a \otimes \rho_b$. -So my two questions: what is multiplication of modular forms representation-theoretically, and what is tensoring of representations on the modular forms side? - -REPLY [5 votes]: With the notation of the question, the tensor product $\rho_a \otimes\rho_b$ is a degree $4$ Galois representation so corresponds to an automorphic representation for ${\rm GL}(4)$. This representation cannot be constructed in an obvious way from the modular forms $A$ and $B$, but its $L$-function can be obtained as the (now classical) Rankin-Selberg convolution of $A$ and $B$.<|endoftext|> -TITLE: How do most people write permutations? -QUESTION [15 upvotes]: I'd like to know how people prefer to write permutations, or elements of the symmetric group $S_n$ for $n\ge0$. -The most natural way to define a permutation in $S_n$ is as a bijection on the set $\{1,2,....,n\}$. Then the set of permutations (bijections) becomes a group under composition of maps. If $f,g\in S_n$ then there are two ways to define the composition $f\circ g$ depending on whether our functions act from the left or the right: -$$(x)(f\circ g) = ((x)f)g\quad\text{and}\quad (f\circ g)(x)=f(g(x)).$$ -I think that the latter is by far the most common these days. Of course, in the first case I could just define $(f\circ g)(x)=g(f(x))$ but it really is a right action so it should be written this way. -In terms of multiplying permutations using cycle notation the two ways of writing composition correspond, respectively, to whether we read the cycles left-to-right or right-to-left. For example: -$$(1,2)(2,3) = \begin{cases} - (3,2,1), &\text{using the $(x)f$ convention},\\ - (1,2,3), &\text{using the $f(x)$ convention}. -\end{cases}$$ -To me it has always seemed more natural to read permutations from left-to-right, as in the first case, but this implicitly uses the less common convention for composition of maps. -So the question: do you prefer to read products of permutations, written as cycles, from left-to-right or right-to-left? - -REPLY [3 votes]: I asked a well-known veteran of finite permutation group theory what specialists in that field did. The answer was that "almost everyone" does it left to right. That is, if $g,h$ are permutations, then $gh$ means "do $g$ then do $h$". Actions are usually written using exponential notation: $x^g$ is the image of $x$ under $g$. The left-to-right convention means that $(x^g)^h=x^{gh}$. As to why the convention is like it is, the widespread influence of Wielandt's book Finite Permutation Groups was proposed as one reason.<|endoftext|> -TITLE: Connected cycles of Shimura curves in $A_{g}$ not contained in larger Shimura subvarieties -QUESTION [5 upvotes]: Is there always a finite family of Shimura curves $(C_{i})$ in $A_{g}$ the moduli space of principally polarized abelian varieties of dimension $g(\geq 2)$, such that the union $\cup C_{i}$ is connected and such that $\cup C_{i}$ is not strictly contained in larger Shimura subvarieties (special subvarieties) except $A_{g}$ itself? i.e. there does not exist Shimura subvarieties $Z$ in $A_{g}$ with $\cup C_{i}\subsetneq Z$ and $Z\neq A_{g}$? I will appreciate any suggestion of results related to this problem and/or suggestions how one can possibly construct such families. - -REPLY [2 votes]: There are countably many special subvarieties because they are defined over $\overline{Q}$ -but I do not see how this is relevant to your question? -I understand you want to know whether any Shimura curve is contained in a special subvariety; -I think the answer is yes, it should be contained in an appropriate Hilbert modular variety.<|endoftext|> -TITLE: How many different numbers can be obtained as product of first $n$ natural numbers? -QUESTION [23 upvotes]: Let m and n be natural numbers, and consider the set of all possible products of m (not necessarily distinct) elements from the set $\{1,2,\ldots,n\}$, that is consider the set -$\{1^{a_1} \cdot 2^{a_2} \cdot \ldots \cdot n^{a_n}\mid a_i\geq 0, a_1+a_2+\ldots+a_n=m \}$. -How many results can be obtained? -Denote this number with $P(m,n)$ (the number of elements of the set defined above). -For example, if $m=1$, then $P(1,n)=n$. -Similarly, we have: -$P(m,1)=1$, -$P(m,2)=m+1$. -Moreover, if $n=p$ is prime then it is not difficult to see that -$$P(m,p)=\sum_{i=0}^{m}P(i,p-1).$$ -(We define $P(0,n)$ to be equal $1$ for any $n$.) -Furhter, using the above property one can obtain the values of $P(m,n)$, for some small values of $n$, for example for $n\leq 10$ we have: -$P(m,3)=\binom{m+2}{2}$; -$P(m,4)=(m+1)^2$; -$P(m,5)=\frac{(m+1)(m+2)(2m+3)}{6}$ -$P(m,6)=(m+1){{m+2}\choose{2}}$; -$P(m,7)=\frac{(m+1)(m+2)(m+3)(3m+4)}{24}$; -$P(m,8)=\frac{(m+1)^2(m+2)(m+3)}{6}$; -$P(m,9)=\frac{(m+1)^2(m+2)^2}{4}$; -$P(m,10)=\frac{(m+1)^2(m+2)(2m+3)}{6}$. -Is there a general method that would give precise value of $P(m,n)$ for any $m$ and $n$? Or at least an approximation of $P(m,n)$? - -REPLY [5 votes]: I add this as a new answer as it has a somewhat different flavor from my previous response. Moreover the bounds given there may still be useful. Thanks also to Edgardo for a useful discussion. -We consider the case when $n$ is fixed and $m$ is large. First we translate the problem into one of counting lattice points in certain polytopes. It will then follow from work of Khovanskii that for $P(m,n)$ here (for $m$ large) is given by a polynomial in $m$ of degree $\pi(n)$ and whose leading coefficient will be described below. It may be that for $P(m,n)$ is in fact a polynomial from the start, and maybe just the Erhart polynomial attached to a convex polytope defined below (this last point now seems unlikely; see the remark at the end of the answer). -To each natural number below $n$ associate a vector in ${\Bbb Z}^{\pi(n)}$ with non-negative coordinates corresponding to the exponents in the prime factorization of that number. Thus when $n=5$ we have the vectors $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(2,0,0)$ and $(0,0,1)$. Let ${\mathcal C}(n)$ denote the convex hull of these $n$ vectors. If a number $N$ is written as a product of $m$ numbers below $n$, then the prime factorization of $N$ gives a vector which is contained in the set ${\mathcal C}(n)$ dilated by $m$. Therefore it follows that the $P(m,n)$ is at most the number of lattice points contained in $m\cdot {\mathcal C}(n)$ which is asymptotically -$$ -m^{\pi(n)} \text{Vol}({\mathcal C}(n)). -$$ -Further note that any vector in $(m-n)\cdot {\mathcal C}(n)$ can be expressed as a sum of at most $m$ of the $n$ vectors corresponding to $1$ to $n$. Therefore $P(m,n)$ is at least as large as the number of lattice points contained in $(m-n) \cdot {\mathcal C}(n)$ and this also is asymptotically -$$ -m^{\pi(n)} \text{Vol}({\mathcal C}(n)). -$$ -Thus we have shown that $P(m,n) \sim m^{\pi(n)} \text{Vol}({\mathcal C}(n))$. My previous answer may be seen as giving upper and lower bounds for the volume of ${\mathcal C}(n)$ by bounding it from above and below by simplices. -In view of the above translation, we see that the problem may be phrased generally as follows: Given a commutative semigroup $G$ and two finite subsets $A$ and $B$ of $G$, consider all elements that are the sum of an element of $B$ and $N$ elements of $A$; call this set $B+N*A$. Theorem 1 of Khovanskii's paper (Newton polyhedron, Hilbert polynomial, and Sums of Finite sets; translation from Russian of his paper in Functional analysis & Applications 1992) then shows that the number of elements in $B+N*A$ is a polynomial in $N$ for large $N$. Khovanskii also considers the special situation when $G$ is ${\Bbb Z}^n$ and $A$ is a finite subset of $G$ which generates all of $G$. This is the situation of the problem at hand, and his work in Section 3 of the cited paper is along the lines of the argument I gave above. Another interesting paper on this topic is the work of Barvinok and Woods (Short rational generating functions for lattice point problems) which led me to Khovanskii's paper. -Edit: See also my related question An integrality question about expressing an integer as a product of numbers below $n$ ; especially Ilya Bogdanov's excellent example which indicates that $P(m,n)$ for larger values of $n$ is perhaps not a polynomial from the start and also probably not the Erhart polynomial for ${\mathcal C}(n)$.<|endoftext|> -TITLE: Faltings height of a CM abelian variety -QUESTION [7 upvotes]: Let A be a CM abelian variety, say simple of dimension g, with $End(A) = O_K$, where $K$ is a CM field -of degree $2g$. -Is there an upper bound for the Faltings height $h(A)$ in terms of the discriminant $d_K$ of $K$? - -REPLY [2 votes]: The way your question is stated, the answer is positive by Faltings's finiteness theorem for abelian varieties: there are only finitely many $K$-isom. classes of g-dimensional abelian varieties over $K$ with good reduction over $O_K$. In particular, the Faltings height of an abelian scheme over $O_K$ of relative dimension $g$ is bounded (ineffectively speaking) in terms of $g$ and $K$ (or $g$ and $d_K$ by Hermite-Minkowski). -Let me say some things about effectivity. I am guessing this is what you are really interested in. -Firstly, there is no effective version (in general) of the above ineffective statement at the moment. You can hope to do some things in particular cases. -For instance, in the case of dimension one, you can look at Chapter 1 of von Kaenel's thesis: -http://e-collection.library.ethz.ch/eserv/eth:2519/eth-2519-02.pdf -Note that the results in von Kaenel's thesis give you (much) more than just bounds on Faltings heights of elliptic curves with good reduction everywhere. -Moreover, when K is of small discriminant, we know by results of Abrashkin-Fontaine, that there are no non-trivial abelian schemes over O_K, thus you can take the bound to be zero when $d_K$ is at most 8. (But that's cheap...) -Finally, Rafael von Kaenel knows how to bound Faltings heights of certain classes of abelian varieties explicitly in terms of their reduction behaviour, and maybe his methods also work to bound stable Faltings heights of CM abelian varieties. You could email him or me about this if you are interested, as these results are not yet available.<|endoftext|> -TITLE: Probabilities in a riddle involving axiom of choice -QUESTION [25 upvotes]: The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question): -The Riddle: -We assume there is an infinite sequence of boxes, numbered $0,1,2,\dots$. Each box contains a real number. No hypothesis is made on how the real numbers are chosen. -You are a team of 100 mathematicians, and the challenge is the following: each mathematician can open as many boxes as he wants, even infinitely many, but then he has to guess the content of a box he has not opened. Then all boxes are closed, and the next mathematician can play. There is no communication between mathematicians after the game has started, but they can agree on a strategy beforehand. -You have to devise a strategy such that at most one mathematician fails. Axiom of choice is allowed. -The Anwser: -If $\vec u=(u_n)_{n\in\mathbb N}$ and $\vec v=(v_n)_{n\in\mathbb N}$ are sequences of real numbers, we say that $\vec u\approx \vec v$ if there is $M$ such that for all $n\geq M$, $u_n=v_n$. Then $\approx$ is an equivalence relation, and we can use the axiom of choice to choose one representant by equivalence class. The strategy is the following: mathematicians are numbered from $0$ to $99$, and the sequence of boxes $(u_n)_{n\in\mathbb N}$ is split into $100$ sequences of the form $\vec u_i=(u_{100n+i})_{n\in\mathbb N}$ with $0\leq i\leq 99$. Mathematician number $i$ will look at all sequences $\vec u_j$ with $j\neq i$, and for each sequence, it will compute the index $M_j$ from which the sequence matches the representant of its $\approx$-class. He then takes $M$ to be the maximum of the $M_j+1$ and looks at the sequence $\vec u_i$ starting at this $M$. He can deduce the $\approx$-class of the sequence $\vec u_i$, and guesses that $u_{M-1}$ matches the representant. At most one mathematician will be wrong: the one who has the number $i$ with $M_i$ maximal. -The Modification: -I would find the riddle even more puzzling if instead of 100 mathematicians, there was just one, who has to open the boxes he wants and then guess the content of a closed box. -He can choose randomly a number $i$ between $0$ and $99$, and play the role of mathematician number $i$. In fact, he can first choose any bound $N$ instead of $100$, and then play the game, with only probability $1/N$ to be wrong. -In this context, does it make sense to say "guess the content of a box with arbitrarily high probability"? I think it is ok, because the only probability measure we need is uniform probability on $\{0,1,\dots,N-1\}$, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up. - -REPLY [3 votes]: The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think). If I have a secret number and you try to guess a larger one, what are your odds of success? is there a way to make them better (if you have $99$ friends). You have only finitely many ways to fail and infinitely many to succeed, but we can't really assign a probability. That is also behind this puzzle -Going back to the given problem, suppose that each box has a guard. At a certain time each guard sees the contents of every other box then guesses the contents of their own. Under AoC with an agreed set of equivalence class representatives and the familiar protocol we know that all but finitely many will be correct. That is strange but we are used to this consequence of the axiom of choice. If you have full knowledge of all this and are required to become the guard for a box of your choice (after looking into as many other boxes as you wish) can we say that you are certain to make a correct guess? Not really. -The situation in the question above seems clearer to me if we stipulate ahead of time that all but finitely many boxes contain $0$ (no assumption , other than finiteness on which the exceptions are). If there are $100$ potential guards who are friends, can they be sure that each can pick a box and guess $0$ (what else can they do?) with at least $99$ correct? yes, as before split the boxes into $100$ subsequences, person $k$ looks at all the boxes in every subsequence other than his own then chooses a box further than the last non-zero box in every other subsequence. -If you are one member of this set of $100$ friends are your odds of being correct (when you guess $0$) improved $100$ fold, or at all? Not really. -Essentially $100$ people are assigned hidden integers and must name an integer greater than their assigned integer (so only finitely ways to fail and infinitely many to succeed!). If each person can see all the other integers then only one will fail. -In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) there is no loss in saying that the chosen sequence is in a certain coset (so why not the coset of the zero sequence)?<|endoftext|> -TITLE: Number of curves over a finite field -QUESTION [16 upvotes]: Let $K$ be a finite field. Is there a formula for the number of isomorphism classes of genus $g$ smooth curves over $K$? -In other words does there exists a formula for the number of rational points of the stack $\mathcal{M}_{g}$ over $K$? If so, what is a standard reference for this? - -REPLY [20 votes]: A formula for the number of isomorphism classes of curves over $\mathbf F_q$ is probably hopeless. As pointed out by Olivier Benoist and Qiaochu Yuan, the much more well behaved number is given by isomorphism classes weighted by their automorphism group, in other words, the groupoid cardinality of the groupoid $\mathcal M_g(\mathbf F_q)$: -$$ \# \mathcal M_g(\mathbf F_q) = \sum_{[C]/\cong}\frac 1 {\# \mathrm{Aut}(C)}$$ -This is also the number of $\mathbf F_q$-points of the coarse moduli space. It has a cohomological interpretation (Grothendieck-Lefschetz trace formula) that reads -$$ \# \mathcal M_g(\mathbf F_q) = \sum_{k}(-1)^k \mathrm{Tr}(\mathrm{Frob}_q \mid H^k_c(\mathcal M_g,\mathbf Q_\ell)).$$ -Thus finding a formula for the number of $\mathbf F_q$-points becomes a question about understanding the cohomology of $\mathcal M_g$ as an $\ell$-adic Galois representation. -For small $g$ all the cohomology will be of Tate type which means that $\# \mathcal M_g(\mathbf F_q)$ is a polynomial in $q$, but for large $g$ this will certainly fail to be true and it's then not really clear what it would mean to write down an explicit formula, as suggested by Joe Silverman. Here are the formulas for $g \leq 4$: -$$ \# \mathcal M_2(\mathbf F_q) = q^3$$ -$$ \# \mathcal M_3(\mathbf F_q) = q^6+q^5+1$$ -$$ \# \mathcal M_4(\mathbf F_q) = q^9+q^8+q^7-q^6$$ -For the first of these, it's a classical fact that $\mathcal M_2$ has the rational cohomology of a point. Also for $\mathcal M_3$ and $\mathcal M_4$ one actually knows the cohomology and its Galois structure in each degree, not just the alternating sum of cohomology groups: see Looijenga's "Cohomology of $\mathcal M_3$ and $\mathcal M_3^1$" and Tommasi's "Rational cohomology of the moduli space of genus 4 curves". I don't know if $\#\mathcal M_5(\mathbf F_q)$ is in the literature but it's probably something one could figure out. But quite soon writing down an explicit formula is going to get hopeless. -One can also start putting in marked points. The number of $\mathbf F_q$-points of $\mathcal M_{1,n}$ and $\mathcal M_{2,n}$ are in a sense known for all $n$, see Getzler's "Resolving mixed Hodge modules on configuration spaces" and my preprint http://arxiv.org/abs/1310.2508 . Here it's also true that for $n$ large it's not a polynomial in $q$, but instead you find a polynomial expression in $q$ and certain traces of Hecke operators on elliptic and Siegel cusp forms. You might also be interested in papers of (various combinations of) Bergström, Tommasi, Faber, van der Geer... - -Addendum, much later: The paper "On the number of curves of genus 2 over a finite field" by Gabriel Cardona determines the actual number of isomorphism classes of genus two curves over a finite field (of odd characteristic). The even characteristic case is treated in a companion paper.<|endoftext|> -TITLE: percolation probability in a hexagonal region -QUESTION [6 upvotes]: Suppose one takes a large hexagonal region in the tiling of the plane by unit hexagons, with $n+1$ hexagons on each side, as seen in the figure below (taken from the COMAP website) for the case $n=5$. - -Starting from any corner cell (call it $C$) and proceeding cyclically, color $n$ consecutive boundary hexagons blue, the next $n$ yellow, the next $n$ blue, the next $n$ yellow, the next $n$ blue, and the next $n$ yellow. Now color the interior hexagons blue and yellow uniformly at random. This determines three (non-intersecting) percolation paths from the boundary of the region to itself, with yellow hexagons on one side of each path and blue hexagons on the other. -What is the probability, in the limit as $n \rightarrow \infty$, that the percolation path that starts next to $C$ will terminate at the diametrically opposite point on the boundary of the region? -I know that Smirnov et al. have proved theorems asserting conformal invariance for percolation on this lattice, but I don't know the technical details, so I don't know if all the hypotheses that those theorems require are satisfied here. Assuming the answer is "yes", then the question I'm asking is a special case of a much more general (and natural) question about a probability model associated with pairings of $2n$ points on the boundary of a disk, and I'd like to learn more about that question as well (though for my purposes, it really is the case $n=3$ with 6 equally-spaced points that I want to know about, and a closed-form expression or close approximation to the probability in question). -[Added after j.c.'s proposed solution: Note that the probability is less than 1/3, since (labeling the corners 1,2,3,4,5,6 in cyclic order) the events 1-is-connected-to-4, 2-is-connected-to-5, and 3-is-connected-to-6 are disjoint, have equal probability (by symmetry), and have total probability less than 1. So .4295722 is not a possible answer.] - -REPLY [5 votes]: The papers "Euler integrals for commuting SLEs" by Julien Dubédat and "Logarithmic operator intervals in the boundary theory of critical percolation" by Jacob JH Simmons contain formulas and equations satisfied by probabilities for different crossing events in hexagons for continuum percolation at criticality. It was not immediately clear to me if due to Smirnov's proof of Cardy's formula these formulas are rigorously proven or if there are still some missing steps. -Here is a picture of the 5 different crossing configurations (from Simmons's paper). - -I will work off of this figure in what follows. Let the top corner of the hexagon (adjacent to the blue-colored side B) be your corner "C", then the event that the percolation interface terminates at the bottom corner is $W^{BC}_A$. (A previous edit of this answer was incorrect as I had misread the question). -Dubédat and Simmons have computed all of these events analytically for the case of a regular hexagon and they are (eqs 27,28 of Simmons paper): -$W^{(ABC)}=W_{ABC}=\frac{1}{2}-\frac{3^{5/2}\Gamma(2/3)^9}{2^{10/3}\pi^5}{}_3F_2\left(1,\frac{5}{6},\frac{5}{6};\frac{3}{2}\frac{3}{2}|1\right)\approx0.288717$ -The equality between events here is because $p_c=1/2$. - -$W^{AB}_C=W^{BC}_A=W^{CA}_B=\frac{3^{3/2}\Gamma(2/3)^9}{2^{7/3}\pi^5}{}_3F_2\left(1,\frac{5}{6},\frac{5}{6};\frac{3}{2}\frac{3}{2}|1\right)\approx0.140855$ - -The equalities between events here are by rotational symmetry. -As for your more general question about $2n$-gons, Steven M Flores and Peter Kleban are currently writing up some results in that direction. See these two papers.<|endoftext|> -TITLE: For which varieties is the natural map from the Chow ring to integral cohomology an isomorphism? -QUESTION [27 upvotes]: My apologies if this question is too naive. -Let $X$ be a smooth projective complex variety. There is a natural map $A^{\bullet}(X) \to H^{2\bullet}(X)$ of graded rings from the Chow ring of $X$ to the integral cohomology of $X$ given by taking Poincaré duals of fundamental classes. (Is there a convenient name for it?) For some particularly nice varieties, e.g. projective spaces and more generally Grassmannians, this map is an isomorphism. - -What are some more general $X$ for which $A^{\bullet}(X) \to H^{2\bullet}(X)$ is an isomorphism? - -I think it suffices that $X$ have a stratification by affine spaces. (Is there a convenient name for such spaces?) In this case there is apparently a difficult theorem of Totaro asserting that $A(X)$ is free abelian on the strata, which I think is also the case for the integral cohomology via cellular cohomology, and degrees and intersections ought to match as well. Are there interesting families of examples where $X$ doesn't admit such a stratification? - -Edit: A somewhat more general sufficient condition, if I've understood my reading correctly, is that the Chow motive of $X$ is a polynomial in the Lefschetz motive. Guletskii and Pedrini showed that this is true (edit: rationally) of the Godeaux surface so this is more general than admitting a stratification by affine spaces. Are there interesting families of examples where this doesn't hold either? - -REPLY [21 votes]: [I've incorporated or addressed comments of Dan Petersen and Daniel Litt into this. My thanks to them.] -One sometimes says that $X$ admits a cellular decomposition if it admits a stratification by affine spaces. The isomorphism of the type you mention was known before Totaro, cf Fulton's Intersection Theory 19.1.11 (in my edition). -This applies to flag varieties. Also such an isomorphism holds for toric varieties for similar reasons. -A surface with $p_g=q=0$ i.e. $\dim H^0(X,\Omega^2)=\dim H^0(X,\Omega^1)=0$, -and $A_0(X)\cong \mathbb{Z}$ satisfies the condition of $A(X)\cong H^*(X)$ by the Lefschetz $(1,1)$ theorem. Note that Bloch has conjectured that the second condition follows from the first. -One has examples of surfaces of Kodaira dimension zero $\ge 0$, -such as Enriques surfaces (Bloch-Kas-Lieberman) and Godeaux surfaces (Voisin) where these conditions hold. -These do not have cellular decompositions. -Finally, assuming the Hodge and Bloch-Beilinson conjectures [thanks Dan], an isomorphism $A(X)\otimes \mathbb{Q}\cong H^{2*}(X,\mathbb{Q})$ holds if and only if the Hodge numbers $h^{pq}=0$ for $p\not= q$.<|endoftext|> -TITLE: A construction of Kähler differentials and Illusie cotangent complex as colimit over embeddings -QUESTION [6 upvotes]: Let $\Bbbk$ be a field, $X$ affine scheme of finite type over $\Bbbk$. Let $\mathcal C_X$ be the category of closed embeddings of $X$ into (say affine) smooth $Y$'s of finite type over $\Bbbk$, morphisms being closed embeddings of $Y$'s (forming commutative triangle). We have a functor $\mathcal C_X^{\mathrm{op}}\to \text{Vect}_\Bbbk$ by $[X\hookrightarrow Y] \mapsto\Omega^1(Y)$ You can also post-compose it with the embedding $\text{Vect}_\Bbbk\to\text{Ch}_\Bbbk$ to the $\infty$-category of chain complexes. -I would like to claim that if you take the direct limit over $\mathcal C^{\mathrm{op}}_X$ of both versions of this functor (the vector-space and the chain-complex ones), what you get is the module of Kähler differentials, resp. the Illusie cotangent complex, of $X$. (There is a natural map in one direction and I claim it's an isomorphism of $\Bbbk$-vector spaces, resp. of objects of the derived category of $\text{Vect}_\Bbbk$.) - -Are such statements known? Or is there a reference for something similar? - -(I think I have an argument at least for the first version, but I couldn't find any references.) -Edit: (some additional thoghts) - -Clearly the first statment follows from the second, since the functor $H_0$ from non-negatively (homologically) graded complexes to vector spaces preserves colimits. -The cotangent complex is defined as a similar colimit for some chosen "cosimplicial resolution" of $X$ by smooth affine varietis. The colimit in question can thus be thought of as "taking all such reslutions at once", but right now I can't formalize this. -It looks like instead of closed embeddings, one could take all maps to affines and (hope to) get the same answer. - -REPLY [2 votes]: This is the approach to the cotangent complex in the Stacks project; there one uses all maps as in your Remark 3 (yes this gives the same answer). See the Stacks project chapter on the cotangent complex. In particular for schemes over a ring in particular see Section Tag 08V7. -Your remark 2 can be made more precise by saying that the homology over your category is computed by taking the usual simplicial resolution of the ring and then applying the functor. The simplest case of this is discussed in Section Tag 08PQ.<|endoftext|> -TITLE: Iterated Automorphism Groups -QUESTION [12 upvotes]: Notation: For each group $G$ define: -$Aut^{(0)}(G):=G$ -$Aut^{(1)}(G):=Aut(G)$ -$\forall n\geq 1~~~Aut^{(n+1)}(G):=Aut(Aut^{(n)}(G))$ -Question: Consider $I\subseteq \omega$. Is there a group $G$ such that: -(1) $\forall i\in I~~~~~Aut^{(i)}(G)$ is an Abelian group. -(2) $\forall i\notin I~~~~~Aut^{(i)}(G)$ is a non-Abelian group. - -REPLY [5 votes]: I have a partial answer, which would be a complete answer (that is to say a complete classification of all such sets $I$) if we assume Dickson's conjecture in number theory. -If $G$ is non-Abelian, then $Aut(G)$ is also non-Abelian. Specifically, for two non-commuting elements $a,b \in G$, the inner automorphisms $x \rightarrow a x a^{-1}$ and $x \rightarrow b x b^{-1}$ do not commute. -Hence, $I$ can only be either: - -The empty set, by letting $G$ be a non-Abelian group; -$\mathbb{N}$, by letting $G$ be the trivial group or some 'ancestor' thereof; -A set of the form $\{ 0, 1, 2, \dots, n \}$ for some $n$. - -In the third case, let $P(n)$ denote the proposition that there exists a group $G$ with the property that $Aut^n(G)$ is Abelian but $Aut^{n+1}(G)$ is non-Abelian. We clearly have $P(n) \implies P(n-1)$, by replacing $G$ with $Aut(G)$. -So, we want to show that $P(n)$ is true for arbitrarily large values of $n$. - -Lemma 1: If $G$ is a finite non-cyclic Abelian group, then $Aut(G)$ is non-Abelian. -Proof: By the Structure Theorem for finitely-generated modules over a PID, we can express $G$ as a product of cyclic groups of prime power orders. Then, $G$ is non-cyclic if and only if the product contains cyclic groups of orders $p^a$ and $p^b$ for the same $p$. If we can find non-commuting automorphisms of $C_{p^a} \times C_{p^b}$, then we are done, since they would extend to non-commuting automorphisms of $G$. -Without loss of generality assume $a \geq b$. and consider these two automorphisms: - -The automorphism $f : C_{p^a} \times C_{p^b} \rightarrow C_{p^a} \times C_{p^b}$ sending $(x,y)$ to $(x,y+x)$. -The automorphism $g : C_{p^a} \times C_{p^b} \rightarrow C_{p^a} \times C_{p^b}$ sending $(x,y)$ to $(x+p^{a-b}y,y)$. - -These do not commute. In particular, $f(g(1,0)) = (1,1)$ whereas $g(f(1,0)) = (1+p^{a-b},1)$. - -Lemma 2: If $G$ is a cyclic group that does not eventually become the trivial group under repeated iteration of $Aut$, then $G$ will eventually become non-Abelian. -Proof: Induction on the order of $G$. If $Aut(G)$ is non-cyclic, then $Aut(Aut(G))$ is non-Abelian by the previous lemma. If $Aut(G)$ is cyclic and $G$ is not the trivial group, then $Aut(G)$ is smaller than $G$ and we are done by induction. - -We know precisely which cyclic groups lead to the trivial group (http://oeis.org/A117729). Any other cyclic group will thus eventually lead to a non-Abelian group. -Assuming Dickson's conjecture, we can find arbitrarily long Cunningham chains of the first kind. These are sequences $(p_1, \dots, p_m)$ of primes where $p_{i+1} = 2p_i + 1$. Note that under the iterated automorphism operator, we get a nice long sequence of cyclic groups: -$C_{2 p_m} \rightarrow C_{2 p_{m-1}} \rightarrow \dots \rightarrow C_{2 p_1}$ -Also, if $m \geq 6$ then these groups do not belong to A117729, so must eventually lead to a non-Abelian group by Lemma 2. -Consequently, we establish proposition $P(n)$ for some $n \geq m$, and (assuming Dickson's conjecture) this gives us $P(n)$ for all $n \in \mathbb{N}$. -It's not hard to show a partial converse: if we restrict ourselves to finite groups, $P(n)$ is true for all $n$ if and only if there exist arbitrarily long Cunningham chains. - -Without Dickson's conjecture, how far can we go? Well, starting with $G$ being the cyclic group of order $361736822347711983585853438$ (twice the largest element of the longest known Cunningham chain), it takes $19$ iterations to reach a non-Abelian group. Hence, $P(n)$ is true for $n \leq 18$.<|endoftext|> -TITLE: Algebraic definition for intermediate Jacobians -QUESTION [14 upvotes]: Let $X$ be a Fano threefold over $\mathbb{C}$ (e.g. a cubic threefold in $\mathbb{P}^4$). Then one may define the intermediate Jacobian $J(X)$ of $X$ via the Hodge decomposition on $H^3(X,\mathbb{C})$. A priori this is just a complex torus, however the vanishing of $H^{3,0}(X)$ implies that this is in fact an abelian variety with -$$\dim J(X) = 1/2 \cdot \dim H^3(X,\mathbb{C}).$$ -Note that we constructed $J(X)$ complex analytically, but it turned out that it was algebraic. This is much the same as with the Jacobian of algebraic curves, but it turns out that there is an algebraic definition for the Jacobians of curves which works over any field (namely Pic^0). My question is whether something similar happens here. - - -Does there exist an algebraic definition for the intermediate Jacobian? - - -More specifically, let $X$ be a Fano threefold over a field $k$ (not necessarily of charactertistic zero). May one define an intermediate Jacobian $J(X)$ here over $k$, which in the case where $k=\mathbb{C}$ recovers the above definition? -More generally, I would like to be able to define the intermediate Jacobian for a family of Fano threefolds parametrised by suitable schemes (any algebraic definition should also hopefully give this). -I have a vague idea how one might go about this for cubic threefolds, but it seems a little ad hoc. Namely, Clemens and Griffiths showed that the Fano variety of lines $F(X)$ in a cubic threefold over $\mathbb{C}$ is a surface of general type, and that the Abel-Jacobian map $F(X) \to J(X)$ induces an isomorphism of abelian varieties $Alb(F(X)) \to J(X)$ (Here $Alb$ means take the Albanese variety). Therefore over general fields, it seems possible that one could simply define $J(X) := Alb(F(X))$. -There are a few reasons I don't like this. Namely: - -I don't know anything about $F(X)$ over general fields (e.g. is it smooth in char. $p$?). -It's not clear that this will work or make sense for general Fano threefolds (I'm not sure that $Alb(F(X)) \to J(X)$ need be an isomorphism in general.) -Ultimately I want a Torelli type theorem to hold, which I am not sure this approach gives. -It is very ad hoc. - -REPLY [13 votes]: There are many answers, but none of them is completely satisfactory. First of all, there is the following book by Gerd Welters. -MR0633157 (84k:14035) Reviewed -Welters, G. E. -Abel-Jacobi isogenies for certain types of Fano threefolds. -Mathematical Centre Tracts, 141. Mathematisch Centrum, Amsterdam, 1981. i+139 pp. -ISBN: 90-6196-227-7 -14K30 (14J30) -In this book, Welters proves that for many Fano threefolds $X$, for some "natural" irreducible component $M$ of the Hilbert scheme / Chow variety / Kontsevich space of effective curves on $X$, the induced Abel-Jacobi map $\alpha:\text{Alb}(M)\to J(X)$ is an isogeny. Thus, if you are willing to work up to isogeny, then you may take $\text{Alb}(M)$ as one algebraic definition. -Of course that seems ad hoc: why use $M$ instead of some other irreducible component? This was one motivation for the study of the geometry of the different irreducible components $M_\beta$ by myself, Harris, Roth, de Jong, Coskun, etc. Harris raised the question whether the Albanese varieties $\text{Alb}(M_\beta)$ might stabilize as the curve class $\beta$ becomes more and more positive. Alas, in her thesis, Ana-Maria Castravet proved that this does not hold, at least for $(2,2)$ complete intersections in $\mathbb{P}^5$, and other Fano manifolds arising as moduli spaces of stable bundles on a curve. Castravet found that there are two different Abelian varieties that both occur infinitely often as $\text{Alb}(M_\beta)$, one of which is $J(X)$, but the other of which is $J(X)\times J(X)$. -MR2039210 (2005i:14038) Reviewed -Castravet, Ana-Maria(1-TX) -Rational families of vector bundles on curves. (English summary) -Internat. J. Math. 15 (2004), no. 1, 13–45. -14H60 (14J45 14M20) -If you really only care about the intermediate Jacobian up to isogeny in a strong sense, i.e., you really only need the functor $T\mapsto J(X)(T)\otimes_{\mathbb{Z}}\mathbb{Q}$ rather than $T\mapsto J(X)(T)$, then you can access this "motivically". Certainly the Bloch-Beilinson conjecture describes this as a certain subquotient of the Chow group of $X$. But I believe there are actually theorems that show how to describe $J(X)\otimes \mathbb{Q}$ purely algebraically, over the same (characteristic 0) base field as $X$. My recollection is that this is unknown in positive characteristic, but I will look into this and revise this post. -Finally, there is also a definition of Yi Zhu for $J(X)$ when $X$ admits a morphism $X\to C$ where $C$ is a smooth, projective curve and the fibers $Y$ of $X$ have sufficiently high "coniveau", i.e., something like $Y$ is rationally connected and $h^{2,1}(Y)=0$. Note, $X$ itself cannot be rationally connected, because $C$ is not rationally connected. However, this sort of fibration does arise in the study of rationally simply connected fibrations, for instance. Zhu constructs $J(X)$ as a certain moduli space of torsors on $C$ for the torus bundle over $C$ that is (Pontrjagin / Cartier) dual to the relative Picard of $X$ over $C$. -Homogeneous Fibrations over Curves -Yi Zhu -http://arxiv.org/abs/1111.2963 -Zhu's definition works over an arbitrary base field, even in positive characteristic.<|endoftext|> -TITLE: How to find counterfeit coins by weighing -QUESTION [9 upvotes]: In one variant of the classic counterfeit coins problem you are given a bag of $n$ numbered but otherwise identical looking coins and a scale and your job is to find which coins are counterfeit. Counterfeit coins all have one weight and the other coins all have another weight. The scale can only tell you if two sets of coins have the same weight, or which one of the two is heavier. -I am interested in a small extension where rather than using scales you can explicitly weigh sets of coins (on a digital scale, say). For simplicity let us also set the weights of the coins to either $1$ or $0$ depending on whether they are counterfeit or not. How many weighings do you need to be sure to determine which of the coins are counterfeit? I am interested in large $n$ answers. Bounds or asymptotic results (or references to them) would be really great. -The first observation is that if $n>4$ you can always get away with fewer than $n$ weighings. -I have the book "Combinatorial group testing and its applications" by Hwang and Du but I can't find any reference to this version of the problem. (See http://bit.ly/1cACC2G for Google books link.) -[Equivalent problem posted to https://math.stackexchange.com/questions/600328/a-whats-my-vector-game before I had reformulated it as this classic looking counterfeit coin problem. I have also posted some explicit solutions for small $n$ there.] - -REPLY [8 votes]: This problem was solved (up to a small multiplicative factor) by Erdos and Renyi: -http://www.renyi.hu/~p_erdos/1963-12.pdf -Ps. Wow, Douglas Zare has a good intuition!<|endoftext|> -TITLE: On discrete version of curve shortening flow -QUESTION [7 upvotes]: One can define an analogous version of the curve shortening flow for polygons in $\mathbb R^2$, namely defined by the differential equation $\dot{p_i}(t)=\frac{v_i(t)}{|v_i(t)|^2}$, where $p_i$ is the $i$-th vertex of the polygon and and $v_i$ is the vector that goes from $p_i$ to the circumcenter of the triangle defined by $p_{i-1}$, $p_i$ and $p_{i+1}$. -The Gage-Hamilton-Grayson theorem states that simple curves remain simple under the curve-shortening flow. -Does this still hold for polygons under this analogous flow? - -REPLY [4 votes]: Not exactly an answer to your question, but Peter Scott and I worked out a polygonal flow that is guaranteed to keep curves embedded in "Shortening Curves on Surfaces", Topology 33, (1994) 25-43. -A version of this is implemented in java at -disk flow applet -The problem with the Birkhoff flow is that long segments move faster than short segments, and can overtake them, creating self-intersections. This can be overcome by getting the length of segments from the surface (using intersections with some sort of grid for example) rather than from a parametrization of the domain, as with Birkhoff.<|endoftext|> -TITLE: Is there a suitably generalized Baire property for topological spaces of arbitrary cardinalities? -QUESTION [10 upvotes]: Is there some suitable generalization to the notion of Baire property for topological spaces of arbitrary cardinalities which satisfies the following condition: - -The meager sets are sets which are union of $\lambda$ nowhere dense sets where $\lambda < \kappa$ and $\kappa$ is the cardinality of the space. -If we consider a model of ZFC $\mathfrak{M}$ in which every (suitably definable) set of reals have the property of Baire, then every (suitably definable) set of a suitably defined toplogical space have the "generalized" Baire property. - -REPLY [11 votes]: I am not sure if this is the kind of answer you were looking for, but since no one has given an answer yet, I think it is a good idea to say what little I know about larger cardinal analogues of the Baire category theorem. -The only spaces that I can currently think of where one would want to consider larger cardinal generalizations of Baire spaces to larger cardinals are the $P_{\kappa}$-spaces, so I will say a few things about how Baire spaces relate to $P_{\kappa}$-spaces. -In order to give motivation for this answer, I will first have to say a few things about $P_{\kappa}$-spaces in general before I talk about the Baire property. -$\textbf{Preliminaries concerning $P_{\kappa}$-spaces}$ -Many of the basic results from general topology hold when one replaces each instance of "finite" with an instance of "less than $\kappa$" for some regular cardinal $\kappa$. In this context, instead of dealing with ordinary topological spaces where the intersection of finitely many open sets is open, one deals with spaces where the intersection of less than $\kappa$ many open sets is open. -If $\lambda$ is a cardinal, then a $P_{\lambda}$-space is a completely regular space where the intersection of less than $\lambda$ many open sets is open. If $\lambda$ is a singular cardinal, then every $P_{\lambda}$ space is automatically a $P_{\lambda^{+}}$-space. Therefore, without loss of generality, we may restrict our attention to $P_{\lambda}$-spaces where $\lambda$ is a regular cardinal. -A space $X$ is said to be $\lambda$-compact if the intersection of less than $\lambda$ many open sets is open. If $\lambda$ is a regular cardinal, then the product of finitely many $\lambda$-compact $P_{\lambda}$-spaces is $\lambda$-compact. -One can also generalize the notion of a metric space and a uniform space to $P_{\kappa}$-spaces. We therefore define a $P_{\kappa}$-uniform space to be a uniform space where the intersection of less than $\kappa$ many entourages is an entourage. If $\kappa$ is uncountable, then it is easy to show that every $P_{\kappa}$-uniform space is generated by equivalence relations. It is well known that a uniform space is induced by a metric if and only if it is generated by a countable sequence of entourages. The notion of a metric space is analogous to the notion of a uniform space generated by linearly ordered descending sequence of entourages of length $\kappa$. -$\textbf{$P_{\kappa}$-spaces and Baire spaces}$ -If $\lambda$ is an infinite cardinal, then a $\lambda$-Baire space is a a topological space $X$ such that the intersection of less than $\lambda$ many dense open sets is dense. With this definition, every topological space is an $\aleph_{0}$-Baire space, and an $\aleph_{1}$-Baire space is simple a Baire space. -Fortunately, the Baire category theorem does hold for some $P_{\kappa}$-spaces where you might expect there to be some form of the Baire category theorem. Let $\kappa$ be a regular cardinal. Suppose that $I$ is a set and $X_{i}$ is a space for $i\in I$. Let -$\prod_{i\in I}^{\kappa}X_{i}$ be the topology with underlying set $\prod_{i\in I}X_{i}$ generated by the basis consisting of products $\prod_{i\in I}U_{i}$ where each $U_{i}$ is open in $X_{i}$ and where $|\{i\in I|U_{i}\neq X_{i}\}|<\kappa$. If each $X_{i}$ is a $P_{\kappa}$-space, then $\prod_{i\in I}^{\kappa}X_{i}$ is also a $P_{\kappa}$-space. - -$\mathbf{Proposition}$ If $I$ is a set, $\kappa$ is a regular - cardinal, and $X_{i}$ is a discrete space for each $i\in I$, then - $\prod_{i\in I}^{\kappa}X_{i}$ satisfies the $\kappa^{+}$-Baire - property. - -$\mathbf{Proof}$ The proof of this result is analogous to the proof of the ordinary Baire category theorem. Let $U_{\alpha}\subseteq\prod_{i\in I}^{\kappa}X_{i}$ be a dense open set for each $\alpha<\kappa$ and let $U\subseteq\prod_{i\in I}^{\kappa}X_{i}$ be a nonempty open set. We shall using transfinite induction construct sets $I_{\alpha}\subseteq I$ with $|I_{\alpha}|<\kappa$ and functions $f_{\alpha}\in\prod_{i\in I_{\alpha}}X_{i}$ such that - -if $\alpha<\beta$, then $I_{\alpha}\subseteq I_{\beta}$ and $f_{\alpha}=f_{\beta}|_{I_{\alpha}}$ -If $f\in\prod_{i\in I}^{\kappa}X_{i}$, and $f|_{I_{\alpha}}=f_{\alpha}$, then -$f\in U\cap\bigcap_{\beta<\alpha}U_{\alpha}$. - -Zero step- Since $U$ is a non-empty open set, there is some $I_{0}\subseteq I$ and some $f_{0}\in\prod_{i\in I_{0}}X_{i}$ where if $f\in\prod_{i\in I}X_{i}$ and $f|_{I_{0}}=f$. -Limit ordinal step- If $\lambda$ is a limit ordinal with $\lambda<\kappa$, then let $I_{\lambda}=\bigcup_{\alpha<\lambda}I_{\alpha}$ and let -$f_{\lambda}=\bigcup_{\alpha<\lambda}f_{\alpha}$. -Successor ordinal step- Suppose that $I_{\alpha},f_{\alpha}$ have been defined already and $|I_{\alpha}|<\kappa$. Then $\{f\in\prod_{i\in I}X_{i}:f|_{I_{\alpha}}=f_{\alpha}\}$ is a non-empty open set, so $U_{\alpha}\cap\{f\in\prod_{i\in I}X_{i}:f|_{I_{\alpha}}=f_{\alpha}\}$ is a non-empty open set. Therefore, there is some set $I_{\alpha+1}$ with $I_{\alpha}\subseteq I_{\alpha+1}$ and some $f_{\alpha+1}\in\prod_{i\in I_{\alpha+1}}X_{i}$ where if $f\in\prod_{i\in I}X_{i}$ and $f|_{I_{\alpha+1}}=f_{\alpha+1}$, then $f\in U_{\alpha}$. -Let $J=\bigcup_{\alpha<\kappa}I_{\alpha}$ and let $f\in\prod_{i\in I}X_{i}$ be a function with $f|_{J}=\bigcup_{\alpha<\kappa}f_{\alpha}$. Then $f\in U\cap\bigcap_{\alpha<\kappa}U_{\alpha}$. Therefore, since $U\cap\bigcap_{\alpha<\kappa}U_{\alpha}$ is non-empty, the set $\bigcap_{\alpha<\kappa}U_{\alpha}$ is dense. $\mathbf{QED}$ -Interestingly, for $P_{\lambda}$-spaces, the property of being a $\lambda$-Baire space is hereditary under taking dense subspaces. - -$\mathbf{Proposition}$ Let $\lambda$ be a cardinal and suppose that - $Y$ is a dense subspace of a space $X$. - -If $Y$ is a $\lambda$-Baire space, then $X$ is also a $\lambda$-Baire space. -If $X$ is a $\lambda$-Baire $P_{\lambda}$-space, then $Y$ is also a $\lambda$-Baire space. - - -One can also construct $P_{\kappa}$-spaces with strong Baire properties using ultraproducts. Recall that an ultrafilter $U$ is $\lambda$-regular if there is some $E\subseteq U$ with $|E|=\lambda$ but where $\bigcap D=\emptyset$ for each infinite $D\subseteq E$. Furthermore, the countably incomplete $\lambda$-good ultrafilters are precisely the ultrafilters where the ultraproducts are always $\lambda$-saturated. -$\mathbf{Proposition}$ (Bankston) - -The ultraproduct of regular spaces by a $\lambda$-regular ultrafilter is a $P_{\lambda^{+}}$-space. -Furthermore, the ultraproduct of topological spaces by a countably incomplete $\lambda$-good ultrafilter is $\lambda$-Baire. - -$\textbf{Ideas of proof}$ A proof of these facts is given in the paper Topological reduced products via good ultrafilters by Paul Bankston. A proof of 1 uses basic facts about regular ultrafilters. A proof of 2 uses the fact that the ultraproduct of models by a $\lambda$-good ultrafilter is saturated. -$\textbf{The failure of the Baire property}$ -One might expect for $\kappa$-compact $P_{\kappa}$-spaces to be Baire spaces, or complete uniform spaces generated by a linearly ordered set of entourages (i.e. generalized metric spaces) to be Baire spaces. Unfortunately, this is not the case. In fact, for most $P_{\kappa}$-spaces it is easy to find a closed subspace that is not even a Baire space. If $X$ is a $P_{\kappa}$-space, and there is a sequence of closed subsets $C_{n}$ where $C_{n}$ is nowhere dense in $C_{n+1}$ for all $n$, then $\bigcup_{n}C_{n}$ is a closed subspace of $X$ and each $C_{n}$ is nowhere dense in $\bigcup_{n}C_{n}$, so $\bigcup_{n}C_{n}$ is not a Baire space. -Let $X_{n}$ be a $P_{\kappa}$-space for all natural numbers $n>0$ and $x_{n}\in X_{n}$ be a non-isolated point for all $n$. Give $X=\prod_{n\in\mathbb{N}}X_{n}$ the box topology. Then $\prod_{n\in\mathbb{N}}X_{n}$ becomes a $P_{\kappa}$-space with this topology. For natural numbers $N$, let $C_{N}\subseteq\prod_{n\in\mathbb{N}}X_{n}$ consist of all sequences $(y_{n})_{n}$ such that $y_{n}=x_{n}$ whenever $n>N$. Then $C_{N}\simeq X_{1}\times...\times X_{N}$. Let $C=\bigcup_{N}C_{N}$. Then $C$ is not a Baire space. On the other hand, if each $X_{n}$ is $\kappa$-compact, then each finite product $X_{1}\times...\times X_{n}\simeq C_{n}$ is $\kappa$-compact, so $C$ is $\kappa$-compact. Furthermore, if each $X_{n}$ can be given a complete uniformity generated by a linearly ordered set of cofinality $\kappa$, then $C$ can also be given a -complete uniformity generated by a linearly ordered set of cofinality $\kappa$ (i.e. $C$ is like a complete metric space), but $C$ is still not a Baire space. -$\textbf{An application of generalized Baire spaces}$ -In this section, I will give motivation for the Baire property for $P_{\kappa}$-spaces in terms of point-free topology and Boolean algebras. The notions in this section come from my own personal research. -Recall that a frame is a complete lattice that satisfies the distributive law $x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i})$. Frames are the main objects of study in point-free topology since if $(X,\mathcal{T})$ is a topological space, then $\mathcal{T}$ is a frame. -If $L,M$ are frames, then a frame homomorphism from $L$ to $M$ is a function $f:L\rightarrow M$ such that $f(0)=0,f(1)=1,f(\bigvee R)=\bigvee f[R],f(x\wedge y)=f(x)\wedge f(y)$ for each $R\subseteq L$ and $x,y\in L$. Frame homomorphisms from $L$ to $M$ correspond roughly to continuous functions from $M$ to $L$. -A frame $L$ shall be called $\kappa$-distributive if whenever $|I|<\kappa$ and $R_{i}\subseteq L$ for $i\in I$, then $\bigwedge_{i\in I}\bigvee R_{i}=\bigvee\{\bigwedge_{i\in I}x_{i}|x_{i}\in R_{i}\,\textrm{for}\,i\in I\}.$ It can be shown that a $T_{1}$-space is $\kappa$-distributive if and only if the intersection of less than $\kappa$ many open sets is open. Therefore, the notion of $\kappa$-distributivity is a way to generalize the notion of a $P_{\kappa}$-space to point-free topology. It turns out that the surjective image of a $\kappa$-distributive frame under a frame homomorphism is not necessarily $\kappa$-distributive. On the other hand, one may characterize the completely regular $\kappa$-distributive frames $L$ such that the image of $L$ under a surjective frame homomorphism is $\kappa$-distributive, and this characterization requires the Baire property. We shall state the results here for topological spaces. -$\mathbf{Proposition}$ (Sikorski) Let $X$ be a $P_{\kappa}$-space. Then the -regular open algebra $\mathrm{Ro}(X)$ is $\kappa$-distributive if and -only if $X$ is a $\kappa$-Baire space. - -$\mathbf{Theorem}.$ Let $\kappa$ be a regular cardinal. Let - $(X,\mathcal{T})$ be a $P_{\kappa}$-space. Then the following are - equivalent. - -Every closed subspace of $X$ satisfies the $\kappa$-Baire property. -Every subspace of $X$ satisfies the $\kappa$-Baire property. -If $f:\mathcal{T}\rightarrow M$ is a surjective frame homomorphism, then $M$ is $\kappa$-distributive. -If $f:\mathcal{T}\rightarrow B$ is a surjective frame homomorphism and $B$ is a complete Boolean algebra, then $B$ is - $\kappa$-distributive.<|endoftext|> -TITLE: Cohomology classes represented by submanifolds -QUESTION [10 upvotes]: Let $Y\subset X$ be a codimension $k$ proper inclusion of submanifolds. If we choose a coorientation of $Y$ inside of $X$ (that is, an orientation of the normal bundle), then we get a class $[Y]\in H^k(X)$. If $X$ and $Y$ are oriented, then $[Y]$ may be defined as the fundamental class of $Y$ in the Borel-Moore homology of $X$, which is isomorphic to the cohomology of $X$. What is the simplest definition of $[Y]$ in the general case (where $X$ and $Y$ are not necessarily oriented)? -Note that a simple generalization of this question would be to ask how to define the pushforward in cohomology along a proper oriented map. (Then $[Y]$ would simply be the pushforward of $1\in H^0(Y)$.) I would be happy to know the answer to this more general question, but I asked the simpler version to be as concrete as possible. - -REPLY [3 votes]: This is an answer to your more general question about how to define $$f_\ast:H^\ast(X;\mathbb{Z})\to H^{\ast+\dim(Y)-\dim(X)}(Y;\mathbb{Z})$$ -when $f: X\to Y$ is a proper oriented map. -For $\ell$ sufficiently large, there is an embedding $g: X\hookrightarrow \mathbb{R}^\ell$ which is unique up to isotopy. The map $f':X\to Y\times \mathbb{R}^\ell$ given by $f'(x)=(f(x),g(x))$ is then an embedding, and has a tubular neighbourhood $N$ as in Dan's answer. Note that the normal bundle $\nu_{f'}$ of $f'$ has dimension $\dim(Y)-\dim(X)+\ell$. There is a Pontrjagin-Thom collapse map $F:(Y\times\mathbb{R}^\ell)_+\to T(\nu_f')$, where $+$ denotes a one-point compactification. After making a standard identification, this gives a pointed map $F: \Sigma^\ell Y_+\to T(\nu_{f'})$, where $\Sigma^\ell$ denotes the $\ell$-fold reduced suspension. -Now we can define the pushforward as the composition -$$ H^\ast(X)\to \widetilde{H}^{\ast+\dim(Y)-\dim(X)+\ell}(T(\nu_{f'}))\stackrel{F^\ast}{\to}\widetilde{H}^{\ast+\dim(Y)-\dim(X)+\ell}(\Sigma^\ell Y_+) \to H^{\ast+\dim(Y)-\dim(X)}(Y). -$$ -Here the first map is a Thom isomorphism and the last is a suspension isomorphism, and all coefficients are integers.<|endoftext|> -TITLE: If two projections are close, then they are unitarily equivalent -QUESTION [13 upvotes]: Given two projections $p,q\in B(H)$, it is well-known that if $\|p-q\|<1$, then there exists a unitary $u\in B(H)$ with $q=upu^*$. -The proof that immediately occurs to me uses comparison of projections (the inequality forces $p\sim q$ and also $p^\perp\sim q^\perp$, and then one constructs the unitary by adding the corresponding partial isometries), and implies that one can even replace $B(H)$ with any von Neumann algebra that contains both $p$ and $q$. - -Questions: - -Does anyone know of a more direct proof? Comparison of projections is a very basic result in von Neumann algebras, but somehow here it feels like overkill. -Does this property hold in a C$^*$-algebra? Proof, or counterexample? - -REPLY [4 votes]: In Two Subspaces (Trans. AMS 1969), Halmos made some useful observations about pairs of subspaces/projections. When $\|p-q\|<1$, this yields in particular that we can assume, up to unitary equivalence, that -$$ -p=\pmatrix{1&0\\0&0} \qquad q=\pmatrix{c^2& cs\\cs& s^2}\qquad c^2+s^2=1 -$$ -where $c, s$ are commuting positive contractions which allow computations just like for pairs of rank one projections in the two-dimensional case where $c=\cos \theta$ and $s=\sin\theta$. Then it is natural to come up with the following candidate for the unitary you want, namely -$$ -u=\pmatrix{c & s\\s&-c} -$$ -Once we have found that, we just have to express this in terms of $p$ and $q$. And we find - -$$ -u=(p+q-1)|p+q-1|^{-1} -$$ - which is a self-adjoint involution, lying obviously in the $C^*$-algebra geberated by $p$ and $q$. - -Note To check that the above formula does the job, the key point is that $(p+q-1)^2=1-(p-q)^2$ is invertible, positive, and commutes with $p$ and $q$. -Remark Another candidate would be the "rotation" -$$ -v=\pmatrix{c& -s\\ s& c}\qquad \mbox{i.e.} \quad v=u(2p-1) -$$ -An advantage of the latter unitary is that we have -$$ -\|v-1\|\leq \sqrt{2}\|p-q\| -$$ -If I recall correctly, this is mentioned in Takesaki I for reference purposes.<|endoftext|> -TITLE: extending elementary embeddings -QUESTION [5 upvotes]: Suppose $j : M \to N$ is an elementary embeddings between transitive models of ZFC. Everyone knows that if $G$ is $\mathbb{P}$-generic over $M$, $H$ is $j(\mathbb{P})$-generic over $N$, and $j[G] \subseteq H$, then the map can be extended to $\hat{j} : M[G] \to N[H]$. -Suppose now that $\mathbb{P}$ is a complete boolean algebra generated by some name $\tau$ for a subset of an ordinal $\kappa$. Suppose $A \subseteq \kappa$ is generic. In $M[A]$, we compute a generic for $\mathbb{P}$ corresponding to $A$. Now assume $B \subseteq j(\kappa)$ is $j(\mathbb{P})$-generic over $N$, and for all $\alpha < \kappa$, $\alpha \in A$ iff $j(\alpha) \in B$. Is it always true that we can extend $j$ to $\hat{j} : M[A] \to N[B]$? In other words, is it necessarily the case that the generic filter $G$ computed from $A$ has the property that $j[G] \subseteq H$, where $H$ is the generic filter computed from $B$? - -REPLY [4 votes]: The answer is no. -First, let's just consider the latter part of your question, whether the property $j[G]\subset H$ follows from your assumption on $A$ and $B$. It does not in general. Here is an example with forcing that is atomic. Let $\mathbb{P}$ be the Boolean algebra arising from the full support $\kappa$-product of the two-atom forcing. Thus, a condition is an element of $2^\kappa$, with these conditions forming an antichain of atoms. Let $\tau$ be the name of the subset of $\kappa$ where the generic condition has its $1$s. The Boolean values $[[\check\alpha\in\tau]]$ generate $\mathbb{P}$, since if you know all those values, you know which atom was selected. Let $G$ be the generic filter selecting the all-ones atom: $\vec 1=\langle 1,1,1,\cdots\rangle$. So $A=\kappa$. In $j(\mathbb{P})$, let $H$ be the filter selecting all ones up to $\kappa$, and then zeros. So $B=\kappa\subset j(\kappa)$. Since $\kappa$ is the critical point of $j$, we have $\alpha\in A\iff j(\alpha)=\alpha\in B$, since this only refers to $\alpha<\kappa$. But we don't have $j[G]\subset H$, since $j(\vec 1)$ is the all-ones condition in $j(\mathbb{P})$, which is not in $H$. So the pull-back property on the filters does not hold. -(Meanwhile, since both $G$ and $H$ are trivial forcing, the embedding does lift, so this shows that the two versions of your question are not equivalent.) -So let's now get a negative answer also to the lifting version of the question. We simply modify the forcing as follows, to get an example where the embedding does not lift. Let $\mathbb{P}$ be the forcing that either selects an eventually constant atom as above, or else adds a Cohen subset to $\kappa$. So the generic filter is determined by a subset $A\subset\kappa$, whose name $\tau$ will generate the Boolean algebra. Now, let $A$ be the eventually constant all-ones sequence, so $G$ is atomic, but this time, let $H$ be the filter adding a Cohen subset to $j(\kappa)$, but extending the condition having all ones up to $\kappa$, and after that being Cohen generic. Now, the argument as above shows that $\alpha\in A\iff j(\alpha)=\alpha\in B$, but this time the embedding does not lift to $j:M[G]\to N[H]$, since $G$ was trivial and $H$ was not.<|endoftext|> -TITLE: The holomorphic version of Galois theory -QUESTION [9 upvotes]: We identify the space of polynomials of degree n with $\mathbb{C}^{n+1}-\mathbb{C}^{n}$, that is an $n+1$ tuple $(a_{n},a_{n-1},\ldots,a_{0})$ with $a_{n} \neq 0$ is identified with $p(z)=a_{n}z^{n} +\ldots a_{1}z+a_{0}$. In this question we search for a holomorphic representation for the roots of P. That is, we search for holomorphic maps which send each polynomial $P$ to one of its roots. Motivating by the special case $n=2$ and the "Radical formula" for the roots, it is natural to search for an appropriate version of "Riemann surface of radical-type functions". So here is our question, precisely: -Question: - -Does there exist an $n+1$ dimensional complex manifold M with a covering space structure $\pi \colon M \rightarrow \mathbb{C}^{n+1}-\mathbb{C}^{n}$ and a holomorphic function $f \colon M \rightarrow \mathbb{C}^{n}$, such that for every $\tilde{P} \in M$ with $\pi(\tilde{P})=P$, all $n$ roots of $P$ are arranged in the $n$-tuple $f(\tilde{P})$? - -Remark: -Since "Galois Theory " is an obstruction for existence of a radical (algebraic) formulation for the roots, it was natural that we search for a holomorphic analogy - -REPLY [11 votes]: The thing you are asking was much studied in connection with Hilbert Problem 13. -The roots of a polynomial of degree exactly $d$ form an unordered $d$-tuple. The set of -unordered $d$-tuples is called the configuration space. It is the factor of $C^d$ - over -the action of permutation group. It is equivalent to the space of -polynomials of degree exactly $d$ modulo multiplication by a non-zero constant. -One recent reference is http://arxiv.org/pdf/math/0403120v3, and it contains many other references. The version of Hilbert problem 13 asks whether this function, mapping a polynomial to its roots, can be represented as a composition of functions of fewer variables.<|endoftext|> -TITLE: How can an extremely mathematically talented young person be helped to fulfill his/her potential? -QUESTION [55 upvotes]: Obviously, this question is not a research level mathematics question at all. But, I've just met an extremely mathematically talented $11$ years old student and I don't know how I can help him. For years I was working at a special school for young gifted and talented students. But, I had never met such talent at such a young age in my life. I talked to a mathematician friend of mine (who himself was an IMO gold medalist at a very young age), but he had no idea unfortunately. Basically, there is no one around with any idea. That is why I came to MO. Just in case that this question gets closed please e-mail me at asghari.amir at gmail if you have any idea. - -REPLY [14 votes]: He is a kid. -Maybe brilliant but just a kid.I believe that the best you can do is to make sure that he is good at almost all lessons in school and in mathematics you should give him as much as he wants. -If he wants to learn more,give him more. This way you will not push him or make him anxious to be the best among all. -Feed him only when he is hungry. -Consider Gauss.He was the best among all, but he did not have anyone to push him.(as far as i know) -Let him push you.If he does,he might be so gifted and he finds interesting mathematics indeed. -If he does not, let him live his life.<|endoftext|> -TITLE: Number of isomorphism types of finite groups -QUESTION [15 upvotes]: Are there some good asymptotic estimations for the number $F(n)$ of non-isomorphic finite groups of size smaller than $n$? - -REPLY [31 votes]: The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets a large jump in the value of $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$. -The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims: - -Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: - $$f(n,p) = p^{(2/27 + o(1))n^3}.$$ - -(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give: - -Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most - $$n^{(2/27+o(1))\mu^2}$$ - as $\mu\to\infty$. - -The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result: - -Pyber, L. Enumerating finite groups of given order. - Ann. of Math. (2) 137 (1993), no. 1, 203–220. - -An interesting extra tidbit from that paper is the following: - -Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$). - -(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result! -$(\dagger)$ I said this was a conjecture earlier - my mistake. - -REPLY [15 votes]: It is proved in -Holt, D. F., -Enumerating perfect groups. -J. London Math. Soc. (2) 39 (1989), no. 1, 67–78 -that -$n^{2l(n)^2/27−dl(n)} \le F(n) \le n^{l(n)^2/6+l(n)}$ -for some constant $d$, where $l(n) = \log_2(n)$. The lower bound is coming from Higman's construction of large numbers of $p$-groups of class 2, and the general belief seems to be that the lower bound is close to being the correct number. -I think that there might be better results known now. You could try searching publications of Laszlo Pyber, but I don't have time right now! -Added later: Nick's answer is much more accurate than mine. But, in case it is of any interest, let me add that the main result of the paper I mentioned was an estimate of the number ${\rm perf}(n)$ of finite perfect groups of order at most $n$, which (perhaps surprisingly) is also large, and satisfies: -$n^{l(n)^2/108−cl(n)} \le {\rm perf}(n) \le n^{l(n)^2/48+l(n)}$, where $c=11/36$.<|endoftext|> -TITLE: Bounding the perimeter of a geodesic triangle in spaces of non-positive curvature -QUESTION [5 upvotes]: This is probably an easy question, but I don't know any Riemannian geometry and a literature search hasn't helped. Any help (e.g. providing a reference) would be greatly appreciated. -For a triangle $ABC$ in euclidean space, denote by $\alpha$ the angle at $A$ and by $a$ its opposite side. Given $\epsilon >0$, consider all triangles $ABC$ with $\alpha>\epsilon$. Then it is clear (e.g. from the sine law) that the perimeter $p(ABC)$ of any such triangle satisfies $p(ABC)< C_\epsilon \cdot a$, with $C_\epsilon$ a positive constant only depending of $\epsilon$. -The same type of bound holds true, for the same reasons, for triangles in hyperbolic space of constant negative curvature $-\rho$, with a constant $C_\epsilon$ depending on $\rho$. (However, note that it doesn't hold for spaces of positive curvature, as an example with two geodesics on the sphere shows). -My question: Let $X$ be a complete Riemannian manifold with sectional curvatures in $[-\rho,0]$ for some $\rho>0$. Is the analogous statement true? That is, is it true that for every $\epsilon>0$, there exists $C_\epsilon>0$ such that $p(ABC)< C_\epsilon \cdot a$ holds for every geodesic triangle $ABC$ with $\alpha>\epsilon$? -In fact, I am only interested in the case where $X$ is a symmetric space of non-compact type, but it seemed more reasonable to ask the question in a more general setting. -Thank you! - -REPLY [4 votes]: If the manifold is complete simply connected and curvature $\le 0$ then it is CAT(0) space. -In particular, all the triangles are thin; i.e., the flat triangle with the same sides has bigger angles. -So your question can be reduced to the question in plane geometry and the answer is yes.<|endoftext|> -TITLE: On the equation $a^n + b^n = c^2$ -QUESTION [7 upvotes]: I am interested in the possible natural solutions of the equation $a^n + b^n = c^2$ where $n \geq 4$ is fixed. I am not sure if it is well-known or not, so any suggestion would be helpful. - -REPLY [12 votes]: To complement Boris Novikov's answer, Darmon and Merel have shown that the equation has no solution with $\gcd(a,b,c)=1$. See J. reine angew. Math. 490 (1997), 81-100. - -REPLY [9 votes]: $2^{2n+1}+2^{2n+1}=( 2^{n+1})^2$. - -REPLY [5 votes]: Assuming $gcd(a,b,c)=1$ there are no solutions (see GH's answer), and this fits well for the hyperbolic case of the generalised Fermat equation - $x^p+y^q=z^r$ with -$1/p+1/q+1/r<1$, and $gcd(x,y,z)=1$, because for $n>4$ above we are in this case. -By Darmon-Granville theorem we know that there are only finitely many solutions, and conjecturally at most $10$ solutions. None of these is for $(p,q,r)=(n,n,2)$ with $n>4$.<|endoftext|> -TITLE: Numbers with disjoint sets of multiples of integer parts -QUESTION [9 upvotes]: Let $\alpha>0$ and define $S(\alpha)=\{\lfloor n \alpha \rfloor: n\in\Bbb Z^+ \}$. (Here $\lfloor x\rfloor$ is the integer part of $x$ and $\mathbb Z^+$ the set of positive integers.) -Is there any known characterization for the pairs of positive irrationals $\alpha$ and $\beta$ for which $S(\alpha)\cap S(\beta)=\varnothing$? - -REPLY [6 votes]: You are right about $S(\alpha)\cap S(\beta)=\emptyset\iff \dfrac{n}{\alpha}+\dfrac{m}{\beta}=1$ for some $n,m\in\mathbb Z^+$ (see Theorem 8 of the cited paper). -The implication $\Longleftarrow$ is easy so let's assume that $\frac{n}{\alpha}+\frac{m}{\beta}\neq 1$ for all $n,m\in\mathbb Z^+$. Then one of the following is true. - -The numbers $1,\dfrac{1}{\alpha},\dfrac{1}{\beta}$ are linearly independent over $\mathbb Q$, -there exist some $n,m,k\in\mathbb Z^+$ such that $\left|\dfrac{n}{\alpha}-\dfrac{m}{\beta}\right|=k$, -there exist some $n,m,k\in\mathbb Z^+$ such that $\dfrac{n}{\alpha}+\dfrac{m}{\beta}=k$ with $k>1$, and $\gcd(n,m,k)=1$. - -It is enough to show that any one of items 1, 2 or 3 implies that $S(\alpha)\cap S(\beta)\neq\emptyset$. Elementary proofs of these can be found in this paper (Theorem 5, Theorem 6 and Theorem 7 respectively).<|endoftext|> -TITLE: Which lenses can be squared? -QUESTION [7 upvotes]: A lune and a lens are both planar figures delineated by two circular arcs; the difference is that a lune has a concave and a convex arc (it is one circle minus another) whereas a lens has two convex arcs (it is the intersection of two circles). -It is well-known that there are exactly five lunes that can be squared; that is, there are five essentially different lunes where you can construct both the lune and a square of the same area using compass and straightedge in finitely many steps. (Proof.) Are there any similar results for lenses? -(I was inspired to ask this question by this Google+ post and this page of fascinating background information, linked already above.) - -REPLY [5 votes]: I suspect a complete proof is not so easy. There is a field called "the constructible numbers," sometimes denoted $E$ for Euclid, which is the smallest subfield of the reals closed under square root of positive numbers. A length $x$ is constructible if $x \in E.$ An angle $\alpha$ is constructible if $\cos \alpha \in E,$ or $\sin \alpha \in E,$ or $\tan \alpha \in E,$ these conditions being equivalent. The constructible angles that are rational multiples of $\pi$ are known precisely. However, we get a mess as soon as we consider, say, $\arctan 2,$ which is a transcendental multiple of $\pi.$ -Using roughly the symbols on your link, we want to show that, for acute constructible angles $\alpha, \beta$ and positive constructible lengths x,y, that -$$ \color{magenta}{ x \alpha + y \beta \; \notin \; E.} $$ -Now, if $x, y \in \mathbb Z,$ this follows by Hermite-Lindemann, page 131, Theorem 9.11 (b) in Niven, Irrational Numbers. Diving by a different integer, it is also true if $x, y \in \mathbb Q.$ Finally, multiplying by a single irrational constructible number, it is also true if the ratio $x/ y \in \mathbb Q.$ -Anyway, a proof of the proposition in color would finish this. Probably true, but... -Now that I compare the two, evidently Cebotarev and Dorodnow proved (I think) that the expression in color is only in the field $E$ if equal to $0,$ where they were concentrating on $x,y$ of opposite signs. So it would seem they had enough technology to finish this. I just don't know how it was done. -It seems Postnikov gave a summary of Cebotarev and Dorodnov in his book on Galois theory, an excerpt was translated in the MAA Monthly, and many of those columns collected in a book. see also Quadrature of the Lune -Hmmm...Postnikov evidently assumed commensurable angles which means he does not show how to deal with the problem above. At this point, and with my experience of Nestorovich and Mordukhai-Boltovskoi in about the same time, I have got to wonder if Cebotarev and his student really gave a complete proof of the five lunes, or made some "commensuability" assumption. As Samuel L. Jackson said in the Long Kiss Goodnight, "When you make an assumption, you make an ass out of you and umption." http://www.imdb.com/title/tt0116908/ and -Mitch Henessey: ...everyone knows, when you make an assumption, you make an ass out of "u" and "umption". -http://www.imdb.com/title/tt0116908/trivia?tab=qt&ref_=tt_trv_qu -NOTE: it is supposed to be When you assume, you make an "ass" out of "u" and "me."<|endoftext|> -TITLE: Pontryagin numbers on manifolds with an $S^1$-action -QUESTION [9 upvotes]: Let $M$ is a smooth compact manifold with an $S^1$-action with isolated fixed points. Suppose the representation of $S^1$ at tangent spaces at all fixed points is known. Can one then find all Pontryagin numbers of the manifold? I would be grateful for some nice reference on this topic. -For four-manifolds such a formula for $p_1$ exists but don't know if it exists for higher dimensions. Indeed, $p_1=3\tau$, where $\tau$ is the signature, and the signature can be calculated by proposition 6.18 here http://www.ma.utexas.edu/users/dafr/Index/asindiii.pdf - -REPLY [6 votes]: One can find the Pontryagin numbers. It is an application of a more general version of the G-signature theorem, when one considers the signature operator on M twisted by a vector bundle. It is Theorem 8.11 respectively formula 8.12 in the paper which proves the G-signature theorem: -Atiyah, M. F.; Singer, I. M. The index of elliptic operators. III. Ann. of Math. (2) 87 1968 546–604. -(Actually you provided a link to an electronic version in your question.)<|endoftext|> -TITLE: Bounds on coefficients of factors of a multivariate polynomial -QUESTION [8 upvotes]: Given a multivariate polynomial $F(x, y, ..)$ what is the smallest bound B that can be quickly found such that $|G|_{\infty} \le B$ for all factors $G$ of $F$. (I'm using $|G|_{\infty}$ to denote the largest coefficient in G.) -I am only aware of the Landau-Mignotte bound for polynomials in $\mathbb{Z}[x]$. - -REPLY [6 votes]: Mahler invented what is now known as Mahler measure to establish inequalities like this. His paper "On some inequalities for polynomials in several variables" (J. London Math. Soc. 37, 341-344, 1962) shows that if $f$ and $g$ are polynomials in $\mathbb{C}[X_1,\ldots,X_n]$ whose degrees in $X_i$ are bounded by $m_i$, then -$$ - |f|_\infty|g|_\infty\le2^{m_1+\cdots+m_n}\{(m_1+1)\cdots(m_n+1)\}^{1/2}|fg|_\infty . -$$ -The arguments are elementary, and build via induction on the single variable case Mahler established in an earlier paper. The basic point is that Mahler measure is multiplicative, and can be related to height and length. -His main interest in this was its use in transcendence theory, and in particular some inequalities used by Gelfond.<|endoftext|> -TITLE: Are There Always Group Generators Which Give Unimodal Growth? -QUESTION [17 upvotes]: Suppose $G$ is a $k$-generated finite group. Is there always a set of $k$ elements which generate the group and have a unimodal counting function? -Background: -The counting function, $f(n)$, is a function whose value at $n$ is the number of elements of $G$ of length $n$. -The length of an element $x$, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to $x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion. -A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence $\{s_1, s_2,\ldots,s_n\}$ is unimodal if there exists a $t$ such that $s_1\leq s_2\leq\cdots\leq s_t$, and $s_t \geq s_{t+1} \geq \cdots\geq s_n$. (Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html) -Discussion: -As pointed out in one of the comments, taking $k$ to be the number of elements in the group will trivially give unimodal growth. The question is more interesting if one considers values of k which are less than the number of elements in the group. In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact, nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function. -Here is an example of a group with non-unimodal growth. The two generators are the permutations: -$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$ -$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$ -In this notation $a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,... -The counting function for this pair of generators is: -1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2. - -REPLY [2 votes]: In response to the bounty question: -By the fundamental theorem of finitely generated abelian groups, any finite $k$-generated abelian group $G$ is isomorphic to the lattice $\bigoplus_{i=1}^{k} \mathbb{Z}_{r_i}$ for prime powers $r_1, \ldots, r_k$. Then $f(n)$ is the number of solutions to $\sum_{i=1}^{k}x_{i} = n$ in the integers $0 \leq x_i \leq r_i-1$. This is equal to the coefficient of $t^n$ in the generating function -$$ -\prod_{i=1}^{k} (1 + t + \ldots + t^{r_i -1}) . -$$ -Any polynomial of this form is unimodal, as (I think) Christian Stump alluded to in his comment. Stanley proves this fact in Proposition 1 of this article: -http://math.mit.edu/~rstan/pubs/pubfiles/72.pdf .<|endoftext|> -TITLE: Does the smooth manifold $\#_{l}CP^{2}\#_{k}(-CP^{2})$ admit a symplectic structure? -QUESTION [9 upvotes]: Let $-CP^{2}$ denote the complex projective surface $CP^{2}$ with the reverse orientation. I have seen some results about the existence of symplectic structures on the connected sums $\#_{l}CP^{2}\#_{k}(-CP^{2})$ for some positive integers l,k. -My question is whether there is a complete result which can decribe the existence of symplectic structures on the connected sums $\#_{l}CP^{2}\#_{k}(-CP^{2})$ for all positive integers l,k. - -REPLY [11 votes]: For $l=1$, these are blowups of the projective plane, which are all Kaehler and hence symplectic. For $l>1$, these do not have symplectic structures. For if $l$ is even, then they don't even have almost complex structures (cf. this MO thread), which symplectic manifolds certainly do. If $l>1$ is odd, then their Seiberg-Witten invariants would vanish, since your manifolds decompose as a connected sum into pieces with positive $b_2^+$. But a famous theorem of Taubes says that there is a non-vanishing Seiberg-Witten invariant. -Finally, if $l=0$, then $b_2^+ = 0$, so your manifold cannot be symplectic, since the cohomology class of the symplectic form has positive square.<|endoftext|> -TITLE: Open cell decomposition after applying a Weyl group element -QUESTION [7 upvotes]: Let $G=\operatorname{GL}(n,\mathbb C)$. What follows can be put into a more general context, but I would like to first understand it for this case, the generalization is a second step. -For Zariski-almost all $x\in G$, there is a unique decomposition $x=lu$ with $l$ lower triangular and $u$ upper unipotent triangular. -Now, I can precompose $x$ with some permutation (matrix) $\pi$ (an element of the Weyl group of $G$), i.e. permute the rows of $x$. For generic $x$, the matrix $\pi x$ will again have some LU-Decomposition $\pi x = l_\pi u_\pi$ with $l_\pi$ lower triangular and $u_\pi$ upper unipotent triangular. -We have an open immersion $\iota:L\times U\hookrightarrow G$ mapping $(l,u)\mapsto lu$, where $L$ is the Borel of lower triangular matrices and $U$ the unipotent radical of its opposite Borel, the variety of upper unipotent triangular matrices. Restricting to some open subset $V\subseteq L\times U$, we can consider the map $(l,u)\mapsto \pi.(l,u) := (l_\pi,u_\pi)$ for all $\pi\in\mathfrak S_n$ such that $\iota(\pi.(l,u))=l_\pi u_\pi= \pi lu$. -Since $\iota(\pi.\sigma.(l,u))= (l_\pi)_\sigma (u_\pi)_\sigma = \sigma l_\pi u_\pi = \sigma\pi lu = l_{\sigma\pi} u_{\sigma\pi} = \iota(\pi\sigma.(l,u))$ and because $\iota$ is an immersion, this defines a Group action on $V$, turning $\iota|_V$ into a $\mathfrak S_n$-equivariant map. Hence, the action of $\mathfrak S_n$ on $V$ is algebraic. -My question is: Is there any way to describe the open sets $L\cap V$ (resp. $U\cap V$) and the action $l\mapsto l_\pi$ (resp. $u\mapsto u_\pi$) of the permutation group $\mathfrak S_n$ on them? I have no intuition what-so-ever what they look like, and calculating examples was not very insightful. -Thanks a lot in advance for any pointers. -Edit: The $n=2$ case might still explain better what is going on. Observe that -$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = -\begin{pmatrix} 1 & 0 \\ ca^{-1} & 1 \end{pmatrix} \cdot -\begin{pmatrix} a & b \\ 0 & d-ba^{-1}c\end{pmatrix} -$$ -Now, switching the rows maps -\begin{align*} -\begin{pmatrix} 1 & 0 \\ ca^{-1} & 1 \end{pmatrix} &\mapsto -\begin{pmatrix} 1 & 0 \\ ac^{-1} & 1 \end{pmatrix} \\ -\begin{pmatrix} a & b \\ 0 & d-ba^{-1}c\end{pmatrix} &\mapsto -\begin{pmatrix} c & d \\ 0 & b-dc^{-1}a\end{pmatrix} -\end{align*} -so on the lower unipotent triangular matrices, in the nonvanishing locus of the lower left entry, we map -\begin{align*} -\begin{pmatrix} 1 & 0 \\ a & 1 \end{pmatrix} &\mapsto -\begin{pmatrix} 1 & 0 \\ a^{-1} & 1 \end{pmatrix} -\end{align*} -I don't see immediately what the action on the upper triangular matrices is. - -REPLY [2 votes]: In lieu of any other answers till now: the question reminded me somewhat vaguely of "cell multiplication" in a Bruhat decomposition, namely, in a Bruhat decomposition $G=\bigsqcup_{w\in W} BwB$ of $G=GL_n(k)$ with $B$ a minimal parabolie (Borel), and $W$ (representatives for) the (spherical) Weyl group, the "cells" $BwB$ have a kind of multiplication rule. For $S\subset W$ the collection of reflections attached to simple roots (determined by choice of $B$), and for $\ell(w)$ the length-of-word function associated to generators $S$ for $W$, for $s\in S$, $w\in W$, -$BsB\cdot BwB=BswB$ when $\ell(sw)>\ell(w)$, and $BsB\cdot BwB=BswB\cup BwB$ when $\ell(sw)<\ell(w)$. -One exposition of such stuff, minimizing reliance upon general Coxeter-group stuff, is at http://www.math.umn.edu/~garrett/m/v/bldgs.pdf -The standard treatment, going back many decades, and abstracted by Bruhat-Tits and others, does rely on the combinatorial group theory of Coxeter groups (and for not-split algebraic groups, on the more serious theory of algebraic groups).<|endoftext|> -TITLE: Commutator Width of a direct limit of hyperbolic groups -QUESTION [6 upvotes]: Is it known if the direct limit of hyperbolic groups can have finite commutator width? Every hyperbolic group has infinite verbal width for any word $w$, so in particular for the commutator word $w=x^{-1}y^{-1}xy$ (http://arxiv.org/abs/1107.3719) -Especially I would like to know if the examples of infinite torsion groups with exactly $n$ conjugacy classes as constructed by Ol'shanksii in the book "Geometry of Defining Relations of Groups" necessarily have to have infinite commutator width. -More generally I am interested in examples of infinite groups having only finitely many conjugacy classes. - -REPLY [7 votes]: Assuming that you refer to Ivanov's construction from Ol'shanskii's book of a $2$-generated infinite group $G$ of exponent $p$, for a large prime $p$, having exactly $p$ conjugacy classes, then the commutator width in this group is bounded above by $p-1$. The reason for this is that for any non-trivial commutator $w$ in $G$, every element of $G$ is conjugate to $w^k$ for some $0 \le k \le p-1$ (because, by construction, $\langle w \rangle$ is a subgroup of order $p$ and intersects each conjugacy class of $G$). It is known, of course, that $G$ is a direct limit of hyperbolic groups. -In fact, the above argument together with a different construction of the group you are looking for, is given in Alexey Muranov's paper "Diagrams with Selection and Method for Constructing Boundedly Generated and Boundedly Simple Groups" (Comm. Algebra 33 (2005), no. 4), http://arxiv.org/abs/math/0404472. -As for the last question, infinitely generated examples of infinite groups with finitely many conjugacy classes are constructed in the book of Lyndon and Schupp. The first finitely generated infinite groups with two conjugacy classes were produced by Denis Osin in "Small cancellations over relatively hyperbolic groups and embedding theorems." (Ann. of Math. (2) 172 (2010), no. 1, 1–39). If I remember correctly, in this paper Denis points out that the latter groups cannot be limits of word hyperbolic groups.<|endoftext|> -TITLE: sets without perfect subset in a non-separable completely metrizable space -QUESTION [5 upvotes]: Suppose $X$ is a completely metrizable (but not separable) space. Suppose $D$ is a Borel (actually $F_{\sigma}$) subset of $X$. Is there any logical relation between the following statements? -[1] $D$ does not contain a non-empty perfect subset of $X$ (There is no $P \subseteq D$ such that $P$ is non-empty, closed in $X$, and has no isolated points). -[2] $D$ is a countable union of discrete subsets of $X$. -If [1] and [2] are not equivalent, what conditions would be equivalent with [1]? -Thanks a lot in advance! - -REPLY [5 votes]: [2]$\Rightarrow$ [1]. It is easy to check that each discrete subset $D’$ of a perfect space $P$ is nowhere dense in $P$. Now suppose that $D$ is a countable union of discrete subsets of $X$ and $D$ contains a non-empty perfect subset $P$ of $X$. By Theorem 4.3.11 from [Eng], $P$ is completely metrizable. By Theorem 2.4 from [Eng], $P$ is a Baire space, therefore it cannot be a countable union of its discrete subsets. -[1]$\Rightarrow$ [2]. I googled a little and found articles [Sto] and [Kou]. In the beginning of the paper [Kou] it is written the following: “A classical theorem of Suslin states that every analytic subset of a Polish (separable complete metric) space is either countable, or contains a copy of the Cantor set. A generalization to the non-separable case has been obtained by El’kin [1]: every absolutely analytic space (i.e. homeomorphic to a Suslin subset of some complete metric space) is either $\sigma$-discrete, or contains a copy of the Cantor set. This theorem had previously been proved by Stone [Sto] for absolutely Borel spaces”. -Concerning descriptive generalizations for Polish spaces, I recall that book [Kech] contains The Perfect Sets Theorems for Borel (13.6) (by Alexandrov and Hausdorff), Analytic (14.13) (by Souslin), Co-Analytic (32.2) and Projective (38.17) (by Davis) Sets: each such subset of a Polish space either is countable or else it contains a copy of Cantor set. -It seems that these results can be trivially generalized to locally separable metrizable spaces (by [Eng, Ex. 4.4.F.C] each such space is a disjoint sum of separable spaces) and to $\sigma$-discrete subsets instead of countable. -I remark that if we drop descriptive conditions for the set $D$ (like Borelness) then [1] does not imply [2]. Let $D$ be a Bernstein subset (see [Cic]) of the real line $\mathbb R$ (for non-separability we can take as $X$ a disjoint sum of uncountably many copies of $\mathbb R$). Each discrete subset of a hereditarily separable space (in particular, of a second countable space) is countable. Therefore each countable union of discrete subsets of $\mathbb R$ is countable. But if $D$ were countable, then $\mathbb R\setminus D$ will be a completely metrizable space as a non-empty $G_\delta$ subset of a completely metrizable space by Theorem 4.3.23 from [Eng]. Therefore $\mathbb R\setminus D$ will contain a copy $C$ of a Cantor set, [Cic] a contradiction. -References -[Cic] Jacek Cichoń. On Bernstein Sets. -[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989. -[HC] R. C. Haworth, R. C. McCoy, Baire spaces, Warszawa, Panstwowe Wydawnictwo Naukowe, 1977. -[Kech] Alexander S. Kechris. Classical Descriptive Set Theory , Springer, 1995. -[Kou] George Koumoullis. Cantor sets in Prokhorov spaces. -[Sto] A. H. Stone. Kernel constructions and Borel sets.<|endoftext|> -TITLE: Rational points and torsion points of CM elliptic curve -QUESTION [6 upvotes]: Let $E$ be a CM elliptic curve defined over a quadratic imaginary field $K$ with maximal order i.e., $\mathrm{End}_K(E)\cong \mathcal{O}=\mathcal{O}_K$. Let $\mathfrak{p}$ be a prime of $K$ such that the map $\mathcal{O}^\times \to(\mathcal{O}/\mathfrak{p})^\times$ is not surjective. With this situation, I tried to prove $E[\mathfrak{p}]\not \subset E(K)$. -My idea is as follows; If $E[\mathfrak{p}]\subset E(K)$, then we can consider $(\mathcal{O}/\mathfrak{p})^\times$ as the set of isomorphisms modulo an equivalence relation `$\sim$' defined by for $f,g\in \mathcal{O}$ we say $f\sim g$ when they coincide on $E[\mathfrak{p}]$. Then it seems contradict to the assumption $\mathcal{O}^\times \to(\mathcal{O}/\mathfrak{p})^\times$ is not surjective. Is it plausible? If it is, how to expand details of proof? - -REPLY [4 votes]: See Cor. 5.18 of Rubin "Elliptic curves with complex multiplication" in LNM 1716 for a proof (that uses the main theorem of complex multiplication).<|endoftext|> -TITLE: Markov processes lacking the Feller property -QUESTION [10 upvotes]: Let $E$ be a LCH second countable topological space and let $\mathcal{E}$ be its Borel $\sigma$-algebra. -Let $(P_t)_{t \geq 0}$ be a conservative transition function on $(E, \mathcal{E})$. -This means by definition that, $\forall t \geq 0$, $P_t : E \times \mathcal{E} \mapsto [0,1]$ such that: - -$\forall t \in [0,\infty)$, $P_t(x,\cdot)$ is a probability measure on $E$ $\forall x \in E$ -$\forall t \in [0,\infty)$, $P_t(\cdot, A)$ is $\mathcal{E}$-measurable $\forall A \in \mathcal{E}$ -$\forall t,s \in [0,\infty)$, $P_{t+s}(x,A) = \int_E P_t(x,dy) P_s(y,A)$, $\forall (x,A) \in E \times \mathcal{E}$ - -Assume further that $P$ is normal, i.e. $P_0(x,\cdot) = \delta_x \quad \forall x \in E$. -The so-called Feller property (as in Rogers and Williams and Revuz and Yor, some authors replace $C_0$ with $C_b$, I will probably ask on this point in another post) now reads as follows: -1 - for all $f \in \mathcal{C}_0(E)$, $P_t f \rightarrow f$ uniformly as $t \rightarrow 0$. -2 - for all $f \in \mathcal{C}_0(E)$, $P_t f \in \mathcal{C}_0(E)$. -I would like to have (many) examples of conservative transition functions that satisfy 1 but not 2. -I would like to have both trivial (if possible) and non-trivial examples. If possible some references to existing literature are welcome, but I noted that classical books on Markov processes do not exhibit examples of processes lacking the Feller property. -Point $2$ can fail to hold in many ways: the most brutal is that $P_t f$ is not even continuous for some continuous $f$ vanishing at infinity. But we can also have that $P_t f$ is continuous (or even continuous and bounded) for each $f \in \mathcal{C}_0(E)$ but fail to vanish at infinity for some $f$. If possible, I would like to have all sort of examples. -Thanks a lot in advance for your answers. - -REPLY [9 votes]: Here is a not so trivial example: take for $P_t$ the transition semigroup for the SDE -$$ -dx = -x^3\,dt + dW\;, -$$ -on $\mathbf{R}$. This is not Feller in your sense because solutions come in from infinity in finite time, so $P_t f$ fails to be in $C_0$ even if $f$ is. It is however Feller in the other definition you've seen in the literature, which causes no end of confusion...<|endoftext|> -TITLE: Is there a McDiarmid-type inequality for sequences with a finite range of dependence? -QUESTION [6 upvotes]: Let $X, X_1, X_2, \ldots, X_N$ be a sequence of identically distributed random variables with $0 \leq X \leq 1$. -We do not assume the sequence is iid, but rather allow the random variables to be dependent on their neighbors: we suppose that there exists $k\geq 1$ such that for every $I,J \subseteq \{ 1,\ldots, N\}$ satisfying $\min\{ |i-j| \,:\, i\in I,\, j\in J \} \geq k$, the collections $\{ X_i\,:\, i\in I\}$ and $\{ X_j \,:\, j\in J \}$ are independent. (It might help to just consider the case $k=2$.) -Let $f: \mathbb{R}^N \to \mathbb R$ be a Lipschitz function with Lipschitz constant one. -Here is my question: is there a good concentration inequality for $f(X_1,\ldots,X_N)$? -Let me recall two well-known facts: - -If $k=1$, then the sequence is i.i.d., and we have McDiarmid's inequality, which states that: -\begin{equation} -\mathbb{P} \left[ \left| f(X_1,\ldots,X_N) - \mathbb E \left[ f(X_1,\ldots,X_N) \right] \right| > \lambda \right] \leq \exp \left( -\frac{2\lambda^2}{N} \right). -\end{equation} -On the other hand, if $f(X_1,\ldots,X_N) = X_1 + \cdots+X_N$, then I can run the proof of Bernstein's inequalities (the usual exponential generating function argument) combined with a neat Holder inequality trick to prove a similar bound. - -However, I am interested in general, non-affine, Lipschitz $f$, and the proof of McDiarmid's inequality uses a martingale argument that doesn't seem to work very well with my weaker independence assumption (the Holder trick doesn't seem to me to work). I have tried googling without success. Does a concentration inequality like McDiarmid's inequality survive? I would even be happy with something not quite as strong. - -REPLY [8 votes]: There are several versions of this type: -1) K. Marton has results for dependent variables. Maybe closest to what you ask (for convex functions $f$) is the paper of Samson: -Samson paper -2) For a martingale difference based argument, see Kontorovich and Ramanan: -Kontorovich-Ramanan paper -The references of these articles give more pointers...<|endoftext|> -TITLE: The cyclic branched covers of "simple" knots in $S^3$ -QUESTION [5 upvotes]: Is there a convenient place in the literature where the geometric decompositions of cyclic branched covers of $S^3$ branched over "small" knots is recorded? -By small knots, I'm referring to things like torus knots, 2-bridge knots and the Rolfsen knot table. -For 2-sheeted coverings the Bonahon-Siebenman paper is a canonical reference, and for 2-bridge knots there's the standard connection with lens spaces. But I'd like to know about $n$-sheeted coverings for larger $n$ (3,4,5,6 is a little larger) and with standard small knots. -I imagine there are a few papers out there where much of this is worked out, perhaps a Montesinos paper? But I'm not having much luck finding this kind of information. -In principle I suppose I could generate tables using SnapPy and Regina but I imagine there must be work of this kind that predates such technology. -Thanks for any insights. - -REPLY [8 votes]: Such a classification follows from the orbifold theorem. The point is that the $n$-fold cyclic branched cover over a knot is an $n$-fold orbifold cover over the orbifold whose singular locus is the knot with a cone angle of $2\pi/n$ (in fact, one only needs the theorem for orbifolds of "cyclic type"). The orbifold theorem states that a good orbifold admits a geometric decomposition. An orbifold whose underlying singular locus is a knot in $S^3$ is automatically good, so satisfies the hypothesis of the orbifold theorem (there can be no teardrop or bad football suborbifolds for straightforward reasons). -There is a version of the prime decomposition for good 3-orbifolds. In the case at hand, the only spherical orbifold that may appear is a football, which corresponds to a connect sum. So the connect sum decomposition of the knot will correspond to the prime decomposition of the cyclic orbifolds. -Thus, assume the knot is prime. Now, the only toroidal orbifolds with the same order singular points are the $2222$ pillowcase and the $333$ turnover. Clearly the turnover cannot occur for a knot orbifold, and thus one only sees the $2222$ pillowcases for the cyclic 2-orbifold. This toroidal decomposition then corresponds to the decomposition along Conway spheres discussed by Bonahon-Siebenmann. Thus, one may reduce to considering tangles with no Conway spheres. The Seifert-fibered pieces will correspond to Montesinos tangles, where one inserts rational tangles in a standard picture. -There are a few exceptional cases, corresponding to the other geometric structures, which were classified by Dunbar. One can browse his tables and discover that the only other case is a Euclidean orbifold which is the 3-fold cover of the figure 8 knot complement.<|endoftext|> -TITLE: Can "syntactic forcing" add ordinals? -QUESTION [9 upvotes]: Kunen, in paragraph VII.9 of his book talks about forcing "via syntactical models", where the we do not use set models of ZFC. Still, the functional $x \mapsto \check x$ can be defined as usual and is, in a sense, an embedding of the ground model $\mathbf V$ in any forcing model $\mathbf V ^\mathbb P$ (obtained, say by completeness theorem). Then, there is also a sense in asking if the $\mathbf V ^\mathbb P$ has more ordinals than $\mathbf V$, i.e. if $\mathbf V ^\mathbb P$ contains an ordinal larger than all $\check \alpha$ for $\alpha$ an ordinal in $\mathbf V$. -I cannot find any formulation of this question in $\mathbf V$ or $\mathbf V ^\mathbb P$. I wonder if any exists. -Also, I suspect that as in usual forcing, $\mathbf V ^\mathbb P$ has the same ordinals as $\mathbf V$, but cannot prove it. I wonder how this, or its negation, can be proved. -Thank you - -REPLY [9 votes]: The answer is that there is one sense in which no new ordinals are added, but there is a second sense in which new ordinals are added. -On the one hand, the sense in which no additional ordinals are added is that for any name $\dot\alpha$, we have an equality in the Boolean values: -$$[\![\dot\alpha\text{ is an ordinal}]\!]=\bigvee_{\beta\in\text{Ord}}[\![\dot\alpha=\check\beta]\!],$$ -which says exactly that $\dot\alpha$ is naming an ordinal exactly to the extent that it is equal to one of the ground model ordinals. So every name for an ordinal is a mixture of ground model check-names for ordinals, and any generic filter will select exactly one of those names from the antichain of possibilities. One can prove the equation by induction on the rank of the name $\dot\alpha$, since the value that $\dot\alpha$ is an ordinal is equal to the value that it is a transitive set of ordinals, and this claim reduces to versions of the equation for smaller rank names. -For any forcing notion $\mathbb{P}$, let $\mathbb{B}$ be the Boolean completion of $\mathbb{P}$ and we may form the structure $V^{\mathbb{B}}$ as a Boolean $\mathbb{B}$-valued structure, defining the Boolean values $[\![\varphi]\!]$ by (meta-theoretic) induction on $\varphi$. You can find full details in my paper Well-founded Boolean ultrapowers as large cardinal embeddings, or in the lecture notes from my tutorial on the Boolean ultrapower at the Hausdorff Center in Bonn, 2011, which are available at that link. -If one continues with your idea of the map $x\mapsto\check x$, you arrive fairly quickly at the Boolean ultrapower. Namely, in a contruction due originally to Vopenka and then studied by Solovay and Scott, let $U\subset\mathbb{B}$ be any ultrafilter (not necessarily generic), and then define the equivalence on names $\sigma=_U\tau\iff[\![\sigma=\tau]\!]\in U$. The quotient structure $V^{\mathbb{B}}/U$ is an actual model of set theory, satisfying $\varphi$ if and only if $[\![\varphi]\!]\in U$, which is a form of the Los theorem. The point is that one may define the ground model $\check V$ of $V^{\mathbb{B}}$, and get a corresponding model $\check V/U$ in the quotient. The Boolean ultrapower map is precisely the map $$x\mapsto [\check x]_U,$$ which is an elementary embedding of $V$ into $\check V/U$. -On the other hand, there is a second sense in which one can think that ordinals are added. Namely, as we explain in the article, the filter $U$ is $V$-generic if and only if the corresponding Boolean ultrapower is an isomorphism. So if you take the Boolean ultrapower by a generic filter, you don't get new ordinals in the Boolean ultrapower, but otherwise you do, since the filter will not always decide the antichains that gave rise to those names for ordinals. Basically, if $U$ misses a maximal antichain, then you can mix distinct ordinals on that antichain to get a name for an ordinal that will not be equivalent to any check name in the Boolean ultrapower. -The previous paragraph describes the sense in which "non-generic" is the right generalization of "non-principal" for ultrafilters on a Boolean algebra. Although it might seem that generic filters are very complicated and principal filters are trivial, nevertheless it is the principal filters that are generic on the atomic Boolean algebras that arise from power sets, and it is because of the genericity that the ultrapower by a principal filter is trivial.<|endoftext|> -TITLE: Fourier-Mukai transform for abelian varieties -QUESTION [5 upvotes]: Let $A$ be an abelian variety over $\mathbb{C}$, $L$ be a very ample line bundle on $A$, then the dual abelian variety is $\hat{A} \cong A/K(L)$ with $K(L)$ the kernel of surjective morphism $A \to Pic^0(A)$. Let $P$ be the Poincare sheaf on $A \times \hat{A}$. It is known that $\Phi_{A \to \hat{A}}^{P}$ (i.e.the Fourier-Muaki functor with kernel $P$ ) is a derived equivalence between $D^b(A)$ and $D^b(\hat{A})$. -Suppose $H \subseteq K(L)$ is a subgroup, and let $\tilde{A} = A/H$. Do we still have derived equivalence between $D^b(A)$ and $D^b(\tilde{A})$? -I feel that one could similarly define a Poincare sheaf $\tilde{P}$ on $A \times \tilde{A}$, but I am not sure if this gives the equivalence. - -REPLY [8 votes]: No. As Will Sawin indicates, every finite subgroup $H$ of $A$ is contained in $K(L)$ for some very ample line bundle on $A$: indeed, let $L_1$ be your favorite very ample line bundle on $A$, and let $n = \# H$; then $H \subset A[n] \subset K(nL_1)$. Thus you are asking whether every abelian variety $B$ which is isogenous to $A$ is a Fourier-Mukai partner of $A$. -This is not true: in order for complex abelian varieties to be Fourier-Mukai partners, it is necessary but not sufficient that they be isogenous. Indeed, Proposition 5.1 of - -MR1827500 (2002a:14017) - Bridgeland, Tom Maciocia, Antony - Complex surfaces with equivalent derived categories. - Math. Z. 236 (2001), no. 4, 677–697. - -shows that any complex abelian surface has only finitely many Fourier-Mukai partners. On the other hand, it is shown in - -Hosono, Shinobu; Lian, Bong H.; Oguiso, Keiji; Yau, Shing-Tung Kummer structures on K3 surface: an old question of T. Shioda. Duke Math. J. 120 (2003), no. 3, 635–647. - -that for every positive integer $n$, there is a complex abelian surface having precisely $n$ Fourier-Mukai partners. Via Hodge theory the question is reduced to the study of positive definite integral quadratic forms....which is quite nontrivial and interesting in its own right, of course.<|endoftext|> -TITLE: The mathematics of tavern puzzles -QUESTION [12 upvotes]: I remember seeing a paper on the arxiv this year (which I cannot now find Edit: This paper: http://arxiv.org/abs/1208.6545, found by j.c.) proposing to study the linkage of rigid bodies such as tavern or carnival puzzles. However, they only worked out one simple example (like showing that two circles of the same diameter can't pass through each other). -This is an interesting idea which I'd like to see more about. I know that such ideas have been studied in symplectic geometry (e.g. the symplectic camel). -It would help to examine a relatively simple case. So my question is: -Given $n$ circles of equal radius in 3-space, how many distinct linkages are there? I.e. how many configuration are there which cannot be made the same by isotopies of 3-space which restrict to a family of isometries on the circles? -I would also accept a solution for $n\leq 4$. - -REPLY [2 votes]: A general useful theory probably does not exist. The reason is that even in dimension $n=2$, and even if all but one of the the bodies are rectangles, the problem is PSPACE-hard. That is because a popular board game Rush Hour is PSPACE-complete. This a result of Flake and Baum. -By taking a product with an interval, and enclosing the puzzle in a box, one can of course transfer the result to three (and also higher) dimensions.<|endoftext|> -TITLE: What fraction of the integer lattice can be seen from the origin? -QUESTION [48 upvotes]: Consider the integer lattice points in the positive quadrant $Q$ of $\mathbb{Z}^2$. -Say that a point $(x,y)$ of $Q$ is visible from the origin if the -segment from $(0,0)$ to $(x,y) \in Q$ passes through no other point of $Q$. -So points block visibility, and the only points visible from the origin are those with -a clear line of sight: -    -Let $\nu(n)$ be the ratio of the number of points visible within the -square with corner $(n,n)$ to $n^2$. For example, for $n=8$, $43$ of the $64$ points -are visible, so $\nu(n) = 43/64 \approx 0.67$. - -Q: What is $\lim_{n \to \infty} \nu(n)$? - -It appears to be approaching $\approx 0.614$: -    -The question bears some similarity to Polya's Orchard Problem -(T.T. Allen, "Polya's orchard problem," -The American Mathematical Monthly -93(2): 98-104 (1986). Jstor link), but I cannot see my question -is answered in that literature. -One could ask the same question for $\mathbb{Z}^d$. - -Answered by Pete Clark: the limit is $\frac{1}{\zeta(d)}$. -Thus in higher -dimensions, almost all the lattice points are visible (because $\zeta(d)$ approaches $1$). - -REPLY [3 votes]: Here is an elementary argument using Euler products. I'm not sure if the initial probability part is rigorous, but it's a heuristic argument at least. The probability of a prime $p$ dividing a positive integer is $\frac{1}{p}$. Then the probability of $p$ dividing all entries of a $d$-tuple of positive integers is $\frac{1}{p^d}$. A $d$-tuple has $\gcd$ equal to $1$ if and only if no prime divides all of the elements. So the probability of a $d$-tuple of positive integers being coprime is $$\prod_{p\text{ prime}}{\left(1-\frac{1}{p^d}\right)}.$$ By the formula for an infinite geometric series, -$$\frac{1}{1-\frac{1}{p^d}}=\sum_{k=0}^{\infty}{\frac{1}{p^{kd}}},$$ so the above product is equal to $$\left[\prod_{p\text{ prime}}{\left(1+\frac{1}{p^{d}}+\frac{1}{p^{2d}}+\frac{1}{p^{3d}}+\cdots\right)}\right]^{-1}=\left[\sum_{n=1}^{\infty}{\frac{1}{n^d}}\right]^{-1}=\frac{1}{\zeta(d)}.$$<|endoftext|> -TITLE: Mazur's torsion theorem on elliptic curves and its generalisations -QUESTION [7 upvotes]: I want to study Mazur's torsion theorem for elliptic curves over $Q$ and its generalizations for number fields, i.e., papers by Kamienny, Kenku & Momose, Filip Najman. So please suggest to me what background I should have before starting to read the relevant papers. If you could offer me some books/papers/articles, I would be glad. - -REPLY [14 votes]: Andrew Snowden has just finished teaching a course on Mazur's torsion theorem--video-taped lectures and extensive notes may be found here. I've watched several of the lectures; they are excellent. - -REPLY [4 votes]: The general case over arbitrary number fields has been treated by Loïc Merel. A good place to start would be Bas Edixhoven's Bourbaki exposé Rational torsion points on elliptic curves over number fields, Séminaire Bourbaki, 36 (1993-1994), Exposé No. 782. -There is also an excellent set of notes by Joseph Œsterlé which were to be found on the ICTP website.<|endoftext|> -TITLE: How to compute the normals to Costa's minimal surface? -QUESTION [6 upvotes]: I am trying to draw Costa's minimal surface in high resolution using the PovRay raytracer. For this I need to compute points on the surface as well as the normals. It is relatively easy to compute the points with Mathematica, because it has Weierstrass functions built in: -g = WeierstrassInvariants[{1/2, I/2}]; -e = WeierstrassP[1/2, g]; - -costa[x_, y_] := - With[ - { - wz1 = WeierstrassZeta[x + I y, g], - wz23 = - WeierstrassZeta[x + I y - 1/2, g] - - WeierstrassZeta[x + I y - I/2, g], - wp = WeierstrassP[x + I y, g] - }, - { - (Pi x + Pi^2/(4 e) - Re[wz1] + Pi/(2 e)*Re[wz23])/2, - (Pi y + Pi^2/(4 e) + Im[wz1] + Pi/(2 e)*Im[wz23])/2, - (Sqrt[2 Pi]/4)*Log[Abs[(wp - e)/(wp + e)]] - } - ] - -To draw Costa's surface you can do ParametricPlot3D[costa[x, y], {x, 0, 1}, {y, 0, 1}]: - -Let me write costa[x,y] as $C(x,y)$. Presumably, to compute the normal at $C(x_0,y_0)$ we take the cross product $(\partial C/\partial x) \times (\partial C/\partial y)$ at $(x_0, y_0)$. This is not so easy to do with Mathematica because it does not know how to compute the derivatives of $\wp$ and $\zeta$. -How can I get explicit formulas for the normal? Do I really have to do the cross product of partial derivatives, or is there an easier way? - -REPLY [3 votes]: I went to the trouble of computing the normals with Mathematica (thanks to my colleague Pavle Saksida for teaching me how to compute derivatives). You can see the result at my GitHub costa-surface repository. The result is not pretty. Just for the fun of it, and to torture MathJax, here is the $x$-component of the normal at a point parametrized by $z \in \mathbb{C}$, as computed and simplified by Mathematica ($g$ is as computed above in the original question): -$$\frac{\sqrt{\pi}}{8\Re(\wp(1/2,g))\sqrt{2}}\cdot\left(\left(\pi\Im(\wp(z-i/2,g)-\wp(z-1/2,g))-2\Re(\wp(1/2,g))\Im(\wp(z,g))\right)\cdot \left(\Im(\wp'(z,g))\cdot\left(\frac{\Re(\wp(1/2,u)-\wp(z,g))}{\Im(\wp(1/2,u)-\wp(z,g))^2+\Re(\wp(1/2,g)-\wp(z,g))^2}+\frac{\Re(\wp(1/2,g)+\wp(z,g))}{\Im(\wp(1/2,u)+\wp(z,g))^2+\Re(\wp(1/2,g)+\wp(z,g))^2}\right)+\Re(\wp'(z,g))\cdot\left(\frac{\Im(-\wp(1/2,u)+\wp(z,g))}{\Im(\wp(1/2,u)-\wp(z,g))^2+\Re(\wp(1/2,g)-\wp(z,g))^2}-\frac{\Im(\wp(1/2,g)+\wp(z,g))}{\Im(\wp(1/2,u)+\wp(z,g))^2+\Re(\wp(1/2,g)+\wp(z,g))^2}\right)\right)-\left(2\Re(\wp(1/2,g))(\pi-\Re(\wp(z,g)))+\pi\Re(\wp(z-i/2,g)-\wp(z-1/2,g))\right) \cdot \left(\frac{\Im(-\wp(1/2,u)+\wp(z,g))\Im(\wp'(z,g))}{\Im(\wp(1/2,g)- -\wp(z,g))^2+\Re(\wp(1/2,g)-\wp(z,g))^2}-\frac{\Im(\wp(1/2,g)+\wp(z,g))\Im(\wp'(z,g))}{\Im(\wp(1/2,g)+\wp(z,g))^2+\Re(\wp(1/2,g)+\wp(z,g))^2}+\Re(\wp'(z,g))\cdot\left(\frac{\Re(-\wp(1/2,u)+\wp(z,g))}{\Im(\wp(1/2,u)-\wp(z,g))^2+\Re(\wp(1/2,g)-\wp(z,g))^2}-\frac{\Re(\wp(1/2,g)+\wp(z,g))}{\Im(\wp(1/2,u)+\wp(z,g))^2+\Re(\wp(1/2,g)+\wp(z,g))^2}\right)\right)\right) -$$ -So I wonder how Weierstraß would have computed this.<|endoftext|> -TITLE: Structures that turn out to exhibit a symmetry even though their definition doesn't -QUESTION [50 upvotes]: Sometimes (often?) a structure depending on several parameters turns out to be symmetric w.r.t. interchanging two of the parameters, even though the definition gives a priori no clue of that symmetry. -As an example, I'm thinking of the Littlewood–Richardson coefficients: If defined by the skew Schur function $s_{\lambda/\mu}=\sum_\nu c^\lambda_{\mu\nu}s_\nu$, where the sum is over all partitions $\nu$ such that $|\mu|+|\nu|=|\lambda|$ and $s_{\lambda/\mu}$ itself is defined e.g. by $ -s_{\lambda/\mu}= \det(h _{\lambda_i-\mu_j-i+j}) _{1\le i,j\le n}$, it is not at all straightforward to see from that definition that $c^\lambda_{\mu\nu} =c^\lambda_{\nu\mu} $. -Granted that this way of looking at it may seem a bit artificial, as I guess that in many of such cases, it is possible to come up with a "higher level" definition that shows the symmetry right away (e.g. in the above example, the usual (?) definition of $c_{\lambda\mu}^\nu$ via $s_\lambda s_\mu =\sum c_{\lambda\mu}^\nu s_\nu$), but showing the equivalence of both definitions may be more or less involved. So I am aware that it might just be a matter of "choosing the right definition". Therefore, maybe it would be better to think of the question as asking especially for cases where historically, the symmetry of a certain structure has been only stated 'later', after defining or obtaining it in a different way first. -Another example that would fit here: the Perfect graph theorem, featuring a 'conceptual' symmetry between a graph and its complement. -What are other examples of "unexpected" or at least surprising symmetries? -(NB. The 'combinatorics' tag seemed the most obvious to me, but I won't be surprised if there are upcoming examples far away from combinatorics.) - -REPLY [2 votes]: The $qt$-catalan numbers, which encode the area and bounce statistic of Dyck paths. -These encode a bigraded Frobenius/Hilbert series in diadonal harmonics, and with this interpretation, there is an obvious qt-symmetry. However, it is still an open problem to find a combinatorial proof of the symmetry, by exhibiting a bijection that interchanges the area and bounce statistics.<|endoftext|> -TITLE: V=HOD & The Height of the Large Cardinal Tree -QUESTION [12 upvotes]: As we know the assumption $V=L$ adds a restriction on the height of the large cardinal tree. Also there is a strict border like $0^{\sharp}$ exists such that all large cardinal axioms which are equivalent or stronger than this axiom are contradictory with $V=L$ and all large cardinal assumptions below it are consistent with $V=L$. -Question: What is the situation for the weaker assumption $V=HOD$ and large cardinal tree? Precisely: -(a) Which one of the known large cardinal axioms (together with $ZFC$) does imply $V\neq HOD$? -(b) Is there any strict border in this case as same as "$0^\sharp$ exists" and "$V=L$"? - -REPLY [15 votes]: $\newcommand\HOD{\text{HOD}}$ -There is no such border, because almost all the large cardinal properties, including the very strongest large cardinal axioms, are relatively consistent with $V=\HOD$. For the larger large cardinals, this is generally proved by forcing, and there are several natural ways to force $V=\HOD$. -One such way to force $V=\HOD$ is to force so as to code every set in the pattern of GCH on the cardinals. Basically, one undertakes a forcing iteration that at each cardinal stage decides generically whether to force the GCH at that cardinal or not; in other words, one uses the lottery sum of two posets, one of which forces the GCH there and the other of which forces a failure of GCH there. In the extension, a simple density argument shows that every set is coded into the GCH pattern, and so we get $V=\HOD$. (The general idea of this kind of coding is due originally to McAloon, although he used a different method. Many authors describe the coding with bookkeeping functions, but this is not actually needed, since the generic coding means that it is dense that any new set is coded.) The forcing is somewhat easier when the $\text{GCH}$ holds in the ground model, and in the general case one wants to space out the cardinals at which coding occurs. The assertion that every set is coded into the GCH pattern is named as the continuum coding axiom (CCA) in the dissertation of Jonas Reitz, who proved that this implies the ground axiom (GA). You can find further uses of the CCA in Set-theoretic geology, which also contains full details of this kind of forcing and variations. -Now, the point is that most of the usual large cardinal axioms are preserved by the $V=\HOD$ forcing, by the usual lifting arguments as for GCH. Thus, all the main large cardinals are relatively consistent with $V=\HOD$. -There are many other coding methods besides coding into the GCH pattern. Andrew Brooke-Taylor, for example, undertook coding in the $\Diamond^*_\kappa$ pattern, which allows one to force $V=\HOD+\text{GCH}$ while preserving all the usual large cardinal notions. His paper Large Cardinals and Definable Well-Orderings of the Universe exactly fits the theme of this question. See also my paper The wholeness axiom and $V=\HOD$, which is about precisely this problem in the case of the Wholeness axiom and related large cardinals.<|endoftext|> -TITLE: What is a Homotopy between $L_\infty$-algebra morphisms II -QUESTION [9 upvotes]: I would like to continue on question I asking, what is a homotopy between -Lie infinity algebras, since I'm not satisfied in two directions: -1.) The naive approach to define a homotopy would be ('naive' in my opinion -of course) the following: -Let $(L,(l)_{k\in\mathbb{N}})$ and $(M,(d)_{k\in\mathbb{N}})$ be two Lie infinity -algebras, let $f_\infty,g_\infty:L \to M$ be two morphism -(in the most general sense) and let $(C(L),Q_L)$ and $(C(M),Q_M)$ be the -appropriate differential graded coalgebras with induced morphism -$F,G:C(L)\to C(M)$. -Then a homotopy between $F$ and $G$ is a degree $+1$ map -$H:C(L)\to C(M)$ such that $F-G = HQ_L \pm Q_MH$ -let signs and additional structure of $H$ (linear, coalgebra, ...) aside for -a moment. -It is strange, however, that I never saw this approach in the literature. Is this definition of homotopy equivalent to the previously mentioned approaches -in question I? -Now for what is more important: -2.) The homotopy theory of Lie infinity algebras as given by Urs Schreiber in -question I is obtained by 'transferring' the homotopy theory of differential graded Lie algebras 'along' an adjunction -$F\dashv G$ -($F:L_\infty Alg \to DGLA$ and $G:DGLA \to L_\infty Alg$) -Now the question that really irritates me for quite some time is: how can we be sure that we get the correct homotopy theory of Lie infinity algebras by transferring its 'shadow' in the category of DG Lie algebra back along the previous mentioned adjunction? To me it looks like we can not rule out that there is a more general definition of weak equivalences in the category of Lie infinity algebras, which just project under $F$ onto those we already know. -Sorry if the second question is vague. -Edit; I used the 'Lie algebra cohomology tag' since the homotopy theory of -Lie infinity algebras affects Lie algebra cohomology, too. - -REPLY [2 votes]: First of all, you might want to look at the MO question How to define the equivalence of Maurer-Cartan elements in an $L_\infty$-algebra?, since of course this is effectively what you need (morphisms can be described as MC elements in a particular convolution $L_\infty$-algebra, I suspect you know that). -Let me also say that some sort of approach to homotopy of the kind you are asking for is discussed by Martin Markl in this paper, and in this paper it is checked that the sort of definition proposed by Martin can be obtained from a Sullivan-type approach to homotopies (outlined in an answer to your first question) using appropriate homotopy transfer theorem for homotopy cooperads. -It is worth remarking, in particular, that the definition of a "derivation homotopy" suggested in another answer to this question does not work: the crucial difference between $A_\infty$ and $L_\infty$ is that $A_\infty$ is a nonsymmetric operad, and therefore instead of the Sullivan's algebra of differential forms on the 1-simplex we can use the dg algebras of chains on the 1-simplex (which is noncommutative) - this gives the "derivation homotopy".<|endoftext|> -TITLE: Universal $C^*$-algebra with generators and relations -QUESTION [14 upvotes]: We say that the $C^*$-algebra $A$ generated by $a_1,...,a_n$ is universal subject to relations $R_1,...,R_m$ if for every $C^*$-algebra $B$ with elements $b_1,...,b_n$ satisfying relations $R_1,...,R_m$ there is $C^*$-epimorphism $\varphi: A \to B$ such that $\varphi(a_i)=b_i$. One of the basic examples is the $C^*$-algebra of complex valued function on three sphere $C(S^3)$ which is the universal commutative unital $C^*$-algebra generated by $a,b$ with relation $a^*a+b^*b=1$. My question is the following: what kind of relations can we impose on our $C^*$-algebra? In all examples which I saw the relations were algebraic and were of the form: $f(a_1,...,a_n,a_1^*,...,a_n^*)=0$ where $f$ was some polynomial. In particular do we admit: - -quantification and referring to other elements not being the generators -order properties of $C^*$-algebras -functions which are no longer polynomials (continuous functions, Borel functions etc.) - -If the answer is positive I would be grateful to know some (known in literature) examples of universal $C^*$-algebras arising in such a way. - -REPLY [18 votes]: Since free C*-algebras don't exist, we can't give a concrete description of all relations that are allowed. Instead, we need to give axioms that determine what collections of n-tuples $(a_1,\dots,a_n)$ in $A$ are allowed, where $A$ varies over all C*-algebras. It is important that the elements not "know about" the ambient C*-algebra, so ``a is in a separable C*-algebra'' is not allowed. -Some of what is allowed is conditions like $0 \leq x \leq 1$ and -$$ -0\leq\left[\begin{array}{cc} -\mathbf{1} & x\\ -x^{*} & \mathbf{1} -\end{array}\right]\leq 1 -$$ -where $\mathbf{1}$ is in the unitization of $A$. Another fun example is ``x is hermitian and has spectrum contained in the Cantor set''. We can't use Borel functional calculus, but we can take a continuous function on $\mathbb R$ that is bounded and use as a relation "$x$ is normal and $f(x)=0$." See section 5 of "From Matrix to Operator Inequalities" by me, Canad. Math. Bull. 55(2012), 339--350 for a use of analytic functional calculus. -I could drone on forever here, but I did already: ``C*-Algebra Relations'' in Mathematica Scandinavica 107, 43--72, 2010.<|endoftext|> -TITLE: Bad subforcings of nice forcing notions -QUESTION [5 upvotes]: Let $\mathbb{P}$ and $\mathbb{Q}$ be two forcing notions. Recall that we say $\mathbb{Q}$ is a subforcing of $\mathbb{P}$ if there exists a regular embedding $\mathbb{Q} \to \text{r.o.}(\mathbb{P}).$ -Question. Let $\mathbb{P} \in \{Add(\omega, \kappa), Col(\omega_1, \kappa) \}.$ -(1) Is there a subforcing of $\mathbb{P}$ which is not Proper? -(2) Is there a subforcing of $\mathbb{P}$ which is not semi-Proper? -Remark. In Subalgebras of Cohen algebras need not be Cohen, Koppelberg and Shelah show that there is a subalgebra of Cohen forcing which is not Cohen. - -REPLY [8 votes]: No. A subforcing of a c.c.c. forcing is c.c.c. A subforcing of a countably closed forcing is countably-strategically-closed, which implies proper. (This is easy to see via countable elementary submodels. Use the strategy to construct a generic condition.) -Furthermore, every subforcing of a proper forcing is proper. Properness is equivalent to preserving stationary subsets of $[\kappa]^\omega$ for all $\kappa$. The stationarity of $X \subseteq [ \kappa ]^\omega$ cannot be restored once killed, since killing it is just adding some $f : [\kappa]^{<\omega} \to \kappa$ such that no $x \in X$ is closed under $f$. (See Abraham's chapter of the Handbook.)<|endoftext|> -TITLE: When is a continuous path stochastic process be representable as diffusion or Ito process? -QUESTION [9 upvotes]: When can a continuous path (Markovian) stochastic process in one dimension be represented as an Ito or a diffusion process? What are the examples when it can not be? - -REPLY [4 votes]: These types of questions are treated in great generality in the book -Rogers-Williams: Diffusions, Markov processes and martingales, Volume 1 -One of the great results is due to Dynkin. Let $(X_t)_{t \ge 0}$ be a continuous Markov process whose semigroup is Feller-Dynkin (that is restricts to a strongly continuous semigroup on continuous functions vanishing at $\infty$). Then if the generator of $X$ contains the space of smooth and compactly supported functions, this generator is necessarily a second-order semi-elliptic differential operator and so $X$ is a diffusion process.<|endoftext|> -TITLE: Can you identify complex conjugations in a number field? -QUESTION [6 upvotes]: Every automorphism of an algebraic number field $F$ extends to an automorphism of $\mathbb{\overline{Q}}$, but an order 2 automorphism of $F$ need not extend to one of order 2 on $\mathbb{\overline{Q}}$, -Is there an algebraic criterion, just looking at the action on $F$, to tell whether a given involution on $F$ will extend to a complex conjugation? -Trivially, when $F$ is quadratic, an automorphism extends to a complex conjugation iff it fixes every number with negative norm. That just says either the automorphism is the identity or the field is imaginary. -Does something like this generalize to non-trivial cases? - -REPLY [6 votes]: An involution $\sigma$ is a complex conjugation if not every element of $F^\sigma$ is a sum of squares of elements of $F^\sigma$ and, if $\sigma$ is nontrivial, some element of $F^\sigma$ which is not a sum of squares of elements of $F^\sigma$ is a sum of squares of elements of $F$. -Proof that every complex conjugation satisfies this: $F^\sigma$ has an embedding into the reals. Since every negative real number is not a sum of squares, some element in $F^\sigma$ is not a sum of squares. If $\sigma$ is nontrivial, there is a complex number in $F$ of the form $a+bi$ with $|b|>|a|$. Then $(a+bi)^2 + (a-bi)^2 = 2a^2-2b^2 <0$, so some negative number is a sum of squares in $F$. Of course, negative numbers are not a sum of squares in $F^\sigma$. -Proof that every $\sigma$ that satisfies this is a complex conjugation: $F^\sigma$ is a formally real field, hence it can be given an ordering, and hence an embedding into $\mathbb R$. Moreover, if $x\in F^\sigma$ is an element that is not a sum of squares in $F^\sigma$, then it can be given an ordering such that $x$ is negative. This is because the preordering consisting of the sums of squares can be extended to a preordering including $x$, and thus by Zorn's lemma to an ordering of the field. For $x\in F^\sigma$ an element which is not a sum of squares in $F^\sigma$ but is a sum of squares in $F$, applying this construction gives a real embedding of $F^\sigma$ that extends to a complex embedding of $F$.<|endoftext|> -TITLE: An integrality question about expressing an integer as a product of numbers below $n$ -QUESTION [12 upvotes]: Let $n\ge 2$ be a natural number. Suppose that $N$ is a natural number, composed only of primes below $n$, and that can be expressed as -$$ -N= \prod_{j=1}^{n} j^{x_j} -$$ -where $x_1$, $\ldots$, $x_n$ are non-negative rational numbers with $\sum_{j}x_j \in {\Bbb N}$. Does there necessarily exist a representation of $N$ as -$$ -N=\prod_{j=1}^{n} j^{a_j} -$$ -where $a_1$, $\ldots$, $a_n$ are non-negative integers with $\sum_{j} a_j=\sum_j x_j$? -Let me also give the interpretation in terms of lattice points in polytopes in case that is easier to think about. For each natural number below $n$ associate a point in ${\Bbb Z}^{\pi(n)}$ corresponding to the exponents in the prime factorization of $n$. So $1$ corresponds to $(0,0,\ldots,0)$, $2$ corresponds to $(1,0,\ldots,0)$ and so on. Let ${\mathcal C}(n)$ denote the convex hull of these lattice points. Suppose we know that a number $N$ corresponds to a lattice point in ${\mathcal C}(n)$ dilated by a factor $m \in {\Bbb N}$. Can that lattice point then be written as a sum of $m$ lattice points from ${\mathcal C}(n)$? In general it is not true that the lattice points in the dilate of a set must fall in the $m$-fold sumset. Is it true in this special situation? -Note: I had originally overlooked the condition that $N$ should be composed only of primes below $n$. This subtlety was pointed out by The Masked Avenger, and at the same time Greg Martin produced an excellent example to show why it is necessary. Thanks to both of them, and with apologies to Greg for the oversight. Greg also provided a solution even when $N$ is composed only of primes below $N$. Here the problem is that the situation for numbers permits more relations than the situation for polytopes which is what I kept thinking about. So I've moved the goalpost one more time (sorry) by asking whether rationality of $x_j$ implies integrality. If this also has a simple counterexample, then I give up! -(Note: the condition that $\sum_j x_j \in {\Bbb N}$ is necessary, else as Christian Elsholtz kindly pointed out one has the counterexample $1^0\times 2^{1/2}\times 3^0\times 4^{1/4}=2$.) -This problem arose in connection with the recent MO question: How many different numbers can be obtained as product of first $n$ natural numbers? If the problem posed here has a positive answer, then it would follow (in the notation of the linked question) that $P(m,n)$ is the Erhart polynomial (in $m$) of a certain convex polytope ${\mathcal C}(n)$. - -REPLY [18 votes]: It seems that here is an example for rational exponents. -Let $n=2209=47^2>3^7>2^{11}$, and $N=4\,385\,664=2^7\cdot 3^6\cdot 47=2048^{7/11}\cdot 2187^{6/7}\cdot 2209^{1/2}\cdot 1^{1/154}$ with $7/11+6/7+1/2+1/154=2$. If $N=ab$ with $a,b\leq 47^2$, then 47 divides one of $a$ and $b$ (say, $a=47k$); since $a,b\leq 47^2$ we have $47\geq k\geq 40$ (the right estimate is quite rough). But there is no product of powers of 2 and 3 in this interval.<|endoftext|> -TITLE: Quotient Metric Space -QUESTION [7 upvotes]: Let $(X,d)$ be a metric space and $R\subseteq X \times X$ an equivalence relation. -Is there a condition for which $X/R$ with the usual quotient topology is a metric space? -Thanks! - -REPLY [6 votes]: I am going to answer this question in the more general setting of uniform spaces rather than metric spaces. Most of the content of this answer is taken from the book Introduction to Uniform Spaces by I.M. James. -If $R$ is an equivalence relation on a set $X$, then write $\pi_{R}$ for the projection from $X$ to $X/R$. -Let $R$ be an equivalence relation on a uniform space $(X,\mathcal{U})$. Then we say that $R$ is weakly compatible with the uniform structure on $X$ if for each $D\in\mathcal{U}$ there is an $E\in\mathcal{U}$ such that $E\circ R\circ E\subseteq R\circ D\circ R$. -Furthermore, we say that $R$ is compatible with the uniform structure on $X$ if for each $D\in\mathcal{U}$ there is an $E\in\mathcal{U}$ where $R\circ E\subseteq D\circ R$. Clearly every compatible equivalence relation is weakly compatible. -If an equivalence relation $R$ is weakly compatible with a uniformity $(X,\mathcal{U})$, then the images $\pi_{R}\times\pi_{R}[D]=\{(\pi_{R}(x),\pi_{R}(y))|(x,y)\in D\}$ of the entourages $D\in\mathcal{U}$ generate a uniformity $\mathcal{U}/R$ on the quotient space $X/R$. The mapping $\pi_{Z}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{U}/R)$ is clearly uniformly continuous. Clearly the mapping $\pi_{R}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{U}/R)$ is uniformly continuous. -It is well known that a uniform space $(X,\mathcal{U})$ is generated by a metric if and only if $\mathcal{U}$ is generated by countably many elements. Therefore, if $(X,d)$ is a metric space that induces a uniformity $\mathcal{U}$ and $R$ is a weakly compatible equivalence relation on $X$, then the uniformity $\mathcal{U}$ and also the quotient uniformity $\mathcal{U}/R$ are generated by countably many elements. Therefore, the quotient uniformity $\mathcal{U}/R$ is induced by some metric (and I am sure there is an easy explicit description of such a metric). -If we assume that $R$ is a compatible equivalence relation on a uniform space rather than just a weakly compatible equivalence relation, then the quotient uniform spaces are essentially the uniformly continuous uniformly open surjections. -If $(X,\mathcal{U}),(Y,\mathcal{V})$ are uniform spaces, then we say that a map $f:X\rightarrow Y$ is uniformly open if for each entourage $D\in\mathcal{U}$ there is an entourage $E\in\mathcal{V}$ such that $E[f(x)]\subseteq\phi[D[x]]$ for all $x\in X$. -Of course, by $D[x]$ we mean $D[x]=\{y\in X|(x,y)\in D\}$. - -$\mathbf{Theorem}:$ Suppose that $R$ is an equivalence relation on a - uniform space $(X,\mathcal{U})$. Then: - -If $R$ is compatible with the uniform structure on $(X,\mathcal{U})$, then the mapping - $\pi_{R}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{U}/R)$ is uniformly - continous and uniformly open. -If $\mathcal{V}$ is a uniformity on the quotient set $X/R$ and the projection $\pi_{R}:(X,\mathcal{U})\rightarrow(X/R,\mathcal{V})$ is - uniformly continuous and uniformly open, then $R$ is compatible with - the uniform structure $(X,\mathcal{U})$ and - $\mathcal{V}=\mathcal{U}/R$.<|endoftext|> -TITLE: Hyperbolic 3-manifolds fibering over the circle -QUESTION [8 upvotes]: Suppose you have the mapping torus $M_\phi$ of some pseudo-Anosov map $\phi.$ Is there some sufficient or necessary condition on $\phi$ to assure that $M_\phi$ has large injectivity radius? I am aware of Jeff Brock et al's results on volume bounds, but this is not quite the same... - -REPLY [8 votes]: A related question is studied by Minsky in the paper "Bounded geometry for Kleinian groups": http://arxiv.org/abs/arXiv:math/0105078 -The main theorem gives a necessary and sufficient condition for the infinite cyclic cover $\tilde{M}_{\phi}$ of a pseudo-Anosov mapping torus to have large injectivity radius, in terms of the geometry of the action of $\phi$ on the curve complex $\mathcal{C}(S)$, for $S$ the fiber of your mapping torus. -It states that given $\epsilon>0$, there is some $K>0$ depending on $\epsilon$ and $S$, such that if $\tilde{M}_{\phi}$ has injectivity radius at least $\epsilon$, then all proper subsurface projections satisfy -$$ d_{Y}(\nu^{-}, \nu^{+}) 0$, there exists some $\epsilon'$ such that if all subsurface projections are uniformly bounded above by $K'$, then your degenerate surface group $\tilde{M}_{\phi}$ has injectivity radius at least $\epsilon'$). In the literature, any pseudo-Anosov $\phi$ satisfying the subsurface projection criterion above for $K$ is said to have "$K$-bounded combinatorics". -This result says something about the injectivity radius of the mapping torus $M_{\phi}$, so long as the loop corresponding to the circle direction (i.e., the generator of the infinite cyclic group of deck transformations) isn't too short. But using work of Brock-Bromberg (see Section 7 of the paper "Geometric inflexibility and 3-manifolds that fiber over the circle: http://www.math.brown.edu/~brock/home/text/papers/inflex/www/inflex.pdf), and also Minsky, we can ensure that this doesn't happen by taking a large power of $\phi$. -The relevant result is that given a pseudo-Anosov $\phi$, its stable translation length in $\mathcal{C}(S)$ is coarsely equal to "thick distance" in $M_{\phi}$, which is the length of the shortest arc traversing the circle direction of $M_{\phi}$, after electrifying all thin regions (adding in an extra point $x$ and declaring all Margulis tubes of $M_{\phi}$ to be within $1/2$ of $x$). -So to summarize, here is a necessary condition on $\phi$ for $M_{\phi}$ to be $\epsilon$-thick: -$\phi$ must have $K(\epsilon,S)$-bounded combinatorics. -The problem is that, at least to my knowledge, none of this is effective. Given $S$ and $\epsilon$, it's completely unknown how small one needs to take $K$ (besides extremely soft statements, such that it's impossible for $K$ to be uniform in either $\epsilon$ or the topological complexity of $S$). -On the other hand, If you choose $\phi$ to have large stable translation length, and $K'$-bounded combinatorics, these results tell you that there is some $\epsilon'$ such that $M_{\phi}$ is $\epsilon'$-thick, but again it's difficult to estimate $\epsilon'$ in terms of $K'$ and $S$. -By the way, another way of thinking about all of this is that large injectivity radius of $\tilde{M}_{\phi}$ is implied by the Teichmuller geodesic connecting the end invariants $\nu^{+}, \nu^{-}$ residing in some sufficiently thick part of Teichmuller space (this interpretation is implied by the work referenced above, and also Rafi's work- see "A characterization of short curves along a Teichmuller geodesic": http://www.math.toronto.edu/~rafi/Papers/Short-Curves.pdf).<|endoftext|> -TITLE: Cominuscule property of nilpotent orbits -QUESTION [7 upvotes]: Let $G$ be a complex reductive Lie group, $G/P$ a flag manifold, -and $\Phi: T^* G/P \to {\mathfrak g}^*$ the moment map. So $\Phi(T^* G/P)$ is the closure of a nilpotent orbit. -Lots of classes of nilpotent orbits have special names. Is there a name for those nilpotent orbits arising as $\Phi(T^* G/P)$, where $G/P$ is cominuscule (also known as, a compact Hermitian symmetric space)? -I'm trying to generalize some type $A$ examples, where every $G/P$ is cominuscule for $P$ maximal, but still one gets very few nilpotent orbits this way. - -REPLY [2 votes]: Would "cominuscule Richardson orbits" be a satisfactory name ? It is basically a stringing together of adjectives that imply the two properties that were outlined in the question. A similar name (w/o the "co-") has also appeared in the literature : Mark Reeder, "Small representations and minuscule Richardson orbits" . Side remark : Actually, Reeder is using minuscule property on G and the Richardson property on Langlands dual of G. So, I think "co-minuscule Richardson" would actually suit his setup better!<|endoftext|> -TITLE: The Sato-Tate conjecture for hypersurfaces? -QUESTION [11 upvotes]: The Sato-Tate conjecture for elliptic curves $E$ predicts the distribution of the eigenvalues of Frobenius at $p$ on the Tate module of $E$ as $p$ varies in terms of the distribution of the eigenvalues of a random element of an associated compact group. There are conjectural generalizations for curves of higher genus and abelian varieties of higher dimension in terms of more complicated compact groups. - -What's conjectured about the corresponding question for hypersurfaces? - -To be more precise, consider a smooth hypersurface of degree $d$ in $\mathbb{P}^n$ defined over $\mathbb{Z}$. Its only interesting $\ell$-adic cohomology should be in the middle degree, and its rank is known. What's conjectured about the distribution of the eigenvalues of Frobenius at $p$ acting on it as $p$ varies? -Edit: If, as ulrich suggests, the expectation here is that it resembles the case of curves / abelian varieties, what about the case of more general varieties? In general it seems like the cup product restricts Frobenius eigenvalues in some a priori complicated way. - -REPLY [6 votes]: The conjecture is that associated to any cohomology group of an algebraic variety, or set of cohomology group on the algebraic variety, there should be a complex Lie group (with representations), which can be described in a number of ways: -The Zariski closure of the image of the Galois group inside $GL_n(\mathbb Q_l)$ is an algebraic group. Base change this group from $\mathbb Q_l$ to $\mathbb C$. -The Mumford-Tate group of the Hodge structure on the cohomology groups. This group fixes all algebraic cycle classes. If the algebraic cycle classes aren't defined over $\mathbb Q$, extend this group by the Galois group of the field they are defined over. Then base change to $\mathbb C$. -The image of the (conjectural) motivic Galois group in the representation associated to the motive in the (conjectural) Tannakian category of motives, base changed to $\mathbb C$. -These should all be the same group. This is "only" the Hodge conjecture and the Tate conjecture! -Then we need to make the weight adjustment on this group, which is parallel to dividing $Frob_p$ by $p^{w/2}$. To do this, consider all homomorphisms from this group to $\mathbb G_m$. These should correspond to characters of the Galois group / 1-dimensional Hodge structures with a finite Galois action / 1-dimensional motives. Take the kernel of all homomorphisms which correspond to powers of the cyclotomic character / Hodge classes with trivial Galois action / powers of the Tate motive. This gives a new, slightly smaller Lie group. -Take the maximal compact subgroup of this group. This is the Sato-Tate group. Take the Haar measure on it, and deduce a distribution for whatever data you care about (traces, eigenvalues, etc.). This is the conjectural distribution for the eigenvalues of Frobenius. -That is the Generalized Sato-Tate conjecture. -This follows, I believe, from holomorphicity of the L functions of the Galois representations corresponding to all the nontrivial representations of the Sato-Tate group, which should follow from the Langlands program.<|endoftext|> -TITLE: Example: Principal G bundle that is not Zariski locally trivial, G not finite and G simply connected -QUESTION [13 upvotes]: Let $G$ be an affine algebraic group over $\mathbb{C}$. It is well known that when working with principal $G$ bundles it is too restrictive to require bundles to be locally trivial in the Zariski topology. Instead $G$-bundles are defined to be locally trivial in the etale topology. The finite map $z \to z^n$ from $\mathbb{C}^\times$ to itself is a $\mu_n$ (= $n$th roots of unity) bundle that is not Zariski locally trivial. Also setting $B = Spec \mathbb{C}[s^\pm,t^\pm]$ then $\{x^2 + s y^2 + t z^2 = 0\} \subset \mathbb{P}^2_B$ is a $\mathbb{P}^1$-bundle that gives rise to a $PGL_2$-bundle that is not Zariski locally trivial. -However for some groups (dubbed special groups) being locally trivial in the etale topology implies locally trivial in the Zariski topology. If $G$ is semisimple then Grothendieck proved the only such special $G$ are products of $SL_n$ and $Sp_{2m}$. -QUESTION -I would like an example of a non Zariski locally trivial principal $G$-bundle for a simple, simply connected group $G$, e.g. $Spin(n)$ or $G_2$. -My motivation is largely out curiosity. For $G$ as in the question, I study the moduli space of $G$-bundles on a curve (and although for a curve over $\mathbb{C}$ a bundle will be Zariski locally trivial it wont be in families) and I would like to have a non Zariski locally trivial bundle in my back pocket. -A vague idea would be to mimic the $PGL_2$ example above by looking at some varying family of $G/P$. You could try this with $G = E_8$ because it is simply connected and adjoint and if you're lucky you might have $Aut(G/P) = G$. I have no idea if this works but even if it does it would be nice if there was a simpler example. - -REPLY [4 votes]: I think a construction like the one for ${\rm PGL}_2$ in the question should work. The bundle in the question corresponds to the quaternion algebra over $\mathbb{C}[s^{\pm},t^{\pm}]$ whose norm form is $T^2-sX^2-tY^2+stZ^2$, i.e., the quaternion algebra where $i$ and $j$ are non-commuting square roots of $s$ and $t$, respectively. -Now we do the same thing for octonion algebras. The underlying ring is going to be $A=\mathbb{C}[s^{\pm},t^{\pm},r^{\pm}]$. Apply the Cayley--Dickson doubling construction to the above quaternion algebra, with parameter $r$. (This works not just over fields, definitions or locally ringed spaces can be found in papers of H.P. Petersson.) This produces an octonion algebra over $A$ whose norm form is the 3-fold Pfister form -$$ -\langle 1,-s\rangle\otimes\langle 1,-t\rangle\otimes\langle 1,-r\rangle= -$$ -$$=(X^2-sY^2-tZ^2+stT^2)-r(U^2-sV^2-tW^2+stS^2). -$$ -This is a non-trivial 3-fold Pfister form over the function field $\mathbb{C}(s,t,r)$ and so gives a non-trivial decomposable element in Galois cohomology ${\rm H}^3(\mathbb{C}(s,t,r),\mathbb{Z}/2)$. (You can think of this topologically, it's the top cohomology of the 3-torus ${\rm Spec}A$.) In particular, the corresponding octonion algebra is going to be nonsplit over the function field. Then we can take the corresponding $G_2$-torsor (of local automorphisms of the octonion algebra); by the above this torsor cannot be Zariski-locally trivial. -An example of a torsor for ${\rm Spin}(7)$ or ${\rm Spin}(8)$ can be obtained by the natural change-of-structure-group, alternatively, these can be explicitly described in terms of the 3-fold Pfister form above. I guess the 3-fold Pfister form gives a family of smooth affine quadrics whose generic fiber is anisotropic, another example of a torsor which is not Zariski-locally trivial. -Edit: If I interpret the description of the projective bundle in the question correctly, the $\mathbb{P}^1$-bundle is exactly the variety cut out by the vanishing of the norm on the projectivization of the quaternions of trace zero. The same can be done for the octonion algebra above: the equation $(sY^2-tZ^2+stT^2)-r(U^2-sV^2-tW^2+stS^2)=0$ should define a hypersurface in $\mathbb{P}^6_A$, defined by vanishing of the norm on the projectivization of the octonions of trace zero. If I'm not messing up something, this should be a homogeneous space bundle with fibers $G_2/P_1$ which is not locally trivial in the Zariski topology. I think this is the appropriate $G_2$-analogue of the $\mathbb{P}^1$-bundle in the question.<|endoftext|> -TITLE: Random walk by simplex vertices -QUESTION [5 upvotes]: I apologize if this question is well-known, but I was unable to find it mentioned anywhere. -There exists a bug which moves around in $r$-space. The bug begins at the origin of this $r$-space. If the bug is at the center of a regular $r$-simplex (all of which are oriented the exact same direction) with radial length of $1$, then the bug moves to one vertex of the simplex chosen at random with equal probabilities for each vertex. Call each instance of the bug moving, a step. -My question is: What is the probability, as a function of $r$, that there exists a number $n>0$ such that just after the $n$th step, the bug is at the origin? -An equivalent question is: -Given an infinite sequence of digits, with any given digit in the sequence being randomly chosen with equal probabilities inclusively between $0$ and $b$, what is the probability, as a function of $b+1$, that there exists a point in the sequence such that there are an equal number of occurrences of all $b$ digits out of the digits up to that point? -Some work I have done has provided me with a solution that is not in closed form: -$$-\sum_{k\in A}\left(\prod_{i=1}^{k_{length}}\frac{-(r k_i)!}{(r^{k_i} (k_i!))^r}\right)$$ -where $A$ is the set of all finite sequences of distinct integers and $k_{length}$ is the length of the sequence $k$. -Unfortunately, I cannot remember how I obtained this result; if I recreate it, I will edit this question. I also have a lower bound of $0.7$ for $r=2$ as calculated by Mathematica. -EDIT: -I'm starting to doubt my expression above as the answer. - -REPLY [4 votes]: Thanks to Ben Barber's answer, one can represent your question as follows. -Let $u_i$ be an i.i.d. sequence of unit vectors of length $r+1$ -with exactly one entry nonzero, and set $S_n=\sum_{i=1}^n u_i$. -Return to the origin at time $n$ -corresponds to the event $A_n=\{S_n(j)=n/(r+1), j=1,\ldots,r\}$. -Once one hits the main diagonal, the process restarts, so -a transience/recurrence criterion is whether $\sum_n P(A_n)<\infty$. -Now, because $S_n(1)$ is binomial(n,1/(r+1)), for $n$ an integer multiple -of $r+1$, -$$P(S_n(1)=n/(r+1))\sim c_1/\sqrt{n}$$ -and, for $j=2,\ldots,r$, -$$P(S_n(j)= n/(r+1)|S_n(j')=n/(r+1), j'=1,\ldots,j-1)\sim c_j/\sqrt{n}.$$ -Therefore, -$$P(A_n)\sim c/n^{r/2}$$ -which implies recurrence for $r=2$ and transience for $r\geq 3$. -The probability in the OP is thus $1$ for $r=2$. As to the probability -for $r>2$, an exact computation seems out of reach, but the asymptotics as -$r\to\infty$ is dominated by the first cycle returning to $0$, i.e. -by $\prod_{j=2}^{r}(1-j/(r+1))\sim c e^{-r}$.<|endoftext|> -TITLE: Are infinite planar graphs still 4-colorable? -QUESTION [16 upvotes]: Imagine you have a finite number of "sites" $S$ in the positive quadrant -of the integer lattice $\mathbb{Z}^2$, -and from each site $s \in S$, one connects $s$ to every lattice point to which it -has a clear line of sight, in the sense used -in my earlier question: No other lattice point lies along that line-of-sight. -This creates a (highly) nonplanar graph; -here $S=\{(0,0),(5,2),(3,7),(11,6)\}$: -    -Now, for every pair of edges that properly cross in this graph, delete the longer edge, -retaining the shorter edge. -In the case of ties, give preference to the earlier site, in an initial sorting of -the sites. -The result is a planar graph, because all edge crossings have been removed: -    - -Q1. Is this graph $4$-colorable? - -Some nodes of this graph (at least those on the convex hull) -have a (countably) infinite degree. More generally, - -Q2. Is every infinite planar graph $4$-colorable? - Which types of "infinite planar graphs" are $4$-colorable? - -The context here is that I am considering a type of "lattice visibility Voronoi diagram." -One can ask many specific questions of this structure, but I'll confine myself -to the $4$-coloring question, which may have broader interest. - -REPLY [13 votes]: Regarding Q1: -The graph is a subgraph of the visibility graph of the integer lattice. Every sublattice $x+2\mathbb{Z} \times 2\mathbb{Z}$ is an independent set in the visibility graph, and the integer lattice can be decomposed into four such sublattices (according to the parity of coordinates). This gives a proper $4$-coloring.<|endoftext|> -TITLE: Holonomy group of Enriques surface -QUESTION [8 upvotes]: I expect that the holonomy group of an Enriques surface $S$ is $SU(2)\times C_2$. I think this can be proven by the fact that its double cover, which is a K3 surface, has the full $SU(2)$ holonomy, but I failed proving it. The holonomy group should be either $SU(2)$ or its $\pi_1(S)=C_2$-extension $SU(2)\times C_2$. Could someone help me prove or disprove this? - -REPLY [14 votes]: Assume that you have endowed an Enriques surface $S$ with a Ricci-flat Kähler metric $g$. The holonomy $H$ of $g$ cannot be contained in $\mathrm{SU}(2)$ because the canonical bundle of $S$ is not trivial (though its square is trivial). Meanwhile, the identity component of $H$ has to be equal to $\mathrm{SU}(2)$ because this is the holonomy of the (simply-connected, non-product) K3 surface that is the double cover of $S$ endowed with one of its Ricci-flat Kähler metrics. The fundamental group of $S$ is $\mathbb{Z}_2$, so $H$ must be a $\mathbb{Z}_2$-extension of $\mathrm{SU}(2)$ that lies inside $\mathrm{U}(2)$. There is only one of these, namely the group consisting of those matrices in $\mathrm{U}(2)$ with determinant $\pm1$, so this must be $H$. -N.B.: $H$ is not isomorphic to $\mathrm{SU}(2)\times\mathbb{Z}_2$. In the latter, the set of elements that satisfy $a^2=1$ has $4$ members: $(I_2,1),(-I_2,1),(I_2,-1),(-I_2,-1)$, while, in the former, the set of matrices satisfying $A^2=I_2$ consists of two points, $\pm I_2$, together with the $2$-sphere consisting of the elements of the form $iJ$ where $J$ lies in $\mathrm{SU}(2)$ and satisfies $J^2 = - I_2$.<|endoftext|> -TITLE: localizing subcategories of $HF_p$-local spectra -QUESTION [14 upvotes]: This entire question takes place in the $HF_p$-local category of $p$-local spectra, i.e. the essential image of $HF_p$-localization on the stable homotopy category. $HF_p$ itself is in there, and of course so is the $HF_p$-local sphere $L_{HF_p}S^0$. Let $loc(X)$ denote the smallest localizing subcategory containing an object $X$. -Question: Is there a way to show that $L_{HF_p}S^0$ is NOT in $loc(HF_p)$? -I'm imagining there might be some property $HF_p$ has that's preserved by triangles and coproducts, but that $L_{HF_p}S^0$ does not have. (But remember that coproducts here are not the same as in the full category of spectra; likewise the smash product). -Here's why I'm interested. If the answer is yes, then this local category has at least three localizing subcategories: $loc(0)$, $loc(HF_p)$, and $loc(L_{HF_p}S^0)$, the last one being the entire category. (I'm pretty sure $loc(HF_p)$ is minimal, but that's irrelevant.) This would be exciting, because I recently found out that the Bousfield lattice of this category has exactly two elements. If the answer to my question above is yes, then there's a localizing subcategory, namely $loc(HF_p)$, that isn't a Bousfield class. (The only other time that this is known to happen is in the derived category of a certain bizarre type of non-Noetherian ring; see Stevenson's http://www.arxiv.org/abs/1210.0399). -If one took a $K(n)$ instead of $HF_p$, the answer to my question is no. Hovey and Strickland ("Morava $K$-theories and localization") classified localizing subcategories in the $K(n)$-local category, and there are only two (ditto with Bousfield classes). But their proof uses that $L_{K(n)}F(n)$ is a small graded weak generator in the $K(n)$-local category. Later in the same paper they show that the $HF_p$-local category has no nonzero small objects, so the same trick won't work. -Or, on the other hand, if anyone knows how to build $L_{HF_p}S^0$ from $HF_p$, that's great too. But that seems harder. - -REPLY [12 votes]: First note that it is equivalent to ask whether the mod p Moore spectrum $M(p)$ is in the localizing subcategory (in the local sense) generated by $HF_p$. Indeed, the fiber $C_0 S$ of the map $S \xrightarrow{} H\mathbb{Q}$ is in the localizing subcategory generated by $M(p)$ (in the usual sense), and $L_{HF_p}C_0 S=L_{HF_p}S$. So if $M(p)=L_{HF_p}M(p)$ is in the localizing subcategory generated by $HF_p$ in the local sense, then so is $L_{HF_p}S$. -Now consider the class of all $HF_p$-local spectra $X$ such that $[X,M(p)]_*=0$. This contains $HF_p$ because $HF_p$ is dissonant and $M(p)$ is harmonic. It is obviously a thick subcategory. I claim it is also closed under $HF_p$-local coproducts. Indeed, suppose $[X_i, M(p)]_*=0$ for all $i$. Then $[\coprod X_i, M(p)]_*=0$. But $M(p)$ is $HF_p$-local, so -$ -[L_{HF_p} (\coprod X_i), M(p)]_+ = [\coprod X_i, M(p)]_* =0. -$ -So we have a localizing subcategory in the local sense that contains $HF_p$ but not $M(p)$, like you wanted. -So there is a localizing subcategory that is not a Bousfield class! Very nice!<|endoftext|> -TITLE: Constructively correct notion of unique factorization domain -QUESTION [12 upvotes]: Recall the well-known proof that a unique factorization domain is a GCD domain: - -Let $x, y \in R \setminus \{ 0 \}$. Factor $x$ and $y$ into pairwise non-associated irreducible elements: $$\begin{align*}x &= p_1^{e_1} \cdots p_n^{e_n}, \\ y &= p_1^{f_1} \cdots p_n^{f_n}.\end{align*} $$ Then one can check that the product $p_1^{r_1} \cdots p_n^{r_n}$ with $r_i := \min\{ e_i, f_i \}$ is a greatest common divisor of $x$ and $y$. - -Here, a unique factorization domain is an integral domain such that every nonzero element admits a factorization into irreducible elements which is unique up to reordering and associatedness, and a GCD domain is an integral domain such that there exists a greatest common divisor for any two elements. -Constructively, the problem with the above proof is that we may not be able to decide for given irreducible elements whether they are associated or not. Using the given (classically standard) definitions, I thus do not see that the implication -$$\text{$R$ is a unique factorization domain} \Longrightarrow \text{$R$ is a GCD domain}$$ -is constructively provable. -Question. Is there an elegant constructive notion of unique factorization domain such that the implication does hold constructively? -Remark 1. Note that in unique factorization domains which happen to be GCD domains, associatedness and even divisibility are indeed decidable: Given elements $x$ and $y$, consider a gcd $d$ of $x$ and $y$. Write $x = ds$. Then $x|y$ iff $s$ is invertible. This is the case iff the number of factors in a factorization of $s$ into irreducible elements is zero. Because the natural numbers are discrete, this can be decided. -Remark 2. In Mines, Richman, Ruitenburg: A Course in Constructive Algebra approximately the same definitions are used and the implication is stated without proof (see theorem IV.2.3 on page 114). Richman wrote elsewhere (section 6) that he considers the treatment of unique factorization domains in this book to be incomplete. - -REPLY [6 votes]: It seems that Lombardi and Quitté in their book "Commutative Algebra: Constructive Methods", Ch. XI, §3, define UFDs as GCD-domains such that every regular element is a product of irreducibles. In classical logic, this coincides with the usual definition. (Ingo sketched the proof in his comment.)<|endoftext|> -TITLE: On the fixed point of automorphism of $\mathbb F_3[[T]]$ -QUESTION [8 upvotes]: Consider the automorphism $\sigma$ on ${\Bbb F}_3[[T]]$ such that $T \mapsto c_1T + f(T)$ with $c_1 = 1$ or $-1$, and $f(T) \not=0$ and the non-zero leading term $c_mT^m$ of $f(T)$ satisfies $m \geq 2$. -Question: Is there any fixed element $t \in {\Bbb F}_3[[T]]$ other than those in the constant field ${\Bbb F}_3$? Namely does such $t$ exist as $\sigma(t) = t$ -but $t \notin {\Bbb F}_3$? -Pierre - -REPLY [2 votes]: Let’s call $K=k((T))$, where $k$ is a field of characteristic $p>0$. Suppose $\Gamma$ is any finite subgroup of the group of $k$-automorphisms of $K$, any one such necessarily sending $T$ to $u(T)=\sum_1^\infty a_iT^i$, and thereby sending a general element of $K$, say $g(T)=\sum_?^\infty b_iT^i$, to $g\circ u=\sum_?^\infty b_iu^i$. Then if $|\Gamma|=n$, of course its fixed field $E\subset K$ has $[K\colon E]=n$. The extension will always be totally ramified, and with a nice generator equal to the norm of $T$, that is, $\prod_{\gamma\in\Gamma}\gamma(T)$. If we call this series $S$, then $E=k((S))$. Now if $u$ is a torsion element of Nottingham, then the group $\langle u\rangle$ is just such a $\Gamma$ as above, and we get a whole lot of fixed elements under right composition by $u$, all of them power series (or Laurent series, if you wish) in $S$.<|endoftext|> -TITLE: Picard number of principally polarized abelian varieties -QUESTION [7 upvotes]: Let $A$ be an abelian variety of dimension $n$. Over $\mathbb{C}$, at least, it is known that the Picard number (that is, the rank of the Neron-Severi group of $A$) is less than or equal to $n^2$, with equality if and only if $A$ is isogenous to the self product of an elliptic curve with complex multiplication. -Is there a bound on the Picard number if $A$ is simple? In other words, can a simple abelian variety have Picard number $n^2-1$? - -REPLY [7 votes]: A tight bound for simple $A/\mathbb{C}$ is $\rho(A) \leq 3n/2$. This follows from Proposition 5.5.7 in Birkenhake-Lange. If $A$ does not have indefinite quaternionic multiplication, the stronger bound $\rho(A) \leq n$ holds.<|endoftext|> -TITLE: Who first identified the universal $C^*$-algebra generated by an idempotent of norm at most $C$? -QUESTION [9 upvotes]: So much is known about hermitian and non-hermitian idempotents in a $C^*$-algebra, that someone must have written down the following. - -Theorem The universal $C^*$-algebra generated by one element $x$ subject to the relations $x^2=x$ and $ \| x \| \leq C$ is isomorphic to - $$ -\mathcal{I}_D= -\left\{ -f\in C\left([0,D],\mathbf{M}_{2}(\mathbb{C})\right) -\left| -f(0)\in\left[\begin{array}{cc} -\mathbb{C} & 0\\ -0 & 0 -\end{array}\right] -\right. -\right\} -$$ - where $D = \sqrt{C-1}$ and the isomorphism is specified by - $$ -x \mapsto -\left[\begin{array}{cc} -1 & 0\\ -t & 0 -\end{array}\right], -$$ - -This is not hard to prove, and makes it easy to see why an idempotent is homotopic to a projection. - -Is there a reference for this result, that explicitly mentions a universal $C^*$-algebra or that spells out the universal property? - -REPLY [5 votes]: I almost found it: Example 4.4 in "Presentations and Tietze transformations -of C*-algebras" by Will Grilliette, New York J. Math. 18 (2012) 121--137. The generator in the concrete algebra is not explicit, and must not be -$$ -\left[\begin{array}{cc} -1 & 0\\ -t & 0 -\end{array}\right] -$$ -as the length of the interval does not come out the same. -I am still interested in any earlier reference that might exist.<|endoftext|> -TITLE: Who first used the cross-ratio to describe shapes in hyperbolic geometry? -QUESTION [5 upvotes]: I was reading this Wikipedia article today:https://en.wikipedia.org/wiki/Shape#Similarity_classes -and I realized that it strongly resembles the use of coss-ratios as "shape parameters" in hyperbolic geometry. -I wanted to add material to the article related to this; however, I would like to add a reference concerning the origin of the cross-ratio in hyperbolic geometry. -Who was the first person to use the cross-ratio to parametrize shapes in hyperbolic geometry? - -REPLY [6 votes]: I've hesitated to attempt an answer to this question because -I do not know about shape parameters. However, in the hope that -what is really wanted is a history of the cross-ratio, here goes. -The cross-ratio, and its invariance under projection, was discovered -by Pappus and rediscovered by Desargues around 1640. It appears in -Proposition 129 of Book VII of Pappus' Mathematical Collection from -around 300 CE. The Desargues version appears in a 1648 book called -Mani`ere universelle de Mr Desargues written by Desargues' disciple -Abraham Bosse. -It became standard in projective geometry when the subject flourished -in the 19th century. It was used by Cayley in his "Sixth memoir on -quantics" of 1859 (which was not his thesis) to introduce a metric -into projective geometry. In 1871, Klein realized that this metric -was in fact the metric for the hyperbolic plane in his paper "Ueber -die sogenannte Nicht-Euclidische Geometrie."<|endoftext|> -TITLE: How many finite loops? -QUESTION [7 upvotes]: How many finite loops of order $n$ are there? -I am interested in the exact values ​​of $n$ if $n <40$ or even reasonable estimates. I am also interested in formulae or bounds for all $n$. -Note that similar questions have been studied for groups and semigroups. Much information is available in the group theory case, but even for semigroups not much is known beyond $n=11$. - -REPLY [2 votes]: Here you can find the number of loops of order n up to isotopy and isomorphism for n at most 10, and the number of quasigroups up to isomorphism. -http://cs.anu.edu.au/~bdm/papers/ls_final.pdf<|endoftext|> -TITLE: Does regular field extension preserve regularity? -QUESTION [6 upvotes]: Let $k$ be an arbitrary field and suppose that $K/k$ is a regular field extension. Let $V$ be regular scheme of finite type over $\text{Spec }k$ (not necessarily smooth). Is it true that $\text{Spec }K\times_{\text{Spec }k}V$ is also regular? - -REPLY [6 votes]: If $k \rightarrow K$ is formally smooth for the discrete topology (i.e. separable), by flat base change $A \rightarrow A\otimes_kK$ is formally smooth for any $k$-algebra $A$ essentially of finite type, and so, if $A$ is regular then $A\otimes_kK$ is regular.<|endoftext|> -TITLE: Recursive ordinals and the minimal standard model of ZF -QUESTION [6 upvotes]: Does the minimal standard model of ZF contain all recursive ordinals or is it limited (probably by the proof theoretic ordinal of ZF as I suspect but cannot prove)? -Paul J. Cohen's definition of the minimal standard model for ZF.(http://www.ams.org/journals/bull/1963-69-04/S0002-9904-1963-10989-1/S0002-9904-1963-10989-1.pdf) is iterated through all ordinals. What is the least ordinal it must be iterated up to generate the minimal standard model? - -REPLY [6 votes]: The minimal transitive (or synonymously, standard) model of ZF is $L_\alpha$, where the ordinal $\alpha$ is chosen to be smallest such that $L_\alpha\models\text{ZF}$. Since ZF proves that $\omega_1^{CK}$ exist, and the recursive ordinals are absolute between $V$ and $L_\alpha$, it follows that $\omega_1^{CK}<\alpha$. So the answer to question 1 is yes. -The ordinal $\alpha$ is precisely the number of times you have to iterate Gödel's definable power set operation in order to build the minimal transitive model, and I don't know any simpler definition of $\alpha$ than: the height of the minimal transitive model. -Meanwhile, $\alpha$ is a fairly small countable ordinal, if it exists, and furthermore countable in $L$. An upper bound for complexity is provided by the fact that there is a $\Delta^1_2$ definable relation on $\mathbb{N}$ with order type $\alpha$, simply by noting that we may define the least $L$-code for such a relation, using at most that complexity.<|endoftext|> -TITLE: Does base extension reflect the property of being isomorphic? -QUESTION [16 upvotes]: Let L/K be a (separable?) field extension, let A be a finite dimensional algebra over K, and let M and N be two A-modules. Let $A' = L \otimes_K A$ be the algebra given by extension of scalars, and let $M' = L \otimes_K M$ and $N' = L \otimes_K N$ be the A'-modules given by extension of scalars. -Does $M' \cong N'$ (as A'-modules) imply that $M \cong N$ (as A-modules)? -(This question is obviously related. Note that just as for that question it is easy to see that base extension reflects isomorphisms in the sense that if a map $f: M \rightarrow N$ has the property that $f' : M' \rightarrow N'$ is an isomorphism then f is an isomorphism. This is asking about the more subtle question of whether it reflects the property of being isomorphic.) -I apologize if this is standard (I have a sinking suspicion that I've seen a theorem along these lines before), but I haven't been able to find it. There's a straightforward proof in the semisimple setting, but I have made no progress in the non-semisimple setting. - -REPLY [10 votes]: I hope I'm not misunderstanding the question. Here goes: -We'll show that if $M,N$ are finite-dimensional over $K$, -then they are isomorphic over $K$. -Think of the linear space $X=\mathrm{Hom}_{A}(M,N)$ -as a variety over $K$. Inside $X$ look at -the $K$-subvariety $X'$ of maps that are not isomorphisms $M \rightarrow N$. -Now $X' \neq X$, -because there is an $L$-point of $X$ not in $X'$. Therefore, -over an infinite field $K$, there will certainly exist a $K$-point of $X$ -that doesnt lie in the proper subvariety $X'$. -If $K$ is finite: -$M,N$ are both $K$-forms of the same module $M'$ over $L$. -The $L$-automorphisms of $M'$ are a connected group, because -they amount to the complement of the hypersurface $X'$ inside the linear space $X$. So its Galois cohomology vanishes, thus the -same conclusion.<|endoftext|> -TITLE: Integral versus real (universal) characteristic classes -QUESTION [9 upvotes]: I'm pretty confused about the precise relation of the integral and the real cohomology of the classifying space $BG$ of a compact Lie group $G$. The natural map $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ certainly kills all torsion, but can it have non-torsion elements in the kernel? For instance, if $G=SU_n$, then there is no torsion in $H^n(BG;\mathbb{Z})$ (see [Hatcher, Algebraic Topology, Theorem 4D.4]), so this map would be an injection for all $n$. -All expositions that I know (for instance [E. Brown, The cohomology of $BSO_n$ and $BO_n$ with integer coefficients]) somehow suggest that $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ is injective, but this is certainly not the case for arbitrary spaces $X$ with torsion free $H^n(X;\mathbb{Z})$. The Corresponding statement for homology is true, since the singular chain complex is free, and so the universal coefficient theorem yiels $$ H_n(X;\mathbb{Z})\otimes \mathbb{R} \xrightarrow{\cong} H_n(X;\mathbb{R}). $$ However, this argument does not carry over to cohomology (or at least I don't see it). -What makes me suspicious is that if $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R})$ is injective for all $n$, then the long exact sequence induced from $\mathbb{Z}\to \mathbb{R} \to S^1$ would split, a thing I don't feel comfortable with. - -REPLY [6 votes]: This is also equivalent to the corresponding statement for homology: Since $Hom(C,\mathbb{Z}) \otimes \mathbb{R} = Hom(C,\mathbb{R})$ for $C$ a free abelian group, the cochain complex with $\mathbb{R}$-coefficients is isomorphic to the cochain complex with $\mathbb{Z}$-coefficients tensored with $\mathbb{R}$. The same homological algebra as in the homology universal coefficient theorem gives you that $H^n(X) \rightarrow H^n(X; \mathbb{R})$ precisely kills torsion. (Which of course comes from $H_{n-1}$ if you invoke the universal coefficient sequence as in Marks answer.)<|endoftext|> -TITLE: Constructing a odd homeomorphism between $A$ and $S^n$ -QUESTION [6 upvotes]: I have asked this question here seven months ago and until now I got no answer. -Let $A\subset\mathbb{R}^N\setminus\{0\}$ be a closed symmetric set ($x\in A$ then $-x\in A$). Suppose that $A$ is homeomorphic to some sphere $S^n$, $n\leq N$ ($n$ is the dimension of the sphere). Is it possible to construct a homeomorphism $F:A\to S^n$ such that $F$ is odd? - -REPLY [7 votes]: The answer is "NO". -It follows from existence of exotic smooth involutions of sphere. -Say, this movie explains a construction of an involution $\iota$ of $\mathbb S^5$ such that -the quotient $\Pi=\mathbb S^5/\iota$ is homotopy equivalent, but not not homeomorphic to $\mathbb R\mathrm P^5$. -Consider composition -$$\Pi\longrightarrow^{\!\!\!\!\!h}\mathbb R\mathrm P^5\longrightarrow^{\!\!\!\!\!e}\mathbb R\mathrm P^{N-1},$$ -where $h$ is the homotopy equivalence and $e$ is the standard embedding. -If $N$ is large, Whitney embedding theorem says that you can perterb this composition into smooth embedding $\Pi\to\mathbb R\mathrm P^{N-1}$. -Passing to the double cover of $\Pi$ and $\mathrm P^{N-1}$, -you get a funny embedding $\mathbb S^5\hookrightarrow \mathbb S^{N-1}\subset \mathbb R^N$, -which gives a counterexample. -(If the image admits is an odd homeomorphism to $\mathbb S^5$ then -$\Pi$ has to be homeomorphic to $\mathbb R\mathrm P^5$.)<|endoftext|> -TITLE: Equivariant smoothing of PL structures on $S^3$ -QUESTION [10 upvotes]: Suppose $S^3$ is PL sphere on which a finite group $G$ acts by PL homeomorphisms. Is it always possible to find a compatible smooth structure such that $G$ acts by diffeomorphisms? -I am not quite sure how smoothing works for trivial $G$, but maybe this can be generalized to the situation described above? - -REPLY [6 votes]: The answer is "always", but it is not proven in the paper by Kwasik and Lee. They only show the existence of an equivariant smooth structure which is not necessarily compatible with the PL structure. The statement holds for manifolds up to dimension four. I tried to write a readable proof and put it on the arXiv.<|endoftext|> -TITLE: Law of the $L^2$ norm of a Brownian motion and related -QUESTION [6 upvotes]: Let $B_t$ be a Brownian motion with variance 1. We know that $\int_0^1 B(t) \mathrm{d} t \sim \mathcal{N}(0,1/3)$. I am interested to know what we can say about the law of the two random variables -$X = \int_{0}^1 B(t)^2 \mathrm{d}t = \langle B,B\rangle_{L^2([0,1])}$ and $Y = \int_{0}^{1} \left( B(t) - \int_0^1 B(s)\mathrm{d}s \right)^2 \mathrm{d}t = X - \left( \int_0^1 B(s)\mathrm{d}s \right)^2$. -Is there a well-known probability law hidden behind them? -Thank you for attention. - -REPLY [4 votes]: The distribution of Y has been studied in great details by Donati-Martin and Yor in their work: Fubini's theorem for double Wiener integrals and the variance of the Brownian motion path -http://archive.numdam.org/ARCHIVE/AIHPB/AIHPB_1991__27_2/AIHPB_1991__27_2_181_0/AIHPB_1991__27_2_181_0.pdf -In particular it is proved that in distribution -$Y=^{law}\int_0^1 | W_s |^2 ds$ -where W is a standard complex Brownian bridge.<|endoftext|> -TITLE: Psi operator on Phi-Gamma modules -QUESTION [5 upvotes]: This is a question about the base-rings appearing in the the theory of $(\varphi, \Gamma)$-modules in $p$-adic Hodge theory. -Let $p$ be prime, $n \ge 1$, and let -$$ \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n} = \left\{ \sum_{k \in \mathbf{Z}} a_k T^k : a_k \in \mathbf{Z}_p, v_p(a_k) + \frac{k}{(p-1)p^{n-1}} \ge 0\, \forall{k} \text{ and $\to \infty$ as $k \to -\infty$}\right\}.$$ -(This ring also goes by the name of $\mathscr{O}_{\mathscr{E}}^{\dagger, n}$ in some works). There is a Frobenius map $\varphi: \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n} \to \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n+1}$ given by $T \mapsto (1 + T)^p - 1$. -There is also a map $\psi: \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n+1} \to \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n}[1/T]$, satisfying $\psi \circ \varphi = 1$ and -$$ (\varphi \circ \psi)(f) = \frac{1}{p} \sum_{\zeta: \zeta^p = 1} f( \zeta(1 + T) - 1). $$ - -Does the map $\psi$ send $\mathbf{A}_{\mathbf{Q}_p}^{\dagger, n+1}$ to $\mathbf{A}_{\mathbf{Q}_p}^{\dagger, n}$? - -There is a sentence in the proof of Lemma V.1.4 of the paper "Theorie d'Iwasawa des representations p-adiques d'un corps local" by Cherbonnier and Colmez (JAMS 12(1), 1999) implying that this is the case, but the proof is only very briefly sketched, and I can't see how to make it work. On the other hand, Colmez's massive paper "Représentations de $\operatorname{GL}_2(\mathbf{Q}_p)$ et $(\varphi, \Gamma)$-modules" (in Asterisque 330, 2010) seems to avoid using this fact at points where it would seem (to me!) natural to do so, e.g. Lemma V.1.1, leading me to suspect that maybe it's not actually true. What's the correct statement here? - -REPLY [5 votes]: It seems to me that the statement follows from some formulas appearing in other papers of Colmez. Lemma I.9 (page 224) of "La série principale unitaire de $GL_2(Q_p)$" has some very precise estimates for the coefficients of $\psi(T^k)$ with $k<0$. It says that if $k<0$, then $\psi(T^k)= \sum_{i=k}^{\lfloor k/p \rfloor} b_{k,i} T^i$ with: -$$ v_p(b_{k,i}) \geq \frac{k-pi}{p-1}-1. $$ -Note also that $v_p(b_{k,i}) \geq 0$. If you use this, then it seems to me that you can prove that $a_k T^k \in A^{\dagger,n+1}_{Q_p}$ implies $\psi(a_k T^k) \in A^{\dagger,n}_{Q_p}$. Don't forget to use the fact that all the $v_p(\cdot)$'s are integers: in particular I'm not sure that the claim would be true if you were to look at rings "with coefficients in $E/Q_p$" if $E$ is ramified - which is maybe why Colmez avoids it? -EDIT: even if the above works, I'd personally try a more direct proof along the following line: -$$ A^{\dagger,n+1}_{Q_p} = Z_p[[T]] \{ \frac{p}{T^{p^n(p-1)}} \} = Z_p[[T]] \{ \frac{p}{\phi(T)^{p^{n-1}(p-1)}} \}, $$ -and your claim then becomes clear.<|endoftext|> -TITLE: Is a semigroup always an exponential? -QUESTION [5 upvotes]: Let $H$ be a Banach space, $\mathscr{B}(H)=\{T:H\to H: \text{where $T$ is a bounded linear operator}\}$, and $S:[0,\infty)\to \mathscr{B}(H)$, a map with the following properties: -$$ -S(0)=I, \quad S(s+t)=S(s)S(t), \quad \text{for all $s,t\in[0,\infty)$}, -$$ -and -$$ -\lim_{t\to s^+}S(t)x=S(s)x, \quad \text{for all $x\in H$.} -$$ -Is there a general theorem which allows us to express $S(t)=\exp(tA)$, for a suitable unbounded operator $A$? If yes, what kind of operator is $A$? In particular, what can we tell about its spectrum? - -REPLY [3 votes]: To expand the answer of Michael Renardy: your question concerns the important question when can we "insert" an operator into a function and what properties can we deduce from the properties of the function only. This is the main idea behind various "functional calculus concepts", and I recommend the excellent monograph and papers (for example, see this or this) by Markus Haase. -The most widely used is the so-called Hille-Phillips calculus, which is based on Laplace transform techniques. This and the Post-Widder inversion formula makes the connection Michael mentions.<|endoftext|> -TITLE: Varieties invariant under affine transformations -QUESTION [6 upvotes]: This question stems from this question on Mathematics stack exchange. The top answer there provides a geometric solution; I'm curious though about an algebro-geometric solutions. -Let me rephrase that question a bit: Let $V$ be a variety in affine $n$-space $A^n_k$, over field $k$, given by an ideal $I$. How one would derive from $I$ the subgroup $G$ of affine transformations that leave $V$ invariant? -In particular, if $V$ is a hypersurface and $I$ is a principal ideal $(f)$ what properties of $f$ determine $G$? - -REPLY [4 votes]: Probably, the easiest method is this (at least in characteristic zero, which I will assume henceforth): Suppose that $V\subset \mathbb{A}^n_k$ is the set of zeros of a polynomial ideal $I\subset k[x^1,\ldots,x^n]$, say, generated by some finite set $\{f_1,\ldots,f_m\}\subset k[x^1,\ldots,x^n]$. Let $\frak{a}$ be the (finite dimensional) Lie algebra of affine vector fields on $\mathbb{A}^n_k$ (i.e., the set of derivations of $k[x^1,\ldots,x^n]$ that preserve the subspace of polynomials of degree at most $1$). Consider the linear map $\Phi:{\frak{a}}\to k^m\otimes k[x^1,\ldots,x^n]/I$ defined as -$$ -\Phi(X) = \bigl([X(f_1)]_I,\ldots,[X(f_m)]_I\bigr). -$$ -The kernel of $\Phi$, say ${\frak{g}}_I\subset{\frak{a}}$, is the space of affine symmetry vector fields of the ideal $I$, and it is the Lie algebra of the connected (in the appropriate sense) subgroup $G^0_I$ of the set of affine transformations of $\mathbb{A}^n_k$ that preserve $I$, i.e., it is the irreducible component containing the identity of the algebraic group $G_I$ that consists of all affine symmetries of $I$. In particular, $G^0_I$ is an algebraic subgroup of the group of affine transformations of $\mathbb{A}^n_k$, and ${\frak{g}}_I$ is its (Zariski) tangent space at the identity. (Computing the full symmetry group $G_I$ is not a linear problem and is much harder in general, though one does know that $G^0_I$ is a normal subgroup of $G_I$ and that the quotient group $G_I/G^0_I$ is finite.) -In practice, if $I$ is complicated, computing $[X(p)]_I$ in an efficient way requires using Gröbner bases or some such tool. In all cases, though, the computation of ${\frak{g}}_I$ reduces to the computation of the kernel of a linear map between finite-dimensional vector spaces. -When $I$ is generated by a single polynomial $f$, one is just asking whether the 'remainder' of $X(f)$ 'divided by' $f$ is zero. (Here, I am intending the use of a so-called 'multivariate division algorithm' à la Buchberger's algorithm using a Gröbner basis, though a more naïve degree approach will work, too.) This is fairly easy to compute explicitly using a Gröbner basis, one, say, that uses some total order on the monomials in $k[x^1,\ldots,x^n]$, so the 'continuous' symmetries of a polynomial are not hard to compute using just about any symbolic algebra program. -For example, given $f$ of degree $d>0$, you can always choose coordinates $y^1,\ldots,y^n$ so that -$$ -f = (y^n)^d + c_1(y^1,\ldots,y^{n-1})(y^n)^{d-1} + \cdots + c_d(y^1,\ldots,y^{n-1}) -$$ -where each $c_i$ is a polynomial of degree at most $i$. Clearly, there is a unique linear map $q:{\frak{a}}\to k$ such that -$$ -X(f) = q(X)\,f + R(X,f) -$$ -where $R(X,f)$ has degree at most $d$ in the $y^i$ and at most degree $d{-}1$ in $y^n$. -Now, $X(f)$ is a multiple of $f$ if and only if $R(X,f)=0$, and, for a given $f$, this is a finite number of linear equations for $X$.<|endoftext|> -TITLE: High dimensional generalized Poincare hypothesis without the h-cobordism theorem? -QUESTION [7 upvotes]: The generalized PL Poincare hypothesis states that in dimension $n$ there is a unique PL manifold that has the homotopy-type of $S^n$. It's known to be true in all dimensions except perhaps $n=4$. -For $n \geq 5$ the standard argument uses rather explicit handle manipulations to come up with the proof (proving the h-cobordism theorem in the process), which is somewhat spiritually-similar to (although more complicated than) the proofs in dimension $n=2$. -The proof in dimension $n=3$ is very different than the above two cases. -But I wonder, perhaps there is a proof that avoids the h-cobordism theorem, perhaps there is a more direct proof? Has there been much discussion of this in the literature? -One thought would be to find an appropriately simplified proof of the Farrell fibering theorem (when a manifold fibers over $S^1$), one that perhaps allows you to reduce $S^n$ recognition into a homotopy-unknot recognition problem. -I imagine back in the 60's and 70's there was some discussion of these topics but I wouldn't know where to look. - -REPLY [2 votes]: The Poincare conjecture predates the h-cobordism theorem, and the original proof of it in Smale's paper does not prove the h-cobordism theorem (though the handle manipulations in it are what inspired the h-cobordism theorem, so this is maybe not a good answer). -A very different way of proving the Poincare conjecture is Stallings's proof using "engulfing". A nice modern source for this is Chapter 9 of Ferry's notes here.<|endoftext|> -TITLE: The Eyeball Theorem generalized -QUESTION [27 upvotes]: I have not seen the 2D Eyeball Theorem—that tangents from the centers of two circles, each encompassing the other, intersect each circle in the same segment length—generalized to higher dimensions. It generalizes easily: the radius of the circles of cone/sphere intersections in $\mathbb{R}^3$ (below, red) are equal: -    -What I am wondering is if there is a sense in which some form of -this theorem generalizes to other -objects: axis-aligned cubes, ellipsoids, or other shapes. Or does the theorem in some -sense characterize spheres? If anyone has seen this addressed previously, I'd appreciate a pointer. Thanks! - -(Added). This seems to work for squares/cubes: - -REPLY [2 votes]: I have introduced some variants of the Eyeball theorem and also seems to admit generalizations in 3D. And as if that were not enough the Archimedean twins have been brought together with these theorems. See link below -http://geometriadominicana.blogspot.com/2014/03/praying-eyes-theorem.html<|endoftext|> -TITLE: When taking the fixed points commutes with taking the orbits -QUESTION [13 upvotes]: Let $G$ and $H$ be groups, both acting on a set $X$ on the left, in such a way that the two actions commute. (Equivalently, let $G \times H$ act on $X$.) -The set $\text{Fix}_H(X)$ of $H$-fixed points carries a $G$-action, and we can then take the set $\text{Fix}_H(X)/G$ of orbits. But also, the set $X/G$ of $G$-orbits carries an $H$-action, and we can take the set $\text{Fix}_H(X/G)$ of fixed points. There is a canonical map -$$ -\lambda\colon \text{Fix}_H(X)/G \to \text{Fix}_H(X/G), -$$ -which is easily seen to be injective. -It's a fact that if $G$ and $H$ are finite groups with coprime orders, this map $\lambda$ is bijective. So then, -$$ -\text{Fix}_H(X)/G \cong \text{Fix}_H(X/G). -$$ -(A bit more generally, this is true whenever $G$ and $H$ are possibly-infinite groups with finite coprime exponents.) The proof isn't hard. -I only discovered this yesterday, but I guess it's well-known — maybe as a special case of something more general. Can anyone give me a reference? -Update Perhaps I should have explained the context. In category theory, limits commute with limits and colimits commute with colimits, but limits do not usually commute with colimits. There are various theorems giving restricted conditions under which limits do commute with colimits. This result about group actions, which Peter Johnstone and I came upon, gives a new (?) such theorem. That's why I'm after a reference. For more explanation, see this $n$-Category Café post. - -REPLY [5 votes]: To see when this is possible, we can consider each orbit of $G \times H$ separately. An orbit of $G \times H$ corresponds to a subgroup of $G \times H$. -A subgroup of $G \times H$ corresponds to a triple of $A,B,C,D$ where $A$ is a normal subgroup of $B$ a subgroup of $G$, $C$ is a normal subgroup of $D$ a subgroup of $H$, plus an isomorphism $B/A \cong D/C$. The subgroup corresponding to $A, B, C,D,\cong$ consists of pairs $(g,h) \in G \times H$ where $g \in B$, $h \in D$, and the class of $g$ in $B/A$ is sent by the isomorphism to the class of $h$ in $D/C$. One can also reverse this process. -$X/G$ is a single orbit of $H$. It is the orbit corresponding to the subgroup $D$, that is, the action of $H$ on $H/D$. So $X/G$ has an $H$-fixed point if $D = H$. -On the other hand $Fix_H(X)$ is nonempty if and only if $C= G$. In this case, the map is surjective. So my comment was off slightly - $H$ and $G$ have this property if and only if no nontrivial quotient of $H$ is a subquotient of $G$. This condition is quite messy in general, and there are lots of examples, like $G$ finite and $H$ not residually finite. -But this is certainly a general result that your statements follow from.<|endoftext|> -TITLE: Examples of toposes for analysts -QUESTION [15 upvotes]: I've read that toposes are extremely important in modern mathematics, but I find the definitions and examples given on the nLab page a little too abstract to understand. -Can you provide some examples of toposes which are of use to analysts? e.g., toposes where the objects consist of topological spaces, measure spaces, etc. - -REPLY [4 votes]: Besides the important examples of topoi already mentioned (like SDG), I would argue that the most important topos for any analyst is just the topos of sheaves on some topological space. I assume that the statement "sheaves are important for analysts" is well-known and doesn't require further argumentation. The study of topoi naturally focuses our attention not on model-dependent results, bizarre axioms and exceptional cases, but on general phenomena and parameter-dependent versions of statements, since the objects of a sheaf topos are naturally dependent on a point in the base space. For these reasons topos theory advocates constructive approach to mathematics, not because of some philosophical reasons, but because logical and set-theoretic intricacies simply don't make sense once you start working over a general base. For example, the axiom of choice fails unless our topological space is discrete. Another example: if you want to do some parametrized topology, you really should consider moving from topological spaces to locales, since general topological spaces over a base are very, very badly behaved. -From this point of view topos theory is some abstract machinery that allows you to transform a good enough (i.e. constructive) proof of any theorem into its parametrized version. -Things get even more fun once you become interested in some homotopical phenomena and deformation theory. It can become very tricky to prove something without considering $\infty$-topoi, either explicitly or shyly hiding them.<|endoftext|> -TITLE: Type III factor representation -QUESTION [6 upvotes]: Does there exist any theorem which permits, under suitable hypotheses, to represent a particular complete orthomodular lattice as the projection lattice of a Type III von Neumann factor? - -REPLY [6 votes]: A characterization of projection ortholattices of von Neumann algebras (and more generally JBW algebras) with no type I$_2$ component was given by Bunce and J.D.M. Wright in two papers: [1] and [2] (full references at end). -You obtain an answer since "factorial" and "type III" are expressed in ortholattice terms (see the Loomis - Maeda dimension theory, in particular the last version [3] where equidimensionality is identified with lattice semi-projectivity). -As expected by professor Handelman, a big role in such a characterization is played by (the faces of) the convex set of normal states (exclusion of I$_2$ components is needed to use Gleason's theorem to identify states with completely additive probability measures on the projection ortholattice. It is also needed since, as well known from projective geometry, not every (ortho)lattice automorphism of a projective (ortho)line (i.e. a arbitrary permutation of the points, or half of them in the orthocomplemented case) is semilinearly induced. In particular, order two matrices over real, complex or quaternion numbers all give the same projection ortholattice (as it trivially happens also with order 1 matrices); by artificially restricting to the complex case one has unicity, but only up to a noncanonical isomorphism). -A much better (from the quantum logic point of view) characterization of the projection ortholattices of (real or complex, always excluding type I$_2$ cases) finite factors as "continuous geometries with transition probability" is due to von Neumann [4] (and then unfortunately well forgotten by modern quantum logicians). From this, two generalizations are obtainable with standard methods: - -to the decomposable case, using Boolean valued analysis (a decomposable case is the same as indecomposable object of a boolean valued universe); - -to the semifinite (instead of finite) case, using the fact that in the semifinite case the join-dense ideal of finite elements completely determines $L$ (a standard method "to adjoin 1 to a generalized orthomodular lattice", due to Janowitz, produces the lattice of all finite and cofinite elements; then the Dedekind completion produces $L$). - - -[Digression. In particular, this gives a characteriazion of Hilbertian logics (of type I factors) that is physically much better than the characterization that modern quantum logicians deduce from Soler's theorem (which is however mathematically wonderful). The modern theorem must exclude all finite dimensional factors (why a finite dimensional irreducibly quantum logic should be automatically embeddable in a infinite dimensional one? von Neumann's method instead excludes only the "spin factors", which are not really quantum since they are the only factors with nonclassical logic but with "noncontextual hidden variables", and the nonarguesian planes, which cannot be embedded in any larger logic except by direct product, which means that these exceptional components can have only classical, not quantum, interactions with the other components) and must presuppose together a complete lattice and orthomodularity without physical reasons (orthomodularity is justified by restricting only to certain propositions, and the "complete lattice" property is justified by enlarging using completions, like Dedekind completions. Unfortunately this only produces two possibly different structures, a restricted orthomodular one and a a larger complete lattice; almost no known mathematical theorem produces automatically a orthomodular completion. The only exception is precisely von Neumann's method when applied to type I cases (and, analogoulsy, the metric completion of pre-Hilbert spaces): the only completeness axiom which is not trivially satisfied in the finite dimensional case is used only in the last step, to show that an already constructed Hilbertian representation is surjective; so, were this last completeness axiom not satisfied, one can always take as completion the bicommutant of the algebra in the Hilbertian representation: one has proved that a completion exists, a conceptual case analogous to the well known proof that, assuming the archimedean axiom for the measures of physical quantities, then one can assume that the measures are real numbers: the archimedean axiom, involving two magnitudes and a simple arithmetic progression, is experimentally falsifiable at least ideally, but the completeness axioms for real numbers, with arbitrary infinite sets, is physically hopeless). End digression.] -I know no attempts to concretize the details of a last, third step in the extensions of von Neumann's characterization: using Tomita - Takesaki modular theory to obtain a generic type III factor starting from a type II infinite factor with a suitable automorphism, one has that, in principle, the projection ortholattice $L$ of a type III factor, being equivalent to the factor itself, is somehow obtained from a type II factor with a given automorphism, which is equivalent to a projection ortholattice, with fixed automorphism, of a type II factor. I hope that someone one day will write down the details of this method. - -Concerning the other remark of professor Handelman: -The possibility of characterization of complex AW$^*$-algebras with no type I$_2$ components by their projection otholattices follows from Dye's theorem [5]: each projection ortholattice isomorphism among them extends to one and only one (necessarilly real linear) $*$-ring isomorphism (or equivalently a unique complex linear Jordan isomorphism; however, since there are type II finite factors not anti-isomorphic to themselves, there are cases where a complex linear $*$-ring isomorphism is impossible). -Dye proved his theorem in 1955 for von Neumann algebras, but Yen in 1957 [6] and Berberian in 1982 [7] remarked that the proof works also for AW$^*$-algebras. (Recent interest in Dye's theorem appears in C. Heunen, M. L. Reyes [8]; these authors seem unaware of the concept of orthosymmetric ortholattices introduced by Mayet [9].) -Really, the theorem (but not Dye's own proof) also holds for real AW$^*$ algebras with no abelian or type I$_2$ component (or even more generally for Rickart real $C^*$-algebras of matrix order at least 3 and $C^*$-direct sums of such algebras). An explicit reconstruction of the Rickart $C^*$-algebra $A$ from its projection ortholattice $L$ is the following: - -First note that it is sufficient to reconstruct the $*$-ring $M$ of "affiliated locally measurable operators" (defined by Berberian and Saito using "strongly dense domains" in $L$, but algebraically it is the ring of classical quotients of $A$); in fact, $A$ is the subring of $M$ generated by its projections (or also the $*$-subring of bounded elements in the algebraic sense first used by von Neumann). -Then $M$, being a direct product of matrix rings of order at least 3, is generated as a ring by its idempotents $e,f,\dots$ using (besides idempotence) the relations given by a restriction of the classical "circle operation" to a partial operation on idempotents: $e\circ f=e+f-ef$ is idempotent when $fe=0$; moreover, these generators and relations depend only by lattice theory: idempotents are identified with complementary ordered pairs $(K,I)$ (kernel and image of the idempotent) in the lattice of right ideals of $M$, and the partial circle operation becomes $(K,I)\oplus(K',I')=(K\wedge K',I\vee I')$ when $I\subseteq K'$ (moreover, the join is an independent join and dually for the meet). (All this follows from the easy part of von Neumann's coordinatization, in any ring even without regularity conditions). -Finally: the above pairs $(K,I)$ and the circle partial operation on them only depends upon the lattice $L$ (which is the same for $A$ and $M$; it is the lattice associated to these Rickart rings): these are exactly the complementary and modular pairs in the lattice (by $O$-symmetry of such ortholattices, all known reasonable modularity conditions for pairs of elements are equivalent), with join and meet computed in $L$; lastly, the projections (as opposed to generic idempotents) are the pairs $(K,I)$ which are orthocomplementary (as opposed to only modular complementary) in $L$; the involution in $A$ is the only one that makes such projections (that ring generate $A$) self-adjoint (and then the involution is also unique on the classical quotient ring $M$). - -Note that $A$ contains only some of the idempotents of $M$; precisely, the idempotents corresponding to pairs $(K,I)$ which are "nonasymptotic" (for this classical concept see Topping, Bures [with improvements by S. Maeda in the interaction with lattice theory], and more recently M. Anoussis, A. Katavolos, I. G. Todorov [10]). In the von Neumann algebra case, a (external) lattice description of "nonasymptoticy" is "absolute modularity": for one normal embedding of $L$ in a Hilbert lattice (as its own bicommutant; note that the commuting of projections is ortholattice definable), the pair is modular in the larger lattice (then the same happens for each normal embedding of $L$ in any projection ortholattice of a von Neumann algebra). -[Since the ortholattice $L$ determines $A$, it also determines the orthosymmetric structure that $A$ defines on $L$; really, Mayet's orthosymmetric structure on $L$ is unique since for each element $e$ of $L$ there is only one involutory automorphism of $L$ i.e. of $A$ that fixes exactly the projections that commute with $e$ (see for example lemma 2.4 in euclid.cmp/1103859692 on projecteuclid.org ; it is sufficient the even more folklore case of factors: then apply a subdirect decomposition into factors for the general case)] - -Full references: -[1] Bunce, L. J.; Wright, J. D. Maitland. Quantum logics and convex geometry. -Comm. Math. Phys. 101 (1985), no. 1, 87--96. -[2] Bunce, L. J.; Wright, J. D. Maitland. Quantum logic, state space geometry and operator algebras. -Comm. Math. Phys. 96 (1984), no. 3, 345--348. -[3]Maeda, Shûichirô. Dimension theory on relatively semi-orthocomplemented complete lattices. -J. Sci. Hiroshima Univ. Ser. A-I Math. 25 (1961), no. 2, 369--404. doi:10.32917/hmj/1206139804. -[4] Von Neumann, John. Continuous geometries with a transition probability. Vol. 252. American Mathematical Soc., 1981. -[5] Dye, H. A. “On the Geometry of Projections in Certain Operator Algebras.” Annals of Mathematics, vol. 61, no. 1, 1955, pp. 73–89. JSTOR -[6] Yen, Ti. Isomorphism of AW∗-algebras. Proc. Amer. Math. Soc. 8 (1957), 345–349. -[7] Berberian, S.K. The maximal ring of quotients of a finite Von Neumann algebra. Rocky Mountain J. Math. 12 (1982), no. 1, 149--164. doi:10.1216/RMJ-1982-12-1-149. -[8] Heunen, Chris, and Manuel L. Reyes. "Active lattices determine AW*-algebras." Journal of Mathematical Analysis and Applications 416.1 (2014): 289-313. -[9] Mayet, R. “Orthosymmetric Ortholattices.” Proceedings of the American Mathematical Society, vol. 114, no. 2, 1992, pp. 295–306. JSTOR -[10] Anoussis, M., Aristides Katavolos, and Ivan G. Todorov. "Angles in C*-algebras." arXiv preprint math/0601003 (2005).<|endoftext|> -TITLE: Reference request: colimits of locally presentable categories -QUESTION [8 upvotes]: Consider the 2-category of locally presentable categories, cocontinuous functors, and natural transformations. I believe that this 2-category is 2-cocomplete in the sense of containing all small 2-colimits. One proof is outlined in Mike Shulman's answer to Martin Brandenburg's question 2-colimits in the category of cocomplete categories from a couple years ago. In my paper with Alex Chirvasitu (link), we claim (with a sketch of a proof) that the following works: (i) every cocontinuous functor between presentable categories has a right adjoint, which is continuous and commutes with sufficiently-filtered colimits, and every functor of this type has a left adjoint; (ii) take the diagram whose colimit you want to compute, and consider the corresponding diagram in the category of locally presentable categories, right adjoints, and natural transformations; it suffices to compute the limit of that diagram; (iii) compute the limit of that diagram in the 2-category of all categories, and check that the limit is presentable, and that the functors involved in the limit are continuous and commute with sufficiently filtered colimits. -My question is simply a reference request: is there a paper in the published literature that provides a careful proof that the 2-category of locally presentable categories, cocontinuous functors, and natural transformations is 2-cocomplete? I'd rather cite this fact in my current project than reproduce the above argument, and I don't want to cite a paper (even my own) that only provides a "sketch of proof", if the fact is one I plan to rely on later. -The claim is not in the standard reference by Adamek and Rosicky. The closest there is the fact that the 2-category of accessible categories and functors that commute with sufficiently filtered colimits is 2-complete (and in fact 2-limits can be computed in Cat). - -REPLY [7 votes]: I'm a bit late to the party, but I believe there is a canonical reference for this fact: Greg Bird's 1984 thesis Limits in 2-Categories of Locally Presentable Categories. Although apparently unpublished, Google Scholar lists 20 citations to it. According to this post by Steve Lack to the categories mailing list, Bird proves that the 2-category of locally presentable categories, left adjoint functors, and all natural transformations has all flexible 2-limits and all flexible 2-colimits. This includes all bi(co)limits. -I don't know the history, but I gather that Bird's thesis work was in fact closely related to the very development of the notion of a flexible 2-limit: these are the strict 2-limits whose weights are such that they don't don't actually demand anything too strict, the "homotopically meaningful" strict 2-limits. Lack later showed that in fact they are the cofibrant objects in an appropriate 2-model structure on the 2-category of weights. -Unfortunately, I haven't been able to find a copy of Bird's thesis online. Update: Ross Street has now made it available again (link); see comment below.<|endoftext|> -TITLE: Characterization of Schur's property -QUESTION [7 upvotes]: A Banach space $H$ is said to have Schur's property if weak convergence of a sequence implies converge in norm. The most famous example of such a space is $\ell^1(\mathbb N)$, while $L^1[0,1]$ does not have this property. -My question is the following: -Is there a characterization of such spaces? -Is there a list of known examples, other than examples of the type $\ell^1(X)$? -Is it true that such space are not reflexive, when they are infinite dimensional? - -REPLY [3 votes]: (Turning my comment into an answer here): -Regarding the third question: Yes, reflexive spaces with the Schur property need to be finite-dimensional. To see that, two ingredients suffice, once you notice that the Schur property can be phrased as "weakly sequentially compact sets are sequentially compact": - -You need to show that in an infinite-dimensional space $X$, the unit ball is never (sequentially) compact. This follows from Riesz's lemma, which guarantees the existence of a sequence $(x_n) \in X^{\mathbb N}$ with $|x_n - x_m| \ge \frac 12$ for $n \ne m$. -You need to show that the unit ball in a reflexive space is weakly sequentially compact. - -The proofs for both claims are elementary, unlike the Eberlein–Šmulian theorem; they can be found e.g. (in German, I'm afraid) in - -Werner, Dirk. Funktionalanalysis. (German) [Functional analysis] Third, revised and extended edition. Springer-Verlag, Berlin, 2000. xii+501 pp. ISBN: 3-540-67645-7 MR1787146<|endoftext|> -TITLE: The status of automorphic induction -QUESTION [15 upvotes]: Background: Let $K/F$ be a degree $r$ extension of number fields. It is conjectured that an automorphic representation of GL$_n$ associated to $K$ induces an automorphic representation of GL$_{rn}$ associated to $F$. -The book by Arthur and Clozel (1989) claims to prove automorphic induction for arbitary $n$ and arbitrary cyclic extensions $K/F$. Work by Lapid and Rogawski in the late 1990s showed that the proof is incorrect and that the result of AC is valid only for cyclic extensions of prime degree (i.e. the same class of fields considered earlier by Langlands in the case of GL$_2$). In the context of automorphic motives this restriction is much too strong because in general the fields that appear are not cyclic, not to mention cyclic of prime degree. This leads to the -Question: -What results are known for automorphic induction of GL$_n$ automorphic forms for various choices of $n$ and various types of extensions $K$? -In the context of automorphic motives even results for the smallest groups GL$_1$ and GL$_2$ would already be of interest. - -REPLY [6 votes]: Here are a few recent results which haven't been mentioned so far (in chronological order). - -Rajan, On the image and fibres of solvable base change (2002) - -This relies on the result of Lapid and Rogawski for $\mathrm{GL}(2)$. The paper works with the generalization for $\mathrm{GL}(n)$ as a hypothesis. Remark 1 there says that, "Granting this, our theorem extends to $\mathrm{GL}_n$, and we present the proof in the general case assuming Statement B of Lapid-Rogawski." - -Henniart, Induction automorphe pour $\mathrm{GL}(n,\mathbb{C})$ (2009) - -The result is the same as that obtained in the non-archimedean case. - -Henniart, Lemaire, Formules de caracteres pour l'induction automorphe (2010) - -If $E/F$ is a degree $d$ cyclic extension of non-archimedean local fields then Henniart and Herb proved that if $\tau$ is a tempered representation of $\mathrm{GL}(m,E)$ then there is a tempered representation $\pi$ of $\mathrm{GL}(md,F)$ such that -\begin{equation} -\mathrm{tr}\,\tau(f)=c\cdot\mathrm{tr}\,\pi(f^G)\circ A, -\end{equation} -whenever $f$ and $f^G$ have matching orbital integrals. The constant $c$ depends on $\tau$ and the operator $A$ which intertwines $\pi$ and $\omega\pi$, where $\omega$ is a character of $F^\times$ determining the extension $E/F$. Henniart and Lemaire show that when one uses the canonical normalization of $A$ via Whittaker models then $c$ does not depend on $\tau$. - -Hiraga and Ichino, On the Kottwitz-Shelstad normalization of transfer factors for automorphic induction for $\mathrm{GL}_n$ (2012) - -They show that this constant $c$ is in fact 1 when using normalized transfer factors. - -Henniart, Induction automorphe globale pour les corps de nombres (2012) - -When $E/F$ is a cyclic extension of number fields of degree $d$ and $\tau$ is an induced from cuspidal representation of $\mathrm{GL}(m,\mathbb{A}_E)$, from the abstract, -"We prove that the representation $\pi$ automorphically induced from $\tau$ exists, and we study the fibres and the image of automorphic induction. For that we use and extend the results of Arthur and Clozel on base change, which corresponds to restricting Galois representations from $F$ to $E$, and we clarify the relations between the two processes. Moreover we prove that global automorphic induction is compatible, at finite places, with the local automorphic induction defined by R. Herb and the author."<|endoftext|> -TITLE: Is $\{ p \alpha \}$ for prime $p$ dense in $[0,1]$? -QUESTION [10 upvotes]: Let $\alpha$ an irrational real number. It is well known that the set $\{ \{n \alpha \}|\,\, n \in \mathbb{N} \}$ is dense in$[0,1]$. -($\{x\}$ denotes the fractional part of $x$) -But how to prove the set $\{ \{p \alpha \}|\,\, prime\,\,p\}$ is dense in $[0,1]$? Also, is it uniformly distributed? -It is somehow a strange generalization of Dirichlet's theorem of primes on arithmetic progressions, so I tried to emulate its proof, by considering sums of the form: -$S(k)=\sum\limits_{prime\,\,p} \frac{f(k p\alpha)}{p}$ -Where $f$ would be a function analogous to a character (then we can "isolate" the $p$'s such that $\{p\alpha\}$ is in some some interval $[a,b]$ by considering an adequate linear combination of the $S(k)$'s). However, I fail to bound the sums when simply choosing $f(x)=e^{2i\pi x}$, because it is not multiplicative as characters are. But, a construction of such function seems hopeless. Any ideas? -Thanks in advance. - -REPLY [17 votes]: Exponential sums over primes can be reduced to exponential sums over arithmetic progressions, as discovered by Vinogradov. The basic idea is the same as in the sieve of Eratosthenes. Modern treatments of this reduction rely on particular decompositions of the von Mangoldt function into convolutions of "simpler" functions, e.g. by the Vaughan identity or the more general Heath-Brown identity. -Along these ideas, Vinogradov proved that $\{p\alpha\}$ is equidistributed in $[0,1]$ as $p$ runs through the prime numbers (in increasing order). Of course the rate of equidistribution depends on how well $\alpha$ can be approximated by rational numbers. For a newer version of this result see Corollary 2.2 in Vaughan: On the distribution of $\alpha p$ modulo 1, Mathematika 24 (1977), 135-141.<|endoftext|> -TITLE: What is the definition of picture changing operation? -QUESTION [9 upvotes]: What is the definition of picture changing operation? -What is a standard reference where it is defined - not just used? - -REPLY [3 votes]: This is a physical description of Picture-Changing Operation: -In the RNS formulation of superstring theory, the worldsheet theory has superconformal gauge invariance. One thus needs to fix the gauge. This means that one can (locally) fix the form of metric and the gravitino field. This will introduce ghost and anti-ghost field into the theory: - -If one fixes diffeomorphism (form of the metric), one gets $(b,c)$ -system (they are fermions because they are ghost fields corresponding -to a gauge fixing of a Bosonic field); -If one fixes supersymmetry (form of gravitino), one gets -$(\beta,\gamma)$ system (they are bosons because they are ghost -fields corresponding to a gauge fixing of a Fermionic field); - -The action for the $(\beta,\gamma)$ system has first-order derivative and due to a general properties of such Bosonic systems, the Hamiltonian is unbounded from below which is not a desirable property. To fix this, Friedan, Shenker and Martinec (FSM) introduced an equivalent description of $(\beta,\gamma)$ system in terms of a new set of fields $(\xi,\eta,\phi)$, a process called Bosonization (In this case, it is a Fermionization though, but in the literature it is called Bosonization). The fields $(\beta,\gamma)$ can be written in terms of derivative of $\xi$ and the $\eta$ and $e^{\pm\phi}$ fields. The equivalence of these two description means that OPE of the operators in the two descriptions match. But there is a new feature in this equivalent description : There is a new degree of freedom in $(\xi,\eta,\phi)$ which was somehow hidden in the $(\beta,\gamma)$ description namely the zero mode of $\xi$ field. -This zero mode has crucial importance in string theory : -If this zero mode didn't exist, all Fermionic processes in superstring theory would vanish. -Using ${\bf SL}(2,\mathbb{C})$ invariant vacuum of string theory $\left|0\rangle\right._{{\bf SL}(2,\mathbb{C})}$ (namely the vacuum annihilated by generators of Virasoro algebra $\hat{L}_n$ for $n=0,\pm 1$) and modes of $\beta$ and $\gamma$ fields, one can construct states with arbitrary low energy (due to conformal dimensions of these fields). This means that: -$\left|0\rangle\right._{{\bf SL}(2,\mathbb{C})}$ is not the correct vacuum of the theory. -The correct vacuum of the theory can be written as follows: -$$\left|\boldsymbol{\Omega}\rangle\right.=\hat{c}(0)e^{-\hat{\phi}(0)}\left|0\rangle\right._{{\bf SL}(2,\mathbb{C})}$$ -These two states, namely $\left|0\rangle\right._{{\bf SL}(2,\mathbb{C})}$ and $\left|\boldsymbol{\Omega}\rangle\right.$ lives in disjoint Hilbert space (due to the exponential operator), i.e. one can not reach the other by applying finite number of ghost fields modes $\beta_k$ and $\gamma_k$. This means that the states of the original $(\beta,\gamma)$ system contains a condensate of modes of these fields which is called Bose Sea in the language of FSM. They interpreted the charge of the $\phi$ field as the filling level of these Bose sea and it is called Picture Number. On the other hand the vertex operators associated to the the Ramond sector are Spin Fields. The form of spin fields in different picture number is different. There are subtleties on how one can associate picture number to R and NS sectors but all allowed picture numbers gives equivalent description of string theory. If the form of spin field in picture number $\alpha$ is $\mathcal{V}_\alpha$ then one can define the Picture-changing Operation as follows: -$$\mathcal{V}_{\alpha+1}(z)\equiv\{{\hat{Q}}_{\bf BRST},\xi(z)\mathcal{V}_{\alpha}\}$$ -One can show that (page 139-140 of Friedan, Shenker and Martinec) two vertex operatos which are related by picture-changing opration give equivalent results for computation of on-shell scattering amplitudes in string theory (section 7 of Friedan, Shenker and Martinec). -For a mathematical description, please check: - -Section 4.2 of Notes On Supermanifolds And Integration -Section 4.3 and 4.4 of Notes On Super Riemann Surfaces And Their Moduli -Section 3.6.2 and 3.6.3 of Perturbative Superstring Theory Revisited<|endoftext|> -TITLE: Shortest curve with given convex hull -QUESTION [7 upvotes]: Suppose $S\subset\mathbb{R}^2$ is compact and convex. Suppose $\Gamma:[0,1]\to S$ is a continuous curve that passes through every extreme point of $S$, i.e., the convex hull of $\Gamma([0,1])$ is $S$. I am interested in obtaining a lower bound on the length $|\Gamma|$ of $\Gamma$. -I conjecture that -$$ -|\Gamma| \ge C(S) - \ell -$$ -where $C(S)$ is the perimeter (circumference) of $S$ and $\ell$ is defined as -$$ -\ell = \sup\{d(e,e'): e,e'\text{ are extreme points of $S$ and $\overline{ee'}\subset \partial S$} \}. -$$ -Here, $\overline{ee'}$ is the line segment joining $e$ and $e'$. Intuitively, $\ell$ is the length of the longest straight segment in $\partial S$. I think the shortest path $\Gamma$ with $\mathrm{ConvexHull}(\Gamma([0,1])) =S$ can be formed by traveling along the circumference of $S$ from one vertex of the longest straight edge in $\partial S$ to the other vertex. (Obviously, the direction of this path is the one giving $\mathrm{ConvexHull}(\Gamma([0,1])) =S$). -Is the conjectured bound on $|\Gamma|$ above correct? Can one construct a counterexample? -Motivation I'm trying to come up with a simple proof that Isbell's curve solves a particular instance of Bellman's "Lost in a Forest" problem. See this question on MSE if you want to know more about my motivation, but that background is not at all necessary to understand and solve this problem. - -REPLY [9 votes]: Let $n$ be large and glue $n$ rectangles, each with the proportions $1\times n$, together end-to-end by their short edges. Then perturb slightly so that all the vertices are in convex position. The perimeter will be roughly $2n^2+2$ (modulo the perturbation), and the longest boundary edge will have length roughly $n$, so $C(S)-\ell$ will be roughly $2n^2-n+2$. However, there is a curve with length roughly $n^2+n+1$ through all the rectangle vertices, that passes once through each short edge of a rectangle and through one of the two long edges of each rectangle, alternating between the top and bottom long sides. So in this case the shortest curve is shorter than your proposed bound by roughly a factor of two. -On the other hand, $C(S)/2$ is an easy lower bound (which this example shows to be close to tight): the shortest (simple) closed curve through all extreme points (their traveling salesman tour) is known to be the same as the convex hull, and is also known to be at most twice the length of any spanning path (or more strongly of any spanning tree) of the extreme points.<|endoftext|> -TITLE: Why did Alonzo Church choose the letter $\lambda$ as the "binding operator"? -QUESTION [12 upvotes]: Is there any known reason why Alonzo Church chose Greek $\lambda$ as the "binding operator" for the Lambda Calculus? - -REPLY [10 votes]: This question has been answered on math.SE (as pointed out by Joel David Hamkins). -With a reference to Lambda-Calculus and Combinators in the 20th Century by -Felice Cardone and J. Roger Hindley, Handbook of the History of Logic -Volume 5, 2009, Pages 723–817, it is stated that “$\lambda x$” comes from “$\hat x$” in Principia Mathematica. -Here is a quote from a preprint of Lambda-Calculus and Combinators in the 20th Century: - -By the way, why did Church choose the notation “$\lambda$”? - In [A. Church, 7 July 1964. Unpublished letter to Harald Dickson, §2] - he stated clearly that it came from the notation - “$\hat x$” used for class-abstraction by Whitehead and Russell, by first - modifying “$\hat x$” to “$\wedge x$” to distinguish function-abstraction from - class-abstraction, and then changing “$\wedge$” to “$\lambda$” for ease of - printing. - This origin was also reported in - [J. B. Rosser. Highlights of the history of the lambda calculus. - Annals of the History of Computing, 6:337—349, 1984, p.338]. - On the other hand, in his later years Church told two enquirers that the - choice was more accidental: a symbol was needed and “$\lambda$” just - happened to be chosen. - -Assuming that “$\lambda x$” comes from “$\hat x$” in Principia Mathematica, let us look how it is used there. -In Principia Mathematica there are two ways the notation “$\hat x$” is used. -The first use is to write “propositional functions,” it is introduced in Volume I, in Chapter I of the Introduction, on page 15. -Here is a quote: - -[...] - When we wish to speak of the propositional function corresponding to - “$x$ is hurt,” we shall write “$\hat x$ is hurt.” - Thus “$\hat x$ is hurt” is the propositional function, and “$x$ is hurt” - is an ambiguous value of that function. - Accordingly though “$x$ is hurt” and “$y$ is hurt” occurring in the same - context can be distinguished, “$\hat x$ is hurt” and “$\hat y$ is hurt” - convey no distinction of meaning at all. - [...] - -The second use is to write classes in a way similar to the modern “$\{\,z\mid\psi(z)\,\}$”, it is introduced in Volume I, in Section C of Part I, in definition *20.01, on page 197. -Here is some quote: - -[...] - But it is convenient to regard $f\{\hat z(\psi z)\}$ as though it had an - argument $\hat z(\psi z)$, which we will call “the class determined by the - function $\psi\hat z$.” - [...]<|endoftext|> -TITLE: Reg the motivation behind Lusztig-Vogan bijection -QUESTION [5 upvotes]: Let $G$ be an algebraic group. Choose a Borel subgroup $B$ and -a maximal Torus $T \subset B$. Let $\Lambda$ be the set of weights wrt $T$ and let $\mathfrak{g}$ be the lie algebra of $G$. -Now, consider the following two sets, -1) $\Lambda^+$, the set of dominant weights wrt $B$, -2) The set $N_{o,r}$ of pairs $(e,r)$ (identified upto $\mathfrak{g}$ conjugacy), where e is a nilpotent element in $\mathfrak{g}$ and $r$ is an irreducible representation of the centralizer (in $\mathfrak{g}$) $Z_e$ of a nilpotent element e. -There is a bijective map between the two sets that plays an important -role in the representation theory of $G$ and this is often called the -Lusztig-Vogan bijection, -$\rho_{LV} : \Lambda^+ \rightarrow N_{o,r}$. -In recent works, this bijection has been studied by Ostrik, -Bezrukovnikov, Chmutova-Ostrik, Achar, Achar-Sommers (Edit: See links to refs below) using various -different tools. My question however pertains to the motivations that -point to the existence of such a bijection in the first place. As I -understand, the component group $A(O)$ where $O$ is the nilpotent orbit associated to $e$ (under the adjoint action) and a quotient of the component -group $\overline{A(O)}$ play important roles in the -algorithmic description of this bijection (say for example in determining -the map for certain $h \in \Lambda^+$, where $h$ is the Dynkin element -of a nilpotent orbit in the dual lie algebra). One of the original motivations -for the existence of such a bijection seems to have emerged from the -study of primitive ideals in the universal enveloping algebra of g. -My questions are the following : - -How does $\overline{A(O)}$ enter the story from the point of view of the -study of primitive ideals ? -Are there other representation theoretic motivations that point to the existence of such a bijection ? Here, I am (somewhat vaguely) counting a motivation to be 'different' if its relation to the theory of primitive ideals is nontrivial. - -[Added in Edit] Refs for some recent works on the bijection (in anti-chronological order) : - -Local systems on nilpotent orbits and weighted Dynkin diagrams (link) - P Achar and E Sommers -Calculating canonical distinguished involutions in the affine Weyl groups (link) - T Chmutova and V Ostrik -Quasi-exceptional sets and equivariant coherent sheaves on the nilpotent cone (link) - R Bezrukavnikov - -REPLY [4 votes]: There are some conversations on the affine Weyl group cells perspective of the bijection. I guess I can contribute a very little bit on the primitive ideals side of the story, if it is not too late to do so. -My first encounter of the Lusztig's quotient comes from the paper of Barbasch and Vogan in 1985 (here). It is mainly about finding the (g,K)-modules of a fixed infinitesimal character whose annihilators are the 'largest' possible primitive ideal, which Barbasch-Vogan called 'special unipotent' as in the title of the paper. -These special unipotent representations have a lot to do with the ring of regular function of a nilpotent orbit, R[O]. This is first hinted in Section 12 of Vogan's work (here). And Vogan's version of the above bijection is about the structure of R[O], as mentioned in the Achar-Sommers paper. I have carried out some computations on the bijection from this perspective (effectively I can prove a conjecture in the paper). But talking about why Vogan's version of the bijection matches with that of Lusztig's, I think Barbasch and Vogan know much more about it.<|endoftext|> -TITLE: Is Euclid dead? -QUESTION [80 upvotes]: Apparently Euclid died about 2,300 years ago (actually 2,288 to be more precise), but the title of the question refers to the rallying cry of Dieudonné, "A bas Euclide! Mort aux triangles!" (see King of Infinite Space: Donald Coxeter, the Man Who Saved Geometry by Siobhan Roberts, p. 157), often associated in the popular mind with Bourbaki's general stance on rigorous, formalized mathematics (eschewing pictorial representations, etc.). See Dieudonné's address at the Royaumont seminar for his own articulated stance. -In brief, the suggestion was to replace Euclidean Geometry (EG) in the secondary school curriculum with more modern mathematical areas, as for example Set Theory, Abstract Algebra and (soft) Analysis. These ideas were influential, and Euclidean Geometry was gradually demoted in French secondary school education. Not totally abolished though: it is still a part of the syllabus, but without the difficult and interesting proofs and the axiomatic foundation. Analogous demotion/abolition of EG took place in most European countries during the 70s and 80s, especially in the Western European ones. (An exception is Russia!) And together with EG there was a gradual disappearance of mathematical proofs from the high school syllabus, in most European countries; the trouble being (as I understand it) that most of the proofs and notions of modern mathematical areas which replaced EG either required maturity or were not sufficiently interesting to students, and gradually most of such proofs were abandoned. About ten years later, there were general calls that geometry return, as the introduction of the alternative mathematical areas did not produce the desired results. Thus EG came back, but not in its original form. -I teach in a University (not a high school), and we keep introducing new introductory courses, for math majors, as our new students do not know what a proof is. [Cf. the rise of university courses in the US that come under the heading "Introduction to Mathematical Proofs" and the like.] -I am interested in hearing arguments both for and against the return of EG to high school curricula. Some related questions: is it necessary for high-school students to be exposed to proofs? If so, is there is a more efficient mathematical subject, for high school students, in order to learn what is a theorem, an axiom and a proof? -Full disclosure: currently I am leading a campaign for the return of EG to the syllabus of the high schools of my country (Cyprus). However, I am genuinely interested in hearing arguments both pro and con. - -REPLY [14 votes]: It might be interesting to point out that the support for removing axiomatically taught Euclidean geometry (not all geometry) from the school education predates Bourbaki. Oliver Heaviside (1850-1925), a British mathematician who also made important contributions to physics wrote in his "Electro-Magnetic Theory", vol. 1 (1893): -"As to the need of improvement there can be no question whilst the reign of Euclid continues. My own idea of a useful course is to begin with arithmetic, and then not Euclid but algebra. Next, not Euclid, but practical geometry, solid as well as plane; not demonstration, but to make acquaintance. Then not Euclid, but elementary vectors, conjoined with algebra, and applied to geometry. Addition first; then the scalar product. Elementary calculus should go on simultaneously, and come into vector algebraic geometry after a bit. Euclid might be an extra course for learned men, like Homer. But Euclid for children is barbarous".<|endoftext|> -TITLE: An explicit IP algorithm for chess? -QUESTION [8 upvotes]: If I have 2 large graphs to be tested for isomorphism, and can communicate with some (powerfull but untrusted) machine, I can choose graph at random, permute vertices, ask machine to guess which one was chosen, and the correct guess (100 times) proves me that the graphs are really non-isomorphic. -A famous IP=PSPACE theorem implies that a similar algorithm EXISTS for n$\times$n chess (with 50-moves rule): if a particular position is a winning one for whites, a machine can prove me this in polynomial time. Is it possible to formulate this algorithm in understandable way, in chess terms? It would be really interesing to see it. Also, is "polynomial time" here really implies "efficient": at least if n=8, is the number of operations I would need to perform can be done using a (standard current) computer in a reasonable time? - -REPLY [3 votes]: This is a good question and a nice way to study some complexity-theorems from the '90's. Reviewing Shamir's famous paper, it is not so much "if a particular position is a winning one for white, a machine can prove me this in polynomial time," it is more "if a particular position is a winning one for white, I can engage with a powerful prover machine in a number of rounds of challenge/response to be convinced by the prover that there is a winning position." In the challenge/response, the prover is pretty abstracted from chess and is merely proving that some polynomial over some finite field is non-zero. -For example, $\mathcal{PSPACE}$ refers to problems running in space polynomial to the input size. Much as $\mathsf{3CNF}$ is the standard example of an $\mathcal{NPC}$ decision problem, True Quantified Boolean Formula, $\mathsf{TQBF}$, is the standard example of a decision problem complete for $\mathcal{PSPACE}$. A $\mathsf{QBF}$ formula might be $\forall x\ \exists y\ \exists z\ ((x \lor z) \land y)$. The relation of $\mathsf{TQBF}$ to games such as chess or checkers is noted, in that a "white-to-mate-in-$n$" problem can also be cast as -$$\exists w_1\forall b_1\exists w_2\forall b_2\cdots\exists w_n:\phi(w_1,b_1,w_2,b_2,\cdots,w_n)$$ -where $\phi$ is a Boolean encoding the rules of chess as applied to the given position. We can read this as "there is a move by white, such that for all moves by black, there is a counter-move by white, such that... white checkmates black. -Turning to $\mathcal{IP}$, interactive proofs involve two parties, say $P$ the prover and $V$ the verifier. $V$ and $P$ engage in rounds of challenges and responses. The "interactive proof" of graph non-isomorphism $\mathsf{GNI}$ given in the question is typically framed as a single round of challenge/response, that is amplified to improve soundness, say, 100 times. However, $\mathcal{IP}$ as used by Shamir involves a polynomial number of rounds of interaction between $V$ and $P$. -The denouement, of course, is that $\mathcal{IP}=\mathcal{PSPACE}$. Shamir and his predecessors showed (1) how to convert any $\mathsf{TQBF}$ problem into a canonical form ("simple QBF"), (2) how to "arithmetize" the simple QBF by replacing each $\forall$ with $\Pi$, each $\exists$ with $\Sigma$, each $\land$ with multiplication, etc. thus converting a simple QBF to a polynomial, and (3) how to have a verifier, $V$, challenge a prover, $P$, with random interactive coin tosses to establish that the polynomial is nonzero. Each round of the interaction simplifies the polynomial, and eventually, as long as $V$ is randomly choosing her coin tosses for each round after $P$ provides her answers, $V$ should be convinced by $P$, -However, $P$ really has to work hard to answer $V$'s challenges. For example, she must actually evaluate very large polynomials for each round. The work that the prover has to do does not seem very enlightening with respect to chess itself.<|endoftext|> -TITLE: Free C^*-algebra -QUESTION [8 upvotes]: Let $A_0$ be a set of all polynomials with complex coefficients of infinitely many noncommuting (free) variables, denoted by $X_1,X_2,...,X_1^*,X_2^*,...$. We equip $A_0$ with the operation $*:A_0 \to A_0$ in such a way that this operation becomes involution. For a polynomial $p(X_1,X_2,...,X_1^*,X_2^*,...)$ we define -$$\|p\|:=sup\|p(T_1,T_2,...,T_1^*,T_2^*,...)\|$$ -where the supremum is taken over all possible collections of contractions $\{T_n\}_n$ in separable infinite dimensional Hilbert space (and $p(T_1,...,T_1^*,...)$ has an obvious meaning). It is not hard to see that this formula defines a seminorm satisfying all conditions for a $C^*$-norm but my question is the following: is it actually the norm? More powerful would be the following: is it possible to find an infinite collection $T_1,T_2,...$ of contractions in separable Hilbert space such that there is no "$C^*$-algebraic" relation relating them? -I never saw a proof that the formula above defines a norm but I've heard that at least one author mentions this fact. - -REPLY [11 votes]: See Lemma 3.7 in my paper C*-relations, Math. Scand., 107(1), 43--72 (2010). I show that this seminorm is actually a norm. This is for the finite or infinite case. For the case of countably many generators, one can embed the "free" C*-algebra in $B(H)$ for a separable Hilbert space and conclude you last assertion. -I can't recall who told me this, nor any reference, but I recall that for any NC $*$-polynomial, there are finite matrices, contractions, not satisfying the results. In terms I am more comfortable with, the C*-algebra generated by $n$ universal contractions is RFD, so such matrices exists. (Recall projective implies RFD.)<|endoftext|> -TITLE: Number of unlabelled planar graphs -QUESTION [8 upvotes]: What are the best known bounds on the number of non-isomorphic (unlabelled) planar graphs on $n$ vertices? Is there a simple proof that this number is at most exponential in $n$? - -REPLY [2 votes]: There is a short proof of a stronger result by Norine, Seymour, Thomas, and Wollan that there is an exponential upper bound for every proper minor-closed class. - -For every proper minor-closed class of graphs $\mathcal G$, there exists a constant $c$, such that for every integer $n$, there are at most $n! c^n$ graphs in $\mathcal G$ with vertex set $\{1, \dots, n\}$. - -This is the labelled version of your question, but as Igor mentions, the typical graph has trivial automorphism group, so we can divide by $n!$. -The paper is entitled Proper minor-closed families are small and can be downloaded here or from Robin Thomas's webpage.<|endoftext|> -TITLE: Are $(\infty,1)$-categories $A_\infty$ categories? -QUESTION [11 upvotes]: Let $X$ be a set. One can define a non-symmetric colored version of the non-unital $A_\infty$ operad as follows. The set of colors is the set of ordered pairs in $X$. Let $(x_1,y_1),\dots,(x_n,y_n)$ be an ordered list of $n\geq 2$ colors. The space of $n$-ary operations with domain $(x_1,y_1),\dots,(x_n,y_n)$ and codomain $(x,y)$ is empty unless $x = x_1$, $y = y_n$, and $y_i = x_{i+1}$ for $i=1,\dots,n-1$. When the colors do satisfy these relations, the space of colors is the finite cell complex whose cells are labeled by planar rooted trees, each vertex having at least two inputs ("rooted" means each vertex has a unique output, namely the half-edge connecting the vertex towards the root), and whose regions are labeled, in left-to-right order along the top (the "leaf" end of the planar tree), by the colors $x=x_1,y_1=x_2,\dots,y_n=y$. The dimension of a cell is $\sum_v (\deg v - 2)$, where the sum ranges over the vertices in the tree, and $\deg v$ is the number of inputs to the vertex $v$. The boundary of a cell consists of all cells that can be formed by blowing up a vertex into two vertices. The combinatorics of this "boundary" operation are precisely described by some Stasheff polytopes, and so is a sphere. Composition in this colored operad is by grafting of trees: ignoring the boundary operation, this colored operad is free on the trees with one vertex. So I believe that this colored operad is cofibrant and has the homotopy type of the colored operad whose algebras Set are categories. -A representation of this colored operad is nothing but an $A_\infty$ category with objects indexed by $X$. -Let $\mathcal S$ be a cartesian-closed presentable $(\infty,1)$-category. I'm pretty sure that every $A_\infty$ category in $\mathcal S$ is also an $\mathcal S$-enriched $(\infty,1)$-category. -Is the opposite true? I.e. is the inclusion of $A_\infty$ categories among all $\mathcal S$-enriched categories a homotopy equivalence? My impression from Bergner, Models for $(∞,n)$-categories and the cobordism hypothesis, Mathematical Foundations of Quantum Field Theory and Perturbative String Theory, 2011 was that at least at the time this was expected but unknown. -But my impression could be off, or there could be new results in the intervening years, or I could have misunderstood what an "$A_\infty$ category" is. -Assuming that the answer is (at least after necessary corrections have been made to my definition) "it's not yet known": where does the difficulty lie? - -REPLY [4 votes]: It seems to me that some progress has been made in the recent paper of Giovani Faonte, but I don't think that the result you are looking for has been proved.<|endoftext|> -TITLE: Trying to Understand Lefschetz Pencils -QUESTION [8 upvotes]: All: -I'm reading on Lefschetz pencils, and I'm trying to understand condition ii) below better, though I would appreciate insights on condition i), and in general. Please forgive if the presentation is somewhat-confused, since I have not found a single source including all details, so I have had to borrow from different places. -A Lefschetz pencil on a simply-connected $4$-manifold $X^4$ is a pair $(B, \pi)$, where $B$ is a finite, discrete subset of $X$ , and $\pi$ is a map $\pi$: $(X-B) \rightarrow \mathbb CP^1 $~$ S^2$ so that: -i) Each point $b$ in $B$ has an orientation-preserving local coordinate map to $(\mathbb C^2,0)$ in which $\pi$ corresponds to the projectivization map (i.e., every line through $0^{2n}$ becomes an equivalence class $[t_0:t_1]; t_0,t_1$ not both $0$, partitioning $\mathbb C^2 -0)$ , and : -ii) Every critical point of $\pi$ has an orientation-preserving chart in which $\pi(z_1,z_2)=z_1^2+z_2^2 $, for some local chart in $\mathbb CP^1$. -iii)The problem fibers, i.e., the fibers at non-regular points are "fishtail fibers", i.e., vanishing cycles, and they are finitely-many, and these are described by their monodromy. The monodromy is described by a Dehn twists at each fiber, and the fibration can be fully described as a word (in the mapping class group of the fiber, given by the Dehn twists) given and the entire fibration can be fully described by this word -Now, part of my confusion has to see with the fact that some sources describe either $\pi$, or $X^4$ as being smooth, and some describe the charts as being holomorphic; other sources do not describe them this way, but do not explicitly exclude these conditions. I'm hoping someone may help me clarify the actual conditions on each of $\pi , X^4$ , and on whether the charts are holomorphic. -In addition , I have a few questions I hope someone can help me with: -1) Given that there is no mention , AFAIK, of any necessary smoothness condition for $\pi$, can the mention of critical point be related to something else? I am not aware of any definition of critical point that does not involve a condition of smoothness, or at least differentiability. Does one just assume $\pi$ is smooth, or at least differentiable, so that critical points are those where $d\pi$ does not have full rank? -2) What is the relevance of having a chart in which $\pi(z_1,z_2)$ equals $z_1^2+ z_2^2 $? I'm aware that these pencils extend, after blowing up each point of the finite, discrete point-set $B$ , into a full-blown (ha-ha) Lefschetz fibration. -The blow up consists, AFAIK, of defining a tangent space at a "problem point" where this tangent space is not defined (moreover: is this an algebraic-geometric tangent space, or a differential-geometric one?) ,somehow patching all possible directions at a point by attaching a $\mathbb CP^n$ containing all directions .But I don't fully get the importance or relevance of these two conditions in ii). Any ideas, refs., please? -EDIT: I'm trying to include the little I understand about the algebraic-geometric perspective. please feel free to correct and comment, since my understanding from this perspective is pretty limited: -1') We start with a complex surface M (meaning Real 4-manifold). -2') We consider a codimension-2 , generic linear subspace $L \subset \mathbb CP^n$. Let $B:= L\cap M $ .By a dimension count (and "genericity"), $|B|=n < \infty$ -3') We consider two generic codimension-1 subspaces $S^1, S^2$, generic other than they contain the linear subspace $L$. We have that $S_1,S_2$ can be represented as $V(p_0)$, $V(p_1)$ respectfully , i.e., as algebraic varieties, i.e., as the zero sets of two polynomials $p_0,p_1$ (not sure why this is possible, i.e., what guarantees we can do this.) -4') We consider the varieties associated to/ generated-by the above subspaces and respective polynomials , variety which is generated by any two points $[r_0:r_1], [s_0:s_1]$ in $\mathbb CP^1$ , i.e., the sets $V(r_0p_0+r_1p_1 )$ and $V(s_0p_0+s_1p_1)$. We show this two varieties intersect $S$ precisely at $B$, as in #2'). This intersection is independent of the choice of points $[s_0,s_1], [r_0,r_1]$ used, i.e., for any two points in $\mathbb CP^1 $ used, the associated varieties will intersect in $B$. -As you see, my understanding from this perspective is minimal, but I would love to understand it better. -Thank You. - -REPLY [7 votes]: As fas as I understand, you are trying to define Lefschetz pencils for smooth 4-manifolds. They are defined in order to satisfy all topological property they satisfy in the algebro-geometric setting, so you should failiarize a bit with that first (and in fact I have the impression, from what you said, that you are). -Regarding your questions: -1) $\pi$ must be smooth. When we talk about smooth manifolds, maps with no otherwise specified regularity are assumed to be smooth. I don't understand Sándor's comment that $\pi$ is not usually smooth in Algebraic geometry. Aren't all maps in Algebraic geometry, at worst, quotient of polynomials? How can they be non smooth in the differential geometric sense? -2) Around the base locus B the fibration looks, from a topological point of view, like the map $\mathbb{C}^2 \setminus \{ 0 \}\to \mathbb{C}$ given by $(x,y) \mapsto x/y$. -3) In order to get an intuition of property (ii) you should work out in detail the simplest case where you have a critical point like this: the map $\pi(x_1, x_2)= x_1^2 + x_2^2$ in $\mathbb{C}^2$. To make things short, if you separate real and imaginary part and rewrite the the equation $\pi(x_1,x_2)=1$ as a real equation, you see that its zero locus is a cotangent bundle of $S^2$. On the other hand, $\pi(x_1,x_2)=0$ defines two complex lines meeting at the origin. What happens when you move $1$ towards $0$ is that the zero section of the cotangent bundle becomes smaller and smaller until it's shrunk to the origin. -Now you might wonder why we care about this type of singularity and not about other ones. The short question is because it has this nice local model which we understand well, but there is also a longer answer. Among real functions on a compact manifold, Morse theory says that those with non-degenerate Hessian at all critical points (Morse functions) are generic. Moreover, one can find coordinates around a critical point with non-degenerate Hessian so that the function can be written as a quadratic form. You should regard Lefschetz pencils as a complex analogue of Morse functions, in the sense that, in a neighbourhood of a critical point, a Lefschetz fibration can be written as a non-degenerate complex quadratic form. The difference between the real case and the complex case is that there is only one non-degenerate complex quadratic form, which is exactly the local model of a singular point in a Lefschetz fibration.<|endoftext|> -TITLE: Examples of major theorems with very hard proofs that have not dramatically improved over time -QUESTION [89 upvotes]: This question complement a previous MO question: Examples of theorems with proofs that have dramatically improved over time. -I am looking for a list of -Major theorems in mathematics whose proofs are very hard but was not dramatically improved over the years. -(So a new dramatically simpler proof may represent a much hoped for breakthrough.) Cases where the original proof was very hard, dramatical improvments were found, but the proof remained very hard may also be included. -To limit the scope of the question: -a) Let us consider only theorems proved at least 25 years ago. (So if you have a good example from 1995 you may add a comment but for an answer wait please to 2020.) -b) Major results only. -c) Results with very hard proofs. -As usual, one example (or a few related ones) per post. -A similar question was asked before Still Difficult After All These Years. (That question referred to 100-years old or so theorems.) -Answers -(Updated Oct 3 '15) -1) Uniformization theorem for Riemann surfaces( Koebe and Poincare, 19th century) -2) Thue-Siegel-Roth theorem (Thue 1909; Siegel 1921; Roth 1955) -3) Feit-Thompson theorem (1963); -4) Kolmogorov-Arnold-Moser (or KAM) theorem (1954, Kolgomorov; 1962, Moser; 1963) -5) The construction of the $\Phi^4_3$ quantum field theory model. -This was done in the early seventies by Glimm, Jaffe, Feldman, Osterwalder, Magnen and Seneor. (NEW) -6) Weil conjectures (1974, Deligne) -7) The four color theorem (1976, Appel and Haken); -8) The decomposition theorem for intersection homology (1982, Beilinson-Bernstein-Deligne-Gabber); (Update: A new simpler proof by de Cataldo and Migliorini is now available) -9) Poincare conjecture for dimension four, 1982 Freedman (NEW) -10) The Smale conjecture 1983, Hatcher; -11) The classification of finite simple groups (1983, with some completions later) -12) The graph-minor theorem (1984, Robertson and Seymour) -13) Gross-Zagier formula (1986) -14) Restricted Burnside conjecture, Zelmanov, 1990. (NEW) -15) The Benedicks-Carleson theorem (1991) -16) Sphere packing problem in R 3 , a.k.a. the Kepler Conjecture(1999, Hales) -For the following answers some issues were raised in the comments. -The Selberg Trace Formula- general case (Reference to a 1983 proof by Hejhal) -Oppenheim conjecture (1986, Margulis) -Quillen equivalence (late 60s) -Carleson's theorem (1966) (Major simplification: 2000 proof by Lacey and Thiele.) -Szemerédi’s theorem (1974) (Major simplifications: ergodic theoretic proofs; modern proofs based on hypergraph regularity, and polymath1 proof for density Hales Jewett.) -Additional answer: -Answer about fully formalized proofs for 4CT and FT theorem. - -REPLY [4 votes]: Chazelle's linear time algorithm for the triangulation of a polygon has not been improved upon since its creation in 1991. -Technically, this is a computer science theorem, but I think it belongs here for a couple reasons. It's complicated. No actual code implementation of the algorithm has ever been made. While it is linear time, the constant factor makes the algorithm useless for any practical purposes, beyond its theoretical use in other papers.<|endoftext|> -TITLE: An alternative proof of the Łojasiewicz inequality -QUESTION [11 upvotes]: Is there a "brute force proof" of the Łojasiewicz inequality? By "brute force" I mean a proof without introducing the machinery of semianalytic sets and so on but only using elementary results (i.e., standard Calculus 1+2). I admit i hadn't time to think about it properly, I thought I could prove it first for homogeneous polynomials (and I got stuck) and then extend the result to analytic functions, but I could be wrong... To avoid misunderstandings I recall here the statement of the Łojasiewicz inequality. -Let $f\in C^{\omega}(B_{1},\mathbb{R})$ (real analytic function on the open unit ball centered at the origin) then there are $\alpha\in (0,1)$, $C>0$ and $\varepsilon$ sufficiently small such that $\forall x\in B_{\varepsilon}$ the following holds true: -$$|f(x)-f(0)|^{\alpha}\leq C|\nabla f(x)|.$$ -Thank you in advance. - -REPLY [2 votes]: No, there is no elementary proof of that result even for polynomials of 2 variables. The proof of this result an be reduced to a more elementary one (the case of monomials) by using the theory of sigularities, but this is not at all elementary, cf. the recent works of Paul Feehan.<|endoftext|> -TITLE: Graphs with many edges avoided by Hamiltonian cycles -QUESTION [10 upvotes]: Let $G$ be a $3$-connected Hamiltonian graph with at least one edge that belongs to each H-cycle of $G$. Some authors (e.g. in the link given here) call such an edge an a-edge and an edge that belongs to no H-cycle of $G$ a b-edge. Let $a(G)$ and $b(G)$ denote the number of a-edges and b-edges, respectively. Define $\rho(G)=\dfrac{b(G)}{a(G)} $. - - -Are there good (or even sharp) upper bounds for $\rho(G)$ ? Maybe $\rho(G)<1$? - - -Such graphs can be constructed e.g. using certain non-Hamiltonian graphs like the so-called Tutte fragment $T$ or the “Petersen fragment” $P$, the latter obtained from the Petersen graph by adding a vertex on each of three consecutive edges and an edge leaving at each of those vertices. These edges are labelled by $1,2,3$. - -It is easy to see that $T$ does not admit a H-path between 1 and 3, whereas $P$ admits no H-path starting at 2, only two H-paths between 1 and 3 (one consists of the violet and the red edges, the other one of the violet and the black edges). $T$ has two a-edges (marked in violet) and no b-edges; $P$ has nine a-edges (violet) and three b-edges (light blue). -The simplest $3$-connected Hamiltonian graphs obtained from these are a prism over the big “triangle” of $T$ with $a(G)=3$ and $b(G)=0$ (note that this one is even planar) and likewise a prism over $P$, which has $a(G)=16$ and $b(G)=5$. -By taking a Hamiltonian graph and replacing certain cubic vertices by $T$ or $P$, we can impose restrictions on the H-cycles and obtain graphs with various values of $a(G)$ and $b(G)$. -If we start with the wheel $W_{n}$ and replace one vertex by a $T$ in such a way that the spike becomes an a-edge, the resulting graph 'keeps' only two of the H-cycles of $W_n$ and has $\rho(G)=\dfrac{n-3}{n}$. I wonder if this is best possible. So if generally $\rho(G)<1$ holds, that would be sharp. -What about the maximum if we consider only cubic graphs ? -If we start with the prism over $K_3$ and make one ‘vertical' edge a b-edge by replacing one vertex with a $P$, we get a graph with a unique$^*$ H-cycle and $\rho(G)=\dfrac{5}{13}\approx.3846 $. -If we start with the dodecahedron and 'block' three well chosen edges (i.e. make them into b-edges) by replacing three vertices with $P$'s, we get a graph with a unique$^*$ H-cycle and $\rho(G)=\dfrac{16}{41}\approx.3902 $, slightly larger. -Starting with a truncated icosahedron (soccerball), we can still do better. It depends on how many edges have to be blocked by using $P$'s to remain with a unique$^*$ H-cycle: if there are $k$ such edges and the resulting graph $G$ is unique$^*$ Hamiltonian, it will have $\rho(G)=\dfrac{30+2k}{60+7k} $, which is equal to $\dfrac{16}{41}$ for $k=9$ and gets bigger as $k$ decreases. I've checked that $k=7$, thus $\rho(G)=\dfrac{44}{109}\approx.4037 $, is possible. (Note that in the drawing given in the link, only the green and black 1-factors yield a H-cycle.) -$^*$ Edit: "unique" is meant w.r.t. the H-cycles of the original graph, because inside the instances of $P$ there are of course several ways of making H-paths through them. -Of course I don't claim that this method yields the best possible results. So the question: - - -Are there cubic, 3-connected graphs with $\rho(G)$ even bigger than that? - -REPLY [4 votes]: The first counterexample given by @joro gives rise to the following family $G_k$ of graphs with $\rho(G_k)=k$ : -$G_k$ has $n=4k+3$ vertices numbered $0,1,\dots,n-1$ forming a H-cycle (imagine a regular n-gon embedded such that the vertex $2k+1$ is on the bottom), plus the ‘horizontal’ chords $(2j-1,4k+3-2j)$ for $j=1,...,k$ (call their set $B$) and the other chords $(j,2k+1+j)$ for $j=0,...,2k+1$. It is not hard to see that $(4k+2,0)$ is an a-edge. As the graph without this edge and without $B$ is bipartite, all edges of $B$ must be b-edges. It remains to check that for each other edge, there is a H-cycle that doesn’t contain it. -For the second question, as @Martin Tancer and @nvcleemp have pointed aout, $\rho(G)\le1/2$ for cubic graphs. The following construction comes arbitrarily close to $1/2$ : -For $n=2k$, start with a H-cycle $0,...,2k-1,0$ and add the ‘horizontal’ chords $(j,2k-j)$ for $j=1,...,k-1$ and the ‘vertical’ chord $(0,k)$. Replace the vertex $k$ with an instance of $P$ that blocks the vertical chord. Then it is easy to see (starting at the vertex $0$) that the horizontal chords cannot be part of a H-cycle. So this graph has $a(G)=2k+7$ (the original cycle plus 7 inner edges of $P$) and $b(G)=k+1$ (the horizontal chords plus 2 inner edges of $P$). -So both original questions are answered, thanks to your input. Remains the interesting question raised by joro what can be said about $\rho(G)$ for 4-regular graphs.<|endoftext|> -TITLE: Do syndetic sets on amenable semigroups have positive upper density? -QUESTION [7 upvotes]: Let $\mathbb{G}$ be a discrete amenable semigroup, and $\left\{ F_{n}\right\} $ -a Folner sequence. -For $S\subset \mathbb{G}$ define the upper density as $D^{\ast -}(S)=\limsup_{n\rightarrow \infty }\frac{\left\vert S\cap F_{n}\right\vert }{% -\left\vert F_{n}\right\vert }.$ -Suppose the exists $m\geq 0$ such that for every $g\in \mathbb{G}$, we have -that $gF_{m}\cap S\neq \varnothing .$ -Is $D^{\ast }(S)>0?$ - -REPLY [6 votes]: This is true if $F_n$ is a right Følner sequence, i.e., $\frac{|F_n \Delta F_ng |}{|F_n|} \to 0$ for all $g \in G$. Indeed, if $F \subset G$ is finite, then the condition $g F \cap S \not= \emptyset$, for every $g \in G$, is equivalent to the condition $S F^{-1} = G$. Hence $D^*(S) = \frac{1}{|F|} \sum_{f \in F} D^* (Sf^{-1}) \geq \frac{1}{|F|} D^*(S F^{-1}) = \frac{1}{|F|}$. -This can fail however if we consider only left Følner sequences. A counter-example is given by letting $G$ be the infinite dihedral group $\mathbb Z \rtimes^\alpha (\mathbb Z / 2 \mathbb Z)$, where $\alpha$ implements the flip automorphism on $\mathbb Z$. If we set $S = \mathbb N \cup (-\mathbb N)\alpha \subset G$, and $F = \{ 0, \alpha \}$, then we have $g F \cap S \not= \emptyset$ for all $g \in G$. However, considering $F_n = [-n^2, n] \cup [- n, n^2]\alpha$ we have that $F_n$ is a left Følner sequence such that $D^*(S) = \limsup_{n \to \infty} \frac{2n + 2}{2n^2 + 2n + 2} = 0$.<|endoftext|> -TITLE: Getting unique ergodicity from minimality -QUESTION [10 upvotes]: It is known that minimality does not imply unique ergodicity (Furstenberg example). I ask whether the implication holds in following particular situation: - -Suppose $X$ is a compact space, $f:X \to X$ is a minimal and uniquely ergodic homeomorphism. Suppose $p: Y \to X$ is a 2-to-1 covering map, and that $g:Y \to Y$ is a homeomorphism covering $f$, i.e., $p \circ g = f \circ p$. Suppose $g$ is minimal. Does it follow that $g$ is uniquely ergodic? - -Trivial example: If $f$ is an irrational rotation of the circle then it is easy to see that so is $g$, and so in this case the question has a positive answer. -A less general but already interesting situation is when the covering is trivial, i.e., $Y = X \times \{0,1\}$. In this case, I think that my question is equivalent to the following coboundary rigidity question: - -Let $f$ be as above. Let $\phi:X \to \mathbb{Z}_2$ be a continuous map on the group with two elements. Suppose it is a measurable-coboundary, i.e., there exists a measurable map $\psi:X \to \mathbb{Z}_2$ such that $\phi = \psi \circ f - \psi$ almost everywhere (w.r.t. the unique $f$-invariant probability measure). Does it follow that $\phi$ is a continuous-coboundary? (I.e., can $\psi$ be chosen continuous?) - -Rem.: To relate the two questions, notice that $g(x,t)=(f(x),t+\phi(x))$ is uniquely ergodic (resp., minimal) iff $\phi$ is not a measurable-coboundary (resp., not a continuous coboundary). - -REPLY [4 votes]: To put Anthony's example in a more general context, if you are familiar with Vershik maps on Bratteli diagrams and dimension groups, then you can find lots of minimal uniquely ergodic actions on Cantor sets that come from the $Z_2$-orbit space of a minimal action action with two ergodic probability measures, with an involution that interchanges the measures. -More explicitly, begin with a Bratteli diagram representing a simple dimension group with unique trace (for example, the 2-adic odometer is represented by $\lim \times 2: Z\to Z$); to be interesting, lots of the multiplicities should exceed $1$, in fact, they should be quite large (which can be arranged by telescoping). Then we can replace the integer multiplicities by characters (not irreducible of course) of the group $Z_2$, and having degrees equalling the multiplicities. Pick a Vershik adic map, e.g., using the left/right ordering. -This yields a new Bratteli diagram, together with a natural action of $Z_2$, and such that the $Z_2$-orbit space is precisely the original action. Except for degenerate cases, this yields a minimal system with a $Z_2$-action, and necessarily a two to one map. Generically, the new minimal system is uniquely ergodic. However, we can easily construct examples where there are two ergodic measures (corresponding to two pure traces on the dimension group). -For example, if we telescope the 2-adic odometer thingy above, so that at the $n$th level, there are $2^n$ edges to the next point, we can realize a character of $Z_2$ whose dimension is $2^n$, corresponding to the matrix -$$ -A_n:= \left( \begin{array}{cc} 2^n-1 & 1 \\ 1 & 2^n-1 \end{array} \right). -$$ This dimension group, $\lim A_n: Z^2 \to Z^2$ is simple and has two pure traces, and there are lots and lots of similar examples. This particular one can be made to sit over the 2-adic odometer by choosing the ordering for the Vershik map appropriately. -This forms part of the classification of actions of $Z_2$ (and other groups) on AF C*-algebras which leave the algebraic direct limit stable. There are a few papers in this area, but I've gone on long enough.<|endoftext|> -TITLE: Do varieties with ample canonical bundle have finite automorphism group in small characteristic? -QUESTION [12 upvotes]: Suppose $X$ is a smooth projective variety over a field $k$, with ample canonical bundle. If $\operatorname{char}(k)=0$ or $\operatorname{char}(k)>\dim(X)$ and $X$ lifts to $W_2(k)$ (thanks Olivier Benoist!), it's not hard to see that $X$ has finite automorphism group. -Proof. Let $X\hookrightarrow \mathbb{P}\Gamma(X, \omega_X^{\otimes n})^\vee$ be a closed embedding. Then the induced map $\operatorname{Aut}(X)\hookrightarrow PGL(\Gamma(X, \omega_X^{\otimes n})^\vee)$ is a closed embedding, so $\operatorname{Aut}(X)$ is finite type. But the tangent space to $\operatorname{Aut}(X)$ at the identity is $\Gamma(X, T_X)$, which is trivial by Kodaira vanishing. So $\operatorname{Aut}(X)$ is a $0$-dimensional finite type group scheme, hence finite. $\blacksquare$ -(An essentially identical argument avoiding the representability of $\operatorname{Aut}(X)$ may be found here.) -This proof breaks down for $\operatorname{char}(k)$ small: $\Gamma(X, T_X)$ need not be trivial. But I don't know of an example of a canonically polarized variety with infinite (thus necessarily positive-dimensional) automorphism group in small characteristic. - -Is the automorphism group of a smooth variety with ample canonical bundle always finite, even if $\operatorname{char}(k)$ is small? - -REPLY [16 votes]: A much more general result is proven in Martin-Deschamps, Lewin-Menegaux : Applications rationnelles séparables dominantes sur une variété de type général. Théorème 2 : if $X$ and $Y$ are smooth and proper, and if $Y$ is of general type, the set of dominant separable rational maps from $X$ to $Y$ is finite ! - What you want follows immediately: it is their Corollaire 4, and they attribute it to Matsumura.<|endoftext|> -TITLE: The fibre product of two quotient stacks -QUESTION [6 upvotes]: My question is to know whether the fibre product of $[X/G]$ by $[Y/H]$ over a base scheme is $S$ is $[X \times_S Y/G \times H]$? And if yes, do you have any reference for it? -Thank you. - -REPLY [8 votes]: This is my attempt at fleshing out the details using the universal property as Scott Carnahan suggested. I think this is correct but I'm not totally sure. Let me know if theres anything off. -Let $\mathscr{X} = [X/G]$, $\mathscr{Y} = [Y/H]$ and $\mathscr{P} = [X \times_S Y/G \times H]$. $\mathscr{X}$ is described by the following pseudofunctor: -$$ -\require{AMScd} -\mathscr{X}(U) = \left\{\begin{CD} P @>>> X \\ @VVV \\ U\end{CD}\right\} -$$ -where $P \to U$ is a $G$-bundle, $P \to X$ is $G$-equivariant, and everything is compatible with the morphisms to $S$. The analogous is true for $\mathscr{Y}$ and $\mathscr{P}$ respectively. -The fiber product $\mathscr{X} \times_S\mathscr{Y}$ represents maps to $\mathscr{X}$ and $\mathscr{Y}$ which are equal with the compositions $S$, that is, the pseudofunctor of pairs -$$ -\left\{\begin{CD} P @>{g}>> X \\ @VVV \\ U\end{CD}, \begin{CD} Q @>{h}>> Y \\ @VVV \\ U\end{CD}\right\} -$$ -such that $g$ is $G$-equivariant, $h$ is $H$-equivariant and everything is compatible over $S$. By the universal property of fiber products, this pair of diagrams is naturally equivalent to the diagram -$$ -\begin{CD} P \times_U Q @>{(g,h)}>> X \times_S Y\\ @VVV \\ U \end{CD}. -$$ -$P \times_U Q \to U$ is a principle $G \times H$ bundle by construction and the map $(g,h)$ is a $G \times H$-equivariant morphism by construction. Therefore, the diagram is an element of $\mathscr{P}(U)$. This defines a fully faithful natural transformation of pseudofunctors $\mathscr{X} \times_S \mathscr{Y} \to \mathscr{P}$. -Now, it remains to show that every object in $\mathscr{P}(U)$ naturally comes from such a pair. That is, we must show that every $G \times H$ principle bundle on $U$ with a $G \times H$ equivariant morphism to $X \times_S Y$ is the product of a $G$-bundle and an $H$-bundle with equivariant morphisms to $X$ and $Y$. -So suppose we have a diagram -$$ -\begin{CD} E @>>> X \times_S Y\\ @VVV \\ U \end{CD} (*) -$$ -in $\mathscr{P}(U)$. $G$ and $H$ are naturally viewed as normal subgroups of $G \times H$. Then from $E$ we get a $G$-principal bundle $E/H$ and an $H$-principal bundle $E/G$. The $G \times H$ equivariant morphism $E \to X \times_S Y$ induces a $G$ equivariant morphism $E/H \to X$ and an $H$ equivariant morphism $E/G \to Y$. These are all compatible with the morphisms to $S$ from naturality of fiber products. Therefore, we get a pair of elements -$$ -\left\{\begin{CD} E/H @>>> X \\ @VVV \\ U\end{CD}, \begin{CD} E/G @>>> Y \\ @VVV \\ U\end{CD}\right\} -$$ -in $\mathscr{X} \times_S \mathscr{Y}(U)$. -There is a natural morphism of principal bundles $E \to E/H \times_U E/G$ over $U$ compatible with the maps to $X$ and $Y$ by the universal property of fiber products. Since any morphism of principal bundles is an isomorphism (equivalently since $\mathscr{P}$ is a stack so it is fibered in groupoids) then the diagram $(*)$ is naturally isomorphic to -$$ -\begin{CD} E/H \times_U E/G @>>> X \times_S Y\\ @VVV \\ U \end{CD}. -$$ -This gives a natural equivalence of categories between $\mathscr{X} \times_S \mathscr{Y}(U)$ and $\mathscr{P}(U)$. Therefore, we get an isomorphism of stacks $\mathscr{X} \times_S \mathscr{Y} \cong \mathscr{P}$ over $S$.<|endoftext|> -TITLE: Formalizations of category theory in proof assistants -QUESTION [40 upvotes]: What are the existing formalizations of category theory in proof assistants? - -I'm primarily interested in public-domain code implementing category theory in a proof assistant (Coq, Agda, Isabelle/HOL, Mizar, NuPRL, Twelf, Lego, Idris, Matita, etc.), though I'm also interested in papers about formalizations of category theory in proof assistants. -I've added answers to this question for all of the papers and formalizations that I know about, and details about the constructions in my own repository as of the date of adding. In addition to adding formalizations that you don't see on here, you should feel free to add details and improve the formatting of the other entries (especially including what language the formalization is in, what category theory it covers, links to papers presenting it and/or publicly available source code, whether or not it's under active development, what the newest version of the proof assistant it compiles with is, etc.). - -REPLY [5 votes]: Wolfram Kahl's RATH-Agda formalisation :: http://relmics.mcmaster.ca/RATH-Agda/ -" -The basic category and allegory theory library of the RATH-Agda project, containing (only sporadically truly) literate theories ranging from semigroupoids, which are “categories without identities”, to “action lattice categories”, which are division allegories that are at the same time Kleene categories (i.e., typed Kleene algebras), including also monoidal categories. -These theories are intended as interfaces for high-level programming; this current collection includes implementations in particular using concrete relations, and a number of constructions, including quotients by (abstractions of) partial equivalence relations. -"<|endoftext|> -TITLE: Number of disjoint simple closed geodesics -QUESTION [10 upvotes]: According to Jairo comment on the first version of this question I revise the question as follows; -Let $g$ be a real analytic Riemannan metric on $S^{2}$. Is it true to say that: -There are at most a finite number of disjoint simple closed geodesics on $S^{2}$. -If the answer is yes put $m$= the sup of the number of such disjoint closed geodesics. -What is a geometric interpretation for this geometric invariant $m$? -For a given $n\in \mathbb{N}$, is there a real analytic Riemannian metric on $S^{2} $ for which $m=n$ - -REPLY [4 votes]: A simpler example seems to be an "accordion surface" (take a sinusoid $y = \sin x,$ rotate around the line $y= 3$) It will have as many parallel simple geodesics as you like.<|endoftext|> -TITLE: Is the ideal membership problem solvable for differential ideals? Is there a good notion of a Gröbner basis? -QUESTION [5 upvotes]: Let $K$ be a field of characteristic zero. Let $\Omega = K[x_1, \dots, x_n, dx_1, \dots, dx_n]$ be the differential ring of algebraic differential forms over $K[X_1, \dots, X_n]$. -Is there an algorithm (e.g. by Gröbner basis-like techniques) that solves the ideal membership problem for $\Omega$? That is: given a finitely generated differential ideal $I \subseteq \Omega$, is it decidable whether $f \in I$ or not for a given $f \in \Omega$? -Is calculating the differential radical of $\Omega$ or eliminating variables like in ordinary commutative algebra with Gröbner bases equally possible? -(The reason why I am interested in these things is, of course, algebraic handling of partial differential equations in exterior form.) - -REPLY [5 votes]: Yes. There is such an algorithm. This ring is a finitely generated free module over a polynomial ring, and it is sufficient to solve the submodule membership problem for these modules. But this is easy - the Grobner basis idea works perfectly. Just order the monomials times generators and find a basis for the initial submodule.<|endoftext|> -TITLE: Does anyone know what is the right reference for the following simple lemma from harmonic analysis? -QUESTION [7 upvotes]: The lemma says that given $\lambda\geq 1$, $p\geq 1$, $a_j\geq 0$, for a collection of balls $\{B_j\}_{j\in\mathbb{N}}$ in $\mathbb{R}^n$, it holds -$$\bigg\|\sum_j a_j\chi_{\lambda B_j}\bigg\|_p\leq C(n,p,\lambda)\bigg\|\sum_j a_j\chi_{B_j}\bigg\|_p.$$ -I saw in some papers that people call this Bojarski's lemma since it appeared in the paper of Bojarski: - Bojarski, B. Remarks on Sobolev imbedding inequalities. Complex analysis, Joensuu 1987, 52–68, Lecture Notes in Math., 1351, Springer, Berlin, 1988. -However, I was informed by people that the above paper is in fact not the first reference on this lemma, at least another mathematcian proved this lemma in an unpublished notes (noticed by my supervisor as well). -As far as I know, in mathematics, we give name to some lemmas to express our respect on the mathematician who proved the corresponding results. But usually for young mathematicians, we are not aware of all the results we cited, in particular if some one add a name of some results and we just follow their name without going to the first reference for the result, it is easy to give a wrong title for some results. This sometimes causes servious problems for some mathematicians since they think the result should "belong to them". -The proof of the lemma was based on a maximal function argument and I do not know whether there are more elementary proofs than the one appeared in Bojarski's paper. If there were, then I would expect that there will be earlier references for this result. Then I will correct the name of this lemma in my paper. - -REPLY [9 votes]: This inequality is also a corollary of the main result of -Fefferman, Charles; Stein, Elias M., Some maximal inequalities, Am. J. Math. 93, 107-115 (1971). ZBL0222.26019. -which asserts that -$$ \| \sum_j |f_j^*|^r)^{1/r} \|_{L^q({\bf R}^n)} \leq C(n,q,r) \| (\sum_j |f_j|^r)^{1/r} \|_{L^q({\bf R}^n)}$$ -whenever $1 < r,q < \infty$, where $f_j^*$ is the Hardy-Littlewood maximal function of $f_j$. Indeed, one takes an arbitrary $1 -TITLE: Recreational mathematics: where to search? -QUESTION [29 upvotes]: I am not sure I can strictly define recreational mathematics. But we all feel what it is about: puzzles, problems you can ask your mathematical friends, problems that will bother them for a couple of hours, and then they will get them (or not?). Problems to distract yourself from research. Sometimes, however, these problems lead to some ideas and concepts connected to "serious mathematics". I adore this kind of problems and, with holidays approaching, I let myself surf on Internet for some new problems to distract myself a little near the Christmas tree. I usually look at some blogs but my search is very chaotic. I'll try to formulate more precisely what kind of problems I'm searching for: there are two kinds of them. -First, good mathematical problems. So good that you can talk about the solution for an hour although the formulation could be really easy. -Example: the old Arnold's question about the perimeter of a banknote. By folding up a banknote, we decrease the area of the polytope obtained. But can we increase the perimeter? The answer is yes, and we can increase it as much as we want. The proof is beautiful and doesn't need any knowledge of higher mathematics although it is not at all trivial. -Second, problems giving some publicity for higher mathematics. I will give an example - hat puzzle. -A sultan decides to give a test to his sages (a countable number of sages actually!). He has the sages stand in a line, one behind the other, so that the person in a line sees everybody before himself. Yes, the sages are clever but also they have a very good vision. He puts the hats on them: white or black. Then, the sages cry (all at one time) the color that they think they are wearing. Everybody who is wrong will be killed. The question is, can the sages achieve the result that only the finite number of them will be killed? -I won't spoil you the pleasure giving the answer but this puzzle in some sort opens a path to some serious mathematics and can be a good pretext to explain it on the seminar for high-school student (or even to undergraduate). -I search for problems that are easy to formulate and not trivial to solve, and that could give a nice pretext to talk about them for a couple of hours for undegraduates. In other words, I search for problems that could make a good advertisement of "serious" mathematics for high-school students that love puzzles. -My question is -- are there any journals (I think, Mathematical Intelligencer can be one of the possible answers..) that publish some kind of research articles on the subject? Maybe some blogs I do not know? -Any links or suggestions will be welcomed. - -REPLY [3 votes]: Miklós Schweitzer is a contest about research level problems, look some problems. Actually there a book about it: Contests in Higher Mathematics.<|endoftext|> -TITLE: How to find faces of polytope defined by a Weyl orbit -QUESTION [5 upvotes]: A few days ago I asked the following question at MSE and received no answer. I thought I would try here. -Let $\xi$ be an integral dominant weight of an irreducible root system $\Delta$, and let $\mathcal{O}_{\xi}$ be its orbit under the action of the Weyl group. The elements of the orbit are the vertices of an n-dimensional convex polytope. -Is there an efficient way to pick out all the subsets of points in $\mathcal{O}_{\xi}$ forming the faces of the convex polytope? I would like to have an algorithm that takes a Weyl orbit of an integral dominant weight as input and gives as output a list of subsets of vertices forming the faces of the polytope. By efficient, I mean that I would like something better than a brute force algorithm. -Let $G$ be the compact connected Lie group corresponding to the root system $\Delta$, and let $T\subset G$ be a maximal torus. We could also view the orbit $\mathcal{O}_{\xi}$ as the moment map image for a Hamiltonian group action of $T$ on $G/T$. In this case, the faces of the moment map image can be viewed as the image of points fixed under a one parameter subgroup of $T$. -Is there a way to understand which 1-parameter subgroups have fixed point sets that get mapped to the faces of the moment map image? - -REPLY [8 votes]: The faces are all of the following form: $w W_P / Stab_W(\xi)$, where $W_P$ varies over the subgroups generated by subsets of the simple reflections. In particular, for $\xi$ regular, the number of them is $\sum_P |W / W_P|$. (Note that $\mathcal O_\xi$ is only $G/T$ when $\xi$ is regular; otherwise it's $G/Stab_G(\xi)$.) -I wrote up a Hamiltonian-geometry-based proof of this, then realized the combinatorics is pretty easy too, assuming $\xi$ is regular as it seems you're doing. -The set of faces ("external" faces, to us Hamiltonians) is obviously $W$-invariant. So it's enough to find the ones nearby the basepoint, then permute them. -The polytope is "simple" -- it has the right number of edges coming out of each vertex, namely the rank of $G$. So the number of faces at the basepoint is $2^{rank(G)}$. Each $W_P$ gives one, so that's all of them. QED. -Here are some related computations in the Hamiltonian formalism. If $S \leq T$ is a connected subgroup, then $ (G/T)^S = C_G(S) N(T)/ T $. Proof: $\subseteq$ is easy. For the reverse, let $s$ be a topological generator of $S$. If $s gT = gT$, then $g^{-1} s g \in T$, so $\exists n \in N(T)$ s.t. $n g^{-1} s g n^{-1} = s$, so $g n^{-1} \in C_G(s) = C_G(S)$. -We're interested in the set $\coprod_S C_G(S) N(T)/T$ of all components, over all $S$, so we get the same set if we look at $\coprod_S N(T) C_G(S)/T$. Since the $T$-moment map is $N(T)$-equivariant, we just need to figure out which components of $\coprod_S C_G(S)/T$, i.e. which $S$, give external walls of the moment polytope. -Every Levi subgroup of $G$ (w.r.t. a fixed Weyl chamber) arises as a $C_G(S)$, where $S$ is a dominant coweight. Each of those gives a wall around the basepoint of the Weyl polyhedron, and there are the same number of each: $2^{rank}$. So all the walls near the basepoint are images of $Levi/T$, i.e. with fixed points $W_P$. - -REPLY [6 votes]: This is not a direct answer to your questions (which I still haven't understood completely from your formulation). But it seems important to place these questions within the extensive theoretical background, since I'm unconvinced that computing large examples (by brute force or otherwise) will provide much insight. -A convenient recent source is a short announcement by A. Khare here. He and various collaborators have posted on arXiv a number of related papers on the faces of weight polytopes. For example, a paper with Ridenour was published in 2012 in Algebras and Representation Theory; the preprint is here. -Some of this work refers back to work of Vinberg and others. Since the combinatorics of weight polytopes gets formidable even when the Weyl group is a symmetric group (presumably the best-behaved family of examples), it's a good idea to explore this literature and also to refine your own questions as far as possible.<|endoftext|> -TITLE: What is the motivation for defining the conductor of an abelian variety? -QUESTION [9 upvotes]: Let $K$ be a $p$-adic field, and let $A$ be an abelian variety over $K$. The conductor of the abelian variety is often defined as $2u+t+\delta$, where $u$, $t$ and $\delta$ are invariants related to the special fiber of the neron model of $A$ over the ring of integers of $K$. (See for example the wikipedia page: https://en.wikipedia.org/wiki/Conductor_of_an_abelian_variety). While I have seen this definition used in several texts, it has never been made clear to me where this definition came from, and why it is helpful. -What is the motivation for this odd definition? In what sense is it related to the conductor in number theory? In particular, is the following assertion correct? -Guess pertaining to motivation -Let $T_l$ be the Tate module of $A$, and let $L$ be the unique minimal field over $K$ such that $Gal(L)$ acts trivially on $T_l$. Let $G=Gal(L/K)$, and let the $G_i$'s be the lower numbering of the ramification of $L/K$. Then is the conductor of $A$ defined above the same as $\sum_{i=0}^{\infty} \frac{|G_i|}{|G|}dim(T_l/(T_l^{G_i}))$? If so, where is this proven? - -REPLY [6 votes]: The reference to Serre is good, but for a somewhat more elementary introduction (for elliptic curves), you could look at Chapter IV Section 10 of my book Advanced Topics in the Arithmetic of Elliptic Curves. It includes an explanation of why the tame part of the conductor can be read off from the reduction type (good, multiplicative, additive) of the elliptic curve, and why the wild part of the conductor is 0 for primes $p\ge5$. As noted in one of the comments, the wild part of the conductor is defined in terms of the action of inertia (and the higher inertia groups) on the $\ell$-torsion $E[\ell]$, so one can use the definition that you gave with $T_\ell(E)$ replaced by the finite Galois module $E[\ell]$. There's also a proof that the exponent of the conductor $f(E/K)$ over a local field $K$ with normalized valuation $v_K$ satisfies -$$ f(E/K)\le 2+3v_K(3)+8v_K(2). $$ -(This was generalized to abelian varieties by Lockhart, Rosen, and me [1], and then Brumer and Kramer [2] gave the best possible upper bounds.) -Again for elliptic curves, there is a also the beautiful formula of Ogg [3] and Saito [4] relating the minimal discriminant $\text{Disc}_{E/K}$, the exponent of the conductor $f(E/K)$, and the number of components on the special fiber of the Neron model $m(E/K)$: -$$ - v_K(\text{Disc}_{E/K}) = f(E/K) + m(E/K) - 1. -$$ -Ogg proved every case except char($K)=0$ and residue characteristic 2, while Saito gave a more conceptual proof that covers all cases -[1] Lockhart, P., Rosen, M. and Silverman, J., An upper bound for the -conductor of an abelian variety J. Alg. Geo. 2 (1993), 569-601. -[2] Brumer, A. and Kramer, K., The conductor of an abelian variety, -Compositio Math. (1994). -[3] Ogg, A., Elliptic curves and wild ramification, Amer. J. Math. -89 (1967), 1-21. -[4] Saito, T., Conductor, discriminant, and the Noether formula of -arithmetic surfaces, Duke Math. J. 57 (1988), 151-173.<|endoftext|> -TITLE: Generalization of Darboux's Theorem -QUESTION [15 upvotes]: Darboux's Theorem. If $f:[a,b]\to\mathbb R$ is differentiable and $f'(a)<\xi -TITLE: which varieties can appear as exceptional divisors? -QUESTION [6 upvotes]: Let $\pi$ be the blow up of $\mathbb{A}^n$ at an ideal -$I_{p}$ which is supported at a close point $p$; let $E$ be the associated exceptional divisor. What are the possible varieties $Y$ that can appear as $E \cong Y$? -$\textbf{Partial Answers:}$ -(1) If $I_p = (x_0, \ldots x_n)$, then $E \cong \mathbb{P}^n$. -In this case $\pi$ is the blow up of $\mathbb{A}^n$ at a smoothing point. -(2) If $I_p = \left( x^{d/w_0}, \ldots, x_n^{d/w_n} \right)$ where $d=lcm(w_0, \ldots, w_n)$. Then $E \cong \mathbb{P}(w_0, \ldots w_n)$. In this case $\pi$ is the weighted blow up with respect the weights $(w_0, \ldots w_n)$. -(3) Kawakita article [K] proves that a divisorial contraction in dimension $n=3$ which contracts its exceptional divisor to a smooth point is a suitable weighted blow-up. -I am wondering if there are similar picture for other varieties? ( The toric case is clarified below by Lev Borisov's answer). I am particularly interested in an explicit description! This means: To obtain the variety $E \cong Y$ we must blow up the ideal $I(Y)_{p}$ which generators are... -Particularly, is it possible to obtain a grassmannian this way? -Thanks! -[K] Divisorial contractions in dimension three which contract divisors to smooth points,Kawakita, Masayuki, Inventiones mathematicae, 145, 1, 105--119,2001 - -REPLY [4 votes]: Assuming that the blowup in question is toric, and exceptional divisor has one component, you are not going to get anything beyond these examples. After all, it would mean that you have a subdivision $\Sigma$ of a unimodular cone $C$ that gives $\mathbb A^n$. If the exceptional locus is an irreducible divisor, it means that there is exactly one new ray in $\Sigma$. Then it is just a matter of looking where this ray is, and you get your weighted projective space picture. -If you don't assume that the divisor has only one component, then you can probably get any projective toric variety as a component of the exceptional locus, as a subdivision -of $C$ can be as messy as you want it to be (and it would still correspond to an ideal under projectivity assumptions).<|endoftext|> -TITLE: Direct sum of Hopf algebras -QUESTION [13 upvotes]: I realise that this question might be rather basic but however I was unable to find the answer in any textbook nor manage to figure out the answer. The question is the following: given two Hopf algebras $H_1,H_2$ is there a canonical way to turn the direct sum $H_1 \oplus H_2$ into a Hopf algebra? I have a problem already at the level of bialgebras: the counit is always a algebra morphism (linear and multiplicative) but the set of all linear multiplicative maps $\omega: H_1 \oplus H_2 \to \mathbb{C}$ is a sum of such sets corresponding to $H_1$ nad $H_2$. So I don't see how to define (in a canonical way) the counit map $\varepsilon_{H_1 \oplus H_2}$. - -REPLY [5 votes]: This is a positive version of the answer "no". -As already mentioned before, the disjoint union of groups is not a group, BUT it is a grupoid! -In a similar way, the direct sum of two Hopf algebras is not a Hopf algebra, but a weak Hopf algebra. There is no problem for multiplication and comultiplication, the problem is, as you saw, the counit, but for weak Hopf algebras is the counit axiom that is precisely different. -You can check in -https://arxiv.org/pdf/2008.00606.pdf -UNIVERSAL QUANTUM SEMIGROUPOIDS, HONGDI HUANG, CHELSEA WALTON, ELIZABETH WICKS, AND ROBERT WON -the definition of weak Hopf (Definition 2.1), Lemma 2.24 is precisely your question.<|endoftext|> -TITLE: Hausdorff gaps and $\mathfrak{p}=\mathfrak{t}$ -QUESTION [13 upvotes]: Recently Malliaris and Shelah proved that $\mathfrak{p}=\mathfrak{t}$ -(see: http://math.uchicago.edu/~mem/Malliaris-Shelah-CST.pdf). Their results are far more general, however for the specify context of this question what they showed is that if $G$ is an ultrafilter on $\omega$, then -the ultra product $\omega^\omega/G$ is a linear order with no -$(\kappa,\lambda)$-gaps with $\kappa,\lambda<\mathfrak{t}$. -They can obtain the desired equality combining it with a result of Shelah on the existence of a certain type of small gap -(i.e. with both sequences in the gap of size smaller than $\mathfrak{t}$) -in structures of the form $\omega^\omega/G$ -in the case $\mathfrak{p}<\mathfrak{t}$. -Recall that - -$\mathfrak{t}$ is the least size of a decreasing chain -in $P(\omega)/Fin$ without a positive lower bound in the boolean algebra -$P(\omega)/Fin$. - -$(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$ is a $\kappa,\lambda$ pregap in a partial order $(X,<^*)$ if $f_\alpha<^*f_\beta<^*g_\eta<^*g_\gamma$ for all -$\alpha<\beta<\kappa$ and $\gamma<\eta<\lambda$. - -A pregap $(f_\alpha:\alpha<\kappa,g_\beta:\beta<\lambda)$ on $(X,<^*)$ -is a gap if for no $f\in X$ we have that $f_\alpha<^*f<^*g_\gamma$ -for all -$\alpha<\kappa$ and $\gamma<\lambda$. - -A gap $(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$ on the partial -order $(\omega^\omega,<^*)$ where $<^*$ is eventual domination is an -Hausdorff gap if for all $\beta<\omega_1$ and $n<\omega$ -$\{\alpha<\beta:\forall m>n \: f_\alpha(m)\omega_1$, -$(f_\alpha:\alpha<\omega_1,g_\beta:\beta<\omega_1)$ is an Hausdorff gap. -Let $X\subset P(\omega)$ be the set of $A$ infinite subsets of $\omega$ -such that -$(f_\alpha\restriction A:\alpha<\omega_1,g_\beta\restriction A:\beta<\omega_1)$ -is no longer a gap in the partial order -$(\omega^A,<^*)$. -Then $X$ is open dense in the Ellentuck topology. - -REPLY [5 votes]: Peter Nyikos and Jerry Vaughan in this paper prove this result, although they don't quite state it that way, and they work in the partial order $(P(\omega),\subseteq^*)$ rather than $(\omega^\omega,<^*)$. -For this partial order, a gap is defined in nearly exactly the same way: just replace the symbol $<^*$ with the symbol $\subseteq^*$ in the definition you gave. Given a gap $(F_\alpha,G_\alpha: \alpha < \omega_1)$, they say that a set $E$ is beside the gap if $(F_\alpha \cap E,G_\alpha \cap E: \alpha < \omega_1)$ is no longer a gap in $(P(E),\subseteq^*)$. The analogy with the situation in your question is (I hope) obvious. They say that a gap is tight if there is no set $E$ beside the gap. They prove (theorem 1.2 of their paper) that -Theorem: There is a tight $(\omega_1,\omega_1)$-gap if and only if $\mathfrak{t} = \aleph_1$. -We'll show in a minute (after proving a lemma below) that this actually implies your statement regarding the Ellentuck topology. -This shows that your proposition (or, at least, a version of it for a different partial order) does depend on the assumption $\mathfrak{t} > \aleph_1$. -As for the proposition in the form you gave it, I don't know whether it had a proof prior to Malliaris and Shelah's work. But it does have a fairly simple proof that doesn't depend on their work. I'll sketch that out now. -Recall that the Ellentuck topology has a basis of sets of the form -$$[s,A] = \{s \cup B : B \in [A]^\omega\},$$ -where $s \in [\omega]^{<\omega}$ and $A \in [\omega]^{\omega}$. -Lemma: Suppose $X$ is a subset of $2^\omega$ such that -$\qquad (1)$ if $A \in X$ and $B \subseteq^* A$, then $B \in X$. -$\qquad (2)$ if $B \in [\omega]^{\omega}$, then there is some infinite $A \in X$ such that $A \subseteq B$. -Then $X$ is open dense in the Ellentuck topology. -(To paraphrase: If you're open dense in the partial order $([\omega]^{\omega},\subseteq^*)$, then you're open dense in the Ellentuck topology.) -Proof: -To see that $X$ is open: Suppose $A \in X$. By $(1)$, $[\emptyset,A] \subseteq X$, so $A$ is in the interior of $X$. -To see that $X$ is dense: Let $[s,A]$ be a basic open neighborhood in the Ellentuck topology. By $(2)$, there is some infinite $B \in X$ with $B \subseteq A$. By $(1)$, $s \cup B \in X$. But $s \cup B \in [s,A]$, so $X \cap [s,A] \neq \emptyset$. -QED -Note: this lemma shows that (in the terminology of Nyikos and Vaughan given above) if $\mathfrak{t} > \aleph_1$ and we have a gap $(F_\alpha,G_\alpha:\alpha<\omega_1)$, then -$$\{E \in [\omega]^{\omega} : E \text{ is beside the gap}\}$$ -is open dense in the Ellentuck topology. Condition $(1)$ is obvious. Condition $(2)$ follows from the aforementioned theorem of Nyikos and Vaughan, since otherwise their theorem I quoted would be false for the gap $(F_\alpha \cap B, G_\alpha \cap B: \alpha < \omega_1)$ (which is a gap in the partial order $(P(B),\subseteq^*)$). -Now let's check that the set $X$ mentioned in your question satisfies the conditions of this lemma. -Condition $(1)$ is fairly obvious. -For condition $(2)$, fix $B \in [\omega]^{\omega}$. For each $\alpha < \omega_1$, let -$$C_\alpha = \{(m,n) : m \in B \text{ and } f_\alpha(m) < n < g_\alpha(m)\}.$$ -Using the definition of a pregap, it's fairly easy to see that $\langle C_\alpha : \alpha < \omega_1 \rangle$ is a decreasing chain in $P(\omega \times \omega)/fin$. Since $\mathfrak{t} > \aleph_1$, there is some infinite $C \subseteq \omega \times \omega$ such that $C \subseteq^* C_\alpha$ for every $\alpha$. Let -$$A = \{m : (m,n) \in C \text{ for some } n\},$$ -$$h = \{(m,n) : m \in A \text{ and } n = \min \{k : (m,k) \in C\}\}.$$ -Clearly $h$ is a function from $A$ to $\omega$, and, for every $\alpha$, $f_\alpha \upharpoonright A <^* h <^* g_\alpha \upharpoonright A$. We know that $A$ must be infinite, because $C$ is infinite and, for each $m \in A$, $C \cap \{m\} \times \omega$ is finite. -Notice that this proof doesn't depend on the gap being Hausdorff, so you can eliminate that assumption from the statement of the proposition.<|endoftext|> -TITLE: $Z_{2}$- graded structures for $C_{red} ^{*} (F_{2})$ -QUESTION [5 upvotes]: Let $F_{2}$ be the free group with two generators. -Then $F_{2}=\{\text{odd words}\}\sqcup\{\text{even words}\}$. This gives us a $Z_{2}$ graded structure for $C^{*}_{red} (F_{2})$, in a natural way. -My question: - -Is there another $Z_{2}$ graded structure for $C^{*}_{red} (F_{2})$ which is not graded isomorphic to the above $Z_{2}$- graded structure? - -Please consider the same question for $C^{*}_{red} (F_{1}) \sim C(\mathbb{S}^{1})$. -Namely: is there a $Z_{2}$ graded structure for $C(\mathbb{S}^{1})$ which is not graded isomorphism to the standard grading structure(decomposition to even and odd continuous functions)? - -REPLY [6 votes]: It is well-known that $K_1(C^*_{\rm red}(F_2))={\mathbb Z^2}$ with generators given by $[u]$ and $[v]$, where $F_2=\langle u,v\rangle$. -Now, the automorphism of order two associated with the even-odd grading is multiplying each generator by $-1$, which is homotopic to the multiplication by $1$. Hence it is trivial on $K_1$. -On the other side, the automorphism of order two that comes mapping $u$ to $v$ and $v$ to $u$ gives rise to a non-trivial action on $K_1$. Thus, the two automorphisms of order two cannot be conjugate - and hence the associated graded algebras cannot be isomorphic.<|endoftext|> -TITLE: Are multicolimits suitable colimits? -QUESTION [5 upvotes]: Today I encountered the notion of multicolimit. -Lacking a standard reference for this notion, let me give a self-contained definition of this gadget. -If $S\colon \cal K\to E$ is a diagram, we define its multicolimit as a small set of cocones $\{S\xrightarrow{\varphi_i} L_{i,S}\}_{i\in I}$ such that for any other cone $S \xrightarrow{\delta} \Delta_X $ in $\cal E$ ($\Delta_X$ the constant functor on $X$) there exists a unique index $i(\delta)\in I$ such that $\delta$ factors uniquely through $\varphi_{i(\delta)}$. This explicit definition can be summarized asking that the functor $\cal E\to Set$ which sends any object $E\in\cal E$ to the set of cocones for $S$ with summit $E$ is isomorphic to a small coproduct of representables: -$$ -[{\cal K,E}](S,\Delta_E)\cong \coprod_{i\in I}{\cal E}(L_{i,S},E) -$$ -I am interested in understanding if this notion in a genuine generalization of the notion of colimit for $S$, and in establishing some formal properties for such an object. In particular: - -Can the notion of multicolimit for $S$ be reduced to a suitable (weighted? weak?) colimit? My sensation is that this notion is utterly different, but ... -...I'm wondering if the notation $\{\varinjlim\!{}^i S\}_i$, chosen as a pure portmanteau, meaningful to denote the multicolimit of $S$ (provided that we are aware that each $\varinjlim\!{}^i S$ is a colimit only on a suitable (possibly empty) restriction of $S$)? -Assume that $S\colon \cal C\times D\to E$ is a functor, and that each $\varinjlim\!{}^i_{\cal D} S(c,-)$, $\varinjlim\!{}^j_{\cal C} S(-,d)$, $\varinjlim\!{}^{(i,j)}_{\cal C\times D} S$ exists. Do multicolimits commute with multicolimits? In other words, is it true that -$$ -\varinjlim\!{}^i_{\cal D}\varinjlim\!{}^j_{\cal C} S(c,d)\cong \varinjlim\!{}^j_{\cal C}\varinjlim\!{}^i_{\cal D} S(c,d)\cong \varinjlim\!{}^{(i,j)}_{\cal C\times D} S(c,d) -$$ -Do left adjoints preserve multicolimits? - -REPLY [8 votes]: Yes, but perhaps in an other sense than you may think. -I have nothing to say :-) -Yes. -Yes. -$\newcommand{\mor}[3]{#1 \colon #2 \rightarrow #3}% -\newcommand{\catl}[1]{\mathbb{#1}}% -\newcommand{\catw}[1] {\mathbf{#1}}$ - - -Here is my elaboration. -Let me first omit some irrelevant details. We shall say that a functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left multiadjoint if for every $X \in \catl{D}$, the hom-functor $\hom(X, F(-))$ is a (small) coproduct of representables: -$$\hom(X, F(-)) \approx \coprod_i\hom(G_i(X), -)$$ -At this point I am not sure yet if I fully understand the above definition --- it makes me wonder if there is anything so special about coproducts: i.e. what if we substituted copoducts with other classes of colimits? Let me try: - -A functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left adjoint if for every $X \in \catl{D}$, the hom-functor $\hom(X, F(-))$ is representable: -$$\hom(X, F(-)) \approx \hom(G(X), -)$$ -I have nothing to add here. -A functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left nothingadjoint if for every $X \in \catl{D}$ the hom-functor $\hom(X, F(-))$, is a (small) colimit of representables: -$$\hom(X, F(-)) \approx \mathit{colim}_i\hom(G_i(X), -)$$ -Nothingadjointness is not an interesting concept, because by Yoneda every $\catw{Set}$-valued functor is a colimit of representables. -A functor $\mor{F}{\catl{C}}{\catl{D}}$ has a finitary approximation to a left adjoint if for every $X \in \catl{D}$, the hom-functor $\hom(X, F(-))$ is a filtered colimit of representables: -$$\hom(X, F(-)) \approx \mathit{fcolim}_i\hom(G_i(X), -)$$ -In the literature, a functor that has a finitary approximation to a left adjoint is called (left) flat. - -Another way of looking at the above three situations is that we want to "represent" objects (apologize for contravariance, but I have not realized that I am describing a contravariant world until now) from $\catw{Set}^\catl{C}$ in $\catl{C}^{op}$, in $\catw{Set}^\catl{C}$ (i.e. a free cocompletion of $\catl{C}^{op}$) and in $\mathit{Ind}(\catl{C}^{op})$ (i.e. a free cofiltered completion of $\catl{C}^{op}$) respectively. A reasonable person, who wants to better understand multiadjunctions, should start looking now for free coproduct completion of a category. -Let $\catl{B}$ be a locally small category. One may associate with it the canonical $\catw{Set}$-indexing functor $\mor{\mathit{fam}(\catl{B})}{\catw{Set}^{op}}{\catw{Cat}}$: -$$X \mapsto \catl{B}^X$$ -One may think of $\catl{B}^X$ as of the category of formal $X$-indexed coproducts of objects from $\catl{B}$. The Grothendieck construction for $\mathit{fam}(\catl{B})$ glues categories $\catl{B}^X$ of formal $X$-indexed coproducts along sets $X$, giving the category: -$$\int \mathit{fam}(\catl{B})$$ -which is a formal (small) coproduct completion of $\catl{B}$ (of course, one needs to carefully check this statement). In fact, the above construction rises to a monad on $\catw{Cat}$, and one may develop a formal theory of multiadjunctions (similar to the formal theory of adjunctions through distributors) in the 2-category of Kleisly resolution of the monad. -Nonetheless, there is a less heavy explanation. I claim that a functor $\mor{F}{\catl{C}}{\catl{D}}$ has a left multiadjoint if $\mor{{F^{op}}^\star}{\int \mathit{fam}(\catl{C}^{op})}{\int \mathit{fam}(\catl{D}^{op})}$ defined as: -$${F^{op}}^\star(\{C_i\}_{i \in I}) = \{F^{op}(C_i)\}_{i \in I}$$ -has right adjoint. First, let me show the trivial direction --- assume that ${F^{op}}^\star$ has right adjoint. In particular, this gives us: -$$\hom_{\int \mathit{fam}(\catl{D}^{op})}({F^{op}}^\star(\{C_i\}_{i \in 1}), \{D_j\}_{j \in 1}) \approx \hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C_i\}_{i \in 1}, G(\{D_j\}_{j \in 1}))$$ -which simplifies to: -$$\hom_{\int \mathit{fam}(\catl{D}^{op})}(F^{op}(C), \{D\}) \approx \hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C\}, G(\{D\}))$$ -A morphism $F^{op}(C) \rightarrow \{D\}$ in $\int \mathit{fam}(\catl{D}^{op})$ is just a morphism $D \rightarrow F(C)$ in $\catl{D}$. Similarly, a single morphism $\{C\} \rightarrow G(\{D\})$ in $\int \mathit{fam}(\catl{C}^{op})$ is a morphism $G(\{D\})_k \rightarrow C$ in $\catl{C}$, for one $k \in K$, where $K$ is the indexing set of $G(\{D\})$. So: -$$\hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C\}, G(\{D\}))\approx \coprod_{k\in K} \hom_\catl{C}(G(\{D\})_k, C)$$ -and we get the formula for left multiadjunction: -$$\hom_\catl{D}(F(C), D) \approx \coprod_{k\in K} \hom_\catl{C}(G(\{D\})_k, C)$$ -In the other direction, let us assume that the above formula holds, and freely extend $G$ to ${G^{op}}^\star$: -$${G^{op}}^\star(\{D_j\}_{j \in J}) = \coprod_{j\in J} G(\{D_j\})$$ -Since ${F^{op}}^\star$ is free, it suffices to show: -$$\hom_{\int \mathit{fam}(\catl{D}^{op})}({F^{op}}^\star(\{C\}), \{D_j\}_{j \in J}) \approx \hom_{\int \mathit{fam}(\catl{C}^{op})}(\{C\}, {G^{op}}^\star(\{D_j\}_{j \in J}))$$ -The left side is isomorphic to: -$$\coprod_{j \in J} \hom_\catl{D}(D_j, F(C))$$ -whereas, the right side is isomorphic to: -$$\coprod_{j \in J} \coprod_{k \in K} \hom_\catl{C}(G(\{D_j\})_k, C)$$ -Thus ${G^{op}}^\star$ is right adjoint to ${F^{op}}^\star$. - -Moving back to your questions: - -Yes, they are colimits in the free coproduct completion of a category. -Still nothing to say :-) -Yes, because it is generally true for (co)limits. -Yes, because the coproduct completion, being a 2-functor, preserves adjunctions.<|endoftext|> -TITLE: Generating primes via composition of polynomials -QUESTION [14 upvotes]: It is well known that no nonconstant polynomial $f\in \mathbb{Z}[x]$ can assume only prime values at integer arguments. Indeed, if $a\in \mathbb{Z}$ is so large that $|f(a)|>1$, and if $p$ is a prime factor of $f(a)$, then all values of $f(a+pf(x))$ are divisible by $p$. -Question: Can a sequence of the form $f(a),f(f(a)),\ldots$ (for some fixed $a\in \mathbb{Z}$) take on only prime values and tend to infinity? -If $f$ is linear, then the answer to the question is no. To prove this, assume on the contrary that $f^n(a)$ tends to infinity and takes only prime values. Choose $n$ so large that $p:=|f^n(a)|$ is a prime greater than any coefficient of $f$. Then $f$ is a permutation polynomial mod $p$, therefore $f^m(p)$ is divisible by $p$ for infinitely many $m$, a contradiction. -If $f$ is not linear, and if the sequence in question consists only of primes, then (as Dietrich Burde pointed out) $f$ would be a non-linear polynomial taking infinitely many prime values, something that is not known to exist. -On the other hand, if the sequence $f(a),f(f(a)),\ldots$ can never take only prime values (which is my guess) then maybe the problem has a simple solution. Does anyone have any ideas or references? -This is an expanded form of a question posted on Math SE here. - -REPLY [3 votes]: Let $d=\deg(f)\ge2$. Since the height of $f^n(\alpha)$ grows like $C^{d^n}$ with $C>1$ (unless the orbit is finite), one would expect the sequence $f^n(\alpha)$ to contain only finitely many primes on probabilistic grounds. So people study other questions. For example, how large is the set of primes that divides at least one term in the sequence? See [1]. Or as joro indicates, one might look at the orbit behavior modulo $p$ for varying $p$; see for example [2] or [3]. Or one might ask how many terms in the sequence have a primitive prime divisor, that is, a prime dividing $f^n(\alpha)$ that does not divide $f^m(\alpha)$ for all $m -TITLE: Why does the first Cech cohomology classify twisted forms? -QUESTION [6 upvotes]: Suppose I have a faithfully flat cover of schemes $\phi:X\to Y$, and a sheaf $F$ on $Y$. I might be interested in so-called ``twisted forms for $F$." That is, sheaves $F'$ on $Y$ such that $\phi^\ast(F)\cong \phi^\ast(F')$. In the case that $\phi$ is a cover by some collection $\{U_i\}$, this is often stated as saying that $F$ and $F'$ are locally isomorphic on this cover. Okay, great. You'll have to trust me that this is something I might be interested in. -It is ``known" that if we compute $\check{H}^1(\phi,Aut(F))$, we can determine all twisted forms for $F$ (up to isomorphism class) along $\phi$. Or at least, that is what I believe people to be saying. -1) Is this the correct statement? -2) If so, can someone at least give me a rough explanation of why this is true, if not a complete proof? -3) Instead of starting out with a sheaf on $Y$, can I start off with an effective descent datum for $\phi$ and determine all twisted forms of that descent datum? In other words, can I start with a sheaf $F''$ on $X$ with effective descent data (which tells us that it comes from something over $Y$) and then ask for sheaves on $Y$ which pull back to (something isomorphic to) $F''$? How is their canonical descent datum (coming from pulling back) related to the descent datum on $F''$? -4) What can I do/say if $\phi$ is not faithfully flat? Can I still make these kinds of statements? What if I am interested in only finding twisted forms for one particular descent datum? -I know this is a ton of questions. Sorry if it's way too much. -Thanks!! --Jon - -REPLY [4 votes]: In general the statement is slightly different. What you typically have is a presheaf $F$ of groupoids (i.e. for every element $X$ you can have $QCoh(X)$ the groupoid of quasicoherent sheaves and isomorphisms, but maybe you are interested in the groupoid of algebras, Azumaya algebras, Hopf algebras etc..) and an object $A$ of $F(Y)$. -Then a twisted form of $A$ is an object $B$ of $F(Y)$ which becomes isomorphic to $A$ when they are pulled back to $X$. In this context you have is a map of pointed sets from the set of isomorphism classes of twisted forms of $A$ to $\check{H}^1(X,Aut(A))$, given by $B\mapsto Iso(A,B)$ where $Iso(A,B)$ is seen as an $Aut(A)$-torsor via precomposition. It is clear that if this torsor is trivial (that is it has a section on $Y$) then $B$ is isomorphic to $A$. Your question can be rephrased to asking conditions under which this map is an isomorphism. -The only case in which I know of a general theorem of this type is when the presheaf of groupoids is obtained by $QCoh$ via some sort of algebraic structure. At example call a signature $S$ a collection of pairs $(n_i,m_i)_{i\in I}$. Then the ''algebraic structures of signature $S$'' are the quasicoherent sheaves $F$ with a collection of maps $\Phi_i:F^{\otimes n_i} \to F^{\otimes m_i}$. Then for any signature $S$ the groupoid of ''algebraic structures of signature $S$'' and isomorphisms is such that the map of the previous paragraph is an isomorphism. -You can find a complete discussion when $X\to Y$ is a field extension in the book by Gille and Szamuely Central simple algebras and Galois cohomology. From that discussion and the general descent theorem is not hard to get the general case. -Sorry I pressed the submit button by mistake, hence the incomplete answer.<|endoftext|> -TITLE: Calculations of nonabelian group cohomology of R^n -QUESTION [9 upvotes]: I am looking at $H^1(\mathbb{R}^n,G)$ where $G$ is a finite 2-group. I'm wondering if such things have been calculated. I'm afraid I can't say I know anything here, past the result that this calculates group extensions of $\mathbb{R}^n$ by $G$. -Any related literature would also be appreciated. - -REPLY [10 votes]: I assume that you are looking at group cohomology. Then the action of $\Bbb{R}^n$ on $G$ is necessarily trivial, so $H^1(\Bbb{R}^n,G)$ is just $\mathrm{Hom}(\Bbb{R}^n,G)$ mod. conjugacy by $G$. But this is of course trivial - $\Bbb{R}^n$ has no finite quotient.<|endoftext|> -TITLE: About Aubin-Lions Lemma -QUESTION [5 upvotes]: I have a question about Aubin-Lions Lemma, the standard Aubin-Lions lemma need those Banach Space be reflexive spaces, are there any version of Aubin-Lions without reflexivity? -Standard aubin-lions:http://en.wikipedia.org/wiki/Aubin-Lions_lemma - -REPLY [4 votes]: I was wondering the same recently, and it seems to my that the answer is yes (you can get rid of reflexivity). Look at the paper of Jacques Simon : -Compact sets in the spaces $L^p(0,T,B)$. -The paper claims to give sharp results in any regard and as far as I can see it only asks the spaces to be banach he gives for example Corollary 4 : -if $\{F\}$ is bounded in $L^q(0,T,X), \{F^\prime\}$ bounded in $ L^1(0,T,Y),$ with the usual assumption :$$X\underset{compact}{\hookrightarrow} B\underset{continous}{\hookrightarrow}Y,$$ then $\{F\}$ is relatively compact in $L^p(0,T,B)$, for $p1$ (gives relative compactness in $\mathcal{C}(0,T,B)$). -I guess this is why it is sometimes mentionned as Aubin-Lions-Simon's lemma ...<|endoftext|> -TITLE: Reducibility of polynomials maps -QUESTION [10 upvotes]: Motivated by this question. -Let $f \in \mathbb{Q}[x]$ or$f \in \mathbb{Z}[x]$ . -Consider the sequence $f(x),f(f(x)), \ldots f^n(x)$. -If some $f^k(x)$ is reducible, the rest iterates will be reducible too. -This happens for $g(x) = x^2 - x - 1$. -$$g(g(g(x))) = (x^{4} - 3 x^{3} + 4 x - 1) \cdot (x^{4} - x^{3} - 3 x^{2} + x + 1)$$ -Fermat numbers are related to similar quadratic map and if -it happens to be reducible for some $k$, this will mean -there are infinitely many Fermat composites. - -Given $f$, is it possible to decide if some $f^k(x)$ is reducible? - -REPLY [13 votes]: I believe that the first person with significant results along these lines was Odoni [1]. There are also papers of Rafe Jones that consider questions of this sort, see for example [2] and [3]. A polynomial is called stable if all of its iterates are irreducible, and more generally, a polynomial is called eventually stable if for all sufficiently large $n$, the factorization of $f^n(x)$ in $\mathbb{Q}[x]$ is completely explained by factorizations of $f^m(x)$ for a fixed finite collection of $m$ values. There is a conjecture that says that all (maybe with a few obvious exceptions) polynomials are eventually stable. -This problem has also been studied over finite fields,. Note that if $f\in\mathbb{Z}[x]$ is monic and if $f^n(x) \bmod{p}$ is irreducible for some prime $p$ and all $n$, then $f$ is also stable over $\mathbb{Q}$. For a recent paper and a recent preprint with finite field stability results, see [4] and [5]. -This reference list isn't meant to be exhaustive, but if you look at these papers and forward and backward reference them, you should be able to find pretty much everything that's known and conjectured regarding (eventual) stability. -[1] Odoni, R. W. K., -The Galois theory of iterates and composites of polynomials, -Proc. London Math. Soc. (3), 51 (1985), 385-414. -[2] Jones, Rafe, -An iterative construction of irreducible polynomials reducible - modulo every prime, J. Algebra 369 (2012), 114-128. -[3] Jones, Rafe and Rouse, Jeremy, -Galois theory of iterated endomorphisms, -Proc. Lond. Math. Soc. (3) 100 (2010),763-794. -[4] Ahmadi, Omran and Luca, Florian and Ostafe, Alina and Shparlinski, Igor E., -On stable quadratic polynomials, Glasg. Math. J. 54 (2102), -359-369. -[5] Gomez-Perez, Domingo and Nicolas, Alejandro and Ostafe, Alina and Sadornil, Daniel, -Stable Polynomials over Finite Fields, 2012, arXiv:1206.4979.<|endoftext|> -TITLE: Periodic orbit property -QUESTION [24 upvotes]: A topological space $X$ satisfies the "Periodic orbit property", briefly POP, if for every continuous map $f:X \to X$, there exist a natural number $n$ and a point $x_{0}\in X$ such that $f^{n}(x_{0})=x_{0}$. -Obviously fixed point property (FPP) implies POP. -For a natural number $n$, a topological space $X$ is called $n$-POP if for every continuous map $f$ on $X$, $f^{n}$ has a fixed point. (Ex: $\mathbb{S}^{2n}$ is a 2-POP manifold, because the degree of a fixed-point-free map on $\mathbb{S}^{2n}$ must be $-1$.) -The question: - -Is there an example of a manifold $M$ which satisfies POP but for every $n\in \mathbb{N}$, there is a continuous map $f$ on $M$ such that $f^{n}$ has no fixed point? - -Namely: we search for a manifold for which every self-map has a periodic orbit, but there is no any control on periods. -Equivalently: - -Is there a manifold $M$ which is POP but not $n$-POP for all $n\in \mathbb{N}$? - -In particular, can we say: - -"every compact POP manifold is necessarily a $n$-POP manifold, for some $n$"? - -Motivated by the Lefschetz fixed-point theorem, we ask that: -What algebraic topological criterion, can be introduced for consideration of this property(POP)? -Edit: According to the very interesting answer of Qiaochu Yuan, in the orientable case, the question is equivalent to the following: - -Let M be a closed orientable manifold. Is it true that $M$ is not POP if and only if $\chi(M)=0$? - -Note1 For a related question see this post and it is natural to ask that "Does $S^{2}\vee S^{2}$ satisfy the periodic orbit property?" -Note2 I think the continuation of the argument of Qiaochu Yuan for his first statement is not easy, for arbitrary manifold. Because for the simplest case $S^{3}$ we had the famous conjecture of "existence of a vector field on $S^{3}$ without periodic orbit. In fact consideration of non vanishing vector fields is necessary but not sufficient. Periodic orbits of vector fields are important, too. Moreover, perhaps an approach which is not based on "vector fields" could be useful, for example consideration of orientation-reversing diffeomorphisms. -Note 3: "pointwise periodic homeomorphism' is a concept which is indirectly similar to the subject of this post. - -REPLY [5 votes]: The following is proved by -F. Brock Fuller in "The Existence of Periodic Points," Annals of Mathematics, Vol. 57, 1953, pp. 229-230: -Theorem. Let X be a compact simplicial complex with Betti numbers $B_i$, and nonzero Euler characteristic. If a continuous map - $f: X\rightarrow X$ induces isomorphisms of homology groups, then $f$ has a point whose period is $\le \max_i \big (∑_i B_{2i},∑_i B_{2i-1} \big)$. Thus every homeomorphism of $X$ has a periodic orbit.<|endoftext|> -TITLE: Sets whose elements are mutually "weakly" coprime? -QUESTION [8 upvotes]: Fix $n$ and $k$. I want a set $S\subseteq\{1,\ldots,n\}$ with the property that for every $x\in S$, -$$\mathrm{gcd}\bigg(x,\prod_{y\in S\setminus\{x\}}y\bigg)<\frac{x}{k}.$$ -How small should a random $S$ be to have this property with high probability? More importantly, what sort of math is this, and where can I learn more? (I only guessed in my tags.) - -REPLY [6 votes]: Let's just consider the case $k=1$ where the problem asks for sets $S$ such that each element of $S$ does not divide the product of the rest of the elements of $S$. I claim that if $S$ has fewer than $\exp(\frac{1}{10} \sqrt{\log n\log \log n})$ elements then with high probability this happens. On the other hand if $S$ has more than $\exp(10\sqrt{\log n \log \log n})$ elements then with high probability some element of $S$ will divide the product of the remaining elements. With more effort this gap can probably be closed. (Also, as will be seen from the proof the argument also works if $k$ is not too big -- that is in the range $k\le \exp(\frac{1}{10} \sqrt{\log n\log \log n})$ say.) -The reasoning relies on some facts about smooth numbers. Let $\Psi(x,y)$ denote the number of integers up to $x$ all of whose prime factors are below $y$. We are interested in the range when $y$ is about $\exp(c\sqrt{\log x\log \log x})$ for some constant $c$, and here $\Psi(x,y)$ is about size $x\exp(-\sqrt{\log x\log \log x} (\frac{1}{2c}+o(1)))$; moreover the number of squarefree smooth numbers is also of this size. -I need one more observation. If $s$ numbers are chosen randomly from $1$ to $n$, then their product will very likely be divisible by all primes up to $s/\log n$ -- if not there is a prime $ps^3$ say. To see this, if we pick a prime $p>s^3$ then the chance that $p^2$ divides the product is $O(s^2/p^2)$ ($p$ could divide two of the $s$ elements; the probability of dividing $3$ elements is even smaller) and summing this over all $p>s^3$ still gives a total probability of $O(1/s)$. -Now we are ready for the proof. Suppose $S$ has fewer than $L_1=\exp(\frac 1{10} \sqrt{\log n\log \log n})$ elements. From our remark on the number of smooth integers, we may see with high probability each element of $S$ is not $L_1^3$ smooth. That is each element of $S$ is likely to be divisible by a prime larger than $|S|^3$. But then as observed above, such a large prime is unlikely to divide two elements of $S$, and therefore with high probability every element of $S$ does not divide the product of the rest. -For the other assertion, suppose now that $S$ has more than $L_2 =\exp(10\sqrt{\log n\log \log n})$ elements. From our observation on smooth numbers, with high probability $S$ contains a squarefree number that is $\exp(\sqrt{\log n\log \log n})$ smooth. But with high probability the product of the remaining elements of $S$ is divisible by all primes up to $(|S|-1)/\log n$ which is a good deal bigger than $\exp(\sqrt{\log n\log \log n})$. In other words, the squarefree smooth number that we are likely to find in $S$ will divide the product of the remaining numbers with high probability. -This completes the proof. Let me add that the shape of the answer $\exp(\sqrt{\log n\log \log n})$ arises in other contexts (e.g. factoring algorithms such as the quadratic sieve) for a similar reason: it is the break even value of $y$ where the probability of being $y$ smooth is roughly $1/y$, and that's roughly what's being used above.<|endoftext|> -TITLE: Fomin-Kirillov algebras and Schubert calculus -QUESTION [11 upvotes]: In -Fomin, Sergey; Kirillov, Anatol N. Quadratic algebras, Dunkl elements, and -Schubert calculus. Advances in geometry, 147--182, Progr. Math., 172, -Birkhäuser Boston, Boston, MA, 1999. MR1667680 (2001a:05152), link -the authors introduce a combinatorial model for modern Schubert calculs. They -define an algebra which contains a commutative subalgebra isomorphic to the -cohomology ring of the flag manifold. More precisely, for $n\geq2$ they define -$\mathcal{E}_n$ as the algebra with generators $x_{(ij)}$, $1\leq -i -TITLE: Correspondence between operads and monads requires tensor distribute over coproduct? -QUESTION [8 upvotes]: In checking the details of the correspondence between operads over a symmetric monoidal category and monads on some associated endofunctor of the category, I cannot make the obvious proof work without assuming that the monoidal product distributes over coproducts. But no such assumption is mentioned in my sources (for example Operads, Algebras, Modules by May (PDF).) Am I missing something in my argument? -Let $\mathcal{C}$ (mathcal C) be an operad (for simplicity take it non-symmetric) over a symmetric monoidal category $\mathcal{V},$ with composition $\gamma\colon \mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes\dotsb\otimes\mathcal{C}(m_n)\to\mathcal{C}(m_1+\dotsb+m_n).$ We define a functor $C\colon \mathcal{V}\to\mathcal{V}$ (Roman C) by $CX = \coprod_i \mathcal{C}(i)\otimes X^{\otimes i}$. -Then one wants to verify that the operad structure $\gamma$ gives a monad on $C$. That is, we need a natural morphism $C^2X\to CX,$ or $\coprod_i \mathcal{C}(i)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes i}\to \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}$. By universal prop of coproducts, it will suffice to exhibit an arrow $\mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\to \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}$ for all $n$. -Clearly we have $\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes\dotsb\otimes \mathcal{C}(m_n) X^{\otimes m_1+\dotsb+m_n}\to \\ \mathcal{C}(m_1+\dotsb+m_n)\otimes X^{\otimes m_1+\dotsb+m_n}\to\coprod_k\mathcal{C}(k)\otimes X^{\otimes k},$ where the first arrow is by symmetry of the monoidal structure, the second arrow is the operad composition $\gamma,$ and the third arrow is the canonical inclusion into the coproduct. -Therefore by universal prop of coproducts, we have $\coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to\coprod_k\mathcal{C}(k)\otimes X^{\otimes k}.$ -In general, again using inclusion morphisms of coproducts, we have arrows $\mathcal{C}(m_\ell)\otimes X^{\otimes m_\ell}\to \coprod_j \mathcal{C}(j)\otimes X^{\otimes j}.$ Then by functorality of the monoidal product, we have $\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$. By universal property of coproducts, we therefore have an arrow from the coproduct $\coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n}\to \mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$. -To summarize, we have the obvious maps $\mathcal{C}(n)\otimes \left(\coprod_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\leftarrow \coprod_{m_1,\dotsc,m_n}\mathcal{C}(n)\otimes \mathcal{C}(m_1)\otimes X^{\otimes m_1}\otimes \dotsb \otimes \mathcal{C}(m_n)\otimes X^{\otimes m_n} \to \\ \coprod_k \mathcal{C}(k)\otimes X^{\otimes k}.$ Unless we know that the arrow on the left is an isomorphism, we do not get the structure map for a monad on $C$. And that arrow on the left will generally not be an isomorphism if the monoidal product in $\mathcal{V}$ does not distribute over the coproduct. For example, if the monoidal product is the coproduct itself. - -REPLY [3 votes]: The correspondence between operads and monads still holds -in the absence of commutation of coproducts and tensor products if one chooses correct definitions. -Operads are monoids in symmetric sequences equipped -with the substitution product, algebras and modules -over operads are (left) modules over monoids -(concentrated in degree 0 in the case of algebras), -and the associated monad is the functor O∘−, -with the multiplication and unit maps -coming from the monoid structure of O. -The substitution product defines a monoidal -category in the usual sense only if the monoidal -product preserves infinite coproducts. -In the general case one obtains a monoidal category -in which the associators need not be invertible. -These are known as (op)lax (normal) monoidal categories -and are defined as (op)lax monoids in the 2-category -of categories equipped with the cartesian product. -(Oplax/lax is chosen according to the direction -in which the noninvertible associator goes.) -See the paper “Lax monoids, pseudo-operads, -and convolution” by Day and Street for more details. -The notions of monoids and left modules over monoids -can be generalized to this setting, as explained -in detail in Ching's paper “A note on the composition product of symmetric sequences”. -If one also generalizes monads to this setting, -one recovers the usual correspondence between operads and monads. -All of this becomes important in the case of cooperads, -for which one would need tensor products to commute -with infinite products to obtain the usual notion of a comonad from a cooperad. -Although monoidal products often preserve infinite coproducts, they rarely preserve infinite products, -unless the monoidal structure is cartesian, which forces one to consider the notions mentioned above.<|endoftext|> -TITLE: Can an infinite number of mathematicians guess the number in a box with only one error? -QUESTION [10 upvotes]: In this question the following observation was made: -Consider a sequence of boxes numbered 0, 1, ... each containing one real number. The real number cannot be seen unless the box is opened. -Define a play to be a series of steps followed by a guess. A step opens a set of boxes. A guess guesses the contents of an unopened box. A strategy is a rule that determines the steps and guess in a play, where each step or guess depends only on the values of the previously opened boxes of that play. -Then for every positive integer $k$, there is a set $S$ of $k$ strategies such that, for any sequence of (closed) boxes, there is at at most one strategy in $S$ that guesses incorrectly. -My question is this: Can $k$ be countably infinite (instead of a positive integer)? If not, is there a proof? -[Edit: the original question also asked whether $k$ can be uncountable; this was answered by Dan Turetsky in the negative in comments]. -The best I have been able to show is that, if the function $f:\mathbb{N}\to\mathbb{R}$ defined by the contents of the initial sequence of boxes is recursive (viewing elements of $\mathbb{R}$ as binary sequences), then $k$ can be countably infinite. To see this, call a subset $X$ of $\mathbb{N}$ signature if two recursive functions on $\mathbb{N}$ that eventually agree on $X$ also eventually agree on $\mathbb{N}$. (Two functions "eventually agree" if they differ in finitely many places). Call two Turing Machines equivalent if their associated functions on $\mathbb{N}$ are equivalent (that is, eventually agree). A diagonalization argument on the class representatives of the Turing Machines yields an infinite partition $U$ of $\mathbb{N}$ into signature subsets. The $i$'th strategy in $S$ first opens all the boxes whose indices are not in the $i$'th element $U_i$ of $U$, determines the class representative Turing Machine T that generates the resulting values on the opened boxes for boxes whose indices are greater than $N$ (for some positive $N$), and guesses that a box with index greater than $N$ and in $U_i$ has a value specified by $T$. -However, I have not been able to modify this for the non-computable case. - -REPLY [8 votes]: It is possible to have every mathematician guess the number in one of the boxes with at most one error. - -Partition the natural numbers into countably many sets, $\{S_i\}_{i=0}^\infty$, where each $S_i=\{n_{i_1},n_{i_2},\dots,\}$ is countably infinite. (There are many ways to do this) Since we have countably many mathematicians, we may list them, and assign $S_i$ to the $i^{th}$ mathematician. -If $u_k$ denotes the real number in the $k^{th}$ box, then the $i^{th}$ mathematician will be assigned the sequence of real numbers $u_{n_{i_j}}$, for $j=1,2,3\dots$. Using the axiom of choice, we may chose a representative for each equivalence class of sequences of real numbers under the equivalence relation $\{u_n\}_{n=1}^\infty\equiv\{v_n\}_{n=1}^\infty$ if there exists $M>0$ such that $v_n=u_n$ for all $n>M$. Thus, for the $i^{th}$ mathematician there will exist an integer $M_i$ such that for all $j>M_i$, the sequence $u_{n_{i_j}}$ is equal to the representative of its equivalence class. The goal is to have mathematician $i$ guess an integer $H_i>M_i$ by looking at every box except those in the set $S_i$. If this happens, then mathematician $i$ may look at all of the elements of $u_{n_{i_j}}$ with $j\geq H_i+1$, determine the equivalence class, and guess the box with $j=H_i$. Since $H_i>M_i$, his guess will be correct. It follows that we need all but possibly one mathematician to guess an integer $H_i>M_i$. If the sequence $M_i$ is bounded, then the problem is easy. The difficulty is handling an unbounded sequence $M_i$. -Under the same system of representatives, the sequence $\{M_i\}_{i=0}^\infty$ lies in some equivalence class of real numbers. Since mathematician $i$ knows the value of $M_l$ for all $l\neq i$, each mathematician can determine the equivalence class of the sequence $\{M_i\}_{i=0}^\infty$. Let $\{v_i\}_{i=0}^\infty$ denote the representative of this equivalence class. Then there exists an integer $N$ such that for every $i>N$, $M_i=v_i$. Mathematician $i$ with $i\leq N$ can determine $N$, however each mathematician with $i>N$ only knows that $N\leq i$. The strategy for guessing is as follows: Assign to mathematician $i$ with $i>N$ the integer $$H_i=1+\max\{v_i,M_{i-1},M_{i-2},\dots,M_1,M_0\},$$ and to each mathematician with $i\leq N$, the integer $$H_i=1+\max\{M_{N},M_{N-1},\dots,M_{i+1},M_{i-1},\dots,M_1,M_0\}.$$ Then we must have $H_i>M_i$ for every $i$ except possibly one. Thus, we have set up a strategy which allows every mathematician except possibly one to guess some box correctly.<|endoftext|> -TITLE: All mapping space between CW complexes is a CW complex? -QUESTION [7 upvotes]: Let $\mathrm{Map}(X,Y)$ denote the (unbased) cellular mapping space from $X$ to $Y$. -If $X$ and $Y$ are finite CW complexes, is $\mathrm{Map}(X,Y)$ a CW complex? -Can we know the cell structure of $\mathrm{Map}(X,Y)$? -For example, what is the cell structure of $\mathrm{Map}(S^n,S^k)$ for $n \geq k$? -Please recommend related papers and textbooks. - -REPLY [4 votes]: I think I should be posting this as a comment, but as i don't have enough reputation to comment I take the liberty to post this as an answer. -I am here just elaborating on Eric Wofsey's answer. -Let's take the case, say, $X=\mathbb{N}$ and $Y=I=[0,1]$. Then -$Map(X,Y)$ is Hilbert cube. -On the other hand, the definition of a CW-complex says that a CW complex has to be the union of finite skelta, and -for reasons explained in Eric Wofsey's answer, this is not the case.<|endoftext|> -TITLE: Reference request: Geodesic flow on a manifold with negative curvature is ergodic -QUESTION [17 upvotes]: I'm reading about the Mostow's rigidity theorem, and the proof uses the following (maybe well-known) result: -The geodesic flow on a manifold with negative curvature is ergodic. -The lecture note that I'm reading does not provide a reference for that. I am aware that Gromov's proof of Mostow's rigidity theorem doesn't need this result, but I'm working on an assignment and I need to learn about Mostow's original proof. I have some background in Riemannian geometry but I know nothing about ergodic theory. Can anyone give me some reference about this result? All kinds of resources are welcomed. -Thank you very much! - -REPLY [6 votes]: Let me write a short scheme of proof that I learned in the course of Ergodic Theory in Warwick University (notes are available on-line. You will find several different versions of this proof). -Let us call $X=SL(2,\mathbb{R})/SL(2,\mathbb{Z})$ and consider the geodesic flow $g_t:=\begin{pmatrix}e^{\frac{t}{2}} & 0\\ 0 & e^{-\frac{t}{2}}\end{pmatrix}$ on $X.$ -Theorem: The geodesic flow $g_t:X\to X$ is ergodic. -Let us define the matrices: -$$h_t:=\begin{pmatrix} 1 & t \\ 0 & 1\end{pmatrix} \mbox{ and } h^-_t:=\begin{pmatrix} 1 & 0 \\ t & 1\end{pmatrix}.$$ -Lemma 1. $g_t h_s g_{-t}=h_{se^t}$ and $g_t h^-_s g_{-t}=h^-_{se^{-t}}.$ -Proof: Direct. -Lemma 2. If $\mu$-as $\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t x)dt$ is constant in $x\in X$ (i.e. it depends on $f,$ but not on $x$) and for all $f$ continuous on $X,$ then the flow $g_t:X\to X$ is ergodic with respect to $\mu.$ -Proof: This a classic exercise in Ergodic theory. -Lemma 3. Almost all the matrices $\gamma\in SL(2,\mathbb{R})$ can be write in the form $\gamma=h_{s_1}g_t h_{s_2}^-.$ -Proof: Direct. -Corollary 4. Given almost all points $x,x'\in X,$ we can chose $\gamma\in SL(2,\mathbb{R})$ such that $x'=h_{s_1}g_t h_{s_2}^-x.$ -Proof: Direct. -Lemma 5. If $y=g_s x,$ then $\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t x)dt=\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t y)dt.$ -Proof: Direct. -Lemma 6. If $y=h_s x,$ then $\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t x)dt=\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t y)dt.$ -Proof. Use $g_t$ invariance of the measure $\mu$ and Lemma 1. -Lemma 7. If $y=h_s^- x,$ then $\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t x)dt=\lim_{T\to\infty}\frac{1}{T}\int_0^T f(g_t y)dt.$ -Proof. Use $g_t$ invariance of the measure $\mu$ and Lemma 1. -Proof of Theorem: Use Lemma 2 to characterise ergodicity. For $\mu$-a.e. $x,x',$ we can write (because of Corollary 4) $y_1=h_{s_2}^-x,$ $y_2=g_t y_1$ and $x'=h_{s_1}y_2.$ Then apply Lemmas 5,6 and 7 to conclude the theorem. $\square$ -Remark: The measure $\mu$ is explicit in the Corollary 4 (and omitted in our statement of the Theorem), corresponds to the Liouville measure.<|endoftext|> -TITLE: Is a function with nowhere vanishing derivatives analytic? -QUESTION [48 upvotes]: My question is the following: Let $f\in C^\infty(a,b)$, such that $f^{(n)}(x)\ne 0$, for every $n\in\mathbb N$, and every $x\in (a,b)$. Does that imply that $f$ is real analytic? -EDIT. According to a theorem of Serge N. Bernstein (Sur les fonctions absolument monotones, Acta Mathematica, 52 (1928) pp. 1–66) if $f\in C^\infty(a,b)$ and $f^{(n)}(x)\ge 0$, for all $x\in(a,b)$, then $f$ extends analytically in a ball centered at $(a,0)\in\mathbb C$ and radius $b-a$! - -REPLY [55 votes]: If $f$ is $C^{\infty}$ every derivative is continuous, so the hypothesis on $f$ implies that each derivative $f^{(n)}$ has constant sign. Such functions were studied by S. Bernstein and called regularly monotonic. In particular he proved in 1926 that a regularly monotonic function is real analytic. -This 1971 AMM article by R.P. Boas provides a proof, more history, and further results along these lines. See also this 1975 PAMS article of J. McHugh.<|endoftext|> -TITLE: Is this system identical to S4.4? -QUESTION [6 upvotes]: Consider the normal modal logic system $\mathbf{TAR1}$ given by $\mathbf{T}$ plus the following axiom: -$$\mathrm{AR1}: \lozenge \square p \rightarrow (\square p \lor \square (p \rightarrow \square p))$$ -Analysis using normal forms shows that it is included in $\mathbf{S4.4}$. However, algebraically I cannot either prove or disprove that it is strictly weaker. I was wondering is someone has a proof or an argument based on Kripke semantics. -Additional info, in case it helps: - -$\mathbf{S4.4}$ is usually defined as $\mathbf{S4}$ plus axiom $\mathrm{R1}$: -$$\mbox{R1}: \lozenge \square p \rightarrow (p \rightarrow \square p)$$ -I find that $\mathbf{S4.4}$ is also $\mathbf{T}$ plus the following axiom: -$$\mathrm{4.4}: \lozenge \square p \rightarrow \square (p \rightarrow \square p)$$ -I also find that $\mathbf{TAR1}$ is equivalent to $\mathbf{TR1}$, which is $\mathbf{T}$ plus $\mathrm{R1}$. But the question now becomes whether $\mathbf{TR1}$ is equivalent to $\mathbf{S4.4}$. -Finally, I am trying to prove $\mathrm{4.4}$ in $\mathbf{TAR1}$, but it does not seem to be possible using only normal forms of degree $\le 3$. So, unless my calculations are wrong, either the proof involves modalities of degree 4, or $\mathbf{TAR1}$ is a distinct system. If distinct, then it would not include $\mathbf{S4}$. - -Philosophically, this particular system may not be particularly relevant; but an answer would help me the algebraic properties of other similar systems that I'm analyzing. - -REPLY [4 votes]: $\mathbf{KT}+\mathbf{AR1}$ is strictly weaker than $\mathbf{KT}+\mathbf{4.4}$. Consider the Kripke frame that is the reflexive closure of the following graph (so that any model built on it is a model of $\mathbf{KT}$): -$$\require{AMScd}\begin{CD} -A @>>> B @>>> C -\end{CD}$$ -Then build a model on it with, say, the valuation such that $p$ holds at each of $A$ and $B$ while $\neg p$ holds at $C$, for every propositional variable $p$. If I'm not mistaken, $\mathbf{AR1}$ holds in the resulting model while $\mathbf{4.4}$ fails. Specifically, to see that $\mathbf{AR1}$ holds, note that the antecedent of it is false at $B$ and $C$, while at $A$ the consequent is true since $\square p$ holds. But $\mathbf{4.4}$ does not hold at $A$ in this model.<|endoftext|> -TITLE: Relating the roots of polynomials to the solution sets of certain functional equations -QUESTION [12 upvotes]: Consider a functional equation of the following form: -$$\sum_{k=0}^n a_k\,\underbrace{f(f(\cdots f}_{k}(x)\cdots )=0\quad \big(f:\,\mathbb{R}\to\mathbb{R},\;a_i\in \mathbb{R},\;\text{and}\;f^0=\text{id}\big)$$ -A well-known strategy for producing continuous solutions to this is to take the real roots $\lambda_1,\,\cdots,\, \lambda_m\;-$ if they exist $-$ of the characteristic polynomial $\mathcal{P}(z)=\sum_{k=0}^n a_kz^k$, and set $f_i(x)=\lambda_i x$ for each $1\leq i\leq m$ (note: this does not necessarily generate a full solution set). -It is natural to then explore the case where $\mathcal{P}$ has no roots in $\mathbb{R}$. After fiddling with a couple problems, it seems conceivable that this would immediately imply an absence of continuous solutions altogether: my question is whether this is indeed the case. - -REPLY [4 votes]: It is known that every polynomial having no positive root has a multiple with nonnegative coefficients only. So we may replace $\mathcal P$ by such multiple $\mathcal Q$ (surely, $f$ satisfies the corresponding equation as well). -Now, since $\mathcal P$ has no real roots, the equality $f(x)=x$ is impossible; thus $f(x)-x$ has constant sign, say $f(x)>x$ for all $x\in\mathbb R$ (otherwise we may replace $f$ by $g(x)=-f(-x)$). Take any positive $x$; then $f^k(x)>0$ as well, so when we substitute this $x$ into our equation constructed for $\mathcal Q$ we get the sum of positive terms only. A contradiction. -For completeness, let me sketch the proof of the fact used at the beginning. It suffices to prove it for polynomials of the forms $z+a$ (for positive $a$, it is trivial) and $z^2+az+b$ (having two non-real roots $z_0$ and $\overline z_0$). [SIMPLIFIED] For the second one, observe that for some positive integer $k$ the number $z_0^k$ has negative real part; then a polynomial $(z^k-z_0^k)(z^k-\overline z_0^k)$ is what we want.<|endoftext|> -TITLE: convergence in $\hat{\mathbb{Z}}$, modulo prime power -QUESTION [7 upvotes]: The following problem appears in Lenstra's Galois Theory for Schemes (p 14, Ex 1.16). - -Let $b\in\mathbb Z_{\ge0}$. Define the sequence $(a_n)_{n=0}^\infty$ - by $a_0=b, a_{n+1}=2^{a_n}$. Prove that $(a_n)_{n=0}^\infty$ - converges in $\hat{\mathbb Z}$, and that $a:=\lim_{n\to\infty}a_n$ is - independent of $b$. -Then, write $a=\sum_{n=1}^\infty c_n n!$. Compute $c_n$ for - $n=1,\cdots, 10$. - -My first question is about finding $c_1, \cdots, c_{10}$. Noting the fact that $ \hat{\mathbb Z} \simeq\varprojlim\mathbb Z/n!\mathbb Z$, I have written down a bijection from $\varprojlim\mathbb Z/n!\mathbb Z$ onto the collection $\{\sum_{n=1}^\infty c_n n!: 0\leq c_n\leq n\}$ of formal series. According to the bijections I need to find out $a\text{ mod } n!$ for $n=2,\cdots, 11$. Thanks to the Chinese remainder theorem, the problem reduces to the computation of $a$ modulo prime-power factors in each $n!$. With the help of Euler's formula, $a\text{ mod }p$ can be computed for every odd prime $p$; and from the definition of $a$ we have $a\equiv 0\pmod{2^k}$. But I do not have any idea about calculating $a\text{ mod }p^k$. In particular, what I need are $a\text{ mod }3^2$, $a\text{ mod }3^4$ and $a\text{ mod }5^2$. -Then, we turn back to a more fundamental question: what does it mean by the sequence to converge? (Lenstra has not defined convergence in his notes.) Intuitively, $(a_n)_{n=1}^\infty$ converges if for every $M\in\mathbb N$, there is an $N\in\mathbb N$ such that for each $m\leq M$, the expression $a_n\text{ mod }m!$ is a constant for all $n\geq N$. Is this "definition" concise enough? And why do $a_n\text{ mod }{m!}$ become stable eventually? Finally, why the limit is independent of $b$? - -REPLY [3 votes]: There's a quite quick recursive algorithm to compute the stable value of $a_n$ modulo $k$. -If I do it for $a_{10}$ (i.e. $k=10!$) and put things upside down to be deductive, I get, successively (independently of the choice of $b$): -$$n\ge 2\Rightarrow a_n\equiv 0 \;[2]$$ -since the group $(\mathbf{Z}/3\mathbf{Z})^*$ has exponent 2 and 2 belongs to it, we deduce -$$n\ge 2\Rightarrow 2^{a_n}\equiv 1\; [3]$$ -$$n\ge 3\Rightarrow a_n\equiv 1 [3];\quad a_n\equiv 0\;[2]$$ -$$n\ge 3\Rightarrow a_n\equiv 4 \;[6]$$ -since the group $(\mathbf{Z}/9\mathbf{Z})^*$ has exponent 6 and 2 belongs to it, we deduce -$$n\ge 3\Rightarrow 2^{a_n}\equiv 16 \;[9]$$ -$$n\ge 4\Rightarrow a_n\equiv 16 [9];\quad 2^{a_n}\equiv 0\; [4]$$ -$$n\ge 5\Rightarrow a_n\equiv 16\; [36]$$ -since the group $(\mathbf{Z}/135\mathbf{Z})^*$ has exponent 36 (being isomorphic to $(\mathbf{Z}/27\mathbf{Z})^*\times (\mathbf{Z}/5\mathbf{Z})^*$ where the factors have cardinal 18 and 4, whose lcm is 36), we get -$$n\ge 5\Rightarrow 2^{a_n}\equiv 2^{16} [135]=196\;[135]$$ -$$n\ge 6\Rightarrow a_n\equiv 196 \;[135];\quad a_n\equiv 0\;[4]$$ -$$n\ge 6\Rightarrow a_n\equiv 196 \;[4\times 135]$$ -if $m=14175=3^4.5^2.7$, the group $(\mathbf{Z}/m\mathbf{Z})^*$ contains 2 and has exponent $4.135=540$, whence -$$n\ge 6\Rightarrow 2^{a_n}\equiv 2^{196} \;[14175]= 1275136\; [14175],$$ -where $1275136=2^8.17.293$; hence -$$n\ge 7\Rightarrow a_n\equiv 1275136 \;[14175],\quad a_n\equiv 1275136 \;[2^8]$$ -$$n\ge 7\Rightarrow a_n\equiv 1275136\; [10!]$$ -since $10!=14175\times 2^8$. -Of course the algorithmic (deterministic) way to proceed goes reverse: if you want to compute the stable value of $a_n$ modulo a certain number $k$, write $k=k'k''$ with $k'$ power of 2 and $k''$ odd. In practice the case of $k'$ is rather trivial since the 2-valuation of $a_n$ grows very fastly. Then to know $a_n$ modulo $k''$, since $k''$ is odd, it is enough to know the exponent $m$ of the group $(\mathbf{Z}/k''\mathbf{Z})^*$, which is easy as soon as you can factor $k''$ into primes (write $k''$ as product of powers $k_i$ of distinct primes, and take the lcm of the orders of the $(\mathbf{Z}/k_i\mathbf{Z})^*$). Then you need to compute of the stable value of $a_n$ modulo $m$, and $m$ is usually much smaller than $k$ (above, for $k=10!$, 5 steps were needed). Thus if $a_n\equiv p_m$ for $n\ge C_m$ independently of $b$, you deduce that $a_n\equiv 2^{p_m}$ for all $n\ge C_m+1$ independently of $b$. Then compute $2^{p_m}$ modulo $k''$ (or modulo $k$) to make it reasonable, and deduce from the Chinese remainder theorem the value of $a_n$ modulo $k$ for $n\ge C_m+1$. -Added: the stable value of $a_n\,[k!]$ immediately entails the computation of $c_n$ for $n\le k-1$, namely $c_n=(a\,\mathrm{mod}\,(n+1)!-a\,\mathrm{mod}\,n!)/n!$. Thus $$(c_1,c_2,c_3,c_4,c_5,c_6,c_7,c_8,c_9)=(0,2,2,0,0,0,5,4,3).$$ To find $c_{10}$, you can run the same method to compute $a\pmod{11!}$.<|endoftext|> -TITLE: Random planar, bipartite graphs -QUESTION [7 upvotes]: I have a need to generate random planar graphs none of which have an odd cycle, -i.e., bipartite graphs. -I know there is a substantial two-decade literature on random planar graphs, little with which I am familiar. -I know there are several models: at least -those that specify an edge probability, -those that depend on a random graph process, and -uniform random planar graphs. -I am wondering if those familiar with these and other -models could -suggest one that might be modified to generate even-cycled planar graphs. -I don't have strict distribution requirements—randomness in any one -of several senses would suffice. Thanks for your advice! - -I should have added the only two ideas I had, neither of which I feel is adequate, -even under a loose definition of "random": - -Generate a random planar graph, and add a vertex in the middle of -each edge. Then all these newly added vertices have degree $2$. -Select out a random subgraph of the grid graph. Then no vertex has degree -greater than $4$. - -REPLY [3 votes]: A further approach starts with a random geometric graph (RGG): Assign a domain $D\subset\mathbb{R}^2$, typically a square. Place points using a Poisson process and make links between pairs less than a distance $r$ apart. -Then, colour the points randomly with two colours and delete links between points of the same colour to make the graph bipartite. Then, delete the longest link that intersects another (drawing straight lines between the points) recursively until there are no intersections and the graph is planar. -My interest in RGGs comes from their application to wireless networks (eg arxiv:1401.7188 and references therein); not sure if planar bipartite RGGs have such an application, but in any case it's an interesting question.<|endoftext|> -TITLE: Inverse limit of Gorenstein local rings is again Gorenstein? -QUESTION [5 upvotes]: If we have the system of surjective ring homomorphisms -$f_{i,i+1}: R_{i+1} \twoheadrightarrow R_i$ -for an arbitrary $i \geq 0$ such that all $R_i$ are Gorenstein local ring. Let us put -$R^{\infty} \colon= \varprojlim_{i \geq 0} R_i$. -Question: Assume that $R^{\infty}$ is also a local ring. Then, is $R^{\infty}$ Gorenstein? - -REPLY [6 votes]: This is not true. In fact, large numbers of complete local rings one encounters in "real life" satisfy this condition. Let me give you some background. -Definition: (Hochster) A local ring $(R, \mathfrak{m})$ is called approximately Gorenstein if there is a decreasing sequence of $\mathfrak{m}$-primary ideals $$I_1 \supseteq I_2 \supseteq \ldots$$ -that are cofinal with the powers of $\mathfrak{m}$ and such that each $R/I_i$ is Gorenstein. -Remark: It's easy to see that Gorenstein rings are approximately Gorenstein, since you can choose $x_1, \ldots, x_d$ a system of parameters and let $I_i = \langle x_1^i, \ldots, x_d^i \rangle$. -Remark: Any complete approximately Gorenstein ring satisfies the condition you mentioned. Obviously $R = \lim_{i} R/I_i$. -Ok, so now you can ask what rings are approximately Gorenstein. It's easy to see that a local ring is approximately Gorenstein if and only if its completion is. -Theorem: (Hochster) If $(R, \mathfrak{m})$ is complete and reduced, it is approximately Gorenstein. Thus if $(R, \mathfrak{m})$ is excellent and reduced, it is approximately Gorenstein. Furthermore, if $R$ has depth at least $2$, it is approximately Gorenstein. -Hence, many common rings are approximately Gorenstein. -Edit: One source for this information is these notes by Mel Hochster<|endoftext|> -TITLE: Pure morphisms which are not faithfully flat -QUESTION [7 upvotes]: Joyal and Tierney proved that morphisms of rings which are of effective descent are exactly those morphisms $\phi:R\to S$ such that $\phi$ presents $S$ as a pure $R$-module. Grothendieck had originally shown that being faithfully flat implied being of effective descent, but had not entirely characterized such morphisms. Is there a specific example that drove this characterization? Is there a good family of examples of morphisms which are pure but not faithfully flat? - -REPLY [6 votes]: Kiran Kedlaya wrote up a detailed exposition of the result you mention. His write-up also describes a bit of the history, mentioning Olivier as one of the first to state it. It is available as Section Tag 08WE in the Stacks project. I think this result as part of the research in the late 60's and early 70's done by a group of mathematicians, including Raynaud, Ferrand, Lazard, Olivier, and others, around flatness, descent, etc. (Sorry, don't have the points to make this a comment.)<|endoftext|> -TITLE: Positive ternary quadratic forms in the same genus that represent the same numbers -QUESTION [11 upvotes]: There are three genera of positive, integral, ternary quadratic forms in which both forms (classes...) are regular, so the paired forms represent the same numbers. These pairs (complete genera) are: -$$ x^2 + y^2 + 9 z^2 + xy, \; \; \; x^2 + 3 y^2 + 3 z^2 + 3 y z, $$ -$$ x^2 + y^2 + 7 z^2 + zx, \; \; \; x^2 + 2 y^2 + 4 z^2 + y z + xy, $$ -$$ x^2 + 4y^2 + 7 z^2 + zx, \; \; \; x^2 + 5 y^2 + 7 z^2 + 5 y z +zx + xy. $$ -In these three cases the relationship of represented numbers is proven. -In response to a question by a master's student in New Zealand, originating with his adviser, Steven Galbraith, I have found, so far, two pairs of irregular forms, in the same genus, that represent the same numbers up to $10^6;$ these are -$$ 3x^2 + 3y^2 + 7 z^2 + yz +2zx +xy, \; \; \; 3x^2 + 5 y^2 + 5 z^2 + 3 y z +zx + 3xy, $$ -$$ 5x^2 + 5y^2 + 8 z^2 +4zx +3xy, \; \; \; 5x^2 + 7 y^2 + 7 z^2 + 6 y z +zx + 5xy. $$ -Cannot prove these relationships, of course. In my accounting, these are discriminant 232 and 648, respectively. The 232 genus has three classes, one more than I displayed, and the 648 genus has 6 classes. I'm just saying. For 232, the genus, collectively, misses $4n+2$ and $4^k (16n+6).$ These are traditionally called the "progressions." The two forms also miss $4^k \{1\}.$ For 648, the progressions are $9n \pm 3, \; \; 81n \pm 27, \;\; 4n+2, \; \; 4^k (16 n + 14).$ The two forms indicated above also miss $4^k \cdot \{1,40\}.$ -NOTE, September 2017: I also checked the two pairs of forms I found for primitive representations. In both cases, the pair of forms in each genus primitively represent the same numbers up to a large bound. If ever proved, these examples violate Conjecture A on page 237, Regular Positive Ternary Quadratic Forms, J. S. Hsia, Mathematika, volume 28, (1981), pages 231-238. - -Anyway, i have written a very fast program and am searching for other pairs. i am disappointed at finding so few. -Note that there are finiteness results about the number of exceptions missed by a ternary form, see Chan and Oh (2004), item 12 at OH_HOME. And, if we count number of representations, no two distinct forms match, i.e. the theta series determines the positive ternary form, proof A. Schiemann. -Meanwhile, note that, if we allow the discriminant to change, there are infinitely many pairs that represent the same numbers, elementary proofs: -$$ A(x^2 + xy+ y^2 ) + B z^2, \; \; \; A(x^2 + 3 y^2 ) + B z^2, $$ -$$ A(x^2 + y^2 + z^2 ) + B (yz + z x + x y), \; \; \; Ax^2 + (2A-B) y^2 + (2A+B) z^2 + 2 B zx. $$ For this one, noticed by Irving Kaplansky, we have discriminant $\Delta = 4A^3 - 3 A B^2 + b^3 = (2A-B)^2 (A+B);$ for this to be positive definite, we need $A > 0$ and $2A > B > -A.$ Given that, the form still might not be "reduced," so it took a while to realize that the different appearances of this were really all the same. -Well, there is the QUESTION, are there only finitely many pairs of forms in the same genus that represent the same numbers? -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - -=====Discriminant 232 ==Genus Size== 3 - Discriminant 232 - Spinor genus misses no exceptions - 232: 1 4 15 2 1 0 vs. s.g. 3 7 11 31 43 - 232: 3 3 7 1 2 1 vs. s.g. 1 single squareclass - 232: 3 5 5 3 1 3 vs. s.g. 1 single squareclass ---------------------------size 3 -The 150 smallest numbers represented by full genus - 1 3 4 5 7 8 9 11 12 13 - 15 16 17 19 20 21 23 25 27 28 - 29 31 32 33 35 36 37 39 40 41 - 43 44 45 47 48 49 51 52 53 55 - 56 57 59 60 61 63 64 65 67 68 - 69 71 72 73 75 76 77 79 80 81 - 83 84 85 87 89 91 92 93 95 97 - 99 100 101 103 104 105 107 108 109 111 - 112 113 115 116 117 119 120 121 123 124 - 125 127 128 129 131 132 133 135 136 137 - 139 140 141 143 144 145 147 148 149 151 - 153 155 156 157 159 160 161 163 164 165 - 167 168 169 171 172 173 175 176 177 179 - 180 181 183 184 185 187 188 189 191 192 - 193 195 196 197 199 200 201 203 204 205 - -The 150 smallest numbers NOT represented by full genus - 2 6 10 14 18 22 24 26 30 34 - 38 42 46 50 54 58 62 66 70 74 - 78 82 86 88 90 94 96 98 102 106 - 110 114 118 122 126 130 134 138 142 146 - 150 152 154 158 162 166 170 174 178 182 - 186 190 194 198 202 206 210 214 216 218 - 222 226 230 234 238 242 246 250 254 258 - 262 266 270 274 278 280 282 286 290 294 - 298 302 306 310 314 318 322 326 330 334 - 338 342 344 346 350 352 354 358 362 366 - 370 374 378 382 384 386 390 394 398 402 - 406 408 410 414 418 422 426 430 434 438 - 442 446 450 454 458 462 466 470 472 474 - 478 482 486 490 494 498 502 506 510 514 - 518 522 526 530 534 536 538 542 546 550 - - ALL ODD -Disc: 232 -================================== - - 232: 1 4 15 2 1 0 -misses, compared with full genus - 3 7 11 12 28 - 31 43 44 48 56 - 79 112 115 120 124 - 141 165 168 172 176 - 184 192 224 295 301 - 309 316 448 456 460 - 471 480 487 496 555 - 564 568 589 616 660 - 672 688 704 736 760 - 768 805 840 896 - - - 232: 3 3 7 1 2 1 -misses, compared with full genus - 1: 1 2: 4 4: 16 8: 64 16: 256 - - ONE ODD - - - 232: 3 5 5 3 1 3 -misses, compared with full genus - 1: 1 2: 4 4: 16 8: 64 16: 256 - - ONE ODD - -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - - -=====Discriminant 648 ==Genus Size== 6 - Discriminant 648 - Spinor genus misses no exceptions - 648: 1 4 41 2 1 0 vs. s.g. 7 11 19 23 31 - 648: 1 5 35 4 1 1 vs. s.g. 8 13 29 31 - 648: 1 11 17 9 1 1 vs. s.g. 5 7 8 65 179 - 648: 4 5 9 3 0 2 vs. s.g. 1 8 - 648: 5 5 8 0 4 3 vs. s.g. 1 40 - 648: 5 7 7 6 1 5 vs. s.g. 1 40 ---------------------------size 6 -The 150 smallest numbers represented by full genus - 1 4 5 7 8 9 11 13 16 17 - 19 20 23 25 28 29 31 32 35 36 - 37 40 41 43 44 45 47 49 52 53 - 55 59 61 63 64 65 67 68 71 72 - 73 76 77 79 80 81 83 85 88 89 - 91 92 95 97 99 100 101 103 104 107 - 109 112 113 115 116 117 119 121 124 125 - 127 128 131 133 136 137 139 140 143 144 - 145 148 149 151 152 153 155 157 160 161 - 163 164 167 169 171 172 173 175 176 179 - 180 181 185 187 188 191 193 196 197 199 - 200 203 205 207 208 209 211 212 215 217 - 220 221 223 225 227 229 232 233 235 236 - 239 241 243 244 245 247 251 252 253 256 - 257 259 260 261 263 265 268 269 271 272 - -The 150 smallest numbers NOT represented by full genus - 2 3 6 10 12 14 15 18 21 22 - 24 26 27 30 33 34 38 39 42 46 - 48 50 51 54 56 57 58 60 62 66 - 69 70 74 75 78 82 84 86 87 90 - 93 94 96 98 102 105 106 108 110 111 - 114 118 120 122 123 126 129 130 132 134 - 135 138 141 142 146 147 150 154 156 158 - 159 162 165 166 168 170 174 177 178 182 - 183 184 186 189 190 192 194 195 198 201 - 202 204 206 210 213 214 216 218 219 222 - 224 226 228 230 231 234 237 238 240 242 - 246 248 249 250 254 255 258 262 264 266 - 267 270 273 274 276 278 282 285 286 290 - 291 294 297 298 300 302 303 306 309 310 - 312 314 318 321 322 326 327 330 334 336 - -Disc: 648 -================================== - - - 648: 1 4 41 2 1 0 -misses, compared with full genus - 7 11 19 23 28 - 31 35 44 76 79 - 88 92 107 112 124 - 140 152 176 280 304 - 316 344 352 368 428 - 448 472 496 536 560 - 608 616 704 811 - - - 648: 1 5 35 4 1 1 -misses, compared with full genus - 8 13 29 31 32 - 128 512 - - - 648: 1 11 17 9 1 1 -misses, compared with full genus - 5 7 8 32 65 - 128 179 512 - - - 648: 4 5 9 3 0 2 -misses, compared with full genus - 1: 1 8 32 128 512 - - - - 648: 5 5 8 0 4 3 -misses, compared with full genus - 1: 1 2: 4 4: 16 40 8: 64 - 160 16: 256 640 - - - 648: 5 7 7 6 1 5 -misses, compared with full genus - 1: 1 2: 4 4: 16 40 8: 64 - 160 16: 256 640 - - -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - -REPLY [2 votes]: Spent a month checking, this is what I suspect is the complete list of 'sporadic' or 'exceptional' pairs. No restriction that they be in the same genus or have the same discriminant. I was able to check discriminant ratio $4$ and discriminant ratio $1$ very high. The other ones just seem sort of random little sets, two quadruples (see the repeated forms). Also some patterns that dry up, discriminants $111,333,999$ but not $2997,$ also on the left $24, 72, 216, 648, 1944,$ but not $5832.$ This last begins with three pair of regular forms, I typed those in. The list of pairs of regular forms that agree is enormous. ---------------------------------------------------------------- - 54 : 2 2 4 1 2 0 216 : 2 4 8 4 1 1 - 75 : 1 4 5 1 1 0 111 : 1 4 7 1 0 0 - 78 : 3 3 3 1 1 3 142 : 3 3 5 2 3 1 - 78 : 3 3 3 1 1 3 158 : 3 3 5 -1 2 1 - 78 : 3 3 3 1 1 3 190 : 3 5 5 5 2 3 - 142 : 3 3 5 2 3 1 158 : 3 3 5 -1 2 1 - 142 : 3 3 5 2 3 1 190 : 3 5 5 5 2 3 - 156 : 3 3 5 2 2 0 284 : 3 5 6 4 2 2 - 156 : 3 3 5 2 2 0 316 : 3 5 6 0 2 2 - 156 : 3 3 5 2 2 0 380 : 3 5 7 2 0 2 - 158 : 3 3 5 -1 2 1 190 : 3 5 5 5 2 3 - 162 : 2 2 14 1 2 2 648 : 2 6 14 3 1 0 - 177 : 2 4 7 4 2 1 213 : 2 4 7 0 1 1 - 225 : 3 4 7 4 3 3 333 : 3 4 7 1 0 0 - 232 : 3 3 7 1 2 1 232 : 3 5 5 3 1 3 - 284 : 3 5 6 4 2 2 316 : 3 5 6 0 2 2 - 284 : 3 5 6 4 2 2 380 : 3 5 7 2 0 2 - 316 : 3 5 6 0 2 2 380 : 3 5 7 2 0 2 - 324 : 4 4 6 0 3 2 567 : 4 6 7 3 2 3 - 486 : 2 2 41 1 2 2 1944 : 2 6 41 3 1 0 - 531 : 5 5 6 0 3 2 639 : 5 5 8 -1 2 4 - 648 : 4 7 7 5 2 2 2592 : 4 7 25 -4 2 2 - 648 : 5 5 8 0 4 3 648 : 5 7 7 6 1 5 - 675 : 5 5 8 -1 4 2 999 : 5 8 8 -5 1 4 -These are pairs of positive quadratic forms that represent the -same numbers, and violate a Kaplansky conjecture. - -Delta : A B C R S T means - -f(x,y,z) = A x^2 + B y^2 + C z^2 + R y z + S z x + T x y, - -and Delta = 4ABC + RST - A R^2 - B S^2 - C T^2. - -The two pair within a genus each are -232 : 3 5 5 3 1 3 232 : 3 3 7 1 2 1 -648 : 5 7 7 6 1 5 648 : 5 5 8 0 4 3 - -The most productive discriminant ratio is 4, -which includes Kap's two infinite families, also - 24 : 1 2 4 2 1 1 6 : 1 1 2 1 1 0 - 72 : 2 2 5 1 1 1 18 : 2 2 2 1 2 2 -216 : 2 5 6 3 0 1 54 : 2 2 5 1 2 2 -648 : 2 6 14 3 1 0 162 : 2 2 14 1 2 2 -1944 : 2 6 41 3 1 0 486 : 2 2 41 1 2 2 -or -48N-24: 2 6 N 3 1 0 12N-6: 2 2 N 1 2 2 -where N = (1+ 3^k)/2, and the pairs for N = 1,2,5 are regular -and have been Schiemann reduced. ------------------------------------------------------------- -Reminder: Kap's two infinite families are equivalent to those -below, which need not be "reduced." For the first, require -gcd(A,C) = 1 and 0 0 and -A < R < 2 A. - -4D : A 3A C 0 0 0 D : A A C 0 0 A - -4D: A 2A-R 2A+R 0 2R 0 D : A A A R R R - -For the first, D = 3 A^2 C, for the second D = (A+R)(2A-R)^2 . - -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -Why not. The complete list of regular pairs, evidently 182 pairs among which various triples, quadruples and so on may be found. Smaller discriminant put first on the line. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - 2 : 1 1 1 1 1 1 18 : 1 2 3 2 1 0 - 2 : 1 1 1 1 1 1 32 : 1 2 4 0 0 0 - 2 : 1 1 1 1 1 1 8 : 1 1 2 0 0 0 - 3 : 1 1 1 0 0 1 12 : 1 1 3 0 0 0 - 3 : 1 1 1 0 0 1 12 : 1 2 2 1 1 1 - 3 : 1 1 1 0 0 1 21 : 1 2 3 0 0 1 - 4 : 1 1 1 0 0 0 16 : 1 2 2 0 0 0 - 4 : 1 1 1 0 0 0 36 : 1 2 5 2 0 0 - 5 : 1 1 2 1 1 1 20 : 1 2 3 1 0 1 - 5 : 1 1 2 1 1 1 20 : 1 2 3 2 0 0 - 6 : 1 1 2 0 0 1 22 : 1 2 3 0 1 0 - 6 : 1 1 2 0 0 1 24 : 1 2 3 0 0 0 - 6 : 1 1 2 1 1 0 15 : 1 1 4 0 1 0 - 6 : 1 1 2 1 1 0 24 : 1 1 6 0 0 0 - 6 : 1 1 2 1 1 0 24 : 1 2 4 2 1 1 - 6 : 1 1 2 1 1 0 33 : 1 2 5 1 1 1 - 8 : 1 1 2 0 0 0 18 : 1 2 3 2 1 0 - 8 : 1 1 2 0 0 0 32 : 1 2 4 0 0 0 - 8 : 1 1 3 1 1 1 32 : 1 3 3 2 0 0 - 8 : 1 1 3 1 1 1 72 : 1 3 7 2 1 1 - 9 : 1 1 3 0 0 1 36 : 1 3 3 0 0 0 - 9 : 1 1 3 0 0 1 36 : 1 3 4 3 1 0 - 9 : 1 1 3 0 0 1 63 : 1 3 6 3 0 0 - 10 : 1 2 2 2 1 1 15 : 1 2 2 1 0 0 - 10 : 1 2 2 2 1 1 40 : 1 2 5 0 0 0 - 12 : 1 1 3 0 0 0 12 : 1 2 2 1 1 1 - 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27 : 1 1 9 0 0 1 108 : 1 3 10 3 1 0 - 27 : 1 1 9 0 0 1 108 : 1 3 9 0 0 0 - 27 : 1 1 9 0 0 1 27 : 1 3 3 3 0 0 - 27 : 1 3 3 3 0 0 108 : 1 3 10 3 1 0 - 27 : 1 3 3 3 0 0 108 : 1 3 9 0 0 0 - 27 : 2 2 2 1 1 1 108 : 2 2 8 2 2 1 - 27 : 2 2 2 1 1 1 108 : 2 3 5 0 2 0 - 27 : 2 2 2 1 1 1 189 : 2 3 8 0 1 0 - 28 : 2 2 3 2 2 2 112 : 2 3 5 2 0 0 - 28 : 2 2 3 2 2 2 60 : 2 3 3 0 0 2 - 28 : 2 2 3 2 2 2 92 : 2 3 5 2 0 2 - 30 : 1 1 10 0 0 1 120 : 1 3 10 0 0 0 - 30 : 1 3 3 1 1 1 46 : 1 3 5 3 1 1 - 30 : 1 3 3 1 1 1 56 : 1 3 5 2 0 0 - 32 : 1 3 3 2 0 0 72 : 1 3 7 2 1 1 - 36 : 1 1 12 0 0 1 144 : 1 3 12 0 0 0 - 36 : 1 3 3 0 0 0 36 : 1 3 4 3 1 0 - 36 : 1 3 3 0 0 0 63 : 1 3 6 3 0 0 - 36 : 1 3 4 3 1 0 63 : 1 3 6 3 0 0 - 36 : 2 2 3 0 0 2 144 : 2 3 6 0 0 0 - 44 : 1 2 6 2 0 0 48 : 1 2 6 0 0 0 - 45 : 2 2 3 0 0 1 72 : 2 2 5 1 1 1 - 45 : 2 2 3 0 0 1 72 : 2 3 3 0 0 0 - 45 : 2 2 3 0 0 1 99 : 2 3 5 3 1 0 - 46 : 1 3 5 3 1 1 56 : 1 3 5 2 0 0 - 48 : 1 4 4 4 0 0 192 : 1 4 12 0 0 0 - 50 : 1 4 4 3 1 1 200 : 1 5 10 0 0 0 - 50 : 1 4 4 3 1 1 75 : 1 4 5 0 0 1 - 54 : 1 1 18 0 0 1 216 : 1 3 18 0 0 0 - 54 : 1 4 4 2 1 1 216 : 1 6 9 0 0 0 - 54 : 1 4 4 2 1 1 297 : 1 6 13 3 1 0 - 54 : 2 2 5 1 2 2 135 : 2 5 5 5 1 2 - 54 : 2 2 5 1 2 2 216 : 2 5 6 0 0 2 - 54 : 2 2 5 1 2 2 216 : 2 5 6 3 0 1 - 54 : 2 2 5 1 2 2 297 : 2 5 8 -2 1 1 - 54 : 2 3 3 3 0 0 216 : 2 3 9 0 0 0 - 60 : 1 4 5 4 1 0 132 : 1 5 7 1 0 1 - 60 : 1 4 5 4 1 0 96 : 1 4 7 4 0 0 - 60 : 2 2 5 0 0 2 240 : 2 5 6 0 0 0 - 60 : 2 3 3 0 0 2 112 : 2 3 5 2 0 0 - 60 : 2 3 3 0 0 2 92 : 2 3 5 2 0 2 - 64 : 3 3 3 -2 2 2 256 : 3 3 8 0 0 2 - 64 : 3 3 3 -2 2 2 576 : 3 3 19 -2 2 2 - 72 : 2 2 5 1 1 1 72 : 2 3 3 0 0 0 - 72 : 2 2 5 1 1 1 99 : 2 3 5 3 1 0 - 72 : 2 3 3 0 0 0 99 : 2 3 5 3 1 0 - 75 : 1 4 5 0 0 1 200 : 1 5 10 0 0 0 - 80 : 3 3 3 2 2 2 320 : 3 4 7 0 2 0 - 90 : 1 1 30 0 0 1 360 : 1 3 30 0 0 0 - 92 : 2 3 5 2 0 2 112 : 2 3 5 2 0 0 - 96 : 1 4 7 4 0 0 132 : 1 5 7 1 0 1 - 98 : 3 3 3 -1 1 1 392 : 3 5 7 0 0 2 - 100 : 2 2 7 -1 1 1 100 : 2 3 5 0 0 2 - 100 : 3 3 3 1 1 1 400 : 3 5 7 0 2 0 - 108 : 1 1 36 0 0 1 432 : 1 3 36 0 0 0 - 108 : 1 3 9 0 0 0 108 : 1 3 10 3 1 0 - 108 : 1 4 7 0 1 0 108 : 1 5 7 5 1 1 - 108 : 1 6 6 6 0 0 432 : 1 6 18 0 0 0 - 108 : 2 2 8 2 2 1 108 : 2 3 5 0 2 0 - 108 : 2 2 8 2 2 1 189 : 2 3 8 0 1 0 - 108 : 2 2 9 0 0 2 432 : 2 6 9 0 0 0 - 108 : 2 3 5 0 2 0 189 : 2 3 8 0 1 0 - 108 : 3 3 5 3 3 3 432 : 3 5 8 4 0 0 - 135 : 2 5 5 5 1 2 216 : 2 5 6 0 0 2 - 135 : 2 5 5 5 1 2 216 : 2 5 6 3 0 1 - 135 : 2 5 5 5 1 2 297 : 2 5 8 -2 1 1 - 144 : 3 4 4 4 0 0 576 : 3 4 12 0 0 0 - 150 : 2 5 5 5 0 0 600 : 2 5 15 0 0 0 - 180 : 2 2 15 0 0 2 720 : 2 6 15 0 0 0 - 180 : 3 5 5 5 3 3 288 : 3 5 5 2 0 0 - 180 : 3 5 5 5 3 3 396 : 3 5 8 2 0 3 - 192 : 1 8 8 8 0 0 704 : 1 8 24 8 0 0 - 192 : 1 8 8 8 0 0 768 : 1 8 24 0 0 0 - 196 : 3 5 5 -4 2 2 784 : 3 5 14 0 0 2 - 216 : 1 6 9 0 0 0 297 : 1 6 13 3 1 0 - 216 : 2 5 6 0 0 2 216 : 2 5 6 3 0 1 - 216 : 2 5 6 0 0 2 297 : 2 5 8 -2 1 1 - 216 : 2 5 6 3 0 1 297 : 2 5 8 -2 1 1 - 240 : 1 4 16 4 0 0 384 : 1 4 24 0 0 0 - 256 : 3 3 8 0 0 2 576 : 3 3 19 -2 2 2 - 270 : 3 3 11 3 3 3 1080 : 3 9 11 6 0 0 - 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-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=<|endoftext|> -TITLE: Convex hulls of families of probability measures -QUESTION [6 upvotes]: Let $X$ be a standard Borel space, so that the space of Borel probability measures on $X$ is also a standard Borel space. We denote it by $\mathcal P(X)$. -In this paper for any family of probability measures $P\subset \mathcal P(X)$ its strong convex hull is defined as -$$ - \operatorname{sco}P:=\left\{\left.\int_{\mathcal P(X)}q\;\nu(\mathrm dq)\;\right|\;\nu\in \mathcal P(\mathcal P(X)): \nu^*(P) = 1\right\} \subset \mathcal P(X). -$$ -which is exactly the set of all convex combinations of elements of $P$: here $\nu^*(P)$ is the outer $\nu$-measure of the set $P$. Since $\operatorname{sco}$ is a monotone map, I guess it admits at least one fixpoint which can be perhaps referred to as a convex closure of $P$. I'm not sure though whether the fixpoint is unique. -My question concerns the literature on convex hulls and convex closures of families of measures e.g. on Borel spaces. The aforementioned paper only contains a single reference to "Probability and Potential" by Dellacherie and Meyer, however I have only volume C available in my library, and there I did not find by any means a comprehensive study of these concepts. Any hint is greatly appreciated. -In particular, besides the uniqueness of the fixpoint I am interested whether the limit -$$ - \operatorname{clo} P := \bigcup_{n\in \Bbb N}\operatorname{sco}^n P -$$ -is a fixpoint of $\operatorname{sco}$. - -REPLY [4 votes]: This is more a comment but I don't have that privilege. If I understand your question correctly, then two very comprehensive papers on the topologies of spaces of probability measures by T. Banakh should be of interest. These originally appeared in russian but are now easily available in english translation (arxiv: 1112.6161 and 1206.1727).<|endoftext|> -TITLE: When is the product $(1+1)(1+4)…(1+n^2)$ a perfect square? -QUESTION [22 upvotes]: This is a modification of an unanswered problem on the math StackExchange. -When is the product $(1+1)(1+4)…(1+n^2)$ a perfect square? -If $(1+1)(1+4)…(1+n^2)=k^2$ then one possibility is $n=3$, $k=10$. Could there be other integer solutions for $(n,k)$? -Surely the answer is NO, but I am unable to prove that $n=3$, $k=10$ is the only possibility. - -REPLY [37 votes]: Javier Cilleruelo has shown that $n=3$ is the only solution: see http://www.uam.es/personal_pdi/ciencias/cillerue/Papers/squares-sinlogo.pdf . -A couple more comments: apparently Chebyshev already showed that the largest prime factor of $\prod_{j=1}^{n} (1+j^2)$ is bigger than $Cn$ for any constant $C$ provided that $n$ is large. This would solve the problem for all large $n$ (taking $C=2$ in Chebyshev's work). The problem of finding the largest prime factor of $\prod_{j=1}^{n} f(j)$ for a polynomial $f$ has been studied by many authors (Erdos, Hooley, Heath-Brown ...); for a first improvement of Chebyshev's work see Erdos's paper http://www.renyi.hu/~p_erdos/1952-07.pdf .<|endoftext|> -TITLE: Wiener measure and Bochner Minlos -QUESTION [10 upvotes]: I am reading probability theory and I have a question. The Bochner-Minlos theorem roughly says that if we have $E \subset H \subset E^*$ where $H$ is a Hilbert space, then there is a unique measure on $E^*$ satisfying some nice properties. -On the other hand, the Wiener measure is supported on the space of continuous functions. Does that mean Wiener measure is not Bochner-Minlos type of measure? In other words, is the Wiener measure the result of the Bochner-Minlos construction? -I apologize if this question is too simple or doesn't even make sense. I am a beginner in this. - -REPLY [5 votes]: Wiener measure can most definitely be characterised as the only probability measure $\mathbf{P}$ on the space $C_0$ of continuous functions starting at the origin and such that the identity -$$ -\int_{C_0} \exp\Big(i \int f(t)\,\mu(dt)\Big)\,\mathbf{P}(df) = \exp \Big(-{1\over 2} \iint (s\wedge t)\,\mu(ds)\,\mu(dt)\Big)\;, -$$ -holds for any compactly supported finite measure $\mu$. This is a consequence of some variant of Bochner's theorem, although it may well be that the one stated in Hida does not exactly fit the bill. (I do not have that book at hand right now.) A good reference with all you ever wanted to know about Gaussian measures (and probably much more) is the book by Bogachev.<|endoftext|> -TITLE: Obstructions for a metric to be conformally equivalent to a product metric -QUESTION [10 upvotes]: Is there a Riemannian metric on $\mathbb{R}^{2}=\mathbb{R} \times \mathbb{R}$ which is not conformally equivalent to a product metric? -More generally, assume that $M$ and $N$ are two manifolds. What obstructions are there for a metric $g$ on $M \times N$ to be conformally equivalent to a product metric for metrics -$g_{1}$ and $g_{2}$ on $M$ and $N$, respectively? - -REPLY [17 votes]: I'll give a partial answer to the OP's second question, which I take to be asking for the obstructions for a metric in dimensions greater than $2$ to be conformal to a product metric. -This is, first of all, a local question, since, even locally, most metrics in dimension $3$ or more are not conformally equivalent to a product metric. The reason is that, up to diffeomorphism, product metrics in dimension $n>1$ depend on arbitrary functions of fewer than $n$ variables while, up to diffeomorphism, conformal classes of metrics in dimension $n$ depend on $\frac12(n{+}1)(n{-}2)$ functions of $n$ variables. (Start with the $\frac12n(n{+}1)$ coefficients of the metric in some coordinate system, subtract $n$ for the action of the diffeomorphism group, and then subtract $1$ more for the conformal scaling factor.) Thus, when $n>2$, the generic metric is not conformal to a nontrivial product metric. -The natural place to look for obstructions, of course, is at the tensors that are conformal invariants. To illustrate the process, let's look at dimension $n=3$. A conformal structure on a $3$-manifold $P$ is a conformal class of metrics $[g]$ on $P$. It is known that the lowest conformal invariant is the Cotton tensor $C\bigl([g]\bigr)$, which is a section of the rank $5$ tensor bundle $S^2_0(T^*,[g])\otimes \bigl|\Lambda^3(T^*)\bigr|^{1/3}$, where $S^2_0(T^*,[g])\subset S^2(T^*)$ is the rank $5$ bundle of quadratic forms that are traceless with respect to $[g]$ and $\bigl|\Lambda^3(T^*)\bigr|^{1/3}$ is the $\frac13$-density bundle. $C\bigl([g]\bigr)$ is a third order tensor in $[g]$ and it vanishes if and only if $[g]$ is conformally flat. -If we consider a metric $g = dt^2 + h$ on $\mathbb{R}\times S$, where $h$ is a metric on an oriented surface $S$ with area form $dA$ and Gauss curvature $K$, we find that -$$ -C\bigl([g]\bigr) = -\bigl(dt\circ({\ast}dK)\bigr)\otimes |dt\wedge dA|^{1/3} -$$ -Thus, $[g]$ is conformally flat if and only if $K$ is constant, and, moreover, when $dK\not=0$, the Cotton tensor in this case is always a (weighted) quadratic form of rank $2$. -In particular, in dimension $3$, if $[g]$ has the property that $C\bigl([g]\bigr)$ has rank $3$, then $[g]$ is not conformal to a product metric in a nontrivial way. This is the first obstruction. (In higher dimensions, this first obstruction will be replaced by algebraic conditions on the Weyl tensor.) -Meanwhile, in dimension $3$, this rank restriction on $C\bigl([g]\bigr)$ is not sufficient to make $[g]$ conformal to a product. To see the further restrictions, suppose that $C\bigl([g]\bigr)$ has rank $2$ everywhere on the $3$-manifold $P$. Then it follows by algebra that there will be a coframing $\eta = (\eta_1,\eta_2,\eta_3)$, unique up to some changes of sign and switching $\eta_1$ and $\eta_2$, so that $[g] = [{\eta_1}^2+{\eta_2}^2+{\eta_3}^2]$ and so that -$$ -C\bigl([g]\bigr) = \bigl(2\ \eta_1{\circ}\eta_2\bigr)\otimes (\eta_1\wedge\eta_2\wedge\eta_3)^{1/3} -$$ -Comparing this with the above formula in the case of a known product, in which the two 'factors' of $C\bigl([g]\bigr)$, namely $dt$ and $*dK$, are integrable, one sees that a further necessary condition is that $\eta_1$ and $\eta_2$ be integrable, i.e., $\eta_i\wedge d\eta_i = 0$ for $i = 1, 2$. Because of the Bianchi identities, this turns out to be only one further condition (this one of $4$th order) on $[g]$. -Indeed, the Bianchi identities for the conformal structure in this coframing, plus the above integrability conditions, can be expressed as follows: There exist functions $p_1,p_2,p_3, q_1, q_2$ such that the following formulae hold -$$ -\begin{aligned} -d\eta_1 &= q_1\,\eta_3\wedge\eta_1 + p_1\,\eta_1\wedge\eta_2\\ -d\eta_2 &= q_2\,\eta_2\wedge\eta_3 + p_2\,\eta_1\wedge\eta_2\\ -d\eta_3 &= 2p_1\,\eta_2\wedge\eta_3 + 2p_2\,\eta_3\wedge\eta_1 - + 2p_3\,\eta_1\wedge\eta_2\\ -\end{aligned} -$$ -Now, again comparing with the known product case, one sees that the only two possible local product structures that could work are, first, the surface foliation defined by $\eta_1 = 0$ paired with the orthogonal curve foliation defined by $\eta_2=\eta_3=0$, or, second, the surface foliation defined by $\eta_2 = 0$ paired with the orthogonal curve foliation defined by $\eta_1=\eta_3=0$. -Now, if the first local product structure is to work, then the quadratic form $Q_1 = {\eta_2}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_1$ dual to $\eta_1$. Since the above formulae imply -$$ -L_{X_1}({\eta_2}^2+{\eta_3}^2) -= 2p_2\ ({\eta_2}^2-2{\eta_3}^2) + 4p_3\ \eta_2{\circ}\eta_3\ , -$$ -one sees that this condition is $p_2 = p_3 = 0$. Note that these conditions imply that $\eta_2$ and $\eta_3$ separately are invariant under the flow of $X_1$. Similarly, if the first local product structure is to work, then the quadratic form $Q_2 = {\eta_1}^2+{\eta_3}^2$ will have to be invariant up to multiples under the flow of the vector field $X_2$ dual to $\eta_2$, which is $p_1 = p_3 = 0$. Either way, this represents two more $4$th order equations on $[g]$. -Finally, suppose that $p_2=p_3=0$. Then the necessary and sufficient local condition -is that $d(q_1\ \eta_3 - p_1\ \eta_2)=0$ (which represents two $5$th order equations on $[g]$), for, then setting $d f = q_1\ \eta_3 - p_1\ \eta_2$, one sees that $\eta_1 = e^f\ dx_1$ for some function $x_1$ and that $f$ is constant on the integral curves of $X_1$, so that -$$ -e^{-2f} \bigl({\eta_1}^2+{\eta_2}^2+{\eta_3}^2\bigr) -= {dx_1}^2 + e^{-2f} \bigl({\eta_2}^2+{\eta_3}^2\bigr) -$$ -where $e^{-2f}\bigl({\eta_2}^2+{\eta_3}^2\bigr)$ is a well-defined metric on the surface that is the space of integral curves of $X_1$. Now, the right hand side of the above equation is visibly a product metric. There is a completely analogous case when $p_1=p_3=0$, for the other possible local product structure. -Just one more remark: When $p_1 = p_2 = p_3 = 0$, one can pursue this analysis and show that the conformal structures that are conformal to a product in two distinct nontrivial ways (but are not conformally flat) can be written in the form -$$ -[g] = \bigl[f_1(x_3)\ {dx_1}^2 + f_2(x_3)\ {dx_2}^2 + {dx_3}^2\bigr] -$$ -where $f_i$ for $i=1,2$ are positive functions of one variable. Thus, the nonflat conformal structures that have more than one 'conformal product structure', even locally, depend only on two arbitrary functions of one variable. -In dimensions above 3: While an exhaustive analysis is probably rather complicated, some general remarks about the dimensions above $3$ will give a sense of what is involved: -First, it is reasonable to ask what one can say about the Weyl curvature of a product metric. First, a few definitions: If $V$ is a vector space endowed with a positive definite inner product, let $R(V)\subset S^2\bigl(\Lambda^2(V^*)\bigr)$ denote the space of Riemann curvature tensors, and let $R(V) = W(V) \oplus Rc(V)$ be the $O(V)$-invariant direct sum decomposition of $R(V)$ into the space $W(V)$ of Weyl curvatures and the $O(V)$-invariant complement $Rc(V)$. If $d\ge 3$ is the dimension of $V$, then -$$ -\dim R(V) = \frac{d^2(d^2{-}1)}{12} -\qquad\text{and}\qquad -\dim W(V) = \frac{d(d{+}1)(d{+}2)(d{-}3)}{12}\ . -$$ -Now, if $V_1$ and $V_2$ are two vector spaces with dimensions $d_1>0$ and $d_2>0$, respectively, with positive definite inner products, and $V_1\oplus V_2$ is given the natural inner product structure, then one has a natural mapping -$$ -R(V_1)\oplus R(V_2)\longrightarrow R(V_1\oplus V_2)\longrightarrow W(V_1\oplus V_2), -$$ -and, except when $d_1=d_2=1$, this mapping has a kernel of dimension $1$. (This corresponds to the fact that a nontrivial Riemannian product of dimension $3$ or more is conformally flat in only two cases: $\mathbb{R}\times M_c$, where $M$ has constant sectional curvature $c$, and $M_{-c}\times N_c$, where $M_{-c}$ has constant sectional curvature $-c$ and $N_c$ has constant sectional curvature $c$.) Let the image of the above mapping be denoted $W(V_1,V_2)\subset W(V_1\oplus V_2)$ -A crude dimension count now shows that, when $\dim V = d > 3$, the union of all the $W(V_1,V_2)$ in $W(V)$ when $V_1$ and $V_2$ are orthogonal complements in $V$ is a proper subvariety of $W(V)$ and that, moreover, usually, the generic element in $W(V_1,V_2)$ does not lie in $W(V'_1, V'_2)$ for any other distinct splitting of $V$. (The reason for the 'usually' is that there is an exception when $2=\dim V_1\le \dim V_2$ and $1 = \dim V'_1<\dim V'_2$). -It follows that, most of the time, in dimensions $d>3$, you'll be able to rule out a Riemannian metric $g$ on a $d$-manifold $M$ being conformal to a product, just by examining the Weyl curvature. Moreover, most of the time that the Weyl curvature of $g$ does actually happen to, pointwise, satisfy the condition of being the Weyl curvature of a product, there will be a unique splitting of the tangent bundle $TM = D_1\oplus D_2$ such that the Weyl curvature of $g$ takes values in the bundle $W(D_1,D_2)$. -For example, when $d=4$, the set of Weyl curvatures in $W(V)$ (a vector space of dimension $10$) that are Weyl curvatures of a product of surfaces is a cone of dimension $5$ that is smooth away from the origin. If you have a Riemannian $4$-manifold whose Weyl curvature is nonvanishing but does take values in this cone, then there is only one splitting of the tangent bundle into $2$-plane bundles that could possibly be the tangents to the leaves of a product structure that could support a product metric conformal to your given metric. -Once you have the only possible splitting $TM = D_1\oplus D_2$ that could work, going the rest of the way is easy: First, the two plane fields $D_i$ must be integrable. Second, when you write $g = g_1 + g_2$ where $g_i$ is positive definite on $D_i$ and has $D_{3-i}$ as null space, then the Lie derivative of $g_i$ with respect to any vector field tangent to $D_{3-i}$ has to be a multiple of $g_i$, or else it cannot work. Third, letting $d_i$ denote the rank of $D_i$, if you now let $\Omega_i$ be a decomposable $d_i$-form that has $D_{3-i}$ as its kernel and is the $g_i$-induced volume form on $D_i$ (this specifies $\Omega_i$ up to a sign), then there will exist a unique $1$-form $\lambda$ such that -$$ -d\Omega_i = d_i\ \lambda\wedge \Omega_i -$$ -(this follows from the integrability that was the first condition); the final local requirement is that this $\lambda$ must be closed, i.e., $d\lambda=0$. If you have this, then, setting $\lambda = d f$ for some function $f$, one sees that -$$ -e^{-2f}g = e^{-2f}g_1 + e^{-2f}g_2 -$$ -and each of the scaled metrics $e^{-2f}g_i$ is well-defined on the $d_i$-dimensional leaf space of the integrable plane field $D_{3-i}$. Thus, $g$ is conformal to a product metric. -Of course, there will be metrics whose Weyl curvatures can be written as split in more than one way (for example, the conformally flat metrics, which have zero Weyl curvature), and a more subtle analysis would need to be done for those, but these will be relatively rare and the analysis is liable to be somewhat messy and dimension dependent. I wouldn't undertake to tackle these special cases without a good reason.<|endoftext|> -TITLE: Is "approximate categoricity" absolute? -QUESTION [11 upvotes]: Let $T$ be a countable first-order theory, and assume that $T$ has exactly one atomic model up to isomorphism in every uncountable cardinality. (By "atomic" I mean a model which omits the non-principal types). -Now let $\mathfrak{M}$ be a countable transitive model of set theory, and assume that $T$ is also (countable) in $\mathfrak{M}$. -Is the above property preserved in $\mathfrak{M}$? i.e. does it hold in $\mathfrak{M}$ that there exists only one atomic model of $T$ in every uncountable cardinality in $\mathfrak{M}$? - -REPLY [16 votes]: This is really a comment, but I need a bit more space. -If $\phi$ is a sentence of ${\cal L}_{\omega_1,\omega}$ that is $\aleph_0$-categorial, there is a complete first order theory $T$ in an expanded vocabulary such that the models of $\phi$ are exactly the reducts atomic models of $\phi$. The expansion is done in such a way that two structures will be isomorphic in the original language if and only if they are isomorphic in the expanded language. -So your question is really the same as: For $\phi$ a sentence of ${\cal}L_{\omega_1,\omega}$ is ``$\phi$ is $\kappa$-categorical for all infinite -$\kappa$" absolute? -This is, as far as I know, still an open question. It is also open if ``$\phi$ is $\aleph_1$-categorical" is absolute. -John Baldwin in his paper ``Amalgamation, Absoluteness, and Categoricity" -addresses some issues around this. Here is a link -http://homepages.math.uic.edu/~jbaldwin/pub/singsep2010rev.pdf<|endoftext|> -TITLE: characterization of commutative Banach algebras -QUESTION [8 upvotes]: Let $A$ be a Banach algebra with the following property: -For every two nets $ x_{\alpha}$ and $y_{\alpha}$ in $A$, $x_{\alpha}y_{\alpha}$ converges if and only if $y_{\alpha}x_{\alpha}$ converges. -Is $A$ necessarily commutative? -1)Note that we do not assume that the above two nets converge to the same value. -2)Note that we do not assume that $A$ has an approximate identity. - -REPLY [10 votes]: No. Some algebras satisfy the identity $xy=-yx$ but are not commutative. Such an algebra clearly satisfies your condition. Obviously, it never has an approximate identity. -To construct one, let $V,W$ be two Banach spaces and let $f: V \times V \to W$ be a continuous symplectic bilinear form. Then define a multiplication on $V + W$ where anything in $W$ times anything is zero, and the multiplication on $V$ is given by $f$. Then this is anticommutative, and not commutative unless $f=0$. So it provides a counterexample.<|endoftext|> -TITLE: Advice for pure-math Phd students -QUESTION [54 upvotes]: Pursuing a Phd in pure math can be a daunting task. A number of students who begin a Phd in pure math don't complete it, and there are high-quality dissertations and those which are not so high quality. -My question is: What advice do you, or would you, give beginning or first-year Phd students early in their studies which will likely increase their possibility of successfully completing a high-quality Phd in pure math? Alternatively: What advice do you wish you were given when you started your Phd? -Are there particular qualities or habits, or is there a particular way of approaching or attitude towards Phd studies, shown by Phd students who complete a high-quality Phd in pure math compared to those who don't? -Is there general principles or advice which can be given to "fit all"? Or do those students who successfully complete a pure-math Phd have "wildly-varying" styles, attitudes and approaches to their studies? -I guess another perspective on this would be: What have you found to be the main reasons for Phd students dropping out or completing a poor-quality pure-math Phd, and what advice could have been given to them early in their studies to prevent these reasons from occuring? -I ask this question because I noticed in my department this year that some pure-math Phd students dropped out in their second or third year of study for various reasons, most of which seemed preventable if they had the right advice early on from their supervisors. Also, the qualities required by pure-math Phd students seem in some ways to be unique compared to other fields. Finally, is it fair to say that professors/supervisors are sometimes not particularly skilled at giving this kind of advice, so many (most?) Phd students are "going without" advice that could really benefit them? - -REPLY [16 votes]: From my personal experience (I am currently in my third year of my PHD), I find that the most valuable thing an advisor can provide is simply insisting that you keep working on your problem. Solving mathematical problems is tough, and your first one even more so. After getting a "result" (it was wrong) fairly quickly after getting my problem and quickly finding out that it was incorrect, I worked essentially fruitlessly at my problem for over a year. This was after dredging through dozens of papers relating to the topic. Many times I felt like I should mention to my advisor that perhaps this problem is not suitable and that I should work on something else. He never waivered, and gently insisted that I keep working. Finally this past summer I was able to make non-trivial progress, albeit not quite in the direction originally anticipated. -In turn, the advice I offer is as follows: there is no replacement for hardwork and perseverance. The return on investment (in terms of monetary value) of a pure math PHD is fairly poor; you can earn a lot more money doing a lot less in many other professions. If the intrinsic value and personal satisfaction of the work is not enough and you are doing it for the money, then start coming up with a plan B. However, if you would rather hammer away at your problem all day than do anything else, then be prepared to work hard often. -Finally, it is important to remember that the role of your advisor is to advise, not to do the work for you. It is important to remember that the key transformation to be undertaken during your PHD is the transition from apprentice to master. After the completion of your PHD you will no longer be your advisor's disciple but your own master, where you will presumably start an academic career where you will primarily fill an instructor or mentorship role. Thus I feel you should expect to go beyond the boundaries naturally set out by your advisor; indeed, it is important for you to be better than your advisor at something (mathematically). If your advisor is a strict upper bound of you in every mathematical aspect, then why should you get a job since presumably your advisor can always do it better than you? So, it is important to work hard on your problem and get good results, but always be mindful of other subject areas, especially if your techniques can be applied to said subject areas.<|endoftext|> -TITLE: Finite dimensional Lie algebra with non-degenerate invariant bilinear forms $\Omega_{ab}$ -QUESTION [6 upvotes]: Firstly, my apology to MO experts that I am in a more science/physics background (a PhD). So please feel free refine/modify/comment my language if I have different math accents than yours. From reading this MO post: complete-classification-of-six-dimensional-non-semi-simple-lie-algebra, I just learned many useful comments and papers in the literature. -From what I had learned in the Reference of this post, we can (strictly?) organize the classes of finite dimensional Lie algebra by: -I. semi-simple. (Killing form is non-degenerate) -II. non-semi-simple (Killing form is degenerate): -$\bullet$ non-solvable. -$\bullet$ solvable and nilpotent. -$\bullet$ solvable but not nilpotent. - -QUESTION: Here I am simply interested in focusing on: -"What are the list of 6 dimensional Lie algebra and 8 dimensional Lie algebra, which allow symmetric non-degenerate invariant bilinear forms $\Omega_{ab}$?" -(hopefully the list can be as complete as possible, but a partial list is welcome.) -If there is a list of corresponding $\Omega_{ab}$ metric, it will be the best. - -I am mostly interested in non-semi-simple case, and in real Lie algebra more than the complex Lie algebra (of course, if providing examples of complex Lie algebra will also be nice). -eg: So far I know only one example in 6-dimension is the nilpotent Lie algebra $A_{6,3}$ with symmetric nondegenerate $\Omega_{ab}$: -The algebra is -$$ -[e_1,e_2]=e_6,\;\;[e_1,e_3]=e_4,\;\;[e_2,e_3]=e_5, -$$ -with other commutators are zeros. One can find the nondegenerate $\Omega_{ab}$ to be: -$$ -\Omega_{ab}={\begin{pmatrix} -q_1 & 0&0&0 &q_2&0 \\ -0& q_3 & 0& -q_2&0 &0\\ -0& 0& q_4 & 0& 0& q_2\\ -0& -q_2 & 0& 0& 0 &0\\ -q_2 &0& 0& 0& 0 &0\\ -0& 0& q_2 & 0& 0 &0 -\end{pmatrix}} -$$ -What are other examples in 6 dimensional Lie algebra and 8 dimensional Lie algebra? (for those semi-simple Lie algebra, I suppose we can use Killing form = $\Omega_{ab}=-{f_{ak}}^l{f_{bl}}^k$. What are the complete lists of semi-simple and non-semi-simple ones of 6 and 8 dimensions?) -Papers/Ref are mostly welcome. (This question is well-motivated by constructing a type of Wess-Zumino-Witten model). Thank you for the concern. - -REPLY [6 votes]: There are several classification lists of solvable and nilpotent quadratic Lie algebras, i.e., having a symmetric, invariant non-degenerate bilinear form. For the classification of nilpotent quadratic Lie algebras of dimension $n\le 7$ over the field of real and complex numbers, see -Piu P., Goze M., Gruppi e Algebre di Lie, appunti per un seminario, Universita degli studi di Cagliari, Dipartimento di Mathematica, 1991. -Gr. Tsagas and P. Nerantzi: Symmetric invariant non-degenerate bilinear forms on nilpotent Lie algebras - see here. -It turns out, that in dimension $6$ there is just one indecomposable nilpotent quadratic algebra, and a decomposable arising from the $5$-dimensional and $1$-dimensional quadratic algebra. In dimension $8$ there are many (two-step nilpotent) examples, but I think, no complete classification. -For the classification of solvable ones in dimension $n\le 6$ see -Tien Dat Pham, Anh vu Le, Minh thanh Duong: Solvable quadratic Lie algebras in low dimension. -In general, see double extension construction and work by Medina and Revoy, Favre and Santaroubane, and many others.<|endoftext|> -TITLE: (Non-)Existence of curves of low degree on affine and projective varieties -QUESTION [12 upvotes]: I am interested in papers that investigates the existence or non-existence of curves of low degree (relative to the degree of the ambient variety). The starting example is that of surfaces and hypersurface of Fermat type, namely the variety $F = 0$ where -$$\displaystyle F(x_1, ..., x_n) = a_1 x_1^d + \cdots + a_n x_n^d$$ -with $a_1, \cdots, a_n \in \mathbb{Z}$. -It was shown by Salberger and Marmon that the only curves of degree less than $(d+1)/3$ that lie on a Fermat surface when $n = 4$ are the 'trivial' lines. It can be shown independently, through methods in Thue equations for example, that these lines (while having low degree) have very few rational points on them. Thus, Fermat surfaces can be proved to have few rational points. -The result stating that Fermat surfaces essentially contains no curves of low degree is rather special and relies heavily on the shape of the equation. I am asking if there exists in the literature any results that attempt to tackle this problem for more general polynomials. Any input would be greatly appreciated. - -REPLY [2 votes]: For curves of degree one you could look at Section 2.4 of Debarre's book "Higher dimensional algebraic geometry". Section 2.14 for Fermat hypersurfaces. -More generally for low degree rational curves: -For hypersurfaces: http://arxiv.org/abs/math/0203088 -For conic connected varieties: http://arxiv.org/abs/math/0701885 -For conics in complete intersections: http://arxiv.org/abs/0804.1627 -For arbitrary projective varieties: http://arxiv.org/abs/1106.0124<|endoftext|> -TITLE: What constant ensures hyperbolicity of Dehn surgery? -QUESTION [9 upvotes]: I am interested in showing that certain knots having a surgery description are hyperbolic. Unfortunately I have not had time yet to read Thurston's work, so my understanding of this is vague. But from reading some papers, it seems that the following is correct (so the first question - is it correct?). Given a hyperbolic link, there is a constant, such that filling a subset of components with slopes with denominators bigger than this given constant, one always gets a hyperbolic manifold (complete, finite volume?). -So my real question is: can we explicitly find this constant? We may or may not be given the volume of the initial link. - -REPLY [13 votes]: Just a remark: there is no universal such constant, even for links with bounded volume complement. -For example, there are infinitely many links with the same complement as the Whitehead link, e.g. - -Then $\frac14$ Dehn filling on the blue component gives the trivial knot, -which is not hyperbolic (at least not with finite volume). Adding twists -to the red component, we may obtain examples with arbitrarily large -denominator non-hyperbolic Dehn fillings.<|endoftext|> -TITLE: Is "being a modular category" a universal or categorical/algebraic property? -QUESTION [11 upvotes]: A semisimple braided category with duals is called modular when a certain matrix $S$ is invertible. The components $S_{AB}$ are indexed by (isomorphism classes of) simple objects of the category and one computes $S_{AB}$ by colouring the Hopf link with (representants of) $A$ and $B$ and evaluates the resulting diagram. One can show that a category is modular iff there are no "transparent" objects (objects that braid trivially with every other object) besides the monoidal unit. -Is being modular a specific property, say in the category of braided categories? Is being a modular category equivalent to being the limit of some diagram or satisfying some diagram? - -REPLY [11 votes]: In general, taking a "center" of a higher category is looking at the endomorphisms of the identity functor. For example, if you think of a monoid as a 1-category, then the endomorphisms of the identity functor are exactly the center of the monoid. Thinking of a tensor category as a 2-category with one object this also gives the Drinfeld center. -If you think of your braided tensor category as a 3 category with one object and one morphism, then this "center" construction yields a symmetric tensor category which is exactly the subcategory of transparent objects! (This is sometimes called the "Mueger center" to distinguish it from the Drinfeld center.) So modularity is just saying that the center of the braided tensor category is trivial.<|endoftext|> -TITLE: Mathematical research published in the form of poems -QUESTION [72 upvotes]: The article -Friedrich Wille: Galerkins Lösungsnäherungen bei monotonen Abbildungen, -Math. Z. 127 (1972), no. 1, 10-16 -is written in the form of a lengthy poem, in a style similar to that -of the works of Wilhelm Busch. -Are there any other examples of original mathematical research published in a similar form? - -REPLY [5 votes]: Here is Samuel Taylor Coleridge on Euclid's first proposition (not original research, but I think the other answers have set ample precedent). Rather than type it all in, I grabbed this from www.poemhunter.com: - -This is now--this was erst, -Proposition the first--and Problem the first. -I. -On a given finite Line -Which must no way incline; -To describe an equi-- ---lateral Tri-- ---A, N, G, L, E. -Now let A. B. -Be the given line -Which must no way incline; -The great Mathematician -Makes this Requisition, -That we describe an Equi-- ---lateral Tri-- ---angle on it: -Aid us, Reason--aid us, Wit! -II. -From the centre A. at the distance A. B. -Describe the circle B. C. D. -At the distance B. A. from B. the centre -The round A. C. E. to describe boldly venture. -(Third Postulate see.) -And from the point C. -In which the circles make a pother -Cutting and slashing one another, -Bid the straight lines a journeying go, -C. A., C. B. those lines will show. -To the points, which by A. B. are reckon'd, -And postulate the second -For Authority ye know. -A. B. C. -Triumphant shall be -An Equilateral Triangle, -Not Peter Pindar carp, not Zoilus can wrangle. -III. -Because the point A. is the centre -Of the circular B. C. D. -And because the point B. is the centre -Of the circular A. C. E. -A. C. to A. B. and B. C. to B. A. -Harmoniously equal for ever must stay; -Then C. A. and B. C. -Both extend the kind hand -To the basis, A. B. -Unambitiously join'd in Equality's Band. -But to the same powers, when two powers are equal, -My mind forbodes the sequel; -My mind does some celestial impulse teach, -And equalises each to each. -Thus C. A. with B. C. strikes the same sure alliance, -That C. A. and B. C. had with A. B. before; -And in mutual affiance, -None attempting to soar -Above another, -The unanimous three -C. A. and B. C. and A. B. -All are equal, each to his brother, -Preserving the balance of power so true: -Ah! the like would the proud Autocratorix do! -At taxes impending not Britain would tremble, -Nor Prussia struggle her fear to dissemble; -Nor the Mah'met-sprung Wight, -The great Mussulman -Would stain his Divan -With Urine the soft-flowing daughter of Fright. -IV. -But rein your stallion in, too daring Nine! -Should Empires bloat the scientific line? -Or with dishevell'd hair all madly do ye run -For transport that your task is done? -For done it is--the cause is tried! -And Proposition, gentle Maid, -Who soothly ask'd stern Demonstration's aid, -Has prov'd her right, and A. B. C. -Of Angles three -Is shown to be of equal side; -And now our weary steed to rest in fine, -'Tis rais'd upon A. B. the straight, the given line.<|endoftext|> -TITLE: Would Elliott-Halberstam conjecture follow from GRH? -QUESTION [8 upvotes]: The Wikipedia article about Elliott-Halberstam (EH for short) conjecture says that the so-called Bombieri-Vinogradov theorem, which is a weaker form of EH conjecture, is in some sense an averaged form of the Generalized Riemann Hypothesis (that is, the analogue of RH for Dirichlet L-functions). So, would the full EH conjecture follow from GRH? -Thanks in advance. -Edit March 29th 2021: denoting by $\Theta_{EH}:=\sup\{\theta\vert EH(\theta)\}$, EH conjecture is equivalent to $\Theta_{EH}=1$. The proven lower bound for it has increased from $1/2$ (Bombieri-Vinogradov) to a greater value. On the other hand, the proven upper bound for the de Bruijn-Newman constant $\Lambda$ has decreased from $1/2$ to $0.2$ (Platt and Trudgian), while the proof of Newman's conjecture by Rodgers and Tao implies $\Lambda\geq 0$, and RH is equivalent to this constant vanishing. So that, it seems that $\Theta_{EH}$ and $\Lambda$ are "dual", the duality relation being the involution $s\mapsto 1-s$. Is it thus likely that $\theta<1-\Lambda\Longrightarrow EH(\theta)$? - -REPLY [11 votes]: The Elliott-Halberstam conjecture is not known to follow from GRH. -Even the weak version of EH (which is with $Q=x^{1/2+\epsilon}$ for any fixed $\epsilon>0$) -does not follow from GRH. On the other hand, it is known that the Elliott-Halberstam conjecture almost implies the twin primes conjecture, i.e., it implies that there are infinitely many pairs of primes at distance $≤ 16$ (now $\le 12$, see Sylvain's comment). Furthermore, the Bombieri-Vinogradov theorem is indeed an amazingly strong unconditional replacement for -the GRH bound (and has as natural strengthening the EH conjecture).<|endoftext|> -TITLE: Can We Decide Whether Small Computer Programs Halt? -QUESTION [26 upvotes]: The undecidability of the halting problem states that there is no general procedure for deciding whether an arbitrary sufficiently complex computer program will halt or not. -Are there some large $n$ and interesting computer languages (say C++, Scheme) for which we have a constructive halting algorithms for programs of length up to $n$? Note that for any fixed bound, there is some finite lists of zero's and one's that stores whether all the programs less than that bound halt or not, so non-constructively for any $n$ there is always a decision procedure for the bounded problem. But I would like to know whether we have some $n$ for which we can constructively write a decision procedure for programs of length less than that bound. By stipulating large $n$, I'd like to avoid $n$ and languages which are trivial in some sense, for instance, decision procedure for languages where the first number of programs all halt and we can easily see that they halt. -Note that the usual proof of the undecidability of the halting program prima facia does not apply, since the decision procedure for programs bounded by $n$ might have length much larger than $n$, so our decision procedure need not correctly apply to our diagonlization program. -What kind of work has been done on this? Are there any good references out there? Thank you! - -REPLY [2 votes]: as noted in other answers this question/problem is similar/nearly equivalent to the Busy beaver problem. the short answer is that even for "small" Turing Machines even with "not that many states", there are some known to be provably universal and therefore the halting problem is uncomputable on them.[1] in fact for "small" TMs it has been shown recently that the halting problem leads to questions about number-theoretic problems similar to the unresolved Collatz conjecture (as noted in another answer).[3] a few useful refs along these lines -[1] Small TMs and the generalized busy beaver competition, Michel (note: could only find .ps fmt for this paper) -[2] Small weakly universal TMs Woods/Neary -[3] Problems in number theory from the busy beaver competition Michel<|endoftext|> -TITLE: Relations between affine Grassmannian and Grassmannian -QUESTION [8 upvotes]: Let $\mathcal K = k((t))$ be the field of formal Laurent series over $k$, and by $\mathcal O = k[[t]]$ the ring of formal power series over $k$. -Let $G$ be an algebraic group over $k$. The affine Grassmannian $Gr_G$ is the functor that associates to a $k$-algebra $A$ the set of isomorphism classes of pairs $(E, \varphi)$, where $E$ is a principal homogeneous space for $G$ over $Spec A[[t]]$ and $\varphi$ is an isomorphism, defined over $Spec A((t))$, of $E$ with the trivial $G$-bundle $G \times Spec A((t))$. -By choosing a trivialization of $E$ over all of $Spec \mathcal O$, the set of $k$-points of $Gr_G$ is identified with the coset space $G(\mathcal K)/G(\mathcal O)$. -Let $V$ be an $n$-dimensional vector space. Then $GL(V)$ acts transitively on the set of all $r$-dimensional subspaces of $V$. Let $H$ be the stabilizer of this action. Then the usual Grassmannian is $GL(V)/H$. -What are the relations between affine Grassmannian and Grassmannian? Why it is important to study affine Grassmannian? Thank you very much. - -REPLY [3 votes]: I am not an expert, but the affine Grassmannian is intimately related to the representation theory of the Langlands dual group $G^{\vee}$. Assume that $G$ is complex semisimple and simply-connected (for simplicity, so to speak). Recall that the dominant weights of $G^{\vee}$ index the finite-dimensional irreducible complex $G^{\vee}$-modules. The dominant weights of $G^{\vee}$ are the dominant coweights of $G$. -Note also that the dominant coweights of $G$ index the strata in a stratification of $Gr_G$. Given a dominant coweight $\lambda$ of $G$, let $Gr^{\lambda}$ denote the corresponding stratum. It turns out that the intersection homology $IH_*(\overline{Gr^{\lambda}})$ is naturally a $G^{\vee}$-module, and is isomorphic to the irrep of $G^{\vee}$ of highest coweight $\lambda$. This description also yields some nice bases for irreps of $G^{\vee}$, called MV cycles. You might read some papers by Kamnitzer and Mirkovic-Vilonen on this subject.<|endoftext|> -TITLE: Totally disconnected locally compact Hausdorff spaces -QUESTION [9 upvotes]: Can any totally disconnected locally compact Hausdorff space be written as a disjoint union of subsets that are both compact and open? -If this is true, does anyone know of a good reference? - -REPLY [8 votes]: I claim that every paracompact locally compact totally disconnected space can be written as a disjoint union of compact open sets. -Every paracompact locally compact space can be partitioned into a collection of $\sigma$-compact open sets (for a proof of this result see my answer to another question here). -Take note that every locally compact totally disconnected space is zero-dimensional (i.e. Every locally compact totally disconnected space has a basis of clopen sets. See this answer for a proof). -Suppose that $X$ is a locally compact paracompact totally disconnected space. Then $X$ can be written as a disjoint union $\bigcup_{i\in I}X_{i}$ of $\sigma$-compact open sets. Since each $X_{i}$ is $\sigma$-compact, the sets $X_{i}$ can be written as countable unions $\bigcup_{n}C_{i,n}$ of compact clopen sets. If $D_{i,n}=C_{i,n}\setminus\bigcup_{m -TITLE: When are generalized Severi-Brauer varieties Grassmannians? -QUESTION [7 upvotes]: Let $F$ be a field and $A$ be an $F$-central simple algebra of degree $n$. Let $0< k< n$ and let $SB_k(A)$ denote the generalized Severi-Brauer variety: if $E/F$ is a field extension, $SB_k(A)(E)$ consists of the right ideals of dimension $kn$ of $A_E=A\otimes_F E$. -If $A$ is split, i.e. $A\simeq M_n(F)$, then $SB_k(A)=Gr(k,n)$, the Grassmannian. - -Is the converse true? If not, can you provide a counterexample? - -The result is true if $k=1$, since $SB_k(A)$ has a rational point over $F$ iff the index of $A$ divides $k$, and Grassmannians have rational points over $F$. -This is cross-posted from Math.SE, where I posted it some days ago, with a few upvotes but no answers or comments. I hope it is ok for MO. -https://math.stackexchange.com/questions/618616/when-are-generalized-severi-brauer-varieties-trivial - -REPLY [4 votes]: Daniel Litt's answer is completely correct, of course. You can also see the result using projective geometry (perhaps this is how Chow proves the result). For every $1$-dimensional subspace $L$ of $F^{\oplus n}$, there is an induced embedding, $$\phi_L:\text{Grass}(k-1,F^{\oplus n}/L) \hookrightarrow \text{Grass}(k,F^{\oplus n}),$$ obtained by associating to every subspace $\overline{V}$ of $F^{\oplus n}/L$ the inverse image $V$ in $F^{\oplus n}$. Denote by $P(t)$ the Hilbert polynomial of $\text{Image}(\phi_L)$ with respect to the ample invertible sheaf $\omega^\vee$, where $\omega$ is the dualizing sheaf of $\text{Grass}(k,F^{\oplus n})$. Then there is a well-defined morphism, $$\Phi:\text{Grass}(1,F^{\oplus n}) \to \text{Hilb}^{P(t)}_{\text{Grass}(k,F^{\oplus n})}, \ \ [L]\mapsto [\text{Image}(\phi_L)].$$ -I claim that $\Phi$ gives an isomorphism of the domain with one of the connected components of the target (when $n\neq 2k$, there is only one connected component, so that $\Phi$ is an $F$-isomorphism). Moreover, it is equivariant for the natural actions of $\textbf{PGL}_n$. Therefore, for every central simple algebra $A$ of "degree" $n$, there is an induced isomorphism of $F$-schemes, -$$ SB_1(A) \to \text{Hilb}^{P(t)}_{SB_k(A)},$$ -where again the Hilbert polynomial is with respect to the very ample invertible sheaf $\omega^\vee$ on $SB_k(A)$. -In particular, if $SB_k(A)$ is isomorphic to $\text{Grass}(k,F^{\oplus n})$, then every connected component of $\text{Hilb}^{P(t)}_{SB_k(A)}$ has an $F$-rational point. Thus, also $SB_1(A)$ has an $F$-rational point. That implies that $A$ is split by the result quoted by the OP. -Of course one still needs to verify that $\Phi$ is an isomorphism. However, since this is a purely geometric result, I imagine that most of us are happier with this claim than the original problem. Also it is fairly easy to prove using some "homogeneity" and cohomology vanishing.<|endoftext|> -TITLE: A question about the Ordinal Definable elements of Power Sets -QUESTION [8 upvotes]: Assume that we are working in a set theory T, formalized in the language of ZFC, whose axioms---in addition to those of ZFC---include also the negation of "V=OD". -For any set X, let CARD(X) denote the cardinal number of X, let ORDEF(X) denote the set of ordinal definable elements of X and let P(X) denote the the set of all subsets of X. As usual, N denotes the set of all non-negative integers and R denotes the set of all real numbers. -It is known that T+"CARD(ORDEF(P(N)))=CARD(N)" is consistent if ZFC is. - -I would like to know whether this phenomenon---that the consistency of ZFC implies the consistency of T+"CARD(ORDEF(P(X)))=CARD(X)"---is true for many sets X? In particular, is it true for X=R? - -My reason for being interested in this question is that, for many uncountable sets X, it becomes difficult to work in P(X) because so many sets---most of which we do not care about---belong to P(X). -If CARD(ORDEF(P(X)))= CARD(X) and if no inconsistency is introduced, we could work in ORDEF(P(X)) instead of in P(X). Most of the sets belonging to P(X) in which we are interested should still belong to ORDEF(P(X)) and we would be working in a set is no greater than that of X. -Just consider how few of the sets belonging to P(R) are of any interest to anybody. Most of them are literally indescribable in any conceivable language. - -REPLY [2 votes]: A marginal comment. -ORDEF(P(N)) can be countable while ORDEF(P(R)) cannot since for any countable ordinal $\alpha$ there is a non-empty OD set of all reals which code $\alpha$. Is there an UNcountable set $X$ such that ORDEF(X) is countable (or just empty?)<|endoftext|> -TITLE: Heronian triangle with two sides that are prime -QUESTION [12 upvotes]: Can any prime number form a Heronian triangle with a second prime as another side? I cannot find a second prime to form a Heronian triangle with either 23 or 167. I have checked up to the 10^7th prime for both with no solution. Also if there is a solution for primes other than 23 or 167 to have a second prime as another side, is the solution set for that prime finite or infinite. See OEIS A230666 and A233232 for primes 3 and 5 where the solution sets are infinite. - -REPLY [15 votes]: For $p=23$, one Heronian triangle with another prime side $q$ has -$q = 5280071830550089$, with third side $q-1$ and area -$60663406817631420 = 2^2 \, 3^4 \, 11 \; 23 \; 37^2 \, 47 \; 71 \; 179 \; 181$. -For each $p$ there are probably infinitely many examples -but very sparse; for $p=167$ I didn't find one with a prime side -among the first few dozen solutions (though there might be a few other -variants to try). [Added later: an example is -$q = 231781748893580717709514473745694370721$, -for which the triangle with sides $167$, $q-25$, $q$ -has area $19135685576510124949571252858502010748400$ -$$ -= 2^4 \, 3 \; 5^2 \, 29 \; 43 \; 71 \; 167 \; 769 \; 29063 \; 250233481 \; 1154762937707.] -$$ -This comes down to a few Fermat-Pell equations, as -Gerhard "insert quote here" Paseman suggested; but the difficulty -is not the size of the fundamental unit (which can be as small as -$p + \sqrt{p^2-1}$) but the rare and unpredictable appearance of primes in -the resulting sequence; -I doubt that anything can be proved about the question. -[The Diophantine equation for a triangle of sides $p,q,q-d$ to be Heronian is -$(p^2-d^2) (2q-p-d) (2q+p-d) = x^2$.]<|endoftext|> -TITLE: Smoothness and Kähler differentials -QUESTION [7 upvotes]: Let $X$ be a complex variety. It is well-known that $X$ is smooth if and only if the sheaf of Kähler differentials $\Omega_X^1$ is locally free (Hartshorne p. 177). -Question: What happens for forms of higher degree? I.e. define $\Omega_X^p := \bigwedge^p \Omega_X^1$ (no reflexive hull or something). -For what values of $p$ and under what additional assumptions on $X$ does $\Omega_X^p$ locally free imply $X$ smooth? - -REPLY [5 votes]: I think that if you do not take reflexive hulls, then all $p\leq \dim X$ should work. Since you said "variety" I assume you mean "reduced". In that case, $\mathscr F$ being locally free is equivalent to $\dim \mathscr F_x\otimes \kappa(x)$ being constant. Since tensor operations commute, it seems to me that a coherent sheaf is locally free if and only if any non-zero exterior power of it is locally free, since you can compute the above value for any exterior power and if that is constant, then the original had to be constant. -If you allow reflexive hulls, then obviously there are varieties with $\omega_X$ being a line bundle. The next question could be whether having all reflexive powers locally free would imply smoothness. Of course, this would follow from the Lipman-Zariski conjecture, so this may be known or easy.<|endoftext|> -TITLE: The geometric-mean factorial -QUESTION [12 upvotes]: Think of the factorial as $f(n) = n \odot (n-1) \odot \cdots \odot 2 \odot 1$, -where $\odot$ is the binary operator for multiplication, $\cdot$. This suggests exploring replacing -$\odot$ with other operators. -$\odot = +$ just results in $(n+1)n/2$. E.g., for $n=10$, the plus-factorial is $55$. -$\odot = -$ results in $-n(n-1)/2$. E.g., for $n=10$, the minus-factorial is $-35$. -Here (and below) I am associating to the left, i.e., $((4-3)-2)-1=-2$. -So I explored the binary operation $ a \odot b = \sqrt{a b}$, a geometric-mean operation. -Then the geometric-mean-factorial $gm!(n)$ looks like this: -$$gm!(2) = \sqrt{2 \cdot 1} \approx 1.41421$$ -$$gm!(3) = \sqrt{(\sqrt{3 \cdot 2}) \cdot 1} = 6^{\frac{1}{4}} \approx 1.56508$$ -$$gm!(4) = \sqrt{(\sqrt{(\sqrt{4 \cdot 3}) \cdot 2}) \cdot 1}=\sqrt{2} \cdot 3^{\frac{1}{8}} \approx 1.62239$$ -$$\ldots$$ -$$gm!(20) \approx 1.66169$$ -The geometric-mean-factorial seems to be approaching a limit that -is unfamiliar to me. -Does anyone recognize this constant from another context? - -REPLY [15 votes]: looks like Somos's quadratic recurrence constant<|endoftext|> -TITLE: Are there some list of the finite subgroups of the mapping class groups of low genus surfaces? -QUESTION [6 upvotes]: We already know the bound of the order of the finite subgroups of the $Mod(S_g)$. If we take a further step, to find all the finite subgroups, then what is the result for low genus cases? For example, are there some list of all types of the finite subgroups of $Mod(S_g)$ for $g \leq 10$? - -REPLY [4 votes]: As pointed out in the preceding answer, what you are looking for are all possible automorphisms of a compact Riemann surface of genus $g$. You'll find some answers at this question.<|endoftext|> -TITLE: Find functions such that f(f(x))=f(x)e^x -QUESTION [9 upvotes]: Are there monotonically increasing functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = e^x f(x)$? - -REPLY [6 votes]: By monotonically increasing, do you mean strictly increasing or non-decreasing? -If you mean non-decreasing, then there is the trivial case $f(x)=0$. However, there may well be other cases that are harder to find. -If you mean strictly increasing, then there are no continuous solutions (there might still be discontinuous ones): -As proven by Carnahan, $f(0)=0$. -Now, consider the following limit: -$$\alpha=\lim_{x \to -\infty}{f(x)}$$ -We shall consider three cases, $\alpha=0$, $-\infty<\alpha<0$, and $\alpha=-\infty$. -In the first case, one of three things occurs: it approaches from above, from below, or from equal to zero. All three of these contradict that $f$ is strictly increasing. -In the second case, we have: -$$f(\alpha)=\lim_{x \to -\infty}{f(f(x))}=\lim_{x \to -\infty}{{e^x}{f(x)}}=\lim_{x \to -\infty}{e^x} \times \lim_{x \to -\infty}{f(x)}=0 \times \alpha=0$$ -So, $f(\alpha)=0=f(0)$ and since $\alpha<0$, we get a contradiction. -In the third case, -$$-\infty=\lim_{x \to -\infty}{f(f(x))}=\lim_{x \to -\infty}{{e^x}{f(x)}}=\lim_{x \to -\infty}\frac{f(x)} {e^x}$$ -Which means that it goes down to negative infinity at a greater than exponential rate. -Let $k=f(-1)$ -Firstly, $k<0$ because $k=f(-1)k$ -Also, since it descends to negative infinity faster than exponentially, there must exist a finite $N$ such that $n -TITLE: Duality between orbifold and quasi-Hopf algebra (twisted quantum doubles) -QUESTION [9 upvotes]: A quick Question: - -Is there some duality known between the quasi Hopf algebra -$D^\omega(H)$ of a finite group $H$ to an orbifold model (such as -SU(2)/$G$ or SO(3)/$G$ orbifold of some group $G$)? What is this duality relation precisely? - - -Background: -It is known (in theoretical physics) that the algebraic framework underlying discrete H gauge theories with 2+1D Chern-Simons term is the quasi Hopf algebra $D^\omega(H)$, i.e. the Chern-Simons term introduces a 3-cocycle $\omega \in H^4(BH,\mathbb{Z}) \simeq H^4(H,\mathbb{Z}) \simeq H^3(H,U(1))$ in the cohomology group on the Hopf algebra $D(H)$. People in theoretical physics also call the quasi Hopf algebra $D^\omega(H)$ as another name: twisted quantum doubles, such as A Kitaev's (of Caltech) Annals of Physics 303, 2 (2003), Annals of Physics 321, 2 (2006). The background understanding of these topics (to me) would go to Dijkgraaf-Witten theory original paper. -My question here is inspired by the observation in this arXiv paper published in Nucl.Phys. B. It stated that: "From the point of view of conformal field theory it is of interest to mention that the -fusion rules of $D^\omega(\mathbb{H}_8)$ for p = 1 coincide with the level 1 SU(2)/($\mathbb{Z}_2 \times \mathbb{Z}_2$)-orbifold (cited a paper by Dijkgraaf, Vafa, Verlinde, Verlinde) after -modding out the appropriate $\mathbb{Z}_2$ generated by 1 (see Table 2 here)). Apparently, the algebraic structure of such non-holomorphic orbifolds is still determined by the ‘holomorphic’ Hopf algebra, be it deformed by a non-trivial 3-cocycle. To our knowledge, this has not been noticed before." - -A detailed Question: - -It seems to me that there may have some duality between: -$$ \text{quasi Hopf algebra } D^\omega(\mathbb{H}_8) \text{ for p = 1} \leftrightarrow \text{level 1 SU(2)/($\mathbb{Z}_2 \times \mathbb{Z}_2$) orbifold} $$ -Here $p = 1$ is the 3-cocycles labeled of $H^3(\mathbb{H}_8,U(1))=\mathbb{Z}_8$ for $p$(mod 8) in $\mathbb{Z}_8$. How about other 7 classes other than $p=1$ in $p$(mod 8)? - - -Are there other some dualities exist for -$$D^\omega(\mathbb{H}_8) \leftrightarrow \text{? orbifold}$$ -$$D^\omega(D_8) \leftrightarrow \text{? orbifold} $$ -$$D^\omega(\mathbb{Z}_2^3) \leftrightarrow \text{? orbifold} $$ -What is the general relation (if any, start with a finite group $H$)? -$$D^\omega(H) \leftrightarrow \text{? orbifold} $$ - - -$D_8$ is a dihedral group with 8 group elements. $D^\omega(D_8)$ should have three labels of $p_1$,$p_2$,$p_3$ from $H^3(D_8,U(1))=\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. And $D^\omega(\mathbb{Z}_2^3)$ should have 7 labels of $p_j$ from $H^3(\mathbb{Z}_2^3,U(1))=\mathbb{Z}_2^7$. -ps. Excuse me that my mathematical background is not equivalent to a math PhD (but trained in physics), but this should be a research level question in mathematical physics. Please feel free giving comments/answers. Thank you for all who reply and support! - -REPLY [2 votes]: Regarding the general question, from the point of view of conformal field theory -there is a rather trivial way to obtain (some) $D^\omega(G)$. Namely, the representation category of the $G$-orbifold of a holomorphic (trivial representation category) rational conformal field theory is $\mathrm{Rep}(D^\omega(G))$ for some $[\omega]$, see Corollary 3.6. in -http://arxiv.org/abs/0909.2537 -But I have now idea if as Scott pointed out as a comment all finite groups $G$ (I suppose yes) can be obtained this way; one has to find a holomorphic theory with an action of $G$, for example the Moonshine CFT for the Monster group etc. Then for a given $G$ I also have no idea which $[\omega]$ arise this way. -I think it also follows conversely, that if a CFT has $\mathrm{Rep}(D^\omega(G))$ as representation category, then it is a $G$-orbifold of a holomorphic theory. -But then I don't understand the non-holomorphic examples the op mentioned. In the non-holomorphic examples you have to mod something out to become (the dual of) $D^\omega(G)$<|endoftext|> -TITLE: Why should I care about topological modular forms? -QUESTION [44 upvotes]: There seems to be a lot of recent activity concerning topological modular forms (TMF), which I gather is an extraordinary cohomology theory constructed from the classical theory of modular forms on the moduli space of elliptic curves. I gather that the homotopy theorists regard it as a major achievement. -My question is as follows : what sorts of more classical topological problems is it useful for solving? It is clear why non-algebraic-topologists should care about the more classical extraordinary cohomology theories (e.g. K-theory and cobordism theory); they allow elegant solutions to problems that are obviously of classical interest. Does TMF also allow solutions to such problems, or is it only of technical interest within algebraic topology? -My background : my research interests are in geometry and topology, but I am not an algebraic topologist (though I have used quite a bit of algebraic topology in my research). - -REPLY [31 votes]: One of the closer connections to geometric topology is likely from invariants of manifolds. The motivating reason for the development of topological modular forms was the Witten genus. The original version of the Witten genus associates power series invariants in $\mathbb{C}[[q]]$ to oriented manifolds, and it was argued that what it calculates on M is an $S^1$-equivariant index of a Dirac operator on the free loop space $Map(S^1,M)$. It is also an elliptic genus, which Ochanine describes much better than I could here. -This is supposed to have especially interesting behavior on certain manifolds. An orientation of a manifold is a lift of the structure of its tangent bundle from the orthogonal group $O(n)$ to the special orthogonal group $SO(n)$, which can be regarded as choosing data that exhibits triviality of the first Stiefel-Whitney class $w_1(M)$. A Spin manifold has its structure group further lifted to $Spin(n)$, trivializing $w_2(M)$. For Spin manifolds, the first Pontrjagin class $p_1(M)$ is canonically twice another class, which we sometimes call "$p_1(M)/2$"; a String manifold has a lift to the String group trivializing this class. Just as the $\hat A$-genus is supposed to take integer values on manifolds with a spin structure, it was argued by Witten that the Witten genus of a String manifold should take values in a certain subring: namely, power series in $\Bbb{Z}[[q]]$ which are modular forms. This is a very particular subring $MF_*$ isomorphic to $\Bbb{Z}[c_4,c_6,\Delta]/(c_4^3 - c_6^2 - 1728\Delta)$. -The development of the universal elliptic cohomology theory ${\cal Ell}$, its refinement at the primes $2$ and $3$ to topoogical modular forms $tmf$, and the so-called sigma orientation were initiated by the desire to prove these results. They produced a factorization of the Witten genus $MString_* \to \Bbb{C}[[q]]$ as follows: -$$ -MString_* \to \pi_* tmf \to MF_* \subset \Bbb{C}[[q]] -$$ -Moreover, the map $\pi_* tmf \to MF_*$ can be viewed as an edge morphism in a spectral sequence. There are also multiplicative structures in this story: the genus $MString_* \to \pi_* tmf$ preserves something a little stronger than the multiplicative structure, such as certain secondary products of String manifolds and geometric "power" constructions. -What does this refinement give us, purely from the point of view of manifold invariants? - -The map $\pi_* tmf \to MF_*$ is a rational isomorphism, but not a surjection. As a result, there are certain values that the Witten genus does not take, just as the $\hat A$--genus of a Spin manifold of dimension congruent to 4 mod 8 must be an even integer (which implies Rokhlin's theorem). Some examples: $c_6$ is not in the image but $2c_6$ is, which forces the Witten genus of 12-dimensional String manifolds to have even integers in their power series expansion; similarly $\Delta$ is not in the image, but $24\Delta$ and $\Delta^{24}$ both are. (The full image takes more work to describe.) -The map $\pi_* tmf \to MF_*$ is also not an injection; there are many torsion classes and classes in odd degrees which are annihilated. These actually provide bordism invariants of String manifolds that aren't actually detected by the Witten genus, but are morally connected in some sense because they can be described cohomologically via universal congruences of elliptic genera. For example, the framed manifolds $S^1$ and $S^3$ are detected, and Mike Hopkins' ICM address that Drew linked to describes how a really surprising range of framed manifolds is detected perfectly by $\pi_* tmf$. - -These results could be regarded as "the next version" of the same story for the relationship between the $\hat A$-genus and the Atiyah-Bott-Shapiro orientation for Spin manifolds. They suggest further stages. And the existence, the tools for construction, and the perspective they bring into the subject have been highly influential within homotopy theory, for entirely different reasons. -Hope this provides at least a little motivation.<|endoftext|> -TITLE: What is the relation between Lefschetz fixed point theorem and Poincare-Hopf theorem on vector fields? -QUESTION [8 upvotes]: In Dubrovin/Fomenko/Novikov Modern geometry--Methods and applications, Part II, the (Poincare-)Hopf theorem is treated in section 15.2 (see theorem 15.2.7 on page 129), while the Lefschetz theorem on fixed points of self-maps is treated in the adjacent section 15.3 (see theorem 5.3.2 on page 131). -Even though the sections are adjacent, no mention is made of the relation between the two theorems, perhaps because this is "obvious" (take the flow of the vector field for small time, then apply Lefschetz, etc). -Is there a framework that would allow one to deduce both results as a consequence of a more general theorem? - -REPLY [9 votes]: The Poincaré-Hopf theorem is a consequence of the Lefschetz fixed point theorem if you accept the fairly standard fact that the Euler characteristic of a compact manifold $M$ is equal to the self-intersection of the diagonal $\Delta \subset M\times M$. Here is why : -Since $M$ is compact, the vector field $X$ gives rise to a flow $(\varphi _t)_{t\in\mathbb{R}}$; the fixed points of $\varphi _t$ are the zeros of $X$. Suppose these zeros are isolated, and let $p$ be one of them. In local coordinates, we have $\varphi _t(x)= x+tX(x)+o(t)$ around $p$, hence $d\varphi _t(p)=I+t\,dX(p)+o(t)$. This implies that for small $t$, the index of $X$ at $p$ is equal to the index of $p$ as a fixed point of $\varphi _t$. Lefschetz theorem tells you that the sum of these indices is the intersection number $(\Delta .\Delta _t)$, where $\Delta _t$ is the graph of $\varphi _t$. This integer depends continuously on $t$, hence is constant, hence equal to $(\Delta .\Delta) =e(M)$.<|endoftext|> -TITLE: Teaching homology via everyday examples -QUESTION [81 upvotes]: What stories, puzzles, games, paradoxes, toys, etc from everyday life are better understood after learning homology theory? - -To be more precise, I am teaching a short course on homology, from chapter two of Hatcher's book. Before diving into the details of Delta-complexes, good pairs, long exact sequences, degrees, and so on, I would like to present a collection of real-life phenomena that are greatly illuminated by actually knowing homology theory. Ideally, I would refer back to these examples as the course progressed and explain them with the new tools the students learn. -Here is what I have thought of so far. - -Tavern puzzles: before trying to solve a tavern puzzle, it is always a good idea to check that the two pieces are topologically unlinked. You can approximate this by computing the linking number, via the degree of the map from the torus to the two-sphere. -Cowlicks: after getting a very short haircut, there is a place on your scalp (typically near the crown) where the hair is standing up. This is "explained" by the hairy ball theorem: a tangent vector field on the two-sphere must vanish somewhere. "Explained" is in quotes because your hair is a vector field on a disk, not on a sphere. -When you knead dough, and then push it back into its original shape, there is always a bit of dough that has returned to its original position. This is the Brouwer fixed-point theorem. This example is a bit tricky, because it is impossible to see the fixed bit of dough. (Now, when you are hiking, you can clearly see that there is a point of the map lying directly above the point it represents. However, this is better explained by the contraction mapping principle.) -There are no draws in the board game Hex. This is equivalent to the Brouwer fixed-point theorem. This isn't a perfect example, because most people don't know the game. - -Noticeably missing are puzzles, etc that rely on homological algebra (diagram chasing, long exact sequences, etc). -$\newcommand{\SO}{\operatorname{SO}} \newcommand{\ZZ}{\mathbb{Z}}$ -EDIT: Let me also give some non-examples, to clarify what I am asking. - -Linking number also arises in discussions of DNA replication; see discussions of topoisomerase. However DNA is not an everyday object, so doesn't work as an example. -The plate (or belt) trick; this is a fancy move that a waiter can make with your plate, but it is more likely to appear in a juggling show. It is explained by knowing $\pi_1(\SO(3)) = \ZZ/2\ZZ$. However, this is a fact about the fundamental group, not about homology. -Impossible objects such as the Penrose tribar that exist locally, but not globally. These can be explained via non-trivial cohomology classes. But, cohomology is in the next class, so if/when I teach that... - -REPLY [2 votes]: Allen Hatcher's book on topology has a "motivating" discussion of cohomology. He says it's a map. -the countries represent elements of the homology. latitude and longitude are functions from the topological space to the reals. The level sets are then also represented by curves on our map (such as isotherms or isobars). and these will be elements of the cohomology<|endoftext|> -TITLE: Metrics on the space of $C^{*}$ algebras -QUESTION [6 upvotes]: I think that there is a metric on the huge space of all $C^{*}$ algebras. What is the explicit -definition of this metric?may you introduce me a reference? -Moreover is the restriction of this metric to commutative $C^{*}$ algebras gives us a discrete metric? by discrete I mean "every commutative $C^{*}$ algebra has a neighborhood, with respect to this metric, which contains no other commutative $C^{*}$ algebra" - -REPLY [3 votes]: Another possibility is the Kadison-Kastler metric on C*-subalgebras of $B(H)$, which is just the Hausdorff distance between their unit balls. This paper gives references to a number of results about stability under perturbation, which include the case where one of the algebras is separable and abelian; this paper generalizes to the case where one of the algebras is separable and nuclear.<|endoftext|> -TITLE: The Bialynicki-Birula Stratification of the Affine Grassmannian -QUESTION [8 upvotes]: Let $G$ be a connected, simply-connected complex semisimple group with affine Grassmannian $\mathcal{G}r$. Fix a maximal torus and Borel $T\subseteq B\subseteq G$. I am reading "Loop Grassmannian Cohomology, the Principal Nilpotent and Kostant Theorem" by Ginzburg. He says that there exists a one-parameter subgroup $\nu:\mathbb{C}^*\rightarrow T$ so that the Bialynicki-Birula stratification of $\mathcal{G}r$ associated with $\mathbb{C}^*$ acting on $\mathcal{G}r$ through $\nu$ coincides with the usual stratification into Iwahori-orbits. I would appreciate a description of this one-parameter subgroup, if possible. I would also appreciate any relevant references. - -REPLY [10 votes]: If I understand what's written Ginzburg correctly, this claim is incorrect, but very easily fixed. -Why is this incorrect: Because there's no cocharacter into $T$ which has finite dimensional BB cells. If there were, then there would be a character whose action on tangent space at the identity coset $[e]$ had a finite dimensional space of positive weights. Since this space is $\mathfrak g((t))/\mathfrak g[[t]]$, and thus infinitely many copies of the adjoint rep of $\mathfrak{g}$, that's impossible. -How it can easily be fixed: While there's no such cocharacter into $T$, there is one into $T\times \mathbb C^*$ where $\mathbb{C}^*$ acts by loop rotation. You want to pick a cocharacter into this group where the positive weight spaces on $\mathfrak{g}((t))$ are given by the Iwahori. There are many such cocharacters (one for each way of assigning positive integers to the simple roots of the corresponding affine Lie algebra). The "most canonical" choice is to take the principal cocharacter (the one that acts with weight 1 on all simple root spaces) in $T$ and the $h+1$st power of the loop rotation (where $h$ is the Coxeter number). I might have that wrong, but anyways, you take a high enough power of loop rotation to assure you have positive weight on $t\mathfrak{g}[[t]]$.<|endoftext|> -TITLE: Infinite blue eyed islanders puzzle -QUESTION [6 upvotes]: Can the well known blue eyed islanders puzzle be extended to an infinite number of islanders? -In that puzzle, a set of $k$ islanders, each with either blue eyes or non-blue eyes, each knows the color of every other islander's eyes but not his own. At time $t=0,1,...$ an islander who knows the color of his eyes raises his hand. Each islander can see which other islanders raised their hands at earlier times and nothing else. At time $t=0$ a visitor tells the islanders that at least one of them has blue eyes. All of the above is common knowledge among the islanders. -Suppose $k$ is finite, as in the initial version of the puzzle. Then, if all the islanders have blue eyes, they will all raise their hands at time $t=k$. -If $k$ is infinite, will they ever raise their hands? -I think it is best to assume here $t$ is an ordinal, and that each islander knows who raised his hand at any lower ordinal. -Certainly, if initially only a finite number of islanders have blue eyes, then after a finite time everyone will know his color. Thus, if initially they all have blue eyes, then after time $t=\omega$, it will be common knowledge that there are an infinite number of islanders with blue eyes. Unfortunately, that gets us nothing because I think this fact was already common knowledge at time $t=0$. I also considered maybe having the islanders collaborate on a strategy somehow - so long as they all know the others are using the strategy, and so long as it is still the case each only raises his hand when he knows his own eye color - but made no headway. -More speculatively, is there some kind of variant of this puzzle that could usefully distinguish between knowledge at transfinite times? I played around with different kinds of finite knowledge - maybe each islander can only see a subset of his fellows - but got nowhere. - -REPLY [2 votes]: Here's my attempt to rigorously show that it cannot generalize. I originally wrote this for my blog, using the following formulation of the puzzle: - -All islanders are immortal, except as explained below. -All islanders know the color of every other islander's eyes, but not their own. -If an islander has enough information to logically deduce the color of their -own eyes, they will perish at midnight of the following day, which will be -noticed by the other islanders. -Any message washed ashore in a bottle can be trusted. - -Now as it happens, at 12:01am on Day 0, a message washes ashore in a bottle, -and is read by all the islanders: - -There is at least one islander with blue eyes. - -What happens? -Finite solution -The following solution works -for infinitely many islanders, as long as only finitely many have blue eyes. I go into more information on my blog post, but since you already know the blue eyes puzzle you can probably prove the following. -Lemma 1 If an islander sees at least $N$ other islanders with blue -eyes, she knows on Day $0$ that she will not be able to deduce her eye -color before Day $N$. -Lemma 2 If an islander sees $N$ other islanders with blue eyes who -are alive on Day $N$, then she can deduce her eye color by Day $N+1$. -Theorem 3 If there are $N+1$ islanders with blue eyes, they will -perish at midnight on Day $N+1$, and all others will survive. -The infinite case -So, we've handled the case where there are an arbitrary number of islanders -(even infinite!), but only finitely many islanders with blue eyes. So, -what if there are infinitely many islanders with blue eyes? -Well, we already have a very nice lemma above, Lemma 1. That allows us -to prove that, since all islanders see infinitely many pairs of blue eyes, -no islander can deduce their eye color by Day $N$ for all -$N\in\{0,1,2,\dots\}$. -So, assuming that our days are limited to the finite ordinals, the case is -closed. But why should we stop there? What if there exists a day for every -ordinal number? -We should consider what happens on Day $\omega$. Will anyone wake up dead? -The answer is of course, no. All islanders were able to deduce from the -beginning that no one could perish after a finite number of days, so -by the time Day $\omega$ begins, no one has gained information they didn't -already have. We shall be able to extend this argument by -transfinite induction to handle every ordinal -(regardless of the cardinality of the ordinal or the set of blue-eyed islanders). -Theorem 4 If there infinitely many islanders with blue eyes, then no islanders -can deduce their eye color by Day $\alpha$, where $\alpha$ is any ordinal. -Proof. If $\alpha$ is finite, we are done by Lemma 1. By transfinite induction, -we may assume the theorem holds for all $\beta\lt\alpha$. In order to -deduce her eye color by Day $\alpha$, $Alice$ requires new information on -a previous day. However, by the induction hypothesis, $Alice$ knows from -the beginning that every islander cannot deduce their eye color by a previous -day. So, if $\alpha$ is a limit ordinal, we are done. The other case -is when $\alpha=\beta+1$ for some $\beta\lt\alpha$. Alice wakes up -on Day $\beta$, and is immediately bored. She already knew that everyone -sees infinitely many islanders with blue eyes, so she already knew that -everyone would be alive. Indeed, they are. So, she has gained no new information -that morning, and therefore cannot deduce anything new by the end of the day, -the beginning of Day $\alpha$. $\square$<|endoftext|> -TITLE: Determinant and eigenvalues of a specific matrix -QUESTION [11 upvotes]: This came up in a conversation with an engineer friend of mine. -Let $c>0$ be a constant. Let $A_{ij}$ be an $n$ by $n$ matrix with entries -$$ -A_{ij} = e^{-c(i-j)^2}. -$$ -Is there a name for this matrix? What is known, perhaps approximately or asymptotically as -$c$ and $n$ change about the determinant and the eigenvalues of it? Are there other functions -of $(i-j)$ for which the answer is more explicit? - -REPLY [6 votes]: To answer the question: "...is there a name for this matrix..." -Beyond the "Gaussian-Toeplitz matrix" mentioned in the comments, the said matrix is a special case of the Gaussian Kernel (which is also the alluded to "heat kernel"): -\begin{equation*} - k(x_i,x_j) = e^{-c(x_i-x_j)^2}\qquad x_1,\ldots,x_n \in \mathbb{R}. -\end{equation*} -Clearly, the matrix $A = [k(x_i,x_j)]$ is semidefinite. If the $x_i$s are unique, then it is strictly positive definite (this requires some more work to prove). -The Gaussian kernel is of course the canonical example of a translation invariant kernel, so searching more for such kernel matrices should bring up more examples, namely, matrices of the form $A_{ij} = k(x_i,x_j)=\varphi(x_i-x_j)$, for suitable $\varphi$. -Incidentally, in the paper mentioned by @Lucia, the smallest eigenvalue for the Gaussian-Toeplitz $A$ is bounded from below by (assuming $A$ is $n\times n$): -\begin{equation*} - \lambda_n \ge \frac{\det(A)}{n!}. -\end{equation*} -It seems possible to improve this to $\lambda_n \ge \frac{\det(A)}{2^{n-1}}$ by using the technique mentioned in my answer here.<|endoftext|> -TITLE: It looks so coKleisli, but it's not. What is it? -QUESTION [5 upvotes]: Fix a symmetric monoidal category $(M,\otimes,I)$ and a small discrete monoidal subcategory $M'\subseteq M$. Define a new symmetric monoidal category $C:=CoKl(M,M')$ as follows: $Ob(C):=Ob(M)$, and for any $X,Y\in Ob(C)$, -$$Hom_C(X,Y):=\sum_{m\in M'} Hom_M(X\otimes m,Y).$$ -There is thus a projection $Hom_C\to M'$. The identity morphisms project to the unit, $m=I$ and use the identity morphisms from $M$. Given morphisms $\phi\colon X\otimes m_1\to Y$ and $\psi\colon Y\otimes m_2\to Z$ the composition $\psi\circ\phi$ projects to $m_1\otimes m_2$ and is given in the obvious way by -$$X\otimes (m_1\otimes m_2)\to Y\otimes m_2\to Z.$$ -The monoidal structure on $C$ is straightforward. -Now doesn't $C$ look like the cokleisli category of some comonad? But of course it's not, in general. There's a functor $M\to C$, but not generally an adjoint. -If $M=\pi_0(M)$ is a discrete monoidal category, then this construction gives something like the quotient, namely -$\pi_0CoKl(M,M')\cong M/M'.$ -Is there a name or reference for this coKleisli-like construction? - -REPLY [4 votes]: This is close to the subject of my (so far unfinished) thesis, so I'll try to explain for the benefit of future readers. -We define an oplax action of a monoidal category $\mathcal X$ upon a category $\mathcal C$ to be an oplax monoidal functor $\mathcal X \to \text{End}(\mathcal C)$; i.e., a functor $\_.\_\colon \mathcal X \times \mathcal C \to \mathcal C$ together with natural transformations $(x\otimes y).a \to x.y.a$ and $I.a \to a$ satisfying some coherences. -It is easy enough to check that if $\mathcal X$ is the unit category, then this is just the same thing as a comonad on $\mathcal C$. Specializing in another direction, if $\mathcal C$ is also a monoidal category, then any oplax monoidal functor $j\colon \mathcal X \to \mathcal C$ gives rise to an oplax action of $\mathcal X$ upon $\mathcal C$ via -$$ -x.a = j(x)\otimes a\,, -$$ -in much the same way that if $b$ is a comonad in a monoidal category $\mathcal C$, then there is a monad on $\mathcal C$ given by $Ta = b\otimes a$ (when $\mathcal C$ is Cartesian, every object has a unique comonad structure given by the diagonal and this is called the reader monad). -It would be nice to generalize the co-Kleisli category to arbitrary lax actions, and this is easy to do. The category we get has the same objects as $\mathcal C$, but the set of morphisms $a$ to $b$ is given by the colimit -$$ -\varinjlim_{x\colon\mathcal X} \mathcal C(x.a,b)\,, -$$ -where the composition of $f\colon a \to x.b$ with $g \colon b \to y.c$ is given by the composite -$$ -a \xrightarrow{f} x.b \xrightarrow{x.g} x.y.c \to (x\otimes y).c\,. -$$ -If $\mathcal X$ is discrete, then the colimit above is of course a sum. -Why is this a natural generalization of the Kleisli category? One answer is Mike's above: just as the co-Kleisli category of a monad is its oplax colimit when the monad is considered as a lax functor $1 \to \mathbf{Cat}$. Similarly, the category we have described above is the oplax colimit of the action, when we consider it as a functor $\mathbf{B}\mathcal X \to \mathbf{Cat}$. -In fact, there is a nice recipe for constructing arbitrary oplax colimits of lax functors in $\mathbf{Cat}$, which is set out nicely on the nLab. We start by taking the Grothendieck construction of the lax functor, which gives us a bicategory, and then we turn the bicategory into a category by identifying $1$-cells that are related by a $2$-cell. -In the particular case of a lax functor $\mathbf{B}\mathcal X \to \mathbf{Cat}$, corresponding to an action of the monoidal category $\mathcal X$ upon a category $\mathcal C$, the bicategory we end up when we take the Grothendieck construction has objects given by the objects of $\mathcal C$, $1$-cells from $a$ to $b$ given by morphisms $a \to x.b$ in $\mathcal C$ and $2$ cells from $f\colon a \to x.b$ to $g\colon a\to y.b$ given by morphisms $h\colon x\to y$ in $\mathcal X$ such that $f;(h.b)=g$. To turn this bicategory into a category, we identify all $f,g$ that are related by such an $h$, which gives us the same thing as the colimit formula above. -Another way to think of the Kleisli category is that it is what we get when we embed functors into profunctors. A comonad on a category $\mathcal C$ is a comonoid in the category of endofunctors on $\mathcal C$, which means that it gives us a monoid in the category of endoprofunctors on $\mathcal C$. But a little thought (see here, for example) tells us that a monoid in the category of endoprofunctors on $\mathcal C$ is the same thing as a category that admits an identity-on-objects functor out of $\mathcal C$; when we start with a comonad, the category we get is that comonad's co-Kleisli category. -If instead of a monoid in the category of endoprofunctors on $\mathcal C$, we consider a lax monoidal functor out of a monoidal category $\mathcal X$, we see that the structure we get is that of an $[\mathcal X,\mathbf{Set}]$-enriched category, where $[\mathcal X,\mathbf{Set}]$ carries the Day convolution product. Our construction can also be understood in this guise. Given an oplax action of a monoidal category $\mathcal X$ on a category $\mathcal C$, we may define an $[\mathcal X,\mathbf{Set}]$-enriched category on the object set of $\mathcal C$ where the object of morphisms from $a$ to $b$ is the functor -$$ -x \mapsto \mathcal C(x.a,b)\,, -$$ -composed in the Kleisli style. If $\mathcal X$ is small, we can get back to our category by change of base along the colimit functor $[\mathcal X,\mathbf{Set}]\to \mathbf{Set}$. -I am not quite sure what to call it. There is a reference to the $[\mathcal X,\mathbf{Set}]$-enriched category, in some unpublished notes of Melliès (see part 3), so the $[\mathcal X,\mathbf{Set}]$-enriched version could be called the Melliès category. Otherwise, I think your idea to call it the 'quotient' is good, or even just to call it the 'co-Kleisli category' associated to the action that you get from your inclusion $M'\to M$.<|endoftext|> -TITLE: Can we force the existence of this function? -QUESTION [12 upvotes]: I want to see if it is possible to force the existence of a function -$F:\aleph_2 \times \aleph_2\rightarrow \aleph_1$ such that: -a) $F(a,b)=F(b,a)$, for all $a,b\in \aleph_2$ and -b) for all distinct $a,b$, the set $\{x|F(a,x)=F(b,x)\}$ is finite. -What is known: -1) Under CH, there is no such function. -2) If such a function exists, let A be a subset of $\aleph_2$ of size $\aleph_1$. Consider the functions $F(a,\cdot)$ restricted to A. We have a family of $\aleph_2$ many such functions from $\aleph_1$ to $\aleph_1$ and any two of them agree only on a finite set. That's a strongly almost disjoint family of size $\aleph_2$. Baumgartner call this $A(\aleph_1,\aleph_2,\aleph_1,\aleph_0)$. It is consistent together with the negation of CH that either $A(\aleph_1,\aleph_2,\aleph_1,\aleph_0)$ or its negation hold. In particular, -$\neg CH$ + "there is no such function" is consistent. -I want to see if anyone knows any result on the positive side. - -REPLY [7 votes]: While this is extremely late, I previously ran into something similar when looking for examples of objects on $\omega_2$ that can be forced with side conditions, and I think the following method has certain advantages. If you want to avoid the use of a Delta-function and force directly, using the poset that you'd imagine, this can be done and is one of the easiest applications of Neeman's method of ``forcing with sequences of models of two types." The initial forcing one might try is to let $\mathbb{P}$ consist of $p=(f_p,A_p)$, where $f_p : \omega_2 \times \omega_2 \rightarrow \omega_1$ is a finite partial symmetric function and $A_p$ is the set of first coordinates in the domain of $f_p$, where $(f_q, A_q) \leq (f_p,A_p)$ if $f_p \subseteq f_q$ and for every $\delta \in A_q \setminus A_p$ and $\alpha \neq \beta \in A_p$, $f_q(\alpha,\delta) \neq f_q(\beta,\delta)$. However, as noted by Péter Komjáth, $F_G '' \omega \times \omega$ will be cofinal in $\omega_1$. Furthermore, because we're also worried about collapsing $\omega_2$, only considering conditions $p = (f_p, A_p, \mathcal{M}_p)$, where $\mathcal{M}_p$ is a finite $\in$-chain of countable elementary substructures of $H_{\omega_2}$, with the interaction requirement that if $\alpha, \beta \in A_p \cap M$, then $f_p(\alpha,\beta) \in M$ for every $M \in \mathcal{M}_p$, will result in a proper poset, but will not work because we cover $H_{\omega_2}$ by an increasing chain of countable structures, so in particular $\omega_2$ is collapsed. -However, if we let $\mathcal{M}_p$ instead be a finite $\in$-chain of elementary submodels of $H_{\omega_2}$ that can be either countable or internally approachable (for example) closed under intersections, with the same interaction requirement (which says nothing for the larger models), then it's not too difficult to show that $\mathbb{P}$ is strongly proper for the set of countable elementary submodels of $H_{\omega_2}$ and for the set of internally approachable elementary submodels of $H_{\omega_2}$, and because the former set is club and the latter is stationary, in $[H_{\omega_2}]^{\omega}$ and $[H_{\omega_2}]^{\omega_1}$, respectively, $\omega_1$ and $\omega_2$ are preserved. By strongly proper here we mean that conditions within models can be extended to strongly generic conditions, where $p$ is $(M, \mathbb{P})$-strongly generic if for every $r \leq p$, there exists $r \restriction M \in M$ such that $q \leq r \restriction M$ with $q \in M$ implies that $q \parallel r$. There are a few details to check, but once you see a few examples of this method, this particular forcing is one of the easiest to consider.<|endoftext|> -TITLE: Realizing homology classes on surfaces with boundary by simple closed curves -QUESTION [5 upvotes]: Let $\Sigma$ be a compact oriented surface with boundary. Assume that the genus of $\Sigma$ is positive. We say that an element $h \in H_1(\Sigma)$ can be realized by a simple closed curve if there exists an oriented simple closed curve $\gamma$ on $\Sigma$ such that $[\gamma] = h$. -If $\Sigma$ has $0$ or $1$ boundary components, then $h \in H_1(\Sigma)$ can be realized by a simple closed curve if and only if $h$ is primitive, that is, if we cannot write $h = n \cdot h'$ for some $n \in \mathbb{Z}$ and $h' \in H_1(\Sigma)$ with $n > 1$. This is a standard fact; for instance, it is contained in Farb and Margalit's Primer on Mapping Class Groups. -This brings me to my question : if $\Sigma$ has more than $1$ boundary component, then what elements of $H_1(\Sigma)$ can be realized by simple closed curves? -One might guess that the answer is still the primitive elements. However, this guess is wrong. Indeed, assume that $\Sigma$ has at least $2$ boundary components. Let $\delta$ be an oriented simple closed nonseparating curve in the interior of $\Sigma$ and let $b$ be one of the boundary components of $\Sigma$. Observe that $[b] \neq 0$, and hence that $2[\delta]+[b]$ is a primitive element of $H_1(\Sigma)$. Assume that $\gamma$ is an oriented simple closed curve in $\Sigma$ such that $[\gamma] = 2[\delta]+[b]$. Let $S$ be the surface obtained by gluing discs to all the boundary components of $\Sigma$. There is then an inclusion map $i : \Sigma \hookrightarrow S$, and we have -$$[i(\gamma)] = 2[i(\delta)] + [i(b)] = 2[i(\delta)],$$ -a contradiction. - -REPLY [2 votes]: This is related to a nontrivial question, address in this paper of Chas and Krongold (there are other related papers of Moira Chas with Fabiana Krongold and Dennis Sullivan, which a google search will bring up). -The original question, however, is trivial, since if we take some curve (think of it as a hyperbolic geodesic) realizing the homology class, we can perform a surgery on each crossing, which removes it, and possibly disconnects the curve, so eventually we will have a multicurve realizing the homology class. Some components of this multicurve will be boundary-parallel. from this multicurve it is pretty easy to see when the class is realizable (unless I am confused, which is quite possible, you need to be realizable in the cupped-off surface, plus something that is not a multiple of a boundary component). -EDIT Firstly, the OP is apparently trying to win friends and influence people for downvoting my answer and Ryan's. Not cool at all. -Secondly, if you want a different answer, knock yourself out (notice that he is solving a more general, thus harder, problem): -MR2335737 (2008e:57015) Reviewed -Granda, Larry M.(1-STL) -Representing homology classes of a surface by disjoint curves. (English summary) -Houston J. Math. 33 (2007), no. 3, 807–813. -57M50 (57M20 57N05) -A more extensive discussion of the same problem solved in Granda's paper (without, however, a complete answer) is given by Allen Edmonds in: -Edmonds, Allan L.(1-IN) -Systems of curves on a closed orientable surface. -Enseign. Math. (2) 42 (1996), no. 3-4, 311–339. -Another edit -The best reference is W. Meeks and J. Patrusky, where Theorem 1 is: -enter link description here -For the link-challenged, it says that a class can be represented by simple closed curve if and only if it is primitive in the capped-off surface OR it is a sum of (some of the) boundary curves, which is what Ryan and I have been saying.<|endoftext|> -TITLE: Small objects in categories -QUESTION [5 upvotes]: I would like to pick out small objects from a category. I would like to find such a notion which -Dream 1. Picks out the schemes of finite type over $k$ from the category of $k$-schemes. Or at least picks out something relevant. -Dream 2. Picks out the top spaces homotopic to finite CW-complexes from the category of topological spaces (equipped with the Quillen model structure). (or something relevant) -Question: What notions exist for "small" objects (other than the compact objects)? -References or comments are welcomed.:) -Thanks! - -REPLY [5 votes]: The subcategory of spaces equivalent to finite CW-complexes is the smallest one containing a point and closed under finite homotopy colimits. (In fact, this is the universal homotopy theory generated under finite hocolims by a single object). -Someone more patient than I could might fill in the details to make the following sentence both (i) sensical and (ii) correct: If $G$ is finite, the subcategory of $G$-spaces equivalent to a finite $G$-CW-complex is the smallest containing a point and closed under finite 'enriched' (weighted is probably the right word) homotopy colimits. (Here I'm enriched in $G$-spaces instead of spaces like above). -The category of schemes is probably too ugly of a place to try and make a similar statement, but perhaps one of the many colimit-populated enlargements would allow you to make the statement that some reasonable enlargement of the category of schemes of finite type over $k$ is generated by a point under some allowed collection of colimits. -(By the way, in the first two examples, honest compact objects are obtained by idempotent completion, i.e. adding retracts.)<|endoftext|> -TITLE: Associators, Grothendieck-Teichmüller group and monoidal categories -QUESTION [14 upvotes]: The standard definition of an associator seems to be that it a a grouplike power series in two variables $x$ and $ y $ satisfying some pentagon and hexagon relations. -In other words, denoting by $ \mathfrak{lie}_2 $ the free Lie algebra on two variables, an associator is a grouplike element in the completed hopf algebra $ \widehat{U}(\mathfrak{lie_2})$ satisfying the above mentionned relations. -In more abstract terms, there is a bijection between the set of associators and a certain morphism of operads. Using Bar-Natan's notations it is a morphism of operads between $ \widehat{PaB_K} $ and $ PaCD $. -In this context, the Grothendieck-Teichmüller group $ \widehat{GT} $ is the group of automorphism of $ \widehat{PaB_K} $. It can also be described as pairs $ (f,\lambda) \in \widehat{F_2(K)} \times K^* $ again satisfying some relations. -These seem to be the "textbook" definitions, however I don't understand their motivation and what these objects do. My questions are the following. (The lack of precision comes from my lack of knowledge of the subject) -0) What is the purpose of associators? What is the purpose of the G-T group in this context (not in algebraic geometry). -The following questions are possible answers to the previous one. -1) Let $ (C, \beta, \gamma) $ be a braided monoidal category with associativity constraint $ \gamma $ and braiding $ \beta $. Suppose we wish to change $ \beta $ and $ \gamma $ into $ \beta' $ and $ \gamma' $ such that $ (C, \beta', \gamma') $ is again a braided monoidal category. Does this procedure have a name? Do elements of the Grothendieck-Teichmüller control this? -2) Let $ (C, \sigma, \gamma) $ be a symmetric monoidal category. Do the associators control how to transform $ C $ into a braided monoidal category? -3) Does the term infinitesimally braided monoidal category exist? -4) Does the Kontsevich integral somehow fit into this story? -As I mentioned above, I know little about the subject: I have skimmed through Bar-Natan's article "On associators and the Grothendieck-Teichmüller group group" and the paper arXiv:0903.4067, therefore any reference is welcome. - -REPLY [8 votes]: 0) This is a wide question. Probably the best answer/definition is precisely Bar Natan's one, which of course is implicit in Drinfeld's work: an associator is a filtered isomorphism between the completion of a naturally filtered, complicated, topological category, and a much more manageable, explicit, naturally graded, combinatorial one. In addition to the obvious interest in braid/knot theory, it turns out that the former is closely related to "quantum" objects, while the latter is related to "classical" objects. Hence associators shows up everywhere in deformation-quantization. Another "conceptual" explanation of this is that they are also responsible for the formality of the little disc operad, which is a close relative of the above isomorphism. -1) yes, if your category is $k[[\hbar]]$-linear for $k$ of char 0, this is basically the motivation behind the definition of GT. This is essentially the same things as the definition you gave, as the automorphism group of the completion of the braid category (as opposed to the automorphism group of the braid category, which is almost trivial). -2),3) yes, if the symmetric category is infinitesimal braided (i.e. if it's a representation of PaCD) then an associator turns it into a braided monoidal category (and even ribbon if it has duals). -4) Yes, it is exactly this: the Kontsevich integral for braids is exactly the above mentionned isomorphism, which extends to a functor from the category of tangles which is again an isomorphism after completion, and whose restriction to links is equal to the Kontsevich integral for any choice of associator. -edit: More precisely: as you say, associators are in bijection with isomorphisms $\widehat{PaB}\rightarrow PaCD$ compatible with the operadic structure. The Kontsevich integral provides such an isomorphism, hence produces a particular associator, namely the image of the trivial braid from $(\cdot \cdot) \cdot$ to $\cdot (\cdot \cdot)$ as explained in Bar-Natan's paper. This is the so-called KZ associator, which is the first (and for some time only) known associator. This is how Drinfeld showed that the set of associators in non-empty (and then deduced that there exists rational associators as well). Then this isomorphism extends to tangles, and by a result of Le-Murakami (implicit in Drinfeld's paper) its restriction to links doesn't depends on the choice of the associator (which proves the rationality of the Kontsevich integral of links). -All of this is explained in a nice, clear, pictorial way in many of Bar Natan's paper, especially of course those on Vassiliev invariants, and in Kassel--Turaev "Chord diagramms invariants of tangles and graphs".<|endoftext|> -TITLE: Floors of rationals to powers: Infinite number of primes? -QUESTION [19 upvotes]: Let $r=a/b$ be a rational number in lowest terms, larger than $1$, -and not an integer (so $b > 1$). - -Q. Does the sequence - $$ \lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor, -\ldots, \lfloor r^n \rfloor, \ldots$$ - always contain an infinite number of primes? - -For example, for $r=13/4$, -$$(r,r^2,r^3,r^4,r^5, \ldots) = (3.25, 10.56, 34.33, 111.57, 362.59, \ldots)$$ -and so the floors are -$$(3, 10, 34, 111, 362, \ldots)$$ -The next prime after $3$ occurs at -$$\lfloor r^{34} \rfloor = 253532870351270971$$ -and I only find four primes up to $\lfloor r^{1000} \rfloor$. - -REPLY [4 votes]: I recently asked a related question (inspired by this one!) . I accepted an answer but also provided an answer of my own. That answer illustrates the fact that for real $r$ one can arrange to have $\lfloor r^n \rfloor$ all even. One can pick the successive terms of $\lfloor r \rfloor, \lfloor r^2 \rfloor, \lfloor r^3 \rfloor, -\ldots, \lfloor r^n \rfloor, \ldots$ fairly freely. By this I mean being slightly vague (or uncommitted) about the value of $r$ and specifying it by providing the sequence of floor values one term at a time. It is easy to preserve freedom of choice enough to have about $\lfloor r \rfloor$ consistent choices at each stage (the smallest of which should be rejected in order to preserve the freedom for the following stage). -Let me relate it to other similar problems and implicitly suggest that for almost all non-integer rationals greater than $1$ there will be infinitely many prime floors. That we might not be able to prove it true for even one rational and that there may not be even one counter-example (at least that we can discover). -Given a positive real $r$ and a real "base" $b \gt 1$ define the base $b$ expansion of $r$ to be the integer sequence $r_b=(x_0,x_1,x_2,\cdots)$ with $x_k=\lfloor rb^k \rfloor$. So this particular integer sequence has $\lim\frac{x_{k+1}}{x_k}=b.$ - -Does the sequence $\pi_{10}=3,31,314,3141,\cdots$ contain infinitely many primes? One would expect so and even that there are about $\frac{\ln{N}}{\ln{10}}$ up to $n=N$ (in an appropriate sense), but one also does not expect to see a proof. -What about the same problem with $b=3$ or $b=4$? Same mutatis-mutandis. -what about $r_{10}=\lfloor 10^n r \rfloor$ for other reals $r?$ Well, we can pick the decimal expansion to avoid primes but otherwise we might predict infinitely many. The case $r=1/9$ is the question "Are there infinitely many prime repunits?" We know that we can only get a prime for $n$ prime, but might guess that something like $\frac{\ln{\ln{N}}}{\ln{10}}$ is the correct growth rate. We could switch this to questions about Mersenne primes by using the non-terminating binary expansion of $c=1$ (or $c=1/2$). -What about $\pi_{\pi}=\lfloor \pi^n \pi \rfloor$ in place of $\lfloor 3^n \pi \rfloor$? Either is a sequence of integers each roughly triple the one before and nothing much should be different. -The main point of my answer to my question is that we can build up a feasible integer sequence $\lfloor r^n \rfloor$ or $\lfloor c r^n \rfloor$ term by term somewhat like a decimal expansion. Eventually typically having about $r$ choices for the next term. At times the next choice can be forced by previous ones but that does not persist. So we can avoid primes if we build $r$ by startiing with the expansion. There are specific reals $r$ with special behavior (see the rest of this answer) but no reason to expect that any non-integer rationals are among them. -In special cases like $r=\sqrt[3]{7}$ we know that every third term of $r_r$ is non-prime, but that still suggests infinitely many prime floors, just at a slower growth rate. -There are certain algebraic integers (Pisot-Vijarayaghavan numbers) $r$ where the distance from $r^n$ to the nearest integer goes to $0$ exponentially fast. Then there is a recurrence relation for $\lfloor r^n \rfloor$ and things can sometimes be cooked to make these all even (for example). With the right definitions ($\lceil rb^n \rceil$ for $r=\frac{1}{\sqrt{5}}, b=\frac{1+\sqrt{5}}2$ ) we get the question "are there infinitely many (odd) prime Fibonacci numbers?" As with repunits, only prime $n$ could give a prime here and the situation should be about the same. -Suppose that we look at possible sequences $\lceil r^n \rceil.$ Then every initial segment of the sequence of Fermat numbers $3,5,9,17\cdots$ is feasible and the nested intervals are specifying a real in $\cap (2,\sqrt[k]{2^k+1}].$ These intervals have empty intersection but $\cap [2,\sqrt[k]{2^k+1}]=\{{2\}}$. In this non-standard expansion only the terms with $n=2^j$ could possibly be prime. Hueristics suggest that we know all the Fermat Primes (just $0 \le j \le 4$) but altering the heuristic arguments can (evidently) support a case that a positive proportion of choices $n=2^j$ give Fermat primes. - -In conclusion: Give a real constant $r$ and a base $b \in \mathbf{R}$, the integer sequence $\lfloor r b^n\rfloor$ gives the base $b$ expansion of $r$. This is a sequence of integers each roughly $b$ times the previous and generally the previous choices give an interval of length $b$ for possible values of the next term. If we start with the expansion we can avoid primes, but a "random" constant $r\ $ should yield a predictable infinite number of primes subject to adjustments we feel that we understand. -Not that much is different if we use $b=r$. There are nameable reals with exceptional expansions $r_r$ (Integers and Pisot numbers), $r=\sqrt[j]{k}$ a root of an integer does certain special things when $j \mid n$. We have a lot of latitude to construct $r_r$, the "base $r$" expansion of $r$ to get an integer sequence enjoying certain properties, but there is no reason to think that this would lead to a rational number.<|endoftext|> -TITLE: Getting the story of Dynkin and Satake diagrams straight -QUESTION [6 upvotes]: I've been trying to teach myself the theory of Lie groups. The sources I've been reading reference Lie algebras in the context of Dynkin and Satake diagrams, but not Lie groups (which I am more interested in). I've been trying to translate these results to Lie groups, but I'm almost sure I'm getting my facts wrong. (I am in particular confused about what is true over $\mathbb{C}$, and what is true over $\mathbb{R}$ because some sources don't mention this explicitly.) -It would be very helpful for me to get some guidance so that I know how to think about the big story before continuing with my study. -Question -Is the following story correct? (And if not, what is the correct story?) - -Semi-simple real Lie algebras correspond to disjoint unions of Satake diagrams. -Semi-simple real Lie algebras are always the Lie algebra of some real compact Lie group. This Lie group is not unique, but it is unique if we assume that it is connected and simply connected. -Any connected compact real Lie group with a semi-simple (real) Lie algebra is semi-simple. Namely, it is reductive with finite center. -Any compact real Lie group is a maximal compact closed subgroup of its complexification. (Although there may be other maximal compact subgroups.) -A complex Lie group is reductive iff it is the complexification of a compact real Lie group. -The maximal closed compact subgroups of a complex reductive Lie group are all real Lie groups. - The maximal closed compact subgroups of a complex reductive Lie group correspond to disjoint unions of Dynkin diagrams. - -Semi-simple complex Lie algebras correspond to disjoint unions of Dynkin diagrams. -$\,\,\,$8. A semi-simple complex Lie algebra is always the Lie algebra of a reductive complex Lie group. (This group may not be unique. I am not sure whether it is unique if we assume that it's simply connected as a manifold.) -EDIT: the point below did not appear in the original question -$\,\,\,$9. Non-isomorphic real (resp. complex) connected and simply connected Lie groups have non-isomorphic real (resp. complex) Lie algebras. -Additional Question -In addition to my confusion about the story above, I am also confused about the way in which this helps classify real Lie groups in general. Namely, if I understand correctly, Satake diagrams only classify connected compact simply connected real Lie groups; and the connected compact real Lie groups are known because they are quotients of connected compact simply connected real Lie groups by central subgroups. But what about non-connected compact real Lie groups? Are they completely unrelated to the Satake diagrams story? Is there any way to classify them? - EDIT: The following is another point of confusion that did not appear in the original question -Let's say we are given a semi-simple complex Lie algebra, and let's say that it's the Lie algebra of some reductive connected and simply connected complex Lie group. Since Dynkin diagrams are a coarser tool than Satake diagrams, I take from that the this complex Lie algebra is potentially the complexification of several non-isomorphic real Lie algebras. However, if I understand correctly, a complex Lie group is the complexification of at most one (up to iso.) real Lie group. So how could it be that the complex Lie algebra has several real forms, but the complex Lie group has only one? What are these real Lie groups that correspond to the various real forms of the complex Lie algebra? -Never mind, I got it. Satake diagrams correspond to not-nec.-compact, connected, simply connected, real Lie groups. - -REPLY [5 votes]: 2 is false. The smallest counterexample is $\mathfrak{sl}_2(\mathbb{R})$. A necessary and sufficient condition for a semisimple real Lie algebra to be the Lie algebra of a compact Lie group is that the Killing form is negative definite (compact Lie algebra). I think it is known that every semisimple complex Lie algebra has a unique compact real form. - -But what about non-connected compact real Lie groups? Are they completely unrelated to the Satake diagrams story? Is there any way to classify them? - -This is at least as hard as classifying finite groups. -Edit: Regarding 9, a much stronger statement is true. Taking tangent spaces at the identity induces an equivalence of categories between the category of connected, simply connected real Lie groups and the category of finite-dimensional real Lie algebras. I think this statement is still true with "real" replaced by "complex." - -Never mind, I got it. Satake diagrams correspond to not-nec.-compact, connected, simply connected, real Lie groups. - -By the above, classifying these is equivalent to classifying finite-dimensional real Lie algebras, and this classification is hopeless already for nilpotent Lie algebras of some specific small dimension that I can't remember right now. Statement 1 is still correct.<|endoftext|> -TITLE: On a sum involving Euler totient function -QUESTION [5 upvotes]: Let -$$S_a(N)=\sum_{n\le N}\frac{\varphi(an)}{n^2}.$$ -The usual machinery gives an asymptotic formula -$$S_a(N)=\frac1{\zeta(2)}\cdot\frac{a^2}{\varphi_+(a)}\log N+C(a)+O(N^{-1+\varepsilon}a^{1+\varepsilon}),$$ -where $C(a)$ some complicated function and -$$\varphi_+(a)=a\prod_{p\mid a}\left(1+\frac1p\right).$$ -Is it possible to give a reference on this asymptotic formula? (My proof is rather long.) - -REPLY [2 votes]: I have found a proof of more general formula in the book Postnikov, A. G. Introduction to analytic number theory American Mathematical Society, 1988, (section 4.2). This proof is simple but it has a small mistake inside. (For arithmetic progression starting from $0$ this mistake vanishes.) -Please give more references if you know ones.<|endoftext|> -TITLE: Subgroups of finite solvable groups with paradoxical properties -QUESTION [24 upvotes]: For which pairs of finite solvable groups H, G, is the following true: - H embeds in two ways into G, say as H1 and H2, where H1 is maximal in - G and H2 is not? Are there any such pairs? -Some comments on the question. -1) Terminology: when H embeds in G maximally and non-maximally, - call H a paradoxical subgroup of G, and say that G has H as - a paradoxical subgroup. -2) For finite groups in general, as opposed to finite solvable, there are such things as paradoxical subgroups, it's not a vacuous concept. If H is a non-abelian simple group, and G is the direct product H x H, take - H1 to be the diagonal subgroup, elements of G of the form (x,x), x in - H, and take - H2 to be a component subgroup, elements of G of the form (x,1), x in H. -Then H1 and H2 are both isomorphic to H, H1 is maximal in G, and H2 - isn't. I owe this example to Yair Glasner. -3) By Sylow's theorems, a p-group is never a paradoxical subgroup - of any group. -4) A standard theorem on super-solvable groups, (that the maximal - subgroups have prime index,) implies that supersolvabe groups have no - paradoxical subgroups. -5) An important special case is when H1and H2 are both core-free, - i e each one contains only the trivial group as a normal subgroup. in - this case G has two faithful permutation representations of the same - degree (namely the index of H in G), with one representation being - primitive and the other imprimitive. Is this case any easier? - -REPLY [13 votes]: One can obtain a little more in Geoff's situation, where $O_p(G) > 1$. Write -$U = O_p(G)$ and $U_i = U \cap H_i$, so $U_i = O_p(H_i)$, as Geoff showed. In particular, $|U_1| = |U_2|$. Note that the $U_i$ are proper in $U$. I claim that -$U = U_1U_2$. -First, observe that $U_1 \triangleleft G$ since $N_H(U_1)$ contains $H_1$ properly because $N_U(U_1) > U_1$. Since $H_1$ is maximal, $U_1 \triangleleft G$, as claimed. Next, $U/U_1$ is $G$-chief by the maximality of $H_1$, and in particular, $U/U_1$ is abelian. Since $G = UH_2$, it follows that $H_2$ acts irreducibly on $U/U_1$. Now $H_2$ normalizes $U_2$, so it normalizes $U_1U_2$. Since $U_1 \subseteq U_1U_2 \subseteq U$, we have either $U_1U_2 = U_1$ or $U_1U_2 = U$. -Since $|U_1| = |U_2|$, the first possibility yields $U_1 = U_2$, and this is a normal subgroup invariant under an isomorphism from $H_1$ to $H_2$, so we have that the $U_i$ are trivial in this case, and $U$ is minimal normal in $G$. Since $G = UH_2$, this implies that $H_2$ is maximal, contrary to assumption. We thus have $U_1U_2 = U$, as claimed. In particular, this yields $G = H_1H_2$. -But where do we go from there? - -OK, now I know where to go. I can finish the proof in the case where $O_p(G)$ is nontrivial. -Assume $G$ is a minimal counterexample. Let $D_2 = H_1 \cap H_2$ and let $D_1$ be the image of $D_2$ under some isomorphism from $H_2$ onto $H_1$. Thus $D_1$ and $D_2$ are isomorphic subgroups of $H_1$ with $p$-power index equal to $|G:H_2|$. I claim that $D_1$ is maximal in $H_1$ but $D_2$ is not, and also $O_p(H_1) = U_1 > 1$. Since $G$ is a minimal counterexample and $H_1 < G$, this will be contradict the minimality of $G$. -First, to see that $D_2$ is not maximal in $H_1$, choose a subgroup $X$ with -$H_2 < X < G$. Since $H_1H_2 = G$, we deduce that $D_2 = H_1 \cap H_2 < H_1 \cap X < H_1$, as wanted. -Next, to show that $D_1$ is maximal in $H_1$, it suffices via the isomorphism to show that $D_2$ is maximal in $H_2$. Note that $H_1 = U_1D_2$ by Dedekind's lemma. Suppose $D_2 < Y < H_2$. Recall that $U_1 \triangleleft G$, so $H_1Y = U_1D_2Y = U_1Y$ is a subgroup. Also, $|H_1Y:H_1| = |Y:D_2|$, and it follows that $H_1 < H_1Y < G$, contradicting the maximality of $H_1$. -Now we need a proof (or counterexample) in the case where $O_p(G) = 1$. - -My argument proves that if $H$ is paradoxical in solvable $G$ with $p$-power index, then $O_p(G) \not\subseteq H$. I did $not$ actually prove that $O_p(G) = 1$, but we do obtain this conclusion in a minimal counterexample. -Robinson has used my contribution and some deep theory to completely solve the problem for groups of odd order. (I am impressed.) We can also obtain some easier consequences. For example: -THEOREM. If H is supersolvable, then it cannot be paradoxical in a solvable group. -PROOF. As usual in a minimal counterexample, there is no nonidentity normal subgroup of $G$ contained in $H_1 \cap H_2$ and invariant under an isomorphism from $H_1$ to $H_2$. Now as $H_1$ is supersolvable, it has a nontrivial normal Sylow $q$-subgroup $Q$, and we know $q \ne p$, so $Q$ is Sylow in $G$. If $Q$ is normal in $G$ it is also Sylow in $H_2$ and is invariant under an isomorphism from $H_1$ to $H_2$, a contradiction. Otherwise, $H_1$ is the full normalizer of $Q$. A Sylow $q$-subgroup of $H_2$ is conjugate to $Q$ so its normalizer in $G$ is conjugate to $H_1$ and contains $H_2$. It follows that $H_2$ is maximal.<|endoftext|> -TITLE: Can you efficiently solve a system of quadratic multivariate polynomials? -QUESTION [12 upvotes]: Given a system of 2nd-degree polynomials, $P=\{p_1,\dots,p_m\}$ where $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$, can you efficiently find a common zero of all of these polynomials? In other words, given $P$, can you find $x_1,\dots,x_n\in\mathbb{R}$ such that $p_i(x_1,\dots,x_n)=0$ for all $i$? I only need to find one solution. -I know that in the general case (arbitrary degree polynomials) this can't be done efficiently. However, I'm wondering if there are any good techniques in the quadratic case, specifically if there is a polynomial time solution. - -REPLY [18 votes]: There is an exact algorithm that needs $n^{O(m)}$ operations (cf. http://arxiv.org/abs/cs/0403008). One cannot expect anything better than that, unless P=NP. Indeed, it is easy to formulate several NP-complete problems as testing solvability of linear equations in 0-1 variables $x_i$, and the latter can be enforced by quadratic equations $x^2_i-x_i=0$. - -REPLY [11 votes]: The quadratic case is as complex as the general case (up to a polynomial-time reduction). Given a set of polynomials $\{p_1,\dots,p_m\}$, you can express each $p_i$ by a straight-line program with instructions of the form $x_i:=c$ ($c\in\mathbb R$ a constant), $x_i:=x_j+x_k$, and $x_i:=x_j\cdot x_k$, which you can in turn translate back to quadratic polynomials (e.g., the last mentioned instruction becomes $x_i-x_jx_k$). The number of variables in the new system will increase (it is linear in the size of the original system), which suggests that you might get an efficient solution if you additionally restrict the number of variables by a constant. - -REPLY [8 votes]: There have been written many books and articles on this subject (see for example the book -"Sturmfels, Bernd, Solving systems of polynomial equations. American Mathematical Society, Providence, RI, 2002"). -It is not the fact that the polynomials are quadratic which helps, but rather other restrictions which will sometimes result in a more effective solution. -If the Groebner basis of the corresponding ideal in $K[x_1,\ldots ,x_n]$ -can be computed, and has triangular form, then we can usually solve the system. -However, Groebner bases have large exponential complexity and cannot solve in practice systems with, say, more than $15$ variables. -For some overdetermined systems there are other techniques than Groebner bases. One of these is called "relinearization". The exact complexity of this algorithm is not -known, but for sufficiently overdetermined systems it is expected to run in -polynomial time (see papers by Nicolas Courtois; Alexander Klimov, Jacques Patarin, and Adi Shamir).<|endoftext|> -TITLE: Approximating an iteratively defined function -QUESTION [6 upvotes]: Let $f_0,f_1,\ldots$ be a sequence of functions $f_n : [0,1] \rightarrow R$ defined as follows: -$$f_0(x) =1+2x$$ -$$f_{n}(x) := \left\{\frac{5+t}{2} : \text{ where t solves } f_{n-1}\left(\frac{x}{t}\right) = \frac{5+t}{2}\right \}$$ -In other words, $f_n(x)$ is equal to the intercept (with respect to $y$) of the functions $f_{n-1}(x/t)$ and $\frac{5+t}{2}$. -It isn't hard to prove by induction that each $f_n$ (for $n>0$) will be a continuous increasing function such that $f_n(0)=5/2$ and $f_n(1)=3$. If we fix $0 -TITLE: Flows of vector fields and diffeomorphisms isotopic to the identity -QUESTION [22 upvotes]: Let $M$ be a compact manifold and $\varphi : M \longrightarrow M$ be a diffeomorphism which is isotopic to the identity. Does there exist a vector field $ X $ on $M$ such that $\varphi$ is the flow at time $1$ of $X$? If that is not always the case, where does the obstruction for such a $\varphi$ to be a flow lives in ? - -REPLY [7 votes]: In fact, for every manifold $M$ with $\dim(M)\ge 2$, the group $Diff_c(M)$ of diffeomorphisms with compact support contains a smooth curve $c:\mathbb R\to Diff_c(M)$ -(smooth in the sense that the associated mapping $\hat c: \mathbb R\times M\to M$ is smooth) -with $c(0)=Id$ -such that the $\{c(t): t\ne 0\}$ are a free set of generators for a free subgroup of the diffeomorphism group such that no diffeomorphism in this free subgroup with the exception of the identity embeds into a flow. -This is proved in - -Grabowski, J., Free subgroups of diffeomorphism groups, Fundam. Math. 131 (1988), 103–121. - -The proof uses clever constructions with Nancy Kopell's diffeomorphisms on $S^1$ (those described in Andy Putman answer) which do not embed into a flow. -This free subgroup is contractible to the identity: slide the generators $c(t)$ to $c(0)$. -So the image of exponential mapping of $Diff_c(M)$ does not meet very many diffeomorphisms near the identity.<|endoftext|> -TITLE: Well-definedness of single-particle smooth billiards flow -QUESTION [7 upvotes]: Single-particle billiards systems in a domain with corners, or multi-particle billiards in a domain with smooth boundary, can exhibit singularities in finite time. (The former phenomenon is well known; for an example of the latter, see e.g. section 1 of Charles Radin's article "Dynamics of Limit Models", available at http://www.ma.utexas.edu/users/radin/papers/limitmodels.pdf.) -Can single-particle billiards exhibit such catastrophes in a domain with smooth boundary? -(If the answer depends on the degree of smoothness, e.g. if the answer is different for $C^1$ vs $C^\infty$, then interpret my question as "For what forms of smoothness is it the case that ..., and for what forms of smoothness is it not the case that ...?") -Note that this is not the same as asking the corresponding question about the discrete billiards map. The discrete map can be iterated unboundedly many times, but if the return-time to the boundary shrinks quickly, "time infinity" under discrete (return-map) dynamics could correspond to a finite-time catastrophe in the flow. -Note also that I am not asking whether billiards flow is well-defined for a set of initial conditions of full measure; I am asking whether it is well-defined for ALL initial conditions. (So please don't add an ergodic-theory tag to my post!) -Note also that the catastrophe phenomenon is not the same as the ill-definedness of the velocity at the moment of rebound. This ill-definedness does indeed make it slightly tricky to define billiard flow, but this is merely a technical problem that is easily surmountable. Extending the dynamics beyond a catastrophe is more problematic; it entails breaking symmetry and/or introducing discontinuities. -Hopefully I have anticipated all misunderstandings of my question that are likely to arise. - -REPLY [6 votes]: I think the paper you want is B. Halpern, "Strange Billiard Tables." Transactions of the AMS Vol 232, 1977. -Thanks to Carl for pointing out that Halpern considers tables with the additional condition of nonvanishing curvature. He constructs a $C^2$ catastrophe (with collision points on the unit circle, but an irregularly shaped table passing through those points) and rules out a $C^3$ catastrophe. -I believe that without the condition of nonvanishing curvature, Halpern's construction can be adapted to produce a smooth catastrophe with infinitely many collision points on $\exp(-1/x^2)$ approaching the origin. However, I haven't verified that the result can be made smooth.<|endoftext|> -TITLE: Does vanishing of cohomology of locally free sheaves imply affiness of scheme -QUESTION [10 upvotes]: We have Serre criterion of affiness of a scheme which states that if a quasi compact scheme has higher cohomology vanishing for all the quasi coherent sheaves,then the scheme is affine. -I wonder whether we have similar statement for locally free sheaves as following: -Let $X$ be a noetherian scheme,let $F$ be arbitrary locally free sheaf on $X$,if higher cohomology of $F$ vanishing(for $i\geq 1$),then $X$ is affine scheme.Is this statement true? -For $X$ be quasi compact scheme,I think it is not true,but for noetherian scheme,I do not know -Maybe it is a stupid question. - -REPLY [16 votes]: There are many versions of Serre's criterion for affineness. One version states that for every quasi-compact, quasi-separated scheme $X$, $X$ is affine if and only if $H^1(X,\mathcal{F})$ vanishes for every quasi-coherent sheaf $\mathcal{F}$ that is locally finitely generated. In particular, if $X$ is Noetherian, then $X$ is affine if and only if $H^1(X,\mathcal{F})$ vanishes for every coherent sheaf. -For this criterion, it does not suffice to consider only locally free sheaves (of finite rank). Let $n>1$ be any integer. Let $X$ be the quasi-compact, quasi-separated, yet non-separated scheme obtained by glueing two copies, $X_1$ and $X_2$, of $\mathbb{A}^n$ along the common open $X_{1,2} = \mathbb{A}^n\setminus\{0\}$. There is a unique morphism of schemes, $f:X\to \mathbb{A}^n$ that restricts to the identity on $X_1$ and $X_2$. Using the S2 property, every locally free sheaf on $X$ is of the form $f^*E$. Thus, by the Quillen-Suslin theorem, every locally free sheaf on $X$ is a direct sum of copies of the structure sheaf. By straightforward computation, $H^1(X,\mathcal{O}_X)$ vanishes. Yet $X$ is not affine, since $X$ is not separated.<|endoftext|> -TITLE: What is the geometric interpretation of this quantity? -QUESTION [5 upvotes]: Let $(M,g)$ be a compact $n$-dimensional Riemannian manifold. Using the metric to identify the tangent and cotangent bundles defines a natural symplectic -structure on the tangent bundle, $(TM, \omega)$, and a natural volume form $\Omega=\omega^{n}$ on $TM$. -For every $r>0$, let $D_{r}(M)$ be the open disc bundle on $M$, with radius $r$. Namely $D_{r}(M)$ is the subset of $TM$ formed by all tangent vectors with length smaller than $r$. -Define: -${C(r)=(\text{Gromov width of $D_{r}(M)$}})^{2n}$ -$V(r)$=The Volum of $D_{r}(M)$ with respect to $\Omega$ -Question : - -1) Does $\lim_{r\to \infty} C(r)/V(r)$ exist? And what is its geometric interpretation? -2) Does $\lim_{r\to 0} C(r)/V(r)$ exist? And what is its geometric interpretation? - -Recall that the Gromov width of a symplectic manifold $N$ of dimension 2n defined as follows: -$\sup\; \{\rho \mid \text{there is a symplectic embedding from $B_{\rho}(0)\subset \mathbb{R}^{2n}$ to $N$}\}$. -By $B_{\rho}(0)$ I mean the disc around the origin with radius $\rho$. - -REPLY [7 votes]: Both limits (1) and (2) are equal to $C(1)/V(1)$ because of the homogeneity of the volume and the symplectic capacity. Namely, the symplectic form is homogeneous of degree $1$ with respect to dilations on the tangent bundle and so it makes no difference what the radius of your disc bundle is. -What is then geometric meaning of $C(1)/V(1)$ ? That's an interesting and difficult question which falls under the topic of capacity-volume inequalities (something that is just taking off). I'd suggest looking at flat tori first. The reason is that if you consider the universal cover, the lattice, and the ellipsoid that defines your unit disc, then if you have a lattice basis inside the ellipsoid, you also have a symplectic unit ball inside the unit tangent bundle of your torus : just look at the moment map of the torus action on the unit ball: it's image is (or can be made to be) the simplex formed by the origin and your lattice basis. You will probably be able to give a number-geometric interpretation to $C(1)/V(1)$ in this case. -By the way, the original capacity-volume problem is -The Viterbo Conjecture. The symplectic capacity (any capacity!) of a convex body $K \subset \mathbb{R}^{2n}$ whose volume is the same as that of the $2n$-dimensional Euclidean unit ball is less than or equal to $\pi$. -Note that this is obviously true for the Gromov width, but it is open for every other capacity. There is a chance this conjecture is true for the Hofer-Zehnder capacity, which can be more easily described as the least of the actions of all closed characteristics on the boundary of $K$. There are some partial results on this conjecture in the first version of this paper, but beware, by a recent result of Artstein-Avidan, Ostrover, and Karasev, a positive answer to Viterbo's conjecture would also prove the Mahler conjecture (corollary: this can't be easy!).<|endoftext|> -TITLE: Restricted isometry -QUESTION [8 upvotes]: Here is a problem that has been bugging me for a while. -Let $\| \|$ be a norm over $\mathbb{R}^n$, let $C$ be a convex subset of $\mathbb{R}^n$ with non-empty interior, and let $f: (C,\|\|) \rightarrow (\mathbb{R}^n,\|\|)$ be a distance-preserving map. -Is it true that there exists an isometry $g$ of $(\mathbb{R}^n,\|\|)$, such that $f = g|_C$? -A few notes: --I don't require an isometry to be surjective, though it follows from the bounded-compactness of $(\mathbb{R}^n,\|\|)$. --There is flexibility regarding the caracteristics of $C$, which may (to serve my purpose) be: $C$ is an open ball or closed ball with non-empty interior. In that case, $f$ is a bijection from $C = B(x,r)$ onto $B(f(x),r)$. --Mazur-Ulam's theorem states that every surjective isometry between two real normed vector-spaces is affine. --Though it may not seem so, I think the easiest way to prove this is to show that $f$ perserves the middles of any two points of $C$. -Unfortunately, this isn't an easy task, I do not think I would have been able to prove it in the case when $C = \mathbb{R}^n$ if I hadn't come across PFDs about Mazur-Ulam. -I have failed to adapt the proof(s) of Mazur-Ulam theorem to this case. -I don't understand the "fundamental reason" why this theorem works, making it difficult to use. Maybe it's just magic. --The norm may not be strictly convex, therefore, given two distinct points, there may be infinitely many points whose distance to each of the original points is half the distance between the original points. --There might be a solution if we assume that $f$ is differentiable on the interior of $C$, but once again I don't think it would be easy to prove that $f$ is differentiable without having established results which directly imply that it is affine. -Any idea how to proceed? - -REPLY [3 votes]: So the problem is that your map $f$ is not surjective. -However, $f$ is locally surjective in the following sence: -If $\Omega$ is the interior of $C$, $x\in \Omega$ and $\varepsilon>0$ is such that $B_\varepsilon(x)\subset \Omega$ then the restriction $f|_{B_\varepsilon(x)}$ is a bijection from $B_\varepsilon(x)\to B_\varepsilon(f(x))$. -This is true since the distance preserving map has to be volume preserving and $\mathrm{vol}\,B_\varepsilon(x)=\mathrm{vol}\,B_\varepsilon(f(x))$. -This local surjectivity can exchange the surjectivity in the proof of Jväisälä given here (the only proof of Mazur–Ulam's theorem I know). -It gives the following: assume $B_{100{\cdot}|x-y|}(x)\subset \Omega$ -then -$$f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}{2},$$ -i.e., the map $f$ is a restriction of affine map to $C$. -Hence everything follows.<|endoftext|> -TITLE: Geometric interpretation of the half-derivative? -QUESTION [46 upvotes]: For $f(x)=x$, the half-derivative of $f$ is -$$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$ -Is there some geometric interpretation of (Q1) this specific derivative, and, (Q2) of the half-derivative more generally? I have read that fractional derivatives are nonlocal, -but it seems strange to me that integral derivatives -can be described in terms of local -geometry only, while fractional derivatives cannot be so described. -This would suggest an odd discontinuity between, say, $d^{1}$ and $d^{1.01}$. -This seems especially at odds with the many applications of fractional derivatives, -which (superficially) suggests continuity should reign. -I'd appreciate someone clearing up my elementary confusions—Thanks in advance! -Addendum (5Jan14). -@AlexR. found this geometric interpretation of -the fractional integral in Richard Herrmann's book, -Fractional Calculus: An Introduction for Physicists,World Scientific, 2011: - -REPLY [6 votes]: A simple perspective for all fractional integro-derivative operators (FID) of this type is that they satisfy the group power sum property (law of exponents) -$$D_x^{\alpha}D_x^{\beta} = D_x^{\alpha+\beta}$$ -and that their actions satisfy the dual translations -$$D_x^{\beta} \; H(x) \; \frac{x^{\alpha}}{\alpha!} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = D_x^{-\alpha-1} \; H(x) \; \frac{x^{-\beta-1}}{(-\beta-1)!} ,$$ -where $H(x)$ is the Heaviside step function. This is consistent with the rep of this action as the convolutional integral rep of the Euler beta function analytically continued via the Pochhammer contour or the methods of generalized functions/distributions. See below and my response to the MO-Q "What's the matrix of logarithm of derivative operator (lnD)? What is the role of this operator in various math fields?" and the numerous links therein. -The Euler beta integral has a variety of physical, geometric, and probabilistic interpretations. See the MO-Q "Connections to physics, geometry, geometric probability theory of Euler's beta integral (function)". - -Sinc function / cardinal series interpolation and a circular Fourier transform: -Let me summarize some old MO and MSE responses and blog notes for easy reference to relate the FIDs here to a Fourier transform on the circle and to hopefully clear up some confusion of this family of FIDs in fractional calculus with pseudo-diff ops/symbols related to Fourier transforms of continuous functions over the whole real line. -As mentioned above one (of several reps) for the action of the FIDs is the Euler beta integral rep -$$D_x^{\beta} \; H(x)\frac{x^{\alpha}}{\alpha!}= H(x)\int_{0}^{x}\frac{z^{\alpha}}{\alpha!} \; \frac{(x-z)^{-\beta-1}}{(-\beta-1)!}\; dz $$ -$$= [ \; \int_{0}^{1}\frac{t^{\alpha}}{\alpha!} \; \frac{(1-t)^{-\beta-1}}{(-\beta-1)!} \; dt \;] \; H(x) \; x^{\alpha-\beta} .$$ -Focusing on the Euler beta integral, -$$ \int_{0}^{1}\frac{t^{\alpha}}{\alpha!} \; \frac{(1-t)^{-\beta-1}}{(-\beta-1)!} \; dt $$ -$$= \int_{0}^{1}\frac{t^{\alpha}}{\alpha!} \; \sum_{n=0} (-1)^n \frac{1}{n!\;(-\beta-n-1)!} \; t^n \; dt$$ -$$ =\frac{1}{a! (-\beta-1)!} \sum_{n=0} (-1)^n \binom{-\beta-1}{n} \; \frac{t^{n+\alpha+1}}{n+\alpha+1} \; |_{t=0}^1$$ -$$ = \frac{(-\alpha-1)!}{(-\beta-1)!}\; \sum_{n \ge 0} \binom{-\beta-1}{n}\; \frac{\sin(\pi \; (n+\alpha+1))}{\pi(n+\alpha+1)}$$ -$$ = \frac{(-\alpha-1)!}{(-\beta-1)!} \; \sum_{n \ge 0} \binom{-\beta-1}{n} \; \binom{0}{n+\alpha+1} = \frac{(-\alpha-1)!}{(-\beta-1)!} \; \binom{-\beta-1}{\alpha-\beta} = \frac{1}{(\alpha-\beta)!} $$ -with the last two lines valid for all complex $\alpha$ and $Re(\beta) < 0$. -A quick change of variables, reflecting the desirable property -$$D_x^{\beta} \; H(x) \; \frac{x^{\alpha}}{\alpha!} = D_x^{-\alpha-1} \; H(x) \; \frac{x^{-\beta-1}}{(-\beta-1)!} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!},$$ -gives -$$ \int_{0}^{1}\frac{t^{\alpha}}{\alpha!}\;\frac{(1-t)^{-\beta-1}}{(-\beta-1)!}\; dt = \int_{0}^{1}\frac{(1-t)^{\alpha}}{\alpha!}\;\frac{t^{-\beta-1}}{(-\beta-1)!}\; dt ,$$ -with the binomial expansion and subsequent sinc function interpolation for the RHS valid for -all complex $\beta$ and $Re(\alpha) > -1.$ -Since the domains of validity of the two different avenues of expansion overlap to give the same result at their intersection, we have via analytic continuation an expression valid for all $\alpha$ and $\beta$, real or complex, -$$D_x^{\beta} \; H(x) \; \frac{x^{\alpha}}{\alpha!} = D_x^{-\alpha-1} \; H(x) \; \frac{x^{-\beta-1}}{(-\beta-1)!} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!},$$ -with the caveat (really an operator interpretation) -$$H(x) \frac{x^{-n-1}}{(-n-1)!} = (-1)^n \delta^{(n)}(x) $$ -with the standard Dirac delta interpretation -$$H(x) \int_0^x f(x-t) \; \delta(t) \; dt = H(x) \int_0^x f(t) \; \delta(x-t) \; dt = H(x) f(x).$$ -A Cauchy integral rep gives the blow-up from the real line of the Euler beta integral to the complex plane, and, with a bit of regularization, -$$D_x^{\beta} \;H(x)\frac{x^{\alpha}}{\alpha!}= H(x)\frac{1}{2\pi i} \; \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!} \; \frac{\beta!}{(z-x)^{\beta+1}} \;dz$$ -$$ = [\; \frac{1}{2\pi i} \; \oint_{|z-1|=1}\frac{z^{\alpha}}{\alpha!} \; \frac{\beta!}{(z-1)^{\beta+1}} \;dz \;] \; H(x) \; x^{\alpha-\beta}$$ -$$ = [\;\frac{1}{2 \pi} \int_{-\pi}^{\pi} (1+e^{i \theta})^{\alpha} \; e^{-i \beta \theta} d\theta \; ] \; \frac{\beta!}{\alpha!} \; H(x) \; x^{\alpha-\beta}$$ -$$ = [ \; \sum_{n \geq 0} \binom{\alpha}{n} \; \frac{\sin(\pi \; (n-\beta))}{\pi(n-\beta)} \;] \; \frac{\beta!}{\alpha!} \; H(x) \; x^{\alpha-\beta} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!}.$$ -Here, we have that the FID operation is equivalent to the Fourier transform over a circle (or an average of the Fourier transform of a periodic, in general piecewise continuous function over the real line) -$$\frac{1}{2 \pi} \int_{-\pi}^{\pi} (1+e^{i \theta})^{\alpha} \; e^{-i \beta \theta} d\theta.$$ -Added 2/3/21: As noted in other posts, this family of FIDs is also derived by Pincherle through an axiomatic approach, praised by Wiener, and presented in eqn. 4 of "The Role of Salvatore Pincherle in the Development of Fractional Calculus" by Mainardi and Pagnini. It is also consistent with Heaviside's operational calculus. -(I recall seeing eons ago that Ramanujan explored this last integral. Anyone have a ref?) \ No newline at end of file